Circuitos Elétricos Nilsson Riedel 10ed - Problemas de Avaliação e Soluções

NILSSON RIEDEL ELECTRIC CIRCUITS 10th Edition Circuit Variables Assessment Problems AP 11 Use a product of ratios to convert twothirds the speed of light from meters per second to miles per second 23 3 10⁸ m1 s 100 cm1 m 1 in254 cm 1 ft12 in 1 mile5280 feet 12427424 miles1 s Now set up a proportion to determine how long it takes this signal to travel 1100 miles 12427424 miles1 s 1100 milesx s Therefore x 110012427424 000885 885 10³ s 885 ms AP 12 To solve this problem we use a product of ratios to change units from dollarsyear to dollarsmillisecond We begin by expressing 10 billion in scientific notation 100 billion 100 10⁹ Now we determine the number of milliseconds in one year again using a product of ratios 1 year36525 days 1 day24 hours 1 hour60 mins 1 min60 secs 1 sec1000 ms 1 year315576 10⁹ ms Now we can convert from dollarsyear to dollarsmillisecond again with a product of ratios 100 10⁹1 year 1 year315576 10⁹ ms 100315576 317ms 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 12 CHAPTER 1 Circuit Variables AP 13 Remember from Eq 12 current is the time rate of change of charge or i dqdt In this problem we are given the current and asked to find the total charge To do this we must integrate Eq 12 to find an expression for charge in terms of current qt ₀ᵗ ix dx We are given the expression for current i which can be substituted into the above expression To find the total charge we let t in the integral Thus we have qtotal ₀ 20e⁵⁰⁰⁰ˣ dx 205000 e⁵⁰⁰⁰ˣ₀ 205000e e⁰ 2050000 1 205000 0004 C 4000 μC AP 14 Recall from Eq 12 that current is the time rate of change of charge or i dqdt In this problem we are given an expression for the charge and asked to find the maximum current First we will find an expression for the current using Eq 12 i dqdt ddt 1α² tα 1α² eᵅᵗ ddt 1α² ddt tα eᵅᵗ ddt 1α² eᵅᵗ 0 1α eᵅᵗ tα αeᵅᵗ α 1α² eᵅᵗ 1α t 1α eᵅᵗ teᵅᵗ Now that we have an expression for the current we can find the maximum value of the current by setting the first derivative of the current to zero and solving for t didt ddt teᵅᵗ eᵅᵗ tαeᵅᵗ 1 αt eᵅᵗ 0 Since eᵅᵗ never equals 0 for a finite value of t the expression equals 0 only when 1 αt 0 Thus t 1α will cause the current to be maximum For this value of t the current is i 1α eᵅᵅ 1α e¹ 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 13 Remember in the problem statement α 003679 Using this value for α i 1 003679e1 10 A AP 15 Start by drawing a picture of the circuit described in the problem statement Also sketch the four figures from Fig 16 a Now we have to match the voltage and current shown in the first figure with the polarities shown in Fig 16 Remember that 4A of current entering Terminal 2 is the same as 4A of current leaving Terminal 1 We get a v 20 V i 4 A b v 20 V i 4 A c v 20 V i 4 A d v 20 V i 4 A b Using the reference system in Fig 16a and the passive sign convention p vi 204 80 W Since the power is greater than 0 the box is absorbing power c From the calculation in part b the box is absorbing 80 W AP 16 a Applying the passive sign convention to the power equation using the voltage and current polarities shown in Fig 15 p vi To find the time at which the power is maximum find the first derivative of the power with respect to time set the resulting expression equal to zero and solve for time p 80000te500t15te500t 120 104t2e1000t dp dt 240 104te1000t 120 107t2e1000t 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Therefore 240 104 120 107 t 0 Solving t 240 104 120 107 2 103 2 ms b The maximum power occurs at 2 ms so find the value of the power at 2 ms p0002 120 104 00022 e2 6496 mW c From Eq 13 we know that power is the time rate of change of energy or p dwdt If we know the power we can find the energy by integrating Eq 13 To find the total energy the upper limit of the integral is infinity wtotal 0 120 104 x2 e1000x dx 120 104 10003 e1000x 10002 x2 21000x 2 0 0 120 104 10003 e0 0 0 2 24 mJ AP 17 At the Oregon end of the line the current is leaving the upper terminal and thus entering the lower terminal where the polarity marking of the voltage is negative Thus using the passive sign convention p vi Substituting the values of voltage and current given in the figure p 800 103 18 103 1440 106 1440 MW Thus because the power associated with the Oregon end of the line is negative power is being generated at the Oregon end of the line and transmitted by the line to be delivered to the California end of the line Problems 15 Chapter Problems P 11 260 106540 109 1044 gigawatthours P 12 480320 pixels 1 frame 2 bytes 1 pixel 30 frames 1 sec 9216 106 bytessec 9216 106 bytessecx secs 32 230 bytes x 32 230 9216 106 3728 sec 62 min 1 hour of video P 13 a 20000 photos 11151 mm3 x photos 1 mm3 x 200001 11151 121 photos b 16 230 bytes 11151 mm3 x bytes 023 mm3 x 16 2300008 11151 832963 bytes P 14 4 cond 845 mi 5280 ft 1 mi 2526 lb 1000 ft 1 kg 22 lb 205 106 kg P 15 Volume area thickness Convert values to millimeters noting that 10 m2 106 mm2 106 10 106thickness thickness 106 10 106 010 mm P 16 a We can set up a ratio to determine how long it takes the bamboo to grow 10 µm First recall that 1 mm 103µm Lets also express the rate of growth of bamboo using the units mms instead of mmday Use a product of ratios to perform this conversion 250 mm 1 day 1 day 24 hours 1 hour 60 min 1 min 60 sec 250 246060 10 3456 mms Use a ratio to determine the time it takes for the bamboo to grow 10 µm 103456 103 m 1 s 10 106 m x s so x 10 106 103456 103 3456 s 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 b 1 cell length 3456 s 3600 s 1 hr 247 hr 1 week 175000 cell lengthsweek P 17 a First we use Eq 12 to relate current and charge i dqdt 0125 e2500t Therefore dq 0125 e2500t dt To find the charge we can integrate both sides of the last equation Note that we substitute x for q on the left side of the integral and y for t on the right side of the integral q0qt dx 0125 0t e2500y dy We solve the integral and make the substitutions for the limits of the integral qt q0 0125 e2500y 2500 0t 50 106 1 e2500t But q0 0 by hypothesis so qt 501 e2500t μC b As t qT 50 μC c q05 103 50 1061 e250000005 35675 μC P 18 First we use Eq 12 to relate current and charge i dqdt 20 cos 5000 t Therefore dq 20 cos 5000 t dt To find the charge we can integrate both sides of the last equation Note that we substitute x for q on the left side of the integral and y for t on the right side of the integral q0qt dx 20 0t cos 5000 y dy We solve the integral and make the substitutions for the limits of the integral remembering that sin 0 0 qt q0 20 sin 5000 y 5000 0t 205000 sin 5000 t 205000 sin 50000 205000 sin 5000 t But q0 0 by hypothesis ie the current passes through its maximum value at t 0 so qt 4 103 sin 5000 t C 4 sin 5000 t mC P 19 a First we use Eq 12 to relate current and charge i dqdt 40 t e500t Therefore dq 40 t e500t dt To find the charge we can integrate both sides of the last equation Note that we substitute x for q on the left side of the integral and y for t on the right side of the integral q0qt dx 40 0t y e500 y dy We solve the integral and make the substitutions for the limits of the integral qt q0 40 e500 y 5002 500 y 1 0t 160 106 e500 t 500 t 1 160 106 160 106 1 500 t e500 t e500 t But q0 0 by hypothesis so qt 160 1 500 t e500 t e500 t μC b q0001 1601 5000001 e5000001 e5000001 144 μC P 110 n 35 106 Cs 16022 1019 Celec 218 1014 elecs P 111 w q V 16022 10196 961 1019 0961 aJ P 112 a circuit diagram with 60 V and current 10 A p vi 4010 400 W Power is being delivered by the box b Entering c Gaining P 113 a p vi 6010 600 W so power is being absorbed by the box b Entering c Losing P 114 Assume we are standing at box A looking toward box B Use the passive sign convention to get p vi since the current i is flowing into the terminal of the voltage v Now we just substitute the values for v and i into the equation for power Remember that if the power is positive B is absorbing power so the power must be flowing from A to B If the power is negative B is generating power so the power must be flowing from B to A a p 306 180 W 180 W from A to B b p 208 160 W 160 W from A to B c p 604 240 W 240 W from B to A d p 409 360 W 360 W from B to A P 115 a In Car A the current i is in the direction of the voltage drop across the 12 V batterythe current i flows into the terminal of the battery of Car A Therefore using the passive sign convention p vi 3012 360 W Since the power is positive the battery in Car A is absorbing power so Car A must have the dead battery b wt integral 0 to t p dx 1 min 60 s w60 integral 0 to 60 360 dx w 360600 36060 21600 J 216 kJ P 116 p vi w integral 0 to t p dx Since the energy is the area under the power vs time plot let us plot p vs t Note that in constructing the plot above we used the fact that 40 hr 144000 s 144 ks p0 159 x 103 135 x 103 W p144 ks 19 x 103 9 x 103 W w 9 x 103144 x 103 12135 x 103 9 x 103144 x 103 1620 J P 117 p 12100 x 103 12 W 4 hr 3600 s 1 hr 14400 s wt integral 0 to t p dt w14400 integral 0 to 14400 12 dt 1214400 1728 kJ P 118 a p vi 15e250t004e250t 06e500t W p001 06e500001 06e5 000404 404 mW b w total integral 0 to infinity px dx integral 0 to infinity 06e500x dx 06 500 e500x from 0 to infinity 00012 einfinity e0 00012 12 mJ P 119 a p vi 005e1000t75 75e1000t 375e1000t 375e2000t W dpdt 3750e1000t 7500e2000t 0 so 2e2000t e1000t 2 e1000t so ln2 1000t thus p is maximum at t 69315 µs p max p69315 µs 9375 mW b w integral 0 to infinity 375e1000t 375e2000t dt 3751000 e1000t 375 2000 e2000t from 0 to infinity 3751000 3752000 1875 mJ P 120 a p vi 025e3200t 05e2000t 025e800t p625 µs 422 mW b wt integral 0 to t 025e3200t 05e2000t 025e800t 140625 78125e3200t 250e2000t 3125e800t µJ w625 µs 1214 µJ c w total 140625 µJ P 121 a p vi 1500t 1e750t004e750t 60t 004e1500t dpdt 60e1500t 1500e1500t60t 004 90000te1500t Therefore dpdt 0 when t 0 so p max occurs at t 0 b p max 600 004e0 004 40 mW c w integral 0 to t pdx w integral 0 to t 60xe1500x dx integral 0 to t 004 e1500x dx 60e1500x 15002 1500x 10t 004 e1500x 1500 0t When t infinity all the upper limits evaluate to zero hence w 60225 x 104 0041500 5333 µJ P 122 a p vi 3200t 32e1000t160t 016e1000t e2000t512000t2 1024t 0512 dpdt e2000t1024000t 1024 2000e2000t512000t2 1024t 0512 e2000t1024 x 106 t2 1024000t Therefore dpdt 0 when t 0 so p max occurs at t 0 b p max e00 0 0512 512 mW c w integral 0 to t pdx w integral 0 to t 512000x2 e2000x dx integral 0 to t 1024 x e2000x dx integral 0 to t 0512 e2000x dx 512000 e2000x 8 x 1094 x 106 x2 4000 x 2 0t 1024 e2000x 4 x 106 2000x 10t 0512 e2000x 2000 0t When t all the upper limits evaluate to zero hence w 5120002 8 109 1024 4 106 0512 2000 w 128 106 256 106 256 106 640 μJ P 123 a We can find the time at which the power is maximum by writing an expression for pt vtit taking the first derivative of pt and setting it to zero then solving for t The calculations are shown below p 0 t 0 p 0 t 40 s p vi t1 0025t4 02t 4t 03t2 0005t3 W 0 t 40 s dpdt 4 06t 0015t2 0015t2 40t 26667 dpdt 0 when t2 40t 26667 0 t1 8453 s t2 31547 s using the polynomial solver on your calculator pt1 48453 0384532 000584533 15396 W pt2 431547 03315472 0005315473 15396 W Therefore maximum power is being delivered at t 8453 s b The maximum power was calculated in part a to determine the time at which the power is maximum pmax 15396 W delivered c As we saw in part a the other maximum power is actually a minimum or the maximum negative power As we calculated in part a maximum power is being extracted at t 31547 s d This maximum extracted power was calculated in part a to determine the time at which power is maximum pmax 15396 W extracted e w 0t p dx 0t 4x 03x2 0005x3dx 2t2 01t3 000125t4 w0 0 J w30 1125 J w10 1125 J w40 0 J w20 200 J To give you a feel for the quantities of voltage current power and energy and their relationships among one another they are plotted below 112 CHAPTER 1 Circuit Variables P 124 a v10 ms 400e1 sin 2 1338 V i10 ms 5e1 sin 2 167 A p10 ms vi 22380 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 b p vi 2000e200t sin2 200t 2000e200t 12 12 cos 400t 1000e200t 1000e200t cos 400t w 0 1000e200t dt 0 1000e200t cos 400t dt 1000 e200t 200 0 1000 e200t 2002 4002 200 cos 400t 400 sin 400t 0 5 1000 200 4 104 16 104 5 1 w 4 J P 125 a p vi 2000 cos 800πt sin800πt 1000 sin1600πt W Therefore pmax 1000 W b pmax extracting 1000 W c pavg 1 25 103 025103 1000 sin1600πt dt 4 105 cos 1600πt 1600π 025103 250π 1 cos 4π 0 d pavg 1 15625 103 015625103 1000 sin1600πt dt 64 103 cos 1600πt 1600π 015625103 40π 1 cos 25π 2546 W P 126 a q area under i vs t plot 12 812000 1612000 12 164000 48000 192000 32000 272000 C b w p dt vi dt v 250 106 t 8 0 t 16 ks 0 t 12000s i 24 66667 106 t p 192 66667 106 t 16667 109 t2 w1 012000 192 66667 106 t 16667 109 t2 dt 2304 48 96103 2256 kJ 12000 s t 16000 s i 64 4 103 t p 512 16 103 t 106 t2 w2 1200016000 512 16 103 t 106 t2 dt 2048 896 78933103 362667 kJ wT w1 w2 2256 362667 2618667 kJ P 127 a 0 s t 10 ms v 8 V i 25t A p 200t W 10 ms t 30 ms v 8 V i 05 25t A p 200t 4 W 30 ms t 40 ms v 0 V i 250 mA p 0 W 40 ms t 60 ms v 8 V i 25t 125 A p 200t 10 W t 60 ms v 0 V i 250 mA p 0 W b Calculate the area under the curve from zero up to the desired time w001 12 2001 10 mJ w003 w001 12 2001 12 2001 10 mJ w008 w003 12 2001 12 2001 10 mJ Problems 115 P 128 a b it 10 05 103t mA 0 t 10 ks it 15 mA 10 ks t 20 ks it 25 05 103t mA 20 ks t 30 ks it 0 t 30 ks p vi 120i so pt 1200 006t mW 0 t 10 ks pt 1800 mW 10 ks t 20 ks pt 3000 006t mW 20 ks t 30 ks pt 0 t 30 ks 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 P 131 a From the diagram and the table we have pa vaia 46166 27696 W pb vbib 1416472 668352 W pc vcic 3264 2048 W pd vdid 22128 2816 W pe veie 336168 56448 W pf vfif 6604 264 W pg vgig 256128 32768 W ph vhih 0404 016 W Pdel 27696 2816 30512 W Pabs 668352 2048 56448 264 32768 016 35792 W Therefore Pdel Pabs and the subordinate engineer is correct b The difference between the power delivered to the circuit and the power absorbed by the circuit is 30512 35792 528 W Onehalf of this difference is 264 W so it is likely that pf is in error Either the voltage or the current probably has the wrong sign In Chapter 2 we will discover that using KCL at the node connecting components f and h the current if should be 04 A not 04 A If the sign of pf is changed from negative to positive we can recalculate the power delivered and the power absorbed as follows Pdel 27696 2816 264 33152 W Pabs 668352 2048 56448 32768 016 33152 W Now the power delivered equals the power absorbed and the power balances for the circuit P 132 a Remember that if the circuit element is absorbing power the power is positive whereas if the circuit element is supplying power the power is negative We can add the positive powers together and the negative powers together if the power balances these power sums should be equal Psup 600 50 600 1250 2500 W Pabs 400 100 2000 2500 W Thus the power balances b The current can be calculated using i pv or i pv with proper application of the passive sign convention c To find the energy calculate the area under the plot of the power w10 ks 120610000 1210000 15 kJ w20 ks w10 ks 1810000 33 kJ w10 ks w20 ks 120610000 1210000 48 kJ P 129 We use the passive sign convention to determine whether the power equation is p vi or p vi and substitute into the power equation the values for v and i as shown below pa vaia 404 103 160 mW pb vbib 244 103 96 mW pc vcic 164 103 64 mW pd vdid 8015 103 120 mW pe veie 4025 103 100 mW pf vfif 12025 103 300 mW Remember that if the power is positive the circuit element is absorbing power whereas is the power is negative the circuit element is developing power We can add the positive powers together and the negative powers togetherif the power balances these power sums should be equal Pdev 120 300 420 mW Pabs 160 96 64 100 420 mW Thus the power balances and the total power absorbed in the circuit is 420 mW P 130 pa vaia 3000250 106 075 W pb vbib 4000400 106 16 W pc vcic 1000400 106 04 W pd vdid 1000150 106 015 W pe veie 4000200 106 08 W pf vfif 400050 106 02 W Therefore Pabs 16 015 02 195 W Pdel 075 04 08 195 W Pabs Thus the interconnection does satisfy the power check ia pava 600400 15 A ib pbvb 50100 05 A ic pcvc 400200 20 A id pdvd 600300 20 A ie peve 100200 05 A if pfvf 2000500 40 A ig pgvg 1250500 25 A P 133 a If the power balances the sum of the power values should be zero ptotal 0175 0375 0150 0320 0160 0120 0660 0 Thus the power balances b When the power is positive the element is absorbing power Since elements a b c e and f have positive power these elements are absorbing power c The voltage can be calculated using v pi or v pi with proper application of the passive sign convention va paia 01750025 7 V vb pbib 03750075 5 V vc pcic 0150005 3 V vd pdid 0320004 8 V ve peie 0160002 8 V vf pfif 0120003 4 V vg pgig 0660055 12 V P 134 pa vaia 12010 1200 W pb vbib 1209 1080 W pc vcic 1010 100 W pd vdid 101 10 W pe veie 109 90 W pf vfif 1005 500 W pg vgig 1204 480 W ph vhih 2205 1100 W Pdel 1200 1080 2280 W Pabs 100 10 90 500 480 1100 2280 W Therefore Pdel Pabs 2280 W Thus the interconnection now satisfies the power check P 135 a The revised circuit model is shown below b The expression for the total power in this circuit is va ia vb ib vf if vg ig vh ih 12010 12010 1203 120ig 2407 0 Therefore 120ig 1200 1200 360 1680 360 so ig 360120 3 A Thus if the power in the modified circuit is balanced the current in component g is 3 A 2 Circuit Elements a Assessment Problems AP 21 a Note that the current ib is in the same circuit branch as the 8 A current source however ib is defined in the opposite direction of the current source Therefore ib 8 A Next note that the dependent voltage source and the independent voltage source are in parallel with the same polarity Therefore their voltages are equal and vg ib 4 8 4 2 V b To find the power associated with the 8 A source we need to find the voltage drop across the source vi Note that the two independent sources are in parallel and that the voltages vg and v1 have the same polarities so these voltages are equal vi vg 2 V Using the passive sign convention ps 8 Avi 8 A2 V 16 W Thus the current source generated 16 W of power 21 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 22 CHAPTER 2 Circuit Elements AP 22 a Note from the circuit that vx 25 V To find α note that the two current sources are in the same branch of the circuit but their currents flow in opposite directions Therefore αvx 15 A Solve the above equation for α and substitute for vx α 15 A vx 15 A 25 V 06 AV b To find the power associated with the voltage source we need to know the current iv Note that this current is in the same branch of the circuit as the dependent current source and these two currents flow in the same direction Therefore the current iv is the same as the current of the dependent source iv αvx 0625 15 A Using the passive sign convention ps iv25 V 15 A25 V 375 W Thus the voltage source dissipates 375 W AP 23 a The resistor and the voltage source are in parallel and the resistor voltage and the voltage source have the same polarities Therefore these two voltages are the same vR vg 1 kV 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 23 Note from the circuit that the current through the resistor is ig 5 mA Use Ohms law to calculate the value of the resistor R vR ig 1 kV 5 mA 200 kΩ Using the passive sign convention to calculate the power in the resistor pR vRig 1 kV5 mA 5 W The resistor is dissipating 5 W of power b Note from part a the vR vg and iR ig The power delivered by the source is thus psource vgig so vg psource ig 3 W 75 mA 40 V Since we now have the value of both the voltage and the current for the resistor we can use Ohms law to calculate the resistor value R vg ig 40 V 75 mA 53333 Ω The power absorbed by the resistor must equal the power generated by the source Thus pR psource 3 W 3 W c Again note the iR ig The power dissipated by the resistor can be determined from the resistors current pR RiR2 Rig2 Solving for ig i2 g pr R 480 mW 300 Ω 00016 so ig 00016 004 A 40 mA Then since vR vg vR RiR Rig 300 Ω40 mA 12 V so vg 12 V AP 24 a Note from the circuit that the current through the conductance G is ig flowing from top to bottom because the current source and the 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 conductance are in the same branch of the circuit so must have the same current The voltage drop across the current source is vg positive at the top because the current source and the conductance are also in parallel so must have the same voltage From a version of Ohms law vg igG 05 A 50 mS 10 V Now that we know the voltage drop across the current source we can find the power delivered by this source psource vg ig 1005 5 W Thus the current source delivers 5 W to the circuit b We can find the value of the conductance using the power and the value of the current using Ohms law and the conductance value pg G vg2 so G pg vg2 9 152 004 S 40 mS ig G vg 40 mS15 V 06 A c We can find the voltage from the power and the conductance and then use the voltage value in Ohms law to find the current pg G vg2 so vg2 pg G 8 W 200 μS 40000 Thus vg 40000 200 V ig G vg 200 μS200 V 004 A 40 mA AP 25 a Redraw the circuit with all of the voltages and currents labeled for every circuit element Write a KVL equation clockwise around the circuit starting below the voltage source 24 V v2 v5 v1 0 Next use Ohms law to calculate the three unknown voltages from the three currents v2 3i2 v5 7i5 v1 2i1 Problems 25 A KCL equation at the upper right node gives i2 i5 a KCL equation at the bottom right node gives i5 i1 a KCL equation at the upper left node gives is i2 Now replace the currents i1 and i2 in the Ohms law equations with i5 v2 3i2 3i5 v5 7i5 v1 2i1 2i5 Now substitute these expressions for the three voltages into the first equation 24 v2 v5 v1 3i5 7i5 2i5 12i5 Therefore i5 2412 2 A b v1 2i5 22 4 V c v2 3i5 32 6 V d v5 7i5 72 14 V e A KCL equation at the lower left node gives is i1 Since i1 i5 is 2 A We can now compute the power associated with the voltage source p24 24is 242 48 W Therefore 24 V source is delivering 48 W AP 26 Redraw the circuit labeling all voltages and currents We can find the value of the unknown resistor if we can find the value of its voltage and its current To start write a KVL equation clockwise around the right loop starting below the 24 Ω resistor 120 V v3 0 Use Ohms law to calculate the voltage across the 8 Ω resistor in terms of its current v3 8i3 Substitute the expression for v3 into the first equation 120 V 8i3 0 so i3 120 8 15 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 26 CHAPTER 2 Circuit Elements Also use Ohms law to calculate the value of the current through the 24 Ω resistor i2 120 V 24 Ω 5 A Now write a KCL equation at the top middle node summing the currents leaving i1 i2 i3 0 so i1 i2 i3 5 15 20 A Write a KVL equation clockwise around the left loop starting below the voltage source 200 V v1 120 V 0 so v1 200 120 80 V Now that we know the values of both the voltage and the current for the unknown resistor we can use Ohms law to calculate the resistance R v1 i1 80 20 4 Ω AP 27 a Plotting a graph of vt versus it gives Note that when it 0 vt 25 V therefore the voltage source must be 25 V Since the plot is a straight line its slope can be used to calculate the value of resistance R v i 25 0 025 0 25 025 100 Ω A circuit model having the same v i characteristic is a 25 V source in series with a 100Ω resistor as shown below 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 27 b Draw the circuit model from part a and attach a 25 Ω resistor To find the power delivered to the 25 Ω resistor we must calculate the current through the 25 Ω resistor Do this by first using KCL to recognize that the current in each of the components is it flowing in a clockwise direction Write a KVL equation in the clockwise direction starting below the voltage source and using Ohms law to express the voltage drop across the resistors in the direction of the current it flowing through the resistors 25 V 100it 25it 0 so 125it 25 so it 25 125 02 A Thus the power delivered to the 25 Ω resistor is p25 25i2 t 25022 1 W AP 28 a From the graph in Assessment Problem 27a we see that when vt 0 it 025 A Therefore the current source must be 025 A Since the plot is a straight line its slope can be used to calculate the value of resistance R v i 25 0 025 0 25 025 100 Ω A circuit model having the same v i characteristic is a 025 A current source in parallel with a 100Ω resistor as shown below b Draw the circuit model from part a and attach a 25 Ω resistor Note that by writing a KVL equation around the right loop we see that the voltage drop across both resistors is vt Write a KCL equation at the top center node summing the currents leaving the node Use Ohms law 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 28 CHAPTER 2 Circuit Elements to specify the currents through the resistors in terms of the voltage drop across the resistors and the value of the resistors 025 vt 100 vt 25 0 so 5vt 25 thus vt 5 V p25 v2 t 25 1 W AP 29 First note that we know the current through all elements in the circuit except the 6 kΩ resistor the current in the three elements to the left of the 6 kΩ resistor is i1 the current in the three elements to the right of the 6 kΩ resistor is 30i1 To find the current in the 6 kΩ resistor write a KCL equation at the top node i1 30i1 i6k 31i1 We can then use Ohms law to find the voltages across each resistor in terms of i1 The results are shown in the figure below a To find i1 write a KVL equation around the lefthand loop summing voltages in a clockwise direction starting below the 5V source 5 V 54000i1 1 V 186000i1 0 Solving for i1 54000i1 186000i1 6 V so 240000i1 6 V Thus i1 6 240000 25 µA b Now that we have the value of i1 we can calculate the voltage for each component except the dependent source Then we can write a KVL equation for the righthand loop to find the voltage v of the dependent source Sum the voltages in the clockwise direction starting to the left of the dependent source v 54000i1 8 V 186000i1 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 29 Thus v 240000i1 8 V 24000025 106 8 V 6 V 8 V 2 V We now know the values of voltage and current for every circuit element Lets construct a power table Element Current Voltage Power Power µA V Equation µW 5 V 25 5 p vi 125 54 kΩ 25 135 p Ri2 3375 1 V 25 1 p vi 25 6 kΩ 775 465 p Ri2 360375 Dep source 750 2 p vi 1500 18 kΩ 750 135 p Ri2 10125 8 V 750 8 p vi 6000 c The total power generated in the circuit is the sum of the negative power values in the power table 125 µW 25 µW 6000 µW 6150 µW Thus the total power generated in the circuit is 6150 µW d The total power absorbed in the circuit is the sum of the positive power values in the power table 3375 µW 360375 µW 1500 µW 10125 µW 6150 µW Thus the total power absorbed in the circuit is 6150 µW AP 210 Given that iφ 2 A we know the current in the dependent source is 2iφ 4 A We can write a KCL equation at the left node to find the current in the 10 Ω resistor Summing the currents leaving the node 5 A 2 A 4 A i10Ω 0 so i10Ω 5 A 2 A 4 A 1 A Thus the current in the 10 Ω resistor is 1 A flowing right to left as seen in the circuit below 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 210 CHAPTER 2 Circuit Elements a To find vs write a KVL equation summing the voltages counterclockwise around the lower right loop Start below the voltage source vs 1 A10 Ω 2 A30 Ω 0 so vs 10 V 60 V 70 V b The current in the voltage source can be found by writing a KCL equation at the righthand node Sum the currents leaving the node 4 A 1 A iv 0 so iv 4 A 1 A 3 A The current in the voltage source is 3 A flowing top to bottom The power associated with this source is p vi 70 V3 A 210 W Thus 210 W are absorbed by the voltage source c The voltage drop across the independent current source can be found by writing a KVL equation around the left loop in a clockwise direction v5A 2 A30 Ω 0 so v5A 60 V The power associated with this source is p v5Ai 60 V5 A 300 W This source thus delivers 300 W of power to the circuit d The voltage across the controlled current source can be found by writing a KVL equation around the upper right loop in a clockwise direction v4A 10 Ω1 A 0 so v4A 10 V The power associated with this source is p v4Ai 10 V4 A 40 W This source thus delivers 40 W of power to the circuit e The total power dissipated by the resistors is given by i30Ω230 Ω i10Ω210 Ω 2230 Ω 1210 Ω 120 10 130 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems P 21 a Yes independent voltage sources can carry the 5 A current required by the connection independent current source can support any voltage required by the connection in this case 5 V positive at the bottom b 20 V source absorbing 15 V source developing delivering 5 A source developing delivering c P20V 205 100 W abs P15V 155 75 W devdel P5A 55 25 W devdel Pabs Pdel 100 W d The interconnection is valid but in this circuit the voltage drop across the 5 A current source is 35 V positive at the top 20 V source is developing delivering the 15 V source is developing delivering and the 5 A source is absorbing P20V 205 100 W devdel P15V 155 75 W devdel P5A 355 175 W abs Pabs Pdel 175 W P 22 The interconnect is valid since the voltage sources can all carry 5 A of current supplied by the current source and the current source can carry the voltage drop required by the interconnection Note that the branch containing the 10 V 40 V and 5 A sources must have the same voltage drop as the branch containing the 50 V source so the 5 A current source must have a voltage drop of 20 V positive at the right The voltages and currents are summarize in the circuit below P50V 505 250 W abs P10V 105 50 W abs P40V 405 200 W dev P5A 205 100 W dev ΣPdev 300 W P 23 The interconnection is valid The 10 A current source has a voltage drop of 100 V positive at the top because the 100 V source supplies its voltage drop across a pair of terminals shared by the 10 A current source The right hand branch of the circuit must also have a voltage drop of 100 V from the left terminal of the 40 V source to the bottom terminal of the 5 A current source because this branch shares the same terminals as the 100 V source This means that the voltage drop across the 5 A current source is 140 V positive at the top Also the two voltage sources can carry the current required of the interconnection This is summarized in the figure below From the values of voltage and current in the figure the power supplied by the current sources is calculated as follows P10A 10010 1000 W 1000 W supplied P5A 1405 700 W 700 W supplied ΣPdev 1700 W P 24 The interconnection is not valid Note that the 3 A and 4 A sources are both connected in the same branch of the circuit A valid interconnection would require these two current sources to supply the same current in the same direction which they do not P 25 The interconnection is valid since the voltage sources can carry the currents supplied by the 2 A and 3 A current sources and the current sources can carry whatever voltage drop from the top node to the bottom node is required by the interconnection In particular note the the voltage drop between the top and bottom nodes in the right hand branch must be the same as the voltage drop between the top and bottom nodes in the left hand branch In particular this means that v1 8 V 12 V v2 Problems 213 Hence any combination of v1 and v2 such that v1 v2 4 V is a valid solution P 26 a Because both current sources are in the same branch of the circuit their values must be the same Therefore v1 50 04 v1 0450 20 V b p v104 2004 8 W absorbed P 27 a The voltage drop from the top node to the bottom node in this circuit must be the same for every path from the top to the bottom Therefore the voltages of the two voltage sources are equal αi 6 Also the current i is in the same branch as the 15 mA current source but in the opposite direction so i 0015 Substituting α0015 6 α 6 0015 400 The interconnection is valid if α 400 VA b The voltage across the current source must equal the voltage across the 6 V source since both are connected between the top and bottom nodes Using the passive sign convention p vi 60015 009 90 mW c Since the power is positive the current source is absorbing power P 28 a Yes each of the voltage sources can carry the current required by the interconnection and each of the current sources can carry the voltage drop required by the interconnection Note that i1 50 mA b No because the voltage drop between the top terminal and the bottom terminal cannot be determined For example define v1 v2 and v3 as shown 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 The voltage drop across the left branch the center branch and the right branch must be the same since these branches are connected at the same two terminals This requires that v1 20 v2 v3 30 But this equation has three unknown voltages so the individual voltages cannot be determined and thus the power of the sources cannot be determined P 29 The interconnection is invalid In the middle branch the value of the current ix must be 50 mA since the 50 mA current source supplies current in this branch in the same direction as the current ix Therefore the voltage supplied by the dependent voltage source in the right hand branch is 1800005 90 V This gives a voltage drop from the top terminal to the bottom terminal in the right hand branch of 90 60 150 V But the voltage drop between these same terminals in the left hand branch is 30 V due to the voltage source in that branch Therefore the interconnection is invalid P 210 First 10va 5 V so va 05 V Then recognize that each of the three branches is connected between the same two nodes so each of these branches must have the same voltage drop The voltage drop across the middle branch is 5 V and since va 05 V vg 05 5 45 V Also the voltage drop across the left branch is 5 V so 20 v9A 5 V and v9A 15 V where v9A is positive at the top Note that the current through the 20 V source must be 9 A flowing from top to bottom and the current through the vg is 6 A flowing from top to bottom Lets find the power associated with the left and middle branches p9A 915 135 W p20V 920 180 W pvg 645 27 W p6A 605 3 W Since there is only one component left we can find the total power ptotal 135 180 27 3 pds 75 pds 0 so pds must equal 75 W Therefore ΣPdev ΣPabs 210 W Problems 215 P 211 a Using the passive sign convention and Ohms law v Ri 30000015 45 V b PR v2 R 452 3000 0675 675 mW c Using the passive sign convention with the current direction reversed v Ri 30000015 45 V PR v2 R 452 3000 0675 675 mW P 212 a Using the passive sign convention and Ohms law i v R 40 2500 0016 16 mA b PR Ri2 250000162 064 640 mW c Using the passive sign convention with the voltage polarity reversed i v R 40 2500 0016 16 mA PR Ri2 250000162 064 640 mW P 213 a b Vbb noload voltage of battery Rbb internal resistance of battery Rx resistance of wire between battery and switch Ry resistance of wire between switch and lamp A Ra resistance of lamp A Rb resistance of lamp B Rw resistance of wire between lamp A and lamp B Rg1 resistance of frame between battery and lamp A Rg2 resistance of frame between lamp A and lamp B S switch 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 216 CHAPTER 2 Circuit Elements P 214 Since we know the device is a resistor we can use Ohms law to calculate the resistance From Fig P214a v Ri so R v i Using the values in the table of Fig P214b R 7200 6 3600 3 3600 3 7200 6 10800 9 12 kΩ Note that this value is found in Appendix H P 215 Since we know the device is a resistor we can use the power equation From Fig P215a p vi v2 R so R v2 p Using the values in the table of Fig P213b R 82 640 103 42 160 103 42 160 103 82 640 103 122 1440 103 162 2560 103 100 Ω Note that this value is found in Appendix H P 216 The resistor value is the ratio of the power to the square of the current R p i2 Using the values for power and current in Fig P216b 825 103 05 1032 33 103 1 1032 7425 103 15 1032 132 103 2 1032 20625 103 25 1032 297 103 3 1032 33 kΩ Note that this is a value from Appendix H P 217 Label the unknown resistor currents and voltages 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 217 a KCL at the top node 002 i1 i2 KVL around the right loop vo v2 5 0 Use Ohms law to write the resistor voltages in the previous equation in terms of the resistor currents 5000i1 2000i2 5 0 5000i1 2000i2 5 Multiply the KCL equation by 2000 and add it to the KVL equation to eliminate i2 2000i1 i2 5000i1 2000i2 2000002 5 7000i1 35 Solving i1 35 7000 0005 5 mA Therefore vo Ri1 50000005 25 V b p20mA 002vo 00225 05 W i2 002 i1 002 0005 0015 A p5V 5i2 50015 0075 W p5k 5000i2 1 500000052 0125 W p2k 2000i2 2 200000152 045 W ptotal p20mA p5V p5k p2k 05 0075 0125 045 0 Thus the power in the circuit balances P 218 a 20ia 80ib ig ia ib 5ib ia 4ib 50 4ig 80ib 20ib 80ib 100ib ib 05 A therefore ia 2 A and ig 25 A b ib 05 A c vo 80ib 40 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 d p4Ω i2g4 6254 25 W p20Ω i2a20 420 80 W p80Ω i2b80 02580 20 W e p50V delivered 50ig 125 W Check ΣPdis 25 80 20 125 W ΣPdel 125 W P 219 a Write a KCL equation at the top node 15 i1 i2 0 so i1 i2 15 Write a KVL equation around the right loop v1 v2 v3 0 From Ohms law v1 100i1 v2 150i2 v3 250i2 Substituting 100i1 150i2 250i2 0 so 100i1 400i2 0 Solving the two equations for i1 and i2 simultaneously i1 12 A and i2 03 A b Write a KVL equation clockwise around the left loop vo v1 0 but v1 100i1 10012 120 V So vo v1 120 V c Calculate power using p vi for the source and p Ri2 for the resistors psource vo15 12015 180 W p100Ω 122100 144 W p150Ω 032150 135 W p250Ω 032250 225 W ΣPdev 180 W ΣPabs 144 135 225 180 W Problems 219 P 220 Label the unknown resistor voltages and currents a ia 35 175 002 A Ohms law i1 ia 002 A KCL b vb 200i1 200002 4 V Ohms law v1 vb 35 0 so v1 35 vb 35 4 75 V KVL c va 00550 25 V Ohms law vg va v1 0 so vg va v1 25 75 10 V KVL d pg vg005 10005 05 W P 221 a Use KVL for the right loop to calculate the voltage drop across the righthand branch vo This is also the voltage drop across the middle branch so once vo is known use Ohms law to calculate io vo 1000ia 4000ia 3000ia 8000ia 80000002 16 V 16 2000io io 16 2000 8 mA b KCL at the top node ig ia io 0002 0008 0010 A 10 mA c The voltage drop across the source is v0 seen by writing a KVL equation for the left loop Thus pg voig 16001 0160 W 160 mW Thus the source delivers 160 mW P 222 a v2 150 501 100V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 i2 v225 4A i3 1 i2 i3 4 1 3A v1 10i3 25i2 103 254 130V i1 v165 13065 2A Note also that i4 i1 i3 2 3 5A ig i4 io 5 1 6A b p4Ω 5²4 100 W p50Ω 1²50 50 W p65Ω 2²65 260 W p10Ω 3²10 90 W p25Ω 4²25 400 W c Pdis 100 50 260 90 400 900 W Pdev 150ig 1506 900 W P 223 Label all unknown resistor voltages and currents Ohms law for 5 kΩ resistor v1 0015000 50 V KVL for lower left loop 80 v2 50 0 v2 80 50 30 V Ohms law for 15 kΩ resistor i2 v21500 301500 20 mA KCL at center node i2 i3 001 i3 i2 001 002 001 001 10 mA Ohms law for 3 kΩ resistor v3 3000i3 3000001 30 V KVL for lower right loop v1 v3 v4 0 v4 v1 v3 50 30 20 V Problems 221 Ohms law for 500 Ω resistor i4 v4500 20500 004 40 mA KCL for right node i3 iR i4 iR i4 i3 004 001 003 30 mA KVL for outer loop 80 vR v4 0 vR 80 v4 80 20 60 V Therefore R vR iR 60 003 2000 2 kΩ P 224 a va 5 104 60 V 240 va vb 0 so vb 240 va 240 60 180 V ie vb14 6 18020 9 A id ie 4 9 4 5 A vc 4id vb 45 180 200 V ic vc10 20010 20 A vd 240 vc 240 200 40 V ia id ic 5 20 25 A R vdia 4025 16 Ω b ig ia 4 25 4 29 A pg supplied 24029 6960 W P 225 a icd 8016 5 A vac 125 80 45 so iac 4515 3 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 iac ibc icd so ibc 5 3 2 A vab 15iac 5ibc 153 52 35 V so iab 357 5 A ibd iab ibc 5 2 3 A Calculate the power dissipated by the resistors using the equation pR RiR² p7Ω 75² 175 W p30Ω 303² 270 W p15Ω 153² 135 W p16Ω 165² 400 W p5Ω 52² 20 W b Calculate the current through the voltage source iad iab iac 5 3 8 A Now that we have both the voltage and the current for the source we can calculate the power supplied by the source pg 1258 1000 W thus pg supplied 1000 W c Pdis 175 270 135 400 20 1000 W Therefore Psupp Pdis P 226 a v2 100 415 160 V v1 160 9 11 102 100 V i1 v14 16 10020 5 A i3 i1 2 5 2 3 A vg v1 30i3 100 303 190 V i4 2 4 6 A ig i4 i3 6 3 9 A b Calculate power using the formula p Ri² p9Ω 92² 36 W p11Ω 112² 44 W p10Ω 102² 40 W p5Ω 56² 180 W p30Ω 303² 270 W p4Ω 45² 100 W p16Ω 165² 400 W p15Ω 154² 240 W c vg 190 V d Sum the power dissipated by the resistors pdiss 36 44 40 180 270 100 400 240 1310 W The power associated with the sources is pvoltsource 1004 400 W pcurrsource vgig 1909 1710 W Thus the total power dissipated is 1310 400 1710 W and the total power developed is 1710 W so the power balances P 227 a Start by calculating the voltage drops due to the currents i1 and i2 Then use KVL to calculate the voltage drop across and 35 Ω resistor and Ohms law to find the current in the 35 Ω resistor Finally KCL at each of the middle three nodes yields the currents in the two sources and the current in the middle 2 Ω resistor These calculations are summarized in the figure below p147top 14728 4116 W p147bottom 14721 3087 W Therefore the top source supplies 4116 W of power and the bottom source supplies 3087 W of power b Pdis 28²1 7²2 21²1 21²5 14²10 7²35 784 98 441 2205 1960 1715 7203 W Psup 4116 3087 7203 W Therefore Pdis Psup 7203 W 224 CHAPTER 2 Circuit Elements P 228 a Plot the v i characteristic From the plot R v i 130 50 10 0 8 Ω When it 0 vt 50 V therefore the ideal voltage source has a voltage of 50 V b When vt 0 it 50 8 625A Note that this result can also be obtained by extrapolating the v i characteristic to vt 0 P 229 a Plot the vi characteristic 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 225 From the plot R v i 180 100 16 0 5 Ω When it 0 vt 100 V therefore the ideal current source must have a current of 1005 20 A b We attach a 20 Ω resistor to the device model developed in part a Write a KCL equation at the top node 20 it i1 Write a KVL equation for the right loop in the direction of the two currents using Ohms law 5i1 20it 0 Combining the two equations and solving 520 it 20it 0 so 25it 100 thus it 4 A Now calculate the power dissipated by the resistor p20 Ω 20i2 t 2042 320 W P 230 a 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 226 CHAPTER 2 Circuit Elements b v 25V i 25 mA R v i 10 kΩ c 10000i1 2500is i1 025is 002 i1 is 125is is 16 mA d vsopen circuit 20 10310 103 200 V e The open circuit voltage can be found in the table of values or from the plot as the value of the voltage vs when the current is 0 Thus vsopen circuit 140 V from the table f Linear model cannot predict the nonlinear behavior of the practical current source P 231 a Begin by constructing a plot of voltage versus current b Since the plot is linear for 0 is 24 mA amd since R vi we can calculate R from the plotted values as follows R v i 24 18 0024 0 6 0024 250 Ω We can determine the value of the ideal voltage source by considering the value of vs when is 0 When there is no current there is no voltage drop across the resistor so all of the voltage drop at the output is due to the voltage source Thus the value of the voltage source must be 24 V The model valid for 0 is 24 mA is shown below 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 227 c The circuit is shown below Write a KVL equation in the clockwise direction starting below the voltage source Use Ohms law to express the voltage drop across the resistors in terms of the current i 24 V 250i 1000i 0 so 1250i 24 V Thus i 24 V 1250 Ω 192 mA d The circuit is shown below Write a KVL equation in the clockwise direction starting below the voltage source Use Ohms law to express the voltage drop across the resistors in terms of the current i 24 V 250i 0 so 250i 24 V Thus i 24 V 250 Ω 96 mA e The short circuit current can be found in the table of values or from the plot as the value of the current is when the voltage vs 0 Thus isc 48 mA from table f The plot of voltage versus current constructed in part a is not linear it is piecewise linear but not linear for all values of is Since the proposed circuit model is a linear model it cannot be used to predict the nonlinear behavior exhibited by the plotted data 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 P 232 Label unknown voltage and current vx vo 2ix 0 KVL vx 6ix Ohms law Therefore 6ix vo 2ix 0 so vo 4ix Thus ix vo 4 Also i1 vo 2 Ohms law 45 ix i1 KCL Substituting for the currents ix and i1 45 vo 4 vo 2 3vo 4 Thus vo 45 43 60 V The only two circuit elements that could supply power are the two sources so calculate the power for each source vx 6ix 6 vo 4 6604 90 V p45V 45vx 4590 4050 W pds 2ix i1 2vo4vo2 2604602 900 W Only the independent voltage source is supplying power so the total power supplied is 4050 W P 233 Label unknown current 20 450i 150i 0 KVL and Ohms law so 600i 20 i 3333 mA vx 150i 15000333 5 V Ohms law vo 300 vx 100 3005100 15 V Ohms law Calculate the power for all components p20V 20i 2000333 0667 W pds vo vx 100 155100 075 W p450 450i2 45000332 05 W p150 150i2 15000332 01667 W p300 vo2 300 152 300 075 W Thus the total power absorbed is pabs 05 01667 075 14167 W P 234 The circuit v1 4000001 40 V Ohms law 230 CHAPTER 2 Circuit Elements v1 2 2000io 6000io 8000io KVL Thus io v12 8000 402 8000 00025 25 mA Calculate the power for all components p10mA 001v1 00140 04 W pds v12io 40225 103 005 W p4k v2 1 4000 402 4000 04 W p2k 2000i2 o 200025 1032 00125 W p6k 6000i2 o 600025 1032 00375 W Therefore ptotal 04 005 04 00125 00375 0 Thus the power in the circuit balances P 235 a io 0 because no current can exist in a single conductor connecting two parts of a circuit b 18 12 6ig ig 1 A v 6ig 6V v2 3 A 10i1 5i2 so i1 2i1 3 A therefore i1 1 A c i2 2i1 2 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 P 236 a 50 20iσ 18iΔ 0 18iΔ 5iσ 40iσ 0 so 18iΔ 45iσ Therefore 50 20iσ 45iσ 0 so iσ 2 A 18iΔ 45iσ 90 so iΔ 5 A vo 40iσ 80 V b ig current out of the positive terminal of the 50 V source vd voltage drop across the 8iΔ source ig iΔ iσ 8iΔ 9iΔ iσ 47 A vd 80 20 60 V Σ Pgen 50 ig 20 iσ ig 5047 20247 4230 W Σ Pdiss 18iΔ2 5iσig iΔ 40iσ2 8iΔ vd 8iΔ20 1825 1047 5 440 4060 4020 4230 W Therefore Σ Pgen Σ Pdiss 4230 W P 237 40i2 540 510 0 i2 15625 mA v1 80i2 125 V 25i1 12520 0015625 0 i1 3125 mA vg 60i1 260i1 320i1 Therefore vg 1 V P 238 iE iB iC 0 iC β iB therefore iE 1 β iB i2 iB i1 Vo iE RE i1 iB R2 0 i1 R1 VCC i1 iB R2 0 or i1 VCC iB R2 R1 R2 232 CHAPTER 2 Circuit Elements Vo iERE iBR2 VCC iBR2 R1 R2 R2 0 Now replace iE by 1 βiB and solve for iB Thus iB VCCR2R1 R2 Vo 1 βRE R1R2R1 R2 P 239 Here is Equation 225 iB VCCR2R1 R2 V0 R1R2R1 R2 1 βRE VCCR2 R1 R2 1060000 100000 6V R1R2 R1 R2 4000060000 100000 24 kΩ iB 6 06 24000 50120 54 30000 018 mA iC βiB 49018 882 mA iE iC iB 882 018 9 mA v3d 0009120 108V vbd Vo v3d 168V i2 vbd R2 168 60000 28 µA i1 i2 iB 28 180 208 µA vab 40000208 106 832 V iCC iC i1 882 0208 9028 mA v13 882 103750 108 10 V v13 2305 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 233 P 240 a b P 241 Each radiator is modeled as a 48 Ω resistor Write a KVL equation for each of the three loops 240 48i1 0 i1 240 48 5 A 48i1 48i2 0 i2 i1 5 A 48i2 48i3 0 i3 i2 5 A Therefore the current through each radiator is 5 A and the power for each radiator is prad Ri2 4852 1200 W There are three radiators so the total power for this heating system is ptotal 3prad 31200 3600 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 234 CHAPTER 2 Circuit Elements P 242 Each radiator is modeled as a 48 Ω resistor Write a KVL equation for the left and right loops 240 48i1 0 i1 240 48 5 A 48i1 48i2 48i2 0 i2 i1 2 5 2 25 A The power for the center radiator is pcen 48i2 1 4852 1200 W The power for each of the radiators on the right is pright 48i2 2 48252 300 W Thus the total power for this heating system is ptotal pcen 2pright 1200 2300 1800 W The center radiator produces 1200 W just like the three radiators in Problem 241 But the other two radiators produce only 300 W each which is 14th of the power of the radiators in Problem 241 The total power of this configuration is 12 of the total power in Fig 241 P 243 Each radiator is modeled as a 48 Ω resistor Write a KVL equation for the left and right loops 240 48i1 48i2 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 235 48i2 48i3 0 i2 i3 Write a KCL equation at the top node i1 i2 i3 i1 i2 i2 2i2 Substituting into the first KVL equation gives 240 482i2 48i2 0 i2 240 348 167 A Solve for the currents i1 and i3 i3 i2 167 A i1 2i2 2167 333 A Calculate the power for each radiator using the current for each radiator pleft 48i2 1 483332 53333 W pmiddle pright 48i2 2 481672 13333 W Thus the total power for this heating system is ptotal pleft pmiddle pright 53333 13333 13333 800 W All radiators in this configuration have much less power than their counterparts in Fig 241 The total power for this configuration is only 222 of the total power for the heating system in Fig 241 P 244 Each radiator is modeled as a 48 Ω resistor Write a KVL equation for this loop 240 48i 48i 48i 0 i 240 348 167 A Calculate the power for each radiator prad 48i2 481672 13333 W Calculate the total power for this heating system ptotal 3prad 313333 400 W Each radiator has much less power than the radiators in Fig 241 and the total power of this configuration is just 19th of the total power in Fig 241 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 3 Simple Resistive Circuits Assessment Problems AP 31 Start from the right hand side of the circuit and make series and parallel combinations of the resistors until one equivalent resistor remains Begin by combining the 6 Ω resistor and the 10 Ω resistor in series 6 Ω 10 Ω 16 Ω Now combine this 16 Ω resistor in parallel with the 64 Ω resistor 16 Ω64 Ω 1664 16 64 1024 80 128 Ω This equivalent 128 Ω resistor is in series with the 72 Ω resistor 128 Ω 72 Ω 20 Ω Finally this equivalent 20 Ω resistor is in parallel with the 30 Ω resistor 20 Ω30 Ω 2030 20 30 600 50 12 Ω Thus the simplified circuit is as shown 31 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 32 CHAPTER 3 Simple Resistive Circuits a With the simplified circuit we can use Ohms law to find the voltage across both the current source and the 12 Ω equivalent resistor v 12 Ω5 A 60 V b Now that we know the value of the voltage drop across the current source we can use the formula p vi to find the power associated with the source p 60 V5 A 300 W Thus the source delivers 300 W of power to the circuit c We now can return to the original circuit shown in the first figure In this circuit v 60 V as calculated in part a This is also the voltage drop across the 30 Ω resistor so we can use Ohms law to calculate the current through this resistor iA 60 V 30 Ω 2 A Now write a KCL equation at the upper left node to find the current iB 5 A iA iB 0 so iB 5 A iA 5 A 2 A 3 A Next write a KVL equation around the outer loop of the circuit using Ohms law to express the voltage drop across the resistors in terms of the current through the resistors v 72iB 6iC 10iC 0 So 16iC v 72iB 60 V 723 384 V Thus iC 384 16 24 A Now that we have the current through the 10 Ω resistor we can use the formula p Ri2 to find the power p10 Ω 10242 576 W AP 32 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 33 a We can use voltage division to calculate the voltage vo across the 75 kΩ resistor vono load 75000 75000 25000200 V 150 V b When we have a load resistance of 150 kΩ then the voltage vo is across the parallel combination of the 75 kΩ resistor and the 150 kΩ resistor First calculate the equivalent resistance of the parallel combination 75 kΩ150 kΩ 75000150000 75000 150000 50000 Ω 50 kΩ Now use voltage division to find vo across this equivalent resistance vo 50000 50000 25000200 V 1333 V c If the load terminals are shortcircuited the 75 kΩ resistor is effectively removed from the circuit leaving only the voltage source and the 25 kΩ resistor We can calculate the current in the resistor using Ohms law i 200 V 25 kΩ 8 mA Now we can use the formula p Ri2 to find the power dissipated in the 25 kΩ resistor p25k 2500000082 16 W d The power dissipated in the 75 kΩ resistor will be maximum at no load since vo is maximum In part a we determined that the noload voltage is 150 V so be can use the formula p v2R to calculate the power p75kmax 1502 75000 03 W AP 33 a We will write a current division equation for the current throught the 80Ω resistor and use this equation to solve for R i80Ω R R 40 Ω 80 Ω20 A 4 A so 20R 4R 120 Thus 16R 480 and R 480 16 30 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 34 CHAPTER 3 Simple Resistive Circuits b With R 30 Ω we can calculate the current through R using current division and then use this current to find the power dissipated by R using the formula p Ri2 iR 40 80 40 80 3020 A 16 A so pR 30162 7680 W c Write a KVL equation around the outer loop to solve for the voltage v and then use the formula p vi to calculate the power delivered by the current source v 60 Ω20 A 30 Ω16 A 0 so v 1200 480 1680 V Thus psource 1680 V20 A 33600 W Thus the current source generates 33600 W of power AP 34 a First we need to determine the equivalent resistance to the right of the 40 Ω and 70 Ω resistors Req 20 Ω30 Ω50 Ω 10 Ω so 1 Req 1 20 Ω 1 30 Ω 1 60 Ω 1 10 Ω Thus Req 10 Ω Now we can use voltage division to find the voltage vo vo 40 40 10 7060 V 20 V b The current through the 40 Ω resistor can be found using Ohms law i40Ω vo 40 20 V 40 Ω 05 A This current flows from left to right through the 40 Ω resistor To use current division we need to find the equivalent resistance of the two parallel branches containing the 20 Ω resistor and the 50 Ω and 10 Ω resistors 20 Ω50 Ω 10 Ω 2060 20 60 15 Ω Now we use current division to find the current in the 30 Ω branch i30Ω 15 15 3005 A 016667 A 16667 mA 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 35 c We can find the power dissipated by the 50 Ω resistor if we can find the current in this resistor We can use current division to find this current from the current in the 40 Ω resistor but first we need to calculate the equivalent resistance of the 20 Ω branch and the 30 Ω branch 20 Ω30 Ω 2030 20 30 12 Ω Current division gives i50Ω 12 12 50 1005 A 008333 A Thus p50Ω 500083332 034722 W 34722 mW AP 35 a We can find the current i using Ohms law i 1 V 100 Ω 001 A 10 mA b Rm 50 Ω5555 Ω 5 Ω We can use the meter resistance to find the current using Ohms law imeas 1 V 100 Ω 5 Ω 0009524 9524 mA AP 36 a 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 36 CHAPTER 3 Simple Resistive Circuits Use voltage division to find the voltage v v 75000 75000 1500060 V 50 V b The meter resistance is a series combination of resistances Rm 149950 50 150000 Ω We can use voltage division to find v but first we must calculate the equivalent resistance of the parallel combination of the 75 kΩ resistor and the voltmeter 75000 Ω150000 Ω 75000150000 75000 150000 50 kΩ Thus vmeas 50000 50000 1500060 V 4615 V AP 37 a Using the condition for a balanced bridge the products of the opposite resistors must be equal Therefore 100Rx 1000150 so Rx 1000150 100 1500 Ω 15 kΩ b When the bridge is balanced there is no current flowing through the meter so the meter acts like an open circuit This places the following branches in parallel The branch with the voltage source the branch with the series combination R1 and R3 and the branch with the series combination of R2 and Rx We can find the current in the latter two branches using Ohms law iR1R3 5 V 100 Ω 150 Ω 20 mA iR2Rx 5 V 1000 1500 2 mA We can calculate the power dissipated by each resistor using the formula p Ri2 p100Ω 100 Ω002 A2 40 mW p150Ω 150 Ω002 A2 60 mW 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 37 p1000Ω 1000 Ω0002 A2 4 mW p1500Ω 1500 Ω0002 A2 6 mW Since none of the power dissipation values exceeds 250 mW the bridge can be left in the balanced state without exceeding the powerdissipating capacity of the resistors AP 38 Convert the three Yconnected resistors 20 Ω 10 Ω and 5 Ω to three connected resistors Ra Rb and Rc To assist you the figure below has both the Yconnected resistors and the connected resistors Ra 510 520 1020 20 175 Ω Rb 510 520 1020 10 35 Ω Rc 510 520 1020 5 70 Ω The circuit with these new connected resistors is shown below From this circuit we see that the 70 Ω resistor is parallel to the 28 Ω resistor 70 Ω28 Ω 7028 70 28 20 Ω Also the 175 Ω resistor is parallel to the 105 Ω resistor 175 Ω105 Ω 175105 175 105 15 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 38 CHAPTER 3 Simple Resistive Circuits Once the parallel combinations are made we can see that the equivalent 20 Ω resistor is in series with the equivalent 15 Ω resistor giving an equivalent resistance of 20 Ω 15 Ω 35 Ω Finally this equivalent 35 Ω resistor is in parallel with the other 35 Ω resistor 35 Ω35 Ω 3535 35 35 175 Ω Thus the resistance seen by the 2 A source is 175 Ω and the voltage can be calculated using Ohms law v 175 Ω2 A 35 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 39 Problems P 31 a From Ex 31 i1 4 A i2 8 A is 12 A at node b 12 4 8 0 at node d 12 4 8 0 b v1 4is 48 V v3 3i2 24 V v2 18i1 72 V v4 6i2 48 V loop abda 120 48 72 0 loop bcdb 72 24 48 0 loop abcda 120 48 24 48 0 P 32 a p4Ω i2 s4 1224 576 W p18Ω 4218 288 W p3Ω 823 192 W p6Ω 826 384 W b p120Vdelivered 120is 12012 1440 W c pdiss 576 288 192 384 1440 W P 33 a The 5 kΩ and 7 kΩ resistors are in series The simplified circuit is shown below b The 800 Ω and 1200 Ω resistors are in series as are the 300 Ω and 200 Ω resistors The simplified circuit is shown below 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 310 CHAPTER 3 Simple Resistive Circuits c The 35 Ω 15 Ω and 25 Ω resistors are in series as are the 10 Ω and 40 Ω resistors The simplified circuit is shown below d The 50 Ω and 90 Ω resistors are in series as are the 80 Ω and 70 Ω resistors The simplified circuit is shown below P 34 a The 36 Ω and 18 Ω resistors are in parallel The simplified circuit is shown below b The 200 Ω and 120 Ω resistors are in parallel as are the 210 Ω and 280 Ω resistors The simplified circuit is shown below c The 100 kΩ 150 kΩ and 60 kΩ resistors are in parallel as are the 75 kΩ and 50 kΩ resistors The simplified circuit is shown below 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 311 d The 750 Ω and 500 Ω resistors are in parallel as are the 15 kΩ and 3 kΩ resistors The simplified circuit is shown below P 35 Always work from the side of the circuit furthest from the source Remember that the current in all seriesconnected circuits is the same and that the voltage drop across all parallelconnected resistors is the same a Circuit in Fig P33a Req 7000 50006000 8000 120006000 8000 4000 8000 12 kΩ Circuit in Fig P33b Req 500800 1200 300 200 5002000 300 200 400 300 200 900 Ω Circuit in Fig P33c Req 35 15 2510 40 7550 30 Ω Circuit in Fig P33d Req 70 80100 50 90300 150100 50 90300 60 50 90300 200300 120 Ω b Note that in every case the power delivered by the source must equal the power absorbed by the equivalent resistance in the circuit For the circuit in Fig P33a P V 2 s Req 182 12000 0027 27 mW 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 312 CHAPTER 3 Simple Resistive Circuits For the circuit in Fig P33b P V 2 s Req 272 900 081 810 mW For the circuit in Fig P33c P V 2 s Req 902 30 270 W For the circuit in Fig P33d P I2 sReq 0032120 0108 108 mW P 36 Always work from the side of the circuit furthest from the source Remember that the current in all seriesconnected circuits is the same and that the voltage drop across all parallelconnected resistors is the same a Circuit in Fig P34a Req 3618 24 12 24 36 Ω Circuit in Fig P34b Req 200120210280 180 200120120 180 200120300 60 Ω Circuit in Fig P34c Req 75 k50 k 100 k150 k60 k 90 k 30 k 30 k 90 k 150 kΩ Circuit in Fig P34d Req 600 900750500 15003000 2000 1500750500 1000 2000 250 1000 2000 3250 325 kΩ b Note that in every case the power delivered by the source must equal the power absorbed by the equivalent resistance in the circuit For the circuit in Fig P34a P V 2 s Req 182 36 9 W For the circuit in Fig P34b P I2 sReq 003260 0054 54 mW For the circuit in Fig P34c P V 2 s Req 602 150000 0024 24 mW For the circuit in Fig P34d P V 2 s Req 652 3250 13 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 313 P 37 a Circuit in Fig P37a Req 1560 3045 2050 25 10 12 18 2050 25 10 5050 25 10 25 25 10 60 Ω Circuit in Fig P37b begin by simplifying the 75 Ω resistor and all resistors to its right 18 1260 3075 3060 3075 20 3075 5075 30 Ω Now simplify the remainder of the circuit Req 30 2050 206040 5050 1540 25 1540 4040 20 Ω Circuit in Fig P37c begin by simplifying the left and right sides of the circuit Rleft 1800 12002000 300 30002000 300 1200 300 1500 Ω Rright 500 25001000 750 30001000 750 750 750 1500 Ω Now find the equivalent resistance seen by the source Req RleftRright 250 3000 15001500 250 3000 750 250 3000 4000 4 kΩ Circuit in Fig P37d Req 750 2501000 100150 600500 300 10001000 100750500 300 500 100300 300 600600 300 Ω b Note that in every case the power delivered by the source must equal the power absorbed by the equivalent resistance in the circuit For the circuit in Fig P37a P V 2 s Req 302 60 15 W For the circuit in Fig P37b P I2 sReq 008220 0128 128 mW For the circuit in Fig P37c P V 2 s Req 202 4000 01 100 mW For the circuit in Fig P37d P I2 sReq 0052300 075 750 mW 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 314 CHAPTER 3 Simple Resistive Circuits P 38 a Rab 24 9060 12 24 36 12 72 Ω b Rab 4 k 6 k 2 k8 k 52 k 12 k8 k 52 k 48 k 52 k 10 kΩ c Rab 1200720320 480 1200720800 288 Ω P 39 Write an expression for the resistors in series and parallel from the right side of the circuit to the left Then simplify the resulting expression from left to right to find the equivalent resistance a Rab 26 1018 636 3618 636 12 636 1836 12 Ω b Rab 12 18101520 1630 4 14 30101520 1630 4 14 4 1630 4 14 2030 4 14 12 4 14 30 Ω c Rab 5001500750 2502000 1000 250 2502000 1000 5002000 1000 400 1000 1400 Ω d Note that the wire on the far right of the circuit effectively removes the 60 Ω resistor Rab 30 1816 2840 2024 25 1050 4816 2840 2024 25 1050 12 2840 2024 25 1050 4040 2024 25 1050 20 2024 25 1050 4024 25 1050 15 25 1050 5050 25 Ω P 310 a R R 2R b R R R R nR c R R 2R 3000 so R 1500 15 kΩ This is a resistor from Appendix H d nR 4000 so if n 4 R 1 kΩ This is a resistor from Appendix H P 311 a Req RR R2 2R R 2 b Req RRR R n Rs R R n 1 R2n 1 R Rn 1 R2 nR R n c R 2 5000 so R 10 kΩ This is a resistor from Appendix H 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 d Rn 4000 so R 4000n If n 3 r 40003 12 kΩ This is a resistor from Appendix H So put three 12k resistors in parallel to get 4kΩ P 312 a vo 16033004700 3300 66 V b i 1608000 20 mA PR1 400 10647 103 188 W PR2 400 10633 103 132 W c Since R1 and R2 carry the same current and R1 R2 to satisfy the voltage requirement first pick R1 to meet the 05 W specification iR1 160 66R1 Therefore 94R12 R1 05 Thus R1 94205 or R1 17672 Ω Now use the voltage specification R2R2 17672160 66 Thus R2 12408 Ω P 313 4 20R2R2 40 so R2 10 Ω 3 20Re40 Re so Re 12017 Ω Thus 12017 10RL10 RL so RL 24 Ω P 314 a vo 40R2R1 R2 8 so R1 4R2 Let Re R2RL R2RLR2 RL vo 40ReR1 Re 75 so R1 433 Re Then 4R2 433Re 4333600R23600 R2 Thus R2 300 Ω and R1 4300 1200 Ω 316 CHAPTER 3 Simple Resistive Circuits b The resistor that must dissipate the most power is R1 as it has the largest resistance and carries the same current as the parallel combination of R2 and the load resistor The power dissipated in R1 will be maximum when the voltage across R1 is maximum This will occur when the voltage divider has a resistive load Thus vR1 40 75 325 V pR1 3252 1200 8802 m W Thus the minimum power rating for all resistors should be 1 W P 315 Refer to the solution to Problem 316 The voltage divider will reach the maximum power it can safely dissipate when the power dissipated in R1 equals 1 W Thus v2 R1 1200 1 so vR1 3464 V vo 40 3464 536 V So 40Re 1200 Re 536 and Re 18568 Ω Thus 300RL 300 RL 18568 and RL 48726 Ω The minimum value for RL from Appendix H is 560 Ω P 316 Req 106 58 12 106 520 106 4 5 Ω v10A v10Ω 10 A5 Ω 50 V Using voltage division v5Ω 58 12 6 58 1250 4 6 450 20 V Thus p5Ω v2 5Ω 5 202 5 80 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 317 P 317 a Req 10 2012 9010 3015 10 Ω v24A 1024 24 V vo v20Ω 20 10 2024 16 V v90Ω 9010 6 901024 9 1524 144 V io 144 90 016 A b p6Ω v24A v90Ω2 6 24 1442 6 1536 W c p24A 2424 576 W Thus the power developed by the current source is 576 W P 318 Begin by using KCL at the top node to relate the branch currents to the current supplied by the source Then use the relationships among the branch currents to express every term in the KCL equation using just i2 005 i1 i2 i3 i4 06i2 i2 2i2 4i1 06i2 i2 2i2 406i2 6i2 Therefore i2 0056 000833 833 mA Find the remaining currents using the value of i2 i1 06i2 06000833 0005 5 mA i3 2i2 2000833 001667 1667 mA i4 4i1 40005 002 20 mA Since the resistors are in parallel the same voltage 25 V appears across each of them We know the current and the voltage for every resistor so we can use Ohms law to calculate the values of the resistors R1 25i1 250005 5000 5 kΩ 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 318 CHAPTER 3 Simple Resistive Circuits R2 25i2 25000833 3000 3 kΩ R3 25i3 25001667 1500 15 kΩ R4 25i4 25002 1250 125 kΩ The resulting circuit is shown below P 319 242 R1 R2 R3 80 Therefore R1 R2 R3 72 Ω R1 R224 R1 R2 R3 12 Therefore 2R1 R2 R1 R2 R3 Thus R1 R2 R3 2R3 72 R3 36 Ω R224 R1 R2 R3 5 48R2 R1 R2 36 72 Thus R2 15 Ω R1 72 R2 R3 21 Ω P 320 a 20 kΩ 40 kΩ 60 kΩ 30 kΩ60 kΩ 20 kΩ vo1 20000 10000 20000180 120 V vo 40000 60000vo1 80 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 b Circuit image with components 10kΩ 30kΩ 20kΩ 40kΩ 180V and current labelled i i 18040000 45 mA 30000i 135 V vo 4000060000 135 90 V c It removes the loading effect of the second voltage divider on the first voltage divider Observe that the open circuit voltage of the first divider is vo1 3000040000 180 135 V Now note this is the input voltage to the second voltage divider when the currentcontrolled voltage source is used P 321 a At no load vo kvs R2R1 R2 vs At full load vo αvs ReR1 Re vs where Re RoR2Ro R2 Therefore k R2R1 R2 and R1 1 kk R2 α ReR1 Re and R1 1 αα Re Thus 1 αα R2RoRo R2 1 kk R2 Solving for R2 yields R2 k αα1 k Ro Also R1 1 kk R2 R1 k ααk Ro b R1 005068 Ro 25 kΩ R2 005012 Ro 14167 kΩ 320 CHAPTER 3 Simple Resistive Circuits c Maximum dissipation in R2 occurs at no load therefore PR2max 600852 14167 1836 mW Maximum dissipation in R1 occurs at full load PR1max 60 080602 2500 5760 mW d PR1 602 2500 144 W 1440 mW PR2 02 14167 0 W P 322 a Let vo be the voltage across the parallel branches positive at the upper terminal then ig voG1 voG2 voGN voG1 G2 GN It follows that vo ig G1 G2 GN The current in the kth branch is ik voGk Thus ik igGk G1 G2 GN b i5 4002 2 02 0125 01 005 0025 32 A P 323 a The equivalent resistance of the 6 kΩ resistor and the resistors to its right is 6 k5 k 7 k 6 k12 k 4 kΩ 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 321 Using voltage division v6k 4000 8000 400018 6 V b v5k 5000 5000 70006 25 V P 324 a The equivalent resistance of the 100 Ω resistor and the resistors to its right is 10080 70 100150 60 Ω Using current division i50 50 90 60300 50 90 60 003 120 200003 0018 18 mA b v70 80 70100 80 70 0018 60 1500018 00072 72 mA P 325 a The equivalent resistance of the circuit to the right of and including the 50 Ω resistor is 6015 4530 2050 25 Ω Thus by voltage division v25 25 25 25 1030 125 V b The current in the 25 Ω resistor can be found from its voltage using Ohms law i25 125 25 05 A c The current in the 25 Ω resistor divides between two branches one containing 50 Ω and one containing 4530 1560 20 50 Ω Using current division i50 5050 50 i25 25 5005 025 A d The voltage drop across the 50 Ω resistor can be found using Ohms law v50 50i50 50025 125 V e The voltage v50 divides across the equivalent resistance 4530 Ω the equivalent resistance 1560 Ω and the 20 Ω resistor Using voltage division v60 v1560 1560 1560 3045 20125 12 12 18 20125 3 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 322 CHAPTER 3 Simple Resistive Circuits P 326 a The equivalent resistance to the right of the 36 Ω resistor is 6 1826 10 18 Ω By current division i36 3618 36 045 015 150 mA b Using Ohms law v36 36i36 36015 54 V c Before using voltage division find the equivalent resistance of the 18 Ω resistor and the resistors to its right 1826 10 12 Ω Now use voltage division v18 12 12 654 36 V d v10 10 10 2636 1 V P 327 a Begin by finding the equivalent resistance of the 30 Ω resistor and all resistors to its right 12 18101520 1630 12 Ω Now use voltage division to find the voltage across the 4 Ω resistor v4 4 4 12 146 08 V b Use Ohms law to find the current in the 4 Ω resistor i4 v44 084 02 A c Begin by finding the equivalent resistance of all resistors to the right of the 30 Ω resistor 12 18101520 16 20 Ω Now use current division i16 3020 20 02 012 120 mA d Note that the current in the 16 Ω resistor divides among four branches 20 Ω 15 Ω 10 Ω and 12 18 Ω i10 20151012 18 10 012 0048 48 mA e Use Ohms law to find the voltage across the 10 Ω resistor v10 10i10 100048 048 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 323 f v18 18 12 18048 0288 288 mV P 328 a v6k 6 6 218 135 V v3k 3 3 918 45 V vx v6k v3k 135 45 9 V b v6k 6 8Vs 075Vs v3k 3 12Vs 025Vs vx 075Vs 025Vs 05Vs P 329 Use current division to find the current in the branch containing the 10 k and 15 k resistors from bottom to top i10k15k 10 k 15 k3 k 12 k 10 k 15 k 18 675 mA Use Ohms law to find the voltage drop across the 15 k resistor positive at the top v15k 675 m15 k 10125 V Find the current in the branch containing the 3 k and 12 k resistors from bottom to top i10k15k 10 k 15 k3 k 12 k 3 k 12 k 18 1125 mA Use Ohms law to find the voltage drop across the 12 k resistor positive at the top v12k 12 k1125 m 135 V vo v15k v12k 10125 135 3375 V P 330 The equivalent resistance of the circuit to the right of the 90 Ω resistor is Req 15075 4030 60 9090 45 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 324 CHAPTER 3 Simple Resistive Circuits Use voltage division to find the voltage drop between the top and bottom nodes vReq 45 45 903 1 V Use voltage division again to find v1 from vReq v1 15075 15075 401 50 901 5 9 V Use voltage division one more time to find v2 from vReq v2 30 30 601 1 3 V P 331 Find the equivalent resistance of all the resistors except the 2 Ω 5 Ω20 Ω 4 Ω 4 Ω 6 Ω 10 Ω 1015 12 13 8 Ω Req Use Ohms law to find the current ig ig 125 2 Req 125 2 8 125 A Use current division to find the current in the 6 Ω resistor i6Ω 8 6 4125 10 A Use current division again to find io io 520 20 i6Ω 520 20 10 2 A P 332 Use current division to find the current in the 8 Ω resistor Begin by finding the equivalent resistance of the 8 Ω resistor and all resistors to its right Req 2080 430 8 20 Ω i8 60Req Req 025 6020 20 025 01875 1875 mA Use current division to find i1 from i8 i1 304 8020 30 i8 3020 30 01875 0075 75 mA 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 325 Use current division to find i4Ω from i8 i4Ω 304 8020 4 8020 i8 3020 20 01875 01125 1125 mA Finally use current division to find i2 from i4Ω i2 8020 20 i4Ω 8020 20 01125 009 90 mA P 333 The current in the shunt resistor at fullscale deflection is iA ifullscale 3 103 A The voltage across RA at fullscale deflection is always 150 mV therefore RA 150 103 ifullscale 3 103 150 1000ifullscale 3 a RA 150 5000 3 30018 mΩ b Let Rm be the equivalent ammeter resistance Rm 015 5 003 30 mΩ c RA 150 100 3 1546 Ω d Rm 015 01 15 Ω P 334 Original meter Re 50 103 5 001 Ω Modified meter Re 002001 003 000667 Ω Ifs000667 50 103 Ifs 75 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 326 CHAPTER 3 Simple Resistive Circuits P 335 At full scale the voltage across the shunt resistor will be 200 mV therefore the power dissipated will be PA 200 1032 RA Therefore RA 200 1032 10 40 mΩ Otherwise the power dissipated in RA will exceed its power rating of 1 W When RA 40 mΩ the shunt current will be iA 200 103 40 103 5 A The measured current will be imeas 5 0002 5002 A Fullscale reading for practical purposes is 5 A P 336 a The model of the ammeter is an ideal ammeter in parallel with a resistor whose resistance is given by Rm 100 mV 2 mA 50 Ω We can calculate the current through the real meter using current division im 2512 50 2512imeas 25 625imeas 1 25imeas b At full scale imeas 5 A and im 2 mA so 5 0002 4998 mA flows throught the resistor RA RA 100 mV 4998 mA 100 4998 Ω im 1004998 50 1004998 imeas 1 2500imeas c Yes P 337 For all fullscale readings the total resistance is RV Rmovement fullscale reading 103 We can calculate the resistance of the movement as follows Rmovement 20 mV 1 mA 20 Ω Therefore RV 1000 fullscale reading 20 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 a RV 100050 20 49980 Ω b RV 10005 20 4980 Ω c RV 1000025 20 230 Ω d RV 10000025 20 5 Ω P 338 a vmeas 50 10315454980 20 05612 V b vtrue 50 1031545 05625 V error 0561205625 1 100 0224 P 339 The measured value is 60201 1505618 Ω ig 501505618 10 1995526 A imeas 60801 1996 1494768 A The true value is 6020 15 Ω ig 5015 10 2 A itrue 6080 2 15 A error 149476815 1 100 034878 035 P 340 Begin by using current division to find the actual value of the current io itrue 1515 45 50 mA 125 mA imeas 1515 45 01 50 mA 124792 mA error 124792125 1 100 0166389 017 P 341 a Circuit image with resistors 20kΩ 80kΩ 082kΩ 02kΩ voltage source 06 V and 75 V labeled with currents i1 iB and directions 20 103 i1 80 103 i1 iB 75 80 103 i1 iB 06 40iB02 103 100i1 80iB 75 103 80i1 88iB 06 103 Calculator solution yields iB 225 μA b With the insertion of the ammeter the equations become 100i1 80iB 75 103 no change 80 103 i1 iB 103 iB 06 40iB200 80i1 89iB 06 103 Calculator solution yields iB 216 μA c error 216225 1 100 4 P 342 a Since the unknown voltage is greater than either voltmeters maximum reading the only possible way to use the voltmeters would be to connect them in series b Rm1 300900 270 kΩ Rm2 1501200 180 kΩ Rm1 Rm2 450 kΩ i1 max 300270 103 111 mA i2 max 150180 103 0833 mA imax 0833 mA since meters are in series vmax 0833 103270 180103 375 V Thus the meters can be used to measure the voltage c im 320450 103 0711 mA vm1 0711270 192 V vm2 0711180 128 V P 343 The current in the seriesconnected voltmeters is im 2052270000 1368180000 076 mA v50 kΩ 076 10350000 38 V Vpower supply 2052 1368 38 380 V P 344 Rmeter Rm Rmovement 500 V1 mA 1000 kΩ vmeas 50 kΩ 250 kΩ 1000 kΩ10 mA 40 kΩ10 mA 400 V vtrue 50 kΩ 250 kΩ10 mA 4167 kΩ10 mA 41667 V error 40041667 1 100 4 P 345 a vmeter 180 V b Rmeter 100200 20 kΩ 20 70 15555556 kΩ vmeter 18035555556 15555556 7875 V c 20 20 10 kΩ vmeter 18080 10 225 V d vmeter a 180 V vmeter b vmeter c 10126 V No because of the loading effect P 346 a R1 1002103 50 kΩ R2 102103 5 kΩ R3 12103 500 Ω b Let ia actual current in the movement id design current in the movement Then error iaid 1 100 For the 100 V scale ia 10050000 25 10050025 id 10050000 iaid 5000050025 09995 error 09995 1100 005 For the 10 V scale iaid 50005025 0995 error 0995 10100 04975 For the 1 V scale iaid 500525 09524 error 09524 10100 476 P 347 From the problem statement we have 50 Vs1010 Rs 1 Vs in mV Rs in MΩ 4875 Vs66 Rs 2 a From Eq 1 10 Rs 02Vs Rs 02Vs 10 Substituting into Eq 2 yields 4875 6Vs02Vs 4 or Vs 52 mV b From Eq 1 50 52010 Rs or 50Rs 20 So Rs 400 kΩ P 348 a Rmovement 50 Ω R1 Rmovement 301 103 30 kΩ R1 29950 Ω R2 R1 Rmovement 1501 103 150 kΩ R2 120 kΩ R3 R2 R1 Rmovement 3001 103 300 kΩ R3 150 kΩ b v1 096 m150 k 144 V imove 144 120 2995 005 096 mA i1 144 750 k 0192 mA i2 imove i1 096 m 0192 m 1152 mA vmeas vx 144 150i2 3168 V c v1 150 V i2 1 m 020 m 120 mA i1 150750000 020 mA vmeas vx 150 150 k120 m 330 V P 349 a Rmeter 300 kΩ 600 kΩ200 kΩ 450 kΩ 450360 200 kΩ Vmeter 200 240600 500 V b What is the percent error in the measured voltage True value 360 400600 540 V error 500 540 1 100 741 P 350 Since the bridge is balanced we can remove the detector without disturbing the voltages and currents in the circuit It follows that i1 igR2 Rx R1 R2 R3 Rx igR2 Rx ΣR i2 igR1 R3 R1 R2 R3 Rx igR1 R3 ΣR v3 R3i1 vx i2Rx R3igR2 Rx ΣR RxigR1 R3 ΣR R3R2 Rx RxR1 R3 From which Rx R2R3 R1 P 351 a The condition for a balanced bridge is that the product of the opposite resistors must be equal 500Rx 1000750 so Rx 1000750 500 1500 Ω Problems 333 b The source current is the sum of the two branch currents Each branch current can be determined using Ohms law since the resistors in each branch are in series and the voltage drop across each branch is 24 V is 24 V 500 Ω 750 Ω 24 V 1000 Ω 1500 Ω 288 mA c We can use Ohms law to find the current in each branch ileft 24 500 750 192 mA iright 24 1000 1500 96 mA Now we can use the formula p Ri2 to find the power dissipated by each resistor p500 500001922 18432 mW p750 750001922 27618 mW p1000 1000000962 9216 mW p1500 1500000962 13824 mW Thus the 750 Ω resistor absorbs the most power it absorbs 27648 mW of power d From the analysis in part c the 1000 Ω resistor absorbs the least power it absorbs 9216 mW of power P 352 Note the bridge structure is balanced that is 15 5 3 25 hence there is no current in the 5 kΩ resistor It follows that the equivalent resistance of the circuit is Req 750 15000 300025000 5000 750 11250 12 kΩ The source current is 19212000 16 mA The current down through the branch containing the 15 kΩ and 3 kΩ resistors is i3k 11250 180000016 10 mA p3k 30000012 03 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 334 CHAPTER 3 Simple Resistive Circuits P 353 Redraw the circuit replacing the detector branch with a short circuit 6 kΩ30 kΩ 5 kΩ 12 kΩ20 kΩ 75 kΩ is 75 12500 6 mA v1 00065000 30 V v2 00067500 45 V i1 30 6000 5 mA i2 45 12000 375 mA id i1 i2 125 mA P 354 In order that all four decades 1 10 100 1000 that are used to set R3 contribute to the balance of the bridge the ratio R2R1 should be set to 0001 P 355 Use the figure below to transform the to an equivalent Y R1 4025 40 25 375 9756 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 335 R2 25375 40 25 375 91463 Ω R3 40375 40 25 375 14634 Ω Replace the with its equivalent Y in the circuit to get the figure below Find the equivalent resistance to the right of the 5 Ω resistor 100 9756125 91463 14634 75 Ω The equivalent resistance seen by the source is thus 5 75 80 Ω Use Ohms law to find the current provided by the source is 40 80 05 A Thus the power associated with the source is Ps 4005 20 W P 356 Use the figure below to transform the Y to an equivalent Ra 25100 2540 40100 25 7500 25 300 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 336 CHAPTER 3 Simple Resistive Circuits Rb 25100 2540 40100 40 7500 40 1875 Ω Rc 25100 2540 40100 100 7500 100 75 Ω Replace the Y with its equivalent in the circuit to get the figure below Find the equivalent resistance to the right of the 5 Ω resistor 3001251875 37575 75 Ω The equivalent resistance seen by the source is thus 5 75 80 Ω Use Ohms law to find the current provided by the source is 40 80 05 A Thus the power associated with the source is Ps 4005 20 W P 357 Use the figure below to transform the Y to an equivalent Ra 25125 25375 375125 375 8750 375 23333 Ω Rb 25125 25375 375125 25 8750 25 350 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 337 Rc 25125 25375 375125 125 8750 125 70 Ω Replace the Y with its equivalent in the circuit to get the figure below Find the equivalent resistance to the right of the 5 Ω resistor 35010023333 4070 75 Ω The equivalent resistance seen by the source is thus 5 75 80 Ω Use Ohms law to find the current provided by the source is 40 80 05 A Thus the power associated with the source is Ps 4005 20 W P 358 a Use the figure below to transform the Y to an equivalent Ra 2530 2550 3050 30 3500 30 11667 Ω Rb 2530 2550 3050 50 3500 50 70 Ω Rc 2530 2550 3050 25 3500 25 140 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 338 CHAPTER 3 Simple Resistive Circuits Replace the Y with its equivalent in the circuit to get the figure below Find the equivalent resistance to the right of the 13 Ω and 7 Ω resistors 705011667 20140 30 Ω Thus the equivalent resistance seen from the terminals ab is Rab 13 30 7 50 Ω b Use the figure below to transform the to an equivalent Y R1 5020 50 20 30 10 Ω R2 5030 50 20 30 15 Ω R3 2030 50 20 30 6 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 339 Replace the with its equivalent Y in the circuit to get the figure below Find the equivalent resistance to the right of the 13 Ω and 7 Ω resistors 50 1025 15 6 30 Ω Thus the equivalent resistance seen from the terminals ab is Rab 13 30 7 50 Ω c Convert the delta connection R1R2R3 to its equivalent wye Convert the wye connection R1R3R4 to its equivalent delta P 359 Begin by transforming the connected resistors 10 Ω 40 Ω 50 Ω to Yconnected resistors Both the Yconnected and connected resistors are shown below to assist in using Eqs 344 346 Now use Eqs 344 346 to calculate the values of the Yconnected resistors R1 4010 10 40 50 4 Ω R2 1050 10 40 50 5 Ω R3 4050 10 40 50 20 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 340 CHAPTER 3 Simple Resistive Circuits The transformed circuit is shown below The equivalent resistance seen by the 24 V source can be calculated by making series and parallel combinations of the resistors to the right of the 24 V source Req 15 51 4 20 205 20 4 20 24 Ω Therefore the current i in the 24 V source is given by i 24 V 24 Ω 1 A Use current division to calculate the currents i1 and i2 Note that the current i1 flows in the branch containing the 15 Ω and 5 Ω series connected resistors while the current i2 flows in the parallel branch that contains the series connection of the 1 Ω and 4 Ω resistors i1 4 15 5i 4 201 A 02 A and i2 1 A 02 A 08 A Now use KVL and Ohms law to calculate v1 Note that v1 is the sum of the voltage drop across the 4 Ω resistor 4i2 and the voltage drop across the 20 Ω resistor 20i v1 4i2 20i 408 A 201 A 32 20 232 V Finally use KVL and Ohms law to calculate v2 Note that v2 is the sum of the voltage drop across the 5 Ω resistor 5i1 and the voltage drop across the 20 Ω resistor 20i v2 5i1 20i 502 A 201 A 1 20 21 V P 360 a Convert the upper delta to a wye R1 5050 200 125 Ω R2 50100 200 25 Ω R3 10050 200 25 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 341 Convert the lower delta to a wye R4 6080 200 24 Ω R5 6060 200 18 Ω R6 8060 200 24 Ω Now redraw the circuit using the wye equivalents Rab 15 125 12080 200 18 14 48 18 80 Ω b When vab 400 V ig 400 80 5 A i31 48 805 3 A p31Ω 3132 279 W P 361 a After the 20 Ω100 Ω50 Ω wye is replaced by its equivalent delta the circuit reduces to 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 342 CHAPTER 3 Simple Resistive Circuits Now the circuit can be reduced to i 96 4001000 240 mA io 400 1000240 96 mA b i1 80 400240 48 mA c Now that io and i1 are known return to the original circuit v2 500048 6000096 60 V i2 v2 100 60 100 600 mA d vg v2 2006 0048 60 1296 7296 V pg vg1 7296 W Thus the current source delivers 7296 W P 362 8 12 20 Ω 2060 15 Ω 15 20 35 Ω 35140 28 Ω 28 22 50 Ω 5075 30 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 343 30 10 40 Ω ig 24040 6 A io 650125 24 A i140Ω 6 2435175 072 A p140Ω 0722140 72576 W P 363 a Replace the 6012020 Ω delta with a wye equivalent to get is 750 5 24 3614 6 12 43 750 75 10 A i1 24 3614 6 24 36 10 15 6010 25 A b io 10 25 75 A v 36i1 6io 3625 675 45 V c i2 io v 60 75 45 60 825 A d Psupplied 75010 7500 W P 364 Ga 1 Ra R1 R1R2 R2R3 R3R1 1G1 1G11G2 1G21G3 1G31G1 1G1G1G2G3 G1 G2 G3 G2G3 G1 G2 G3 Similar manipulations generate the expressions for Gb and Gc P 365 a Subtracting Eq 342 from Eq 343 gives R1 R2 RcRb RcRaRa Rb Rc Adding this expression to Eq 341 and solving for R1 gives R1 RcRbRa Rb Rc To find R2 subtract Eq 343 from Eq 341 and add this result to Eq 342 To find R3 subtract Eq 341 from Eq 342 and add this result to Eq 343 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 344 CHAPTER 3 Simple Resistive Circuits b Using the hint Eq 343 becomes R1 R3 RbR2R3Rb R2R1Rb R2R1Rb Rb R2R3Rb RbR1 R3R2 R1R2 R2R3 R3R1 Solving for Rb gives Rb R1R2 R2R3 R3R1R2 To find Ra First use Eqs 344346 to obtain the ratios R1R3 RcRa or Rc R1R3Ra and R1R2 RbRa or Rb R1R2Ra Now use these relationships to eliminate Rb and Rc from Eq 342 To find Rc use Eqs 344346 to obtain the ratios Rb R3R2Rc and Ra R3R1Rc Now use the relationships to eliminate Rb and Ra from Eq 341 P 366 a Rab 2R1 R22R1 RL 2R1 R2 RL RL Therefore 2R1 RL R22R1 RL 2R1 R2 RL 0 Thus R2 L 4R2 1 4R1R2 4R1R1 R2 When Rab RL the current into terminal a of the attenuator will be viRL Using current division the current in the RL branch will be vi RL R2 2R1 R2 RL Therefore vo vi RL R2 2R1 R2 RL RL and vo vi R2 2R1 R2 RL b 3002 4R1 R2R1 22500 R2 1 R1R2 vo vi 05 R2 2R1 R2 300 R1 05R2 150 R2 05R2 R1 150 R2 2R1 300 22500 R2 1 R12R1 300 3R2 1 300R1 R2 1 100R1 7500 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Solving R1 50 Ω R2 250 300 400 Ω c From Appendix H choose R1 47 Ω and R2 390 Ω For these values Rab RL so the equations given in part a cannot be used Instead Rab 2R1 R22R1 RL 247 390247 300 94 390394 290 Ω error 290 300 1 100 333 Now calculate the ratio of the output voltage to the input voltage Begin by finding the current through the top left R1 resistor called ia ia vi Rab Now use current division to find the current through the RL resistor called iL iL R2 R2 2R1 RL ia Therefore the output voltage vo is equal to RLiL vo R2RLvi RabR2 2R1 RL Thus vo vi R2RL RabR2 2R1 RL 390300 290390 247 300 05146 error 05146 05 1 100 292 P 367 a After making the YtoΔ transformation the circuit reduces to Combining the parallel resistors reduces the circuit to Now note 075R 3RRL 3R RL 225R2 375RRL 3R RL Therefore Rab 3R 225R2 375RRL 3R RL 3R 225R2 375RRL 3R RL 3R3R 5RL 15R 9RL If R RL we have Rab 3RL8RL 24RL RL Therefore Rab RL b When R RL the circuit reduces to io ii 3RL 45 RL 115 ii 115 vi RL vo 075RL io 12 vi Therefore vo vi 05 P 368 a 353R RL 3R RL 105R 1050 3R 300 75R 1350 R 180 Ω R2 2 180 3002 3 1802 3002 4500 Ω Problems 347 b vo vi 35 42 35 12 V io 12 300 40 mA i1 42 12 4500 30 4500 667 mA ig 42 300 140 mA i2 140 667 13333 mA i3 40 667 3333 mA i4 13333 3333 100 mA p4500 top 667 10324500 02 W p180 left 13333 1032180 32 W p180 right 3333 1032180 02 W p180 vertical 100 1032180 048 W p300 load 40 1032300 048 W The 180 Ω resistor carrying i2 c p180 left 32 W d Two resistors dissipate minimum power the 4500 Ω resistor and the 180 Ω resistor carrying i3 e They both dissipate 02 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 P 369 a va vin R4 Ro R4 ΔR vb R3 R2 R3 vin vo va vb R4 vin Ro R4 ΔR R3 R2 R3 vin When the bridge is balanced R4 Ro R4 vin R3 R2 R3 vin R4 Ro R4 R3 R2 R3 Thus vo R4 vin Ro R4 ΔR R4 vin Ro R4 R4 vin 1 Ro R4 ΔR 1 Ro R4 R4 vin ΔR Ro R4 ΔR Ro R4 ΔR R4 vin Ro R42 since ΔR R4 b ΔR 003 Ro Ro R2 R4 R3 10005000 500 10000 Ω ΔR 003104 300 Ω vo 30050006 150002 40 mV c vo ΔR R4 vin Ro R4 ΔRRo R4 30050006 1530015000 392157 mV P 370 a approx value ΔR R4 vin Ro R42 true value ΔR R4 vin Ro R4 ΔRRo R4 approx value true value Ro R4 ΔR Ro R4 error Ro R4 Ro R4 ΔR 1 100 ΔR Ro R4 100 Note that in the above expression we take the ratio of the true value to the approximate value because both values are negative But Ro R2 R4 R3 error R3 ΔR R4 R2 R3 b error 500300 50001500 100 2 P 371 ΔRR3100 R2 R3 R4 05 ΔR500100 15005000 05 ΔR 75 Ω change 75 10000 100 075 P 372 a Using the equation for voltage division Vy β Ry β Ry 1 β Ry VS β Ry Ry VS β VS b Since β represents the touch point with respect to the bottom of the screen 1 β represents the location of the touch point with respect to the top of the screen Therefore the ycoordinate of the pixel corresponding to the touch point is y 1 β py Remember that the value of y is capped at py 1 350 CHAPTER 3 Simple Resistive Circuits P 373 a Use the equations developed in the Practical Perspective and in Problem 372 Vx αVS so α Vx VS 1 5 02 Vy βVS so β Vy VS 375 5 075 b Use the equations developed in the Practical Perspective and in Problem 372 x 1 αpx 1 02480 384 y 1 βpy 1 075800 200 Therefore the touch occurred in the upper right corner of the screen P 374 Use the equations developed in the Practical Perspective and in Problem 372 x 1 αpx so α 1 x px 1 480 640 025 Vx αVS 0258 2 V y 1 βpy so β 1 y py 1 192 1024 08125 Vy βVS 081258 65 V P 375 From the results of Problem 374 the voltages corresponding to the touch point 480 192 are Vx1 2 V Vy1 65 V Now calculate the voltages corresponding to the touch point 240 384 x 1 αpx so α 1 x px 1 240 640 0625 Vx2 αVS 06258 5 V y 1 βpy so β 1 y py 1 384 1024 0625 Vy2 βVS 06258 5 V When the screen is touched at two points simultaneously only the smaller of the two voltages in the x direction is sensed The same is true in the y direction Therefore the voltages actually sensed are Vx 2 V Vy 5 V These two voltages identify the touch point as 480 384 which does not correspond to either of the points actually touched Therefore the resistive touch screen is appropriate only for single point touches 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Techniques of Circuit Analysis Assessment Problems AP 41 a Redraw the circuit labeling the reference node and the two node voltages The two node voltage equations are 15 v160 v115 v1 v25 0 5 v22 v2 v15 0 Place these equations in standard form v1 160 115 15 v2 15 15 v1 15 v2 12 15 5 Solving v1 60 V and v2 10 V Therefore i1 v1 v25 10 A b p15A 15 Av1 15 A60 V 900 W 900 Wdelivered c p5A 5 Av2 5 A10 V 50 W 50 Wdelivered 42 CHAPTER 4 Techniques of Circuit Analysis AP 42 Redraw the circuit choosing the node voltages and reference node as shown The two node voltage equations are 45 v11 v1 v26 2 0 v212 v2 v16 2 v2 304 0 Place these equations in standard form v1 1 18 v2 18 45 v1 18 v2 112 18 14 75 Solving v1 6 V v2 18 V To find the voltage v first find the current i through the seriesconnected 6Ω and 2Ω resistors i v1 v26 2 6 188 15 A Using a KVL equation calculate v v 2i v2 215 18 15 V AP 43 a Redraw the circuit choosing the node voltages and reference node as shown The node voltage equations are v1 506 v18 v1 v22 3i1 0 5 v24 v2 v12 3i1 0 Problems 43 The dependent source requires the following constraint equation i1 50 v16 Place these equations in standard form v1 16 18 12 v2 12 i13 506 v1 12 v2 14 12 i13 5 v1 16 v20 i11 506 Solving v1 32 V v2 16 V i1 3 A Using these values to calculate the power associated with each source p50V 50i1 150 W p5A 5v2 80 W p3i1 3i1 v2 v1 144 W b All three sources are delivering power to the circuit because the power computed in a for each of the sources is negative AP 44 Redraw the circuit and label the reference node and the node at which the node voltage equation will be written The node voltage equation is vo40 vo 1010 vo 20iΔ20 0 The constraint equation required by the dependent source is iΔ i10Ω i30Ω 10 vo10 10 20iΔ30 Place these equations in standard form vo 140 110 120 iΔ 1 1 vo 110 iΔ 1 2030 1 1030 Solving iΔ 32 A and vo 24 V AP 45 Redraw the circuit identifying the three node voltages and the reference node Note that the dependent voltage source and the node voltages v and v2 form a supernode The v1 node voltage equation is v175 v1 v25 48 0 The supernode equation is v v125 v10 v225 v2 121 0 The constraint equation due to the dependent source is ix v175 The constraint equation due to the supernode is v ix v2 Place this set of equations in standard form v1175 125 v125 v20 ix0 48 v1125 v125 110 v2125 1 ix0 12 v1175 v0 v20 ix1 0 v10 v1 v21 ix1 0 Solving this set of equations gives v1 15 V v2 10 V ix 2 A and v 8 V AP 46 Redraw the circuit identifying the reference node and the two unknown node voltages Note that the rightmost node voltage is the sum of the 60 V source and the dependent source voltage The node voltage equation at v1 is v1 602 v124 v1 60 6iφ3 0 The constraint equation due to the dependent source is iφ 60 6iφ v13 Place these two equations in standard form v112 124 13 iφ2 30 20 v113 iφ1 2 20 Solving iφ 4 A and v1 48 V AP 47 a Redraw the circuit identifying the three mesh currents The mesh current equations are 80 5i1 i2 26i1 i3 0 30i2 90i2 i3 5i2 i1 0 8i3 26i3 i1 90i3 i2 0 Place these equations in standard form 46 CHAPTER 4 Techniques of Circuit Analysis 31i1 5i2 26i3 80 5i1 125i2 90i3 0 26i1 90i2 124i3 0 Solving i1 5 A i2 2 A i3 25 A p80V 80i1 805 400 W Therefore the 80 V source is delivering 400 W to the circuit b p8Ω 8i2 3 8252 50 W so the 8 Ω resistor dissipates 50 W AP 48 a b 8 n 6 b n 1 3 b Redraw the circuit identifying the three mesh currents The three meshcurrent equations are 25 2i1 i2 5i1 i3 10 0 3vφ 14i2 3i2 i3 2i2 i1 0 1i3 10 5i3 i1 3i3 i2 0 The dependent source constraint equation is vφ 3i3 i2 Place these four equations in standard form 7i1 2i2 5i3 0vφ 15 2i1 19i2 3i3 3vφ 0 5i1 3i2 9i3 0vφ 10 0i1 3i2 3i3 1vφ 0 Solving i1 4 A i2 1 A i3 3 A vφ 12 V pds 3vφi2 3121 36 W Thus the dependent source is delivering 36 W or absorbing 36 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 47 AP 49 Redraw the circuit identifying the three mesh currents The mesh current equations are 25 6ia ib 8ia ic 0 2ib 8ib ic 6ib ia 0 5iφ 8ic ia 8ic ib 0 The dependent source constraint equation is iφ ia We can substitute this simple expression for iφ into the third mesh equation and place the equations in standard form 14ia 6ib 8ic 25 6ia 16ib 8ic 0 3ia 8ib 16ic 0 Solving ia 4 A ib 25 A ic 2 A Thus vo 8ia ic 84 2 16 V AP 410 Redraw the circuit identifying the mesh currents Since there is a current source on the perimeter of the i3 mesh we know that i3 16 A The remaining two mesh equations are 30 3i1 2i1 i2 6i1 0 8i2 5i2 16 4i2 2i2 i1 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 48 CHAPTER 4 Techniques of Circuit Analysis Place these equations in standard form 11i1 2i2 30 2i1 19i2 80 Solving i1 2 A i2 4 A i3 16 A The current in the 2 Ω resistor is i1 i2 6 A p2 Ω 622 72 W Thus the 2 Ω resistors dissipates 72 W AP 411 Redraw the circuit and identify the mesh currents There are current sources on the perimeters of both the ib mesh and the ic mesh so we know that ib 10 A ic 2vφ 5 The remaining mesh current equation is 75 2ia 10 5ia 04vφ 0 The dependent source requires the following constraint equation vφ 5ia ic 5ia 04vφ Place the mesh current equation and the dependent source equation is standard form 7ia 2vφ 55 5ia 3vφ 0 Solving ia 15 A ib 10 A ic 10 A vφ 25 V Thus ia 15 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 49 AP 412 Redraw the circuit and identify the mesh currents The 2 A current source is shared by the meshes ia and ib Thus we combine these meshes to form a supermesh and write the following equation 10 2ib 2ib ic 2ia ic 0 The other mesh current equation is 6 1ic 2ic ia 2ic ib 0 The supermesh constraint equation is ia ib 2 Place these three equations in standard form 2ia 4ib 4ic 10 2ia 2ib 5ic 6 ia ib 0ic 2 Solving ia 7 A ib 5 A ic 6 A Thus p1 Ω i2 c1 621 36 W AP 413 Redraw the circuit and identify the reference node and the node voltage v1 The node voltage equation is v1 20 15 2 v1 25 10 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Rearranging and solving v1115 110 2 2015 2510 v1 35 V p2A 352 70 W Thus the 2 A current source delivers 70 W AP 414 Redraw the circuit and identify the mesh currents There is a current source on the perimeter of the i3 mesh so i3 4 A The other two mesh current equations are 128 4i1 4 6i1 i2 2i1 0 30ix 5i2 6i2 i1 3i2 4 0 The constraint equation due to the dependent source is ix i1 i3 i1 4 Substitute the constraint equation into the second mesh equation and place the resulting two mesh equations in standard form 12i1 6i2 144 24i1 14i2 132 Solving i1 9 A i2 6 A i3 4 A ix 9 4 5 A v4A 3i3 i2 4ix 10 V p4A v4A4 104 40 W Thus the 2 A current source delivers 40 W Problems 411 AP 415 a Redraw the circuit with a helpful voltage and current labeled Transform the 120 V source in series with the 20 Ω resistor into a 6 A source in parallel with the 20 Ω resistor Also transform the 60 V source in series with the 5 Ω resistor into a 12 A source in parallel with the 5 Ω resistor The result is the following circuit Combine the three current sources into a single current source using KCL and combine the 20 Ω 5 Ω and 6 Ω resistors in parallel The resulting circuit is shown on the left To simplify the circuit further transform the resulting 30 A source in parallel with the 24 Ω resistor into a 72 V source in series with the 24 Ω resistor Combine the 24 Ω resistor in series with the 16 Ω resisor to get a very simple circuit that still maintains the voltage v The resulting circuit is on the right Use voltage division in the circuit on the right to calculate v as follows v 8 1272 48 V b Calculate i in the circuit on the right using Ohms law i v 8 48 8 6 A Now use i to calculate va in the circuit on the left va 616 8 576 V Returning back to the original circuit note that the voltage va is also the voltage drop across the series combination of the 120 V source and 20 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 resistor Use this fact to calculate the current in the 120 V source ia ia 120 va20 120 57620 312 A p120V 120ia 120312 37440 W Thus the 120 V source delivers 3744 W AP 416 To find RTh replace the 72 V source with a short circuit Note that the 5Ω and 20Ω resistors are in parallel with an equivalent resistance of 520 4Ω The equivalent 4Ω resistance is in series with the 8Ω resistor for an equivalent resistance of 4 8 12Ω Finally the 12Ω equivalent resistance is in parallel with the 12Ω resistor so RTh 1212 6Ω Use node voltage analysis to find vTh Begin by redrawing the circuit and labeling the node voltages The node voltage equations are v1 725 v120 v1 vTh8 0 vTh v18 vTh 7212 0 Place these equations in standard form v115 120 18 vTh18 725 v118 vTh18 112 6 Solving v1 60 V and vTh 648 V Therefore the Thévenin equivalent circuit is a 648 V source in series with a 6Ω resistor Problems 413 AP 417 We begin by performing a source transformation turning the parallel combination of the 15 A source and 8 Ω resistor into a series combination of a 120 V source and an 8 Ω resistor as shown in the figure on the left Next combine the 2 Ω 8 Ω and 10 Ω resistors in series to give an equivalent 20 Ω resistance Then transform the series combination of the 120 V source and the 20 Ω equivalent resistance into a parallel combination of a 6 A source and a 20 Ω resistor as shown in the figure on the right Finally combine the 20 Ω and 12 Ω parallel resistors to give RN 2012 75 Ω Thus the Norton equivalent circuit is the parallel combination of a 6 A source and a 75 Ω resistor AP 418 Find the Thevenin equivalent with respect to A B using source transformations To begin convert the series combination of the 36 V source and 12 kΩ resistor into a parallel combination of a 3 mA source and 12 kΩ resistor The resulting circuit is shown below Now combine the two parallel current sources and the two parallel resistors to give a 3 18 15 mA source in parallel with a 12 k60 k 10 kΩ resistor Then transform the 15 mA source in parallel with the 10 kΩ resistor into a 150 V source in series with a 10 kΩ resistor and combine this 10 kΩ resistor in series with the 15 kΩ resistor The Thevenin equivalent is thus a 150 V source in series with a 25 kΩ resistor as seen to the left of the terminals AB in the circuit below Now attach the voltmeter modeled as a 100 kΩ resistor to the Thevenin equivalent and use voltage division to calculate the meter reading vAB vAB 100000 125000150 120 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 414 CHAPTER 4 Techniques of Circuit Analysis AP 419 Begin by calculating the open circuit voltage which is also vTh from the circuit below Summing the currents away from the node labeled vTh We have vTh 8 4 3ix vTh 24 2 0 Also using Ohms law for the 8 Ω resistor ix vTh 8 Substituting the second equation into the first and solving for vTh yields vTh 8 V Now calculate RTh To do this we use the test source method Replace the voltage source with a short circuit the current source with an open circuit and apply the test voltage vT as shown in the circuit below Write a KCL equation at the middle node iT ix 3ix vT2 4ix vT2 Use Ohms law to determine ix as a function of vT ix vT8 Substitute the second equation into the first equation iT 4vT8 vT2 vT 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Thus RTh vTiT 1Ω The Thévenin equivalent is an 8 V source in series with a 1Ω resistor AP 420 Begin by calculating the open circuit voltage which is also vTh using the node voltage method in the circuit below The node voltage equations are v60 v vTh 160iΔ20 4 0 vTh40 vTh80 vTh 160iΔ v20 0 The dependent source constraint equation is iΔ vTh40 Substitute the constraint equation into the node voltage equations and put the two equations in standard form v160 120 vTh520 4 v120 vTh140 180 520 0 Solving v 1725 V and vTh 30 V Now use the test source method to calculate the test current and thus RTh Replace the current source with a short circuit and apply the test source to get the following circuit Write a KCL equation at the rightmost node iT vT80 vT40 vT 160iΔ80 The dependent source constraint equation is iΔ vT40 Substitute the constraint equation into the KCL equation and simplify the righthand side iT vT10 Therefore RTh vTiT 10Ω Thus the Thévenin equivalent is a 30 V source in series with a 10Ω resistor AP 421 First find the Thévenin equivalent circuit To find vTh create an open circuit between nodes a and b and use the node voltage method with the circuit below The node voltage equations are vTh 100 vφ4 vTh v14 0 v1 1004 v1 204 v1 vTh4 0 The dependent source constraint equation is vφ v1 20 Place these three equations in standard form vTh14 14 v114 vφ14 25 vTh14 v114 14 14 vφ0 30 vTh0 v11 vφ1 20 Solving vTh 120 V v1 80 V and vφ 60 V Problems 417 Now create a short circuit between nodes a and b and use the mesh current method with the circuit below The mesh current equations are 100 4i1 i2 vφ 20 0 vφ 4i2 4i2 isc 4i2 i1 0 20 vφ 4isc i2 0 The dependent source constraint equation is vφ 4i1 isc Place these four equations in standard form 4i1 4i2 0isc vφ 80 4i1 12i2 4isc vφ 0 0i1 4i2 4isc vφ 20 4i1 0i2 4isc vφ 0 Solving i1 45 A i2 30 A isc 40 A and vφ 20 V Thus RTh vTh isc 120 40 3 Ω a For maximum power transfer R RTh 3 Ω b The Thevenin voltage vTh 120 V splits equally between the Thevenin resistance and the load resistance so vload 120 2 60 V Therefore pmax v2 load Rload 602 3 1200 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 418 CHAPTER 4 Techniques of Circuit Analysis AP 422 Sustituting the value R 3 Ω into the circuit and identifying three mesh currents we have the circuit below The mesh current equations are 100 4i1 i2 vφ 20 0 vφ 4i2 4i2 i3 4i2 i1 0 20 vφ 4i3 i2 3i3 0 The dependent source constraint equation is vφ 4i1 i3 Place these four equations in standard form 4i1 4i2 0i3 vφ 80 4i1 12i2 4i3 vφ 0 0i1 4i2 7i3 vφ 20 4i1 0i2 4i3 vφ 0 Solving i1 30 A i2 20 A i3 20 A and vφ 40 V a p100V 100i1 10030 3000 W Thus the 100 V source is delivering 3000 W b pdepsource vφi2 4020 800 W Thus the dependent source is delivering 800 W c From Assessment Problem 421b the power delivered to the load resistor is 1200 W so the load power is 12003800100 3158 of the combined power generated by the 100 V source and the dependent source 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 419 Problems P 41 a 12 branches 8 branches with resistors 2 branches with independent sources 2 branches with dependent sources b The current is unknown in every branch except the one containing the 25 mA current source so the current is unknown in 11 branches c 10 essential branches R1 R2 forms an essential branch as does R8 2 V The remaining eight branches are essential branches that contain a single element d The current is known only in the essential branch containing the current source and is unknown in the remaining 9 essential branches e From the figure there are 7 nodes three identified by rectangular boxes two identified by triangles and two identified by diamonds f There are 5 essential nodes three identified with rectangular boxes and two identified with triangles g A mesh is like a window pane and as can be seen from the figure there are 6 window panes or meshes P 42 a From Problem 41d there are 9 essential branches where the current is unknown so we need 9 simultaneous equations to describe the circuit b From Problem 41f there are 5 essential nodes so we can apply KCL at 5 1 4 of these essential nodes There would also be a dependent source constraint equation c The remaining 4 equations needed to describe the circuit will be derived from KVL equations 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 420 CHAPTER 4 Techniques of Circuit Analysis d We must avoid using the bottom leftmost mesh since it contains a current source and we have no way of determining the voltage drop across a current source The two meshes on the bottom that share the dependent source must be handled in a special way P 43 a There are eight circuit components seven resistors and the voltage source Therefore there are eight unknown currents However the voltage source and the R1 resistor are in series so have the same current The R4 and R6 resistors are also in series so have the same current The R5 and R7 resistors are in series so have the same current Therefore we only need 5 equations to find the 5 distinct currents in this circuit b There are three essential nodes in this circuit identified by the boxes At two of these nodes you can write KCL equations that will be independent of one another A KCL equation at the third node would be dependent on the first two Therefore there are two independent KCL equations c Sum the currents at any two of the three essential nodes a b and c Using nodes a and c we get i1 i2 i4 0 i1 i3 i5 0 d There are three meshes in this circuit one on the left with the components vs R1 R2 and R3 one on the top right with components R2 R4 and R6 and one on the bottom right with components R3 R5 and R7 We can write KVL equations for all three meshes giving a total of three independent KVL equations 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 421 e vs R1i1 R2i2 R3i3 0 R4i4 R6i4 R2i2 0 R3i3 R5i5 R7i5 0 P 44 a At node a i1 i2 i4 0 At node b i2 i3 i4 i5 0 At node c i1 i3 i5 0 b There are many possible solutions For example adding the equations at nodes a and c gives the equation at node b i1 i2 i4 i1 i3 i5 0 so i2 i3 i4 i5 0 This is the equation at node b with both sides multiplied by 1 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 422 CHAPTER 4 Techniques of Circuit Analysis P 45 a As can be seen from the figure the circuit has 2 separate parts b There are 5 nodes the four black dots and the node betweem the voltage source and the resistor R3 c There are 7 branches each containing one of the seven circuit components d When a conductor joins the lower nodes of the two separate parts there is now only a single part in the circuit There would now be 4 nodes because the two lower nodes are now joined as a single node The number of branches remains at 7 where each branch contains one of the seven individual circuit components P 46 Use the lower terminal of the 25 Ω resistor as the reference node vo 24 20 80 vo 25 004 0 Solving vo 4 V P 47 a From the solution to Problem 46 we know vo 4 V therefore p40mA 004vo 016 W p40mA developed 160 mW b The current into the negative terminal of the 24 V source is ig 24 4 20 80 02 A p24V 2402 48 W p24V developed 4800 mW 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 c p20Ω 02220 800 mW p80Ω 02280 3200 mW p25Ω 4225 640 mW pdev 4800 mW pdis 160 800 3200 640 4800 mW P 48 a v0 24 20 80 vo 25 004 0 vo 4 V b Let vx voltage drop across 40 mA source vx vo 50004 2 V p40mA 2004 80 mW so p40mA developed 80 mW c Let ig be the current into the positive terminal of the 24 V source ig 424100 02 A p24V 0224 4800 mW so p24V developed 4800 mW d pdis 02220 02280 4225 004250 008 4800 mW e vo is independent of any finite resistance connected in series with the 40 mA current source P 49 2 vo50 vo 451 4 0 vo 50 V p2A 502 100 W delivering The 2 A source extracts 100 W from the circuit because it delivers 100 W to the circuit P 410 a vo v1R vo v2R vo v3R vo vnR 0 n vo v1 v2 v3 vn vo 1n v1 v2 v3 vn 1n k1n vk b vo 13 100 80 60 40 V P 411 a v1 1285 v160 v1 v24 0 v2 v14 v280 v2 32010 0 In standard form v1 15 160 14 v2 14 1285 v1 14 v2 14 180 110 32010 Solving v1 162 V v2 200 V ia 128 1625 68 A ib 16260 27 A ic 162 2004 95 A id 20080 25 A ie 200 32010 12 A b p128V 12868 8704 W abs p320V 32012 3840 W dev Therefore the total power developed is 3840 W P 412 v1 1444 v110 v1 v280 0 so 29v1 v2 2880 3 v2 v180 v25 0 so v1 17v2 240 Solving v1 100 V v2 20 V P 413 6 v140 v1 v28 0 v2 v18 v280 v2120 1 0 Solving v1 120 V v2 96 V CHECK p40Ω 120240 360 W p8Ω 120 9628 72 W p80Ω 96280 1152 W p120Ω 962120 768 W p6A 6120 720 W p1A 196 96 W pabs 360 72 1152 768 96 720 W pdev 720 W CHECKS 426 CHAPTER 4 Techniques of Circuit Analysis P 414 a v1 40 v1 40 4 v1 v2 2 0 so 31v1 20v2 0v3 400 v2 v1 2 v2 v3 4 28 0 so 2v1 3v2 v3 112 v3 2 v3 v2 4 28 0 so 0v1 v2 3v3 112 Solving v1 60 V v2 73 V v3 13 V b ig 40 60 4 5 A pg 405 200 W Thus the 40 V source delivers 200 W of power P 415 a v1 125 1 v1 v2 6 v1 v3 24 0 v2 v1 6 v2 2 v2 v3 12 0 v3 125 1 v3 v2 12 v3 v1 24 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 In standard form v₁11 16 124 v₂16 v₃124 125 v₁16 v₂16 12 112 v₃112 0 v₁124 v₂112 v₃11 112 124 125 Solving v₁ 10124 V v₂ 1066 V v₃ 10657 V Thus i₁ 125 v₁1 2376 A i₄ v₁ v₂6 1510 A i₂ v₂2 533 A i₅ v₂ v₃12 977 A i₃ v₃ 1251 1843 A i₆ v₁ v₃24 866 A b ΣPdev 125i₁ 125i₃ 527309 W ΣPdis i₁²1 i₂²2 i₃²1 i₄²6 i₅²12 i₆²24 527309 W P 416 Circuit diagram with 40 V source resistors 12Ω 20Ω 25Ω 40Ω and current sources 5A and 75A Equations v₁ 4012 v₁25 v₁ v₂20 5 0 v₂ v₁20 5 v₂ v₁40 75 0 v₃40 v₃ v₂40 75 0 Solving v₁ 10 V v₂ 132 V v₃ 84 V i40V 10 4012 25 A p₅A 5v₁ v₂ 510 132 710 W del p₇5A 84 13275 1620 W del p₄₀V 4025 100 W del p₁₂Ω 25²12 75 W p₂₅Ω v₁²25 10²25 4 W p₂₀Ω v₁ v₂²20 142²20 10082 W p₄₀Ωlower v₃²40 84²40 1764 W p₄₀Ωright v₂ v₃²40 216²40 11664 W Σpdiss 75 4 10082 1764 11664 2430 W Σpdev 710 1620 100 2430 W CHECKS The total power dissipated in the circuit is 2430 W P 417 3 vo200 vo 5iΔ10 vo 8020 0 iΔ vo 8020 a Solving vo 50 V b ids vo 5iΔ10 iΔ 50 8020 15 A ids 425 A 5iΔ 75 V pds 5iΔids 31875 W c p₃A 3vo 350 150 W del p₈₀V 80iΔ 8015 120 W del Σpdel 150 120 270 W CHECK p₂₀₀Ω 2500200 125 W p₂₀Ω 80 50²20 90020 45 W p₁₀Ω 425²10 180625 W Σpdiss 31875 180625 125 45 270 W Problems 429 P 418 vo 160 10 vo 100 vo 150iσ 50 0 iσ vo 100 Solving vo 100 V iσ 1 A io 100 1501 50 5 A p150iσ 150iσio 750 W The dependent voltage source delivers 750 W to the circuit P 419 a 002 v1 1000 v1 v2 1250 0 v2 v1 1250 v2 4000 v2 2000 v2 2500i 200 0 i v2 v1 1250 Solving v1 60 V v2 160 V i 80 mA P20mA 002v1 00260 12 W absorbed ids v2 2500i 200 160 2500008 200 02 A Pds 2500iids 250000802 40 W 40 W developed Pdeveloped 40 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 b P₁ₖ v₁²1000 60²1000 36 W P₁₂₅₀ 1250 iΔ² 1250008² 8 W P₄ₖ v₂²4000 160²4000 64 W P₂ₖ v₂²2000 160²2000 128 W P₂₀₀ 200 ids² 20002² 8 W Pabsorbed P20mA P₁ₖ P₁₂₅₀ P₄ₖ P₂ₖ P₂₀₀ 12 36 8 64 128 8 40 W check P 420 Circuit diagram with 50 V source resistors 500Ω 1 kΩ 2 kΩ 200Ω dependent current source vΔ750 nodes vΔ and vo marked Equations a vΔ 50500 vΔ1000 vΔ vo2000 0 vo vΔ2000 vΔ750 vo200 0 Solving vΔ 30 V vo 10 V b i50V vΔ 50500 30 50500 004 A P50V 50 i50V 50004 2 W 2 W supplied Pds vovΔ750 1030750 04 W 04 W supplied Ptotal 2 04 24 W supplied P 421 a io v240 5 io v120 v1 v25 0 so 10v1 13v2 0v3 0 v2 v15 v240 v2 v310 so 8v1 13v2 4v3 0 v3 v210 v3 115 io5 v3 964 0 so 0v1 63v2 220v3 9600 Solving v1 156 V v2 120 V v3 78 V b io v240 12040 3 A i3 v3 115 io5 78 11535 87 A ig 78 964 45 A p5io 5 io v1 53156 2340 W dev p115io 115 io i3 115387 30015 W abs p96v 9645 432 W dev pdev 2340 432 2772 W CHECK pdis 156220 156 12025 120240 120 78250 8725 4524 30015 2772 W pdev pdis 2772 W P 422 a This circuit has a supernode includes the nodes v1 v2 and the 25 V source The supernode equation is 2 v150 v2150 v275 0 The supernode constraint equation is v1 v2 25 Place these two equations in standard form v1150 v21150 175 2 v11 v21 25 Solving v1 375 V and v2 625 V so vo v1 375 V p2A 2 vo 2375 75 W The 2 A source delivers 75 W b This circuit now has only one nonreference essential node where the voltage is not known note that it is not a supernode The KCL equation at v1 is 2 v150 v1 25150 v1 2575 0 Solving v1 375 V so vo v1 375 V p2A 2 vo 2375 75 W The 2 A source delivers 75 W c The choice of a reference node in part b resulted in one simple KCL equation while the choice of a reference node in part a resulted in a supernode KCL equation and a second supernode constraint equation Both methods give the same result but the choice of reference node in part b yielded fewer equations to solve so is the preferred method P 423 The two node voltage equations are v1 5080 v150 v1 v240 0 v2 v140 075 v2200 v2 50800 0 Place these equations in standard form v1180 150 140 v2140 5080 v1140 v2140 1200 1800 075 50800 Solving v1 34 V v2 532 V Thus vo v2 50 532 50 32 V POWER CHECK ig 50 3480 50 532800 196 mA p50V 500196 98 W p80Ω 50 34280 32 W p800Ω 50 5322800 128 mW p40Ω 532 34240 9216 W p50Ω 34250 2312 W p200Ω 5322200 141512 W p075A 532075 399 W pabs 32 0128 9216 2312 141512 497 W pdel 98 399 497 check 434 CHAPTER 4 Techniques of Circuit Analysis P 424 v1 30000 v1 v2 5000 v1 20 2000 0 so 22v1 6v2 300 v2 1000 v2 v1 5000 v2 20 5000 0 so v1 7v2 20 Solving v1 15 V v2 5 V Thus io v1 v2 5000 2 mA P 425 a v2 230 1 v2 v4 1 v2 v3 1 0 so 3v2 1v3 1v4 0v5 230 v3 v2 1 v3 1 v3 v5 1 0 so 1v2 3v3 0v4 1v5 0 v4 v2 1 v4 230 6 v4 v5 2 0 so 12v2 0v3 20v4 6v5 460 v5 v3 1 v5 6 v5 v4 2 0 so 0v2 12v3 6v4 20v5 0 Solving v2 150 V v3 80 V v4 140 V v5 90 V i2Ω v4 v5 2 140 90 2 25 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 p2Ω 2522 1250 W b i230V v1 v21 v1 v46 230 1501 230 1406 80 15 95 A p230V 23095 21850 W Check Pdis 8021 7021 8021 1526 1021 1021 2522 1526 21850 W P 426 Place 5vΔ inside a supernode and use the lower node as a reference Then vΔ 1510 vΔ2 vΔ 5vΔ20 vΔ 5vΔ40 0 12vΔ 60 vΔ 5 V vo vΔ 5vΔ 45 20 V P 427 a There is only one node voltage equation va 305000 va500 va 801000 001 0 Solving va 30 10va 5va 400 50 0 so 16va 320 va 20 V Calculate the currents i1 30 205000 10 mA i2 20500 40 mA i4 804000 20 mA i5 80 201000 60 mA i3 i4 i5 10 mA 0 so i3 001 002 006 007 70 mA b p30V 30001 03 W p10mA 20 80001 06 W p80V 80007 56 W p5k 00125000 05 W p500Ω 0042500 08 W p1k 80 2021000 36 W p4k 8024000 16 W pabs 05 08 36 16 65 W pdel 03 06 56 65 W checks P 428 The two node voltage equations are 7 vb3 vb vc1 0 2vx vc vb1 vc 42 0 The constraint equation for the dependent source is vx vc 4 Place these equations in standard form vb13 1 vc1 vx0 7 vb1 vc1 12 vx2 42 vb0 vc1 vx1 4 Solving vc 9 V vx 5 V and vo vb 15 V P 429 a The leftmost node voltage is 75 1250iϕ The rightmost node voltage is 75 V Write KCL equations at the essential nodes labeled 1 and 2 b From the values given iϕ v1 v2500 105 85500 004 A vΔ v2 75 85 75 10 V vL 75 1250004 25 V ix vL v11000 vL2500 25 1051000 252500 007 A iy v2 75500 ix 85 75500 007 009 A Calculate the total power Pdstop 1250iϕix 1250004007 35 W Pdsbot v112 103vΔ 10512 10310 126 W P75V 75iy 75009 675 W 438 CHAPTER 4 Techniques of Circuit Analysis P1k vL v12 1000 25 1052 1000 64 W P25k v2 L 2500 252 2500 025 W P500mid 500i2 φ 5000042 08 W P500right v2 500 102 500 02 W P425k v2 2 4250 852 4250 17 W c Psupplied 35 126 161 W Pabsorbed 675 64 025 08 02 17 161 W Psupplied Therefore the analyst is correct P 430 Node equations v1 20 v1 20 2 v3 v2 4 v3 80 3125v 0 v2 40 v2 v3 4 v2 20 1 0 Constraint equations v 20 v2 v1 35iφ v3 iφ v240 Solving v1 2025 V v2 10 V v3 29 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 440 CHAPTER 4 Techniques of Circuit Analysis P 433 a The three mesh current equations are 128 5ia 60ia ic 0 4ic 80ic ie 60ic ia 0 320 80ie ic 10ie 0 Place these equations in standard form ia5 60 ic60 ie0 128 ia60 ic4 80 60 ie80 0 ia0 ic80 ie80 10 320 Solving ia 68 A ic 95 A ie 12 A Now calculate the remaining branch currents ib ia ic 27 A id iv ie 25 A b p128V 12868 8704 W abs p320V 32012 3840 W dev Therefore the total power developed is 3840 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 441 P 434 a The three mesh current equations are 125 1i1 6i1 i6 2i1 i3 0 24i6 12i6 i3 6i6 i1 0 125 2i3 i1 12i3 i6 1i3 0 Place these equations in standard form i11 6 2 i32 i66 125 i16 i312 i624 12 6 0 i12 i32 12 1 i612 125 Solving i1 2376 A i3 1843 A i6 866 A Now calculate the remaining branch currents i2 i1 i3 533 A i4 i1 i6 1510 A i5 i3 i6 977 A b psources ptop pbottom 1252376 1251843 296958 230351 5273 W Thus the power developed in the circuit is 5273 W Now calculate the power absorbed by the resistors p1top 237621 56439 W p2 53322 5679 W p1bot 184321 33959 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 442 CHAPTER 4 Techniques of Circuit Analysis p6 151026 136764 W p12 977212 114522 W p24 866224 179947 W The power absorbed by the resistors is 56439 5679 33959 136764 114522 179947 5273 W so the power balances P 435 The three mesh current equations are 20 2000i1 i2 30000i1 i3 0 5000i2 5000i2 i3 2000i2 i1 0 1000i3 30000i3 i1 5000i3 i2 0 Place these equations in standard form i132000 i22000 i330000 20 i12000 i212000 i35000 0 i130000 i25000 i336000 0 Solving i1 55 mA i2 3 mA i3 5 mA Thus io i3 i2 2 mA 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 444 CHAPTER 4 Techniques of Circuit Analysis P 437 a The four mesh current equations are 230 1i1 i2 1i1 i3 1i1 i4 0 6i2 1i2 i3 1i2 i1 0 2i3 1i3 i4 1i3 i1 1i3 i2 0 6i4 1i4 i1 1i4 i3 0 Place these equations in standard form i13 i21 i31 i41 230 i11 i28 i31 i40 0 i11 i21 i35 i41 0 i11 i20 i31 i48 0 Solving i1 95 A i2 15 A i3 25 A i4 15 A The power absorbed by the 5 Ω resistor is p5 i2 32 2522 1250 W b p230 230i1 23095 21850 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 445 P 438 160 10i1 100i1 io 0 30io 150iσ 20io 100iσ 0 iσ io i1 Solving i1 6 A io 5 A iσ 1 A Pds 150iσio 15015 750 W Thus 750 W is delivered by the dependent source P 439 65 4i1 5i1 i2 6i1 0 8i2 3v 15i2 5i2 i1 0 v 4i1 Solving i1 4 A i2 1 A v 16 V p15Ω 1215 15 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 447 P 441 a 40 125i1 300i1 i 0 75i 25i 300i i1 0 50i2 200i2 500i Solving i1 02 A i2 03 A i 015 A vo 200i2 20003 60 V b pds 500ii2 50001503 225 W pds delivered 225 W P 442 Mesh equations 53i 8i1 3i2 5i3 0 0i 3i1 30i2 20i3 30 0i 5i1 20i2 27i3 30 Constraint equations i i2 i3 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 449 P 444 Mesh equations 128i1 80i2 240 80i1 200i2 120 Solving i1 3 A i2 18 A Therefore v1 406 3 120 V v2 12018 1 96 V P 445 a Mesh equations 65i1 40i2 0i3 100io 0 40i1 55i2 5i3 115io 0 0i1 5i2 9i3 115io 0 1i1 1i2 0i3 1io 0 Solving i1 72 A i2 42 A i3 45 A io 3 A Therefore v1 2053 72 156 V v2 4072 42 120 V v3 542 45 1153 78 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 455 P 451 a Supermesh equations 1000ib 4000ic id 500ic ia 0 ic ib 001 Two remaining mesh equations 5500ia 500ic 30 4000id 4000ic 80 In standard form 500ia 1000ib 4500ic 4000id 0 0ia 1ib 1ic 0id 001 5500ia 0ib 500ic 0id 30 0ia 0ib 4000ic 4000id 80 Solving ia 10 mA ib 60 mA ic 50 mA id 70 mA Then i1 ia 10 mA i2 ia ic 40 mA i3 id 70 mA b psources 30001 1000006001 80007 65 W presistors 10000062 50000012 5000042 4000005 0072 65 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 458 CHAPTER 4 Techniques of Circuit Analysis b The mesh current equations 2500i1 001 2000i1 1000i1 i2 0 5000i2 001 1000i2 i1 1000i2 0 Place the equations in standard form i12500 2000 1000 i21000 25 i11000 i25000 1000 1000 50 Solving i1 6 mA i2 8 mA Find the power in the 1 kΩ resistor i1k i1 i2 2 mA p1k 000221000 4 mW c No the voltage across the 10 A current source is readily available from the mesh currents and solving two simultaneous meshcurrent equations is less work than solving three node voltage equations d vg 2000i1 1000i2 12 8 20 V p10mA 20001 200 mW Thus the 10 mA source develops 200 mW P 455 a There are three unknown node voltages and three unknown mesh currents so the number of simultaneous equations required is the same for both methods The node voltage method has the advantage of having to solve the three simultaneous equations for one unknown voltage provided the connection at either the top or bottom of the circuit is used as the reference node Therefore recommend the node voltage method 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 460 CHAPTER 4 Techniques of Circuit Analysis voltage method it is the preferred approach b Node voltage equations v1 100 0003v v2 250 02 0 02 v3 100 v4 200 0003v 0 Constraints v2 va v3 v v4 v3 04va v2 v1 20 Solving v1 24 V v2 44 V v3 72 V v4 54 V io 02 v2 250 24 mA p20V 200024 480 mW Thus the 20 V source absorbs 480 mW P 457 a The meshcurrent method does not directly involve the voltage drop across the 40 mA source Instead use the nodevoltage method and choose the reference node so that a node voltage is identical to the voltage across the 40 mA source b 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 461 Since the 40 mA source is developing 0 W v1 must be 0 V Since v1 is known we can sum the currents away from node 1 to find v2 thus 0 70 v2 500 0 v2 250 0 750 004 0 v2 30 V Now that we know v2 we sum the currents away from node 2 to find v3 thus v2 70 0 500 v2 0 250 v2 v3 1250 0 v3 80 V Now that we know v3 we sum the currents away from node 3 to find idc thus v3 1000 v3 v2 1250 idc 0 idc 012 120 mA P 458 a If the meshcurrent method is used then the value of the lower left mesh current is io 0 This shortcut will simplify the set of KVL equations The nodevoltage method has no equivalent simplifying shortcut so the meshcurrent method is preferred b Write the mesh current equations Note that if io 0 then i1 0 23 5i2 10i3 46 0 30i2 15i2 i3 5i2 0 Vdc 25i3 46 10i3 15i3 i2 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 463 vo 125000001 125 V b 5000i1 40000i2 30000i3 35 i2 i1 0008 30000i2 70000i3 25 Solving i1 533 mA i2 2667 mA i3 15 mA vo 25000i3 0001 2500000005 125 V P 460 a 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 464 CHAPTER 4 Techniques of Circuit Analysis io 75 25000 03 mA b va 1500000003 45 V ia va 20000 225 µA ib 1 0225 03 1525 mA vb 50000525 103 45 7125 V vc 30001525 103 7125 60001525 103 2085 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 466 CHAPTER 4 Techniques of Circuit Analysis Solving v1 165 V v2 135 V io v2 v1 150 20 mA P 462 a Applying a source transformation to each current source yields Now combine the 12 V and 5 V sources into a single voltage source and the 6 Ω 6 Ω and 5 Ω resistors into a single resistor to get Now use a source transformation on each voltage source thus which can be reduced to io 85 10 1 085 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 467 b 34ia 17ib 12 5 34 51 17ia 185ib 34 Solving ib 085 A io P 463 a First remove the 16 Ω and 260 Ω resistors Next use a source transformation to convert the 1 A current source and 40 Ω resistor which simplifies to vo 250 300480 400 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 469 v1 16667 V so voc 3000 5000v1 100 V Find the shortcircuit current isc 40002000 2000 0075 50 mA Thus IN isc 50 mA RN voc isc 100 005 2 kΩ P 466 50i1 40isc 60 40 40i1 48iscs 32 Solving isc 7 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 470 CHAPTER 4 Techniques of Circuit Analysis RTh 8 1040 50 16 Ω P 467 After making a source transformation the circuit becomes 500 20i1 8i2 300 8i1 432i2 i1 30 A and i2 125 A vTh 12i1 52i2 425 V RTh 812 5230 75 Ω P 468 First we make the observation that the 10 mA current source and the 10 kΩ resistor will have no influence on the behavior of the circuit with respect to the terminals ab This follows because they are in parallel with an ideal voltage source Hence our circuit can be simplified to 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 471 or Therefore the Norton equivalent is P 469 i1 10020 5 A 100 vTh 5RTh vTh 100 5RTh i2 20050 4 A 200 vTh 4RTh vTh 200 4RTh 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 472 CHAPTER 4 Techniques of Circuit Analysis 100 5RTh 200 4RTh so RTh 100Ω vTh 100 500 600 V P 470 125 vTh 2RTh 117 vTh 18RTh Solving the above equations for VTh and RTh yields vTh 126 V RTh 50 mΩ IN 252 A RN 50 mΩ P 471 First find the Thevenin equivalent with respect to Ro 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 473 RoΩ iomA voV 100 1364 1364 120 125 15 150 1111 1667 180 10 18 P 472 a First find the Thevenin equivalent with respect to ab using a succession of source transformations vTh 54 V RTh 45 kΩ 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 475 P 474 OPEN CIRCUIT 100 2500i1 625i1 103v2 v2 6000 100005000i1 Solving i1 002 A v2 voc 60 V SHORT CIRCUIT v2 0 isc 5000 4000i1 i1 100 2500 625 0032 A Thus isc 5 4i1 004 A RTh 60 004 15 kΩ 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 477 280 2000i 2000i isc 0 2000isc 02i 2000isc i 0 Put these equations in standard form i4000 isc2000 280 i2400 isc4000 0 Solving i 01 A isc 006 A RTh 112 006 186667 Ω P 476 a Find the Thevenin equivalent with respect to the terminals of the ammeter This is most easily done by first finding the Thevenin with respect to the terminals of the 48 Ω resistor Thevenin voltage note iφ is zero vTh 100 vTh 25 vTh 20 vTh 16 2 0 Solving vTh 20 V Shortcircuit current isc 12 2isc isc 12 A RTh 20 12 53 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 478 CHAPTER 4 Techniques of Circuit Analysis Rtotal 20 6 333 Ω Rmeter 333 313 02 Ω b Actual current iactual 20 313 638 A error 6 638 638 100 6 P 477 a Replace the voltage source with a short circuit and find the equivalent resistance from the terminals ab RTh 1030 25 10 Ω b Replace the current source with an open circuit and the voltage source with a short circuit Find the equivalent resistance from the terminals ab RTh 1040 8 16 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 479 P 478 a Open circuit v2 9 20 v2 70 18 0 v2 35 V vTh 60 70v2 30 V Short circuit v2 9 20 v2 10 18 0 v2 15 V ia 9 15 20 03 A isc 18 03 15 A RTh 30 15 20 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 481 Solving v1 75 V vt 150 V i 05 A RTh vt 1 A 150 Ω P 480 Since there is no independent source VTh 0 Now apply a test source at the terminals ab to find the Thevenin equivalent resistance v1 100 v1 13ix 10 v1 v2 20 0 v2 v1 20 v2 50 1 0 ix v2 50 Solving v1 24 V v2 16 V ix 032 A vt 241 v2 so vt 16 24 40 V RTh vt 1 A 40 Ω The Thevenin equivalent is simply a 40 Ω resistor 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 482 CHAPTER 4 Techniques of Circuit Analysis P 481 VTh 0 since there are no independent sources in the circuit Thus we need only find RTh v 250ix 500 15ix v 750 1 0 ix v 750 Solving v 1500 V ix 2 A RTh v 1 A 1500 15 kΩ P 482 a RTh 50001600 24004800 1800 25 kΩ Ro RTh 25 kΩ 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 484 CHAPTER 4 Techniques of Circuit Analysis P 483 Write KVL equations for the left mesh and the supermesh place them in standard form and solve At i1 60 2400i1 4800i1 i2 0 Supermesh 4800i2 i1 1600i3 50002500i3 1800i3 0 Constraint i2 i3 0015 0 Standard form i17200 i24800 i30 60 i14800 i24800 i3506667 0 i10 i21 i31 0015 Calculator solution i1 19933 mA i2 174 mA i3 24 mA Calculate voltage across the current source v15mA 4800i1 i2 1216 V Calculate power delivered by the sources p15mA 00151216 1824 mW abs p60V 60i1 600019933 1196 W del pdelivered 1196 W Calculate power absorbed by the 25 kΩ resistor and the percentage power i25k 50002500 2500 i3 16 mA p25k 0001622500 64 mW delivered to Ro 00064 1196 100 0535 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 485 P 484 a From the solution to Problem 471 we have RoΩ pomW 100 18595 120 1875 150 1852 180 18 The 120 Ω resistor dissipates the most power because its value is equal to the Thevenin equivalent resistance of the circuit b c Ro 120 Ω po 1875 mW which is the maximum power that can be delivered to a load resistor P 485 a Since 0 Ro maximum power will be delivered to the 6 Ω resistor when Ro 0 b P 302 6 150 W P 486 a From the solution of Problem 475 we have RTh 186667 Ω and VTh 112 V Therefore Ro RTh 186667 Ω b p 562 186667 168 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 488 CHAPTER 4 Techniques of Circuit Analysis Node voltage equations v1 110 15 01v v1 v2 5 0 v2 v1 5 v2 3i 8 0 Constraint equations i v2 v1 5 v 110 v1 Solving v2 55 V vTh Thevenin resistance using a test source v1 15 01v v1 vt 5 0 vt v1 5 vt 3i 8 1 0 i vt v1 5 v v1 Solving vt 4 V RTh vt 1 4 Ω Ro RTh 4 Ω b pmax 2752 4 1890625 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 491 P 489 a First find the Thevenin equivalent with respect to Ro Open circuit voltage iφ 0 184φ 0 v1 16 v1 180 20 v1 180 10 v1 10 01v 0 v v1 180 10 2 02v1 36 v1 80 V v 20 V VTh 180 v 180 20 160 V Short circuit current v1 16 v1 180 20 v2 8 v2 10 01180 0 v2 184iφ v1 iφ 180 2 v2 8 90 0125v2 v2 640 V v1 1200 V iφ isc 10 A RTh VThisc 16010 16 Ω Ro 16 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 492 CHAPTER 4 Techniques of Circuit Analysis b pmax 80216 400 W c v1 16 v1 180 20 v2 80 8 v2 10 0180 180 0 v2 184iφ v1 iφ 8016 5 A Therefore v1 640 V and v2 280 V thus ig 180 80 2 180 640 20 27 A p180V dev 18027 4860 W P 490 a Find the Thevenin equivalent with respect to the terminals of RL Open circuit voltage The mesh current equations are 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 493 3600 45i1 i2 300i1 i3 30i1 0 30i2 60i2 i3 45i2 i1 0 150iβ 15i3 300i3 i1 60i3 i2 0 The dependent source constraint equation is iβ i1 i2 Place these equations in standard form i145 300 30 i245 i3300 iβ0 3600 i145 i230 60 45 i360 iβ0 0 i1300 i260 i315 300 60 iβ150 0 i11 i21 i30 iβ1 0 Solving i1 996 A i2 78 A i3 1008 A iβ 216 A VTh 300i1 i3 360 V Shortcircuit current The mesh current equations are 3600 45i1 i2 30i1 0 30i2 60i2 i3 45i2 i1 0 150iβ 15i3 60i3 i2 0 The dependent source constraint equation is iβ i1 i2 Place these equations in standard form i145 30 i245 i30 iβ0 3600 i145 i230 60 45 i360 iβ0 0 i10 i260 i360 15 iβ150 0 i11 i21 i30 iβ1 0 Solving i1 92 A i2 7333 A i3 96 A iβ 1867 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 494 CHAPTER 4 Techniques of Circuit Analysis isc i1 i3 4 A RTh VTh isc 360 4 90 Ω RL RTh 90 Ω b pmax 1802 90 360 W P 491 a We begin by finding the Thevenin equivalent with respect to the terminals of Ro Open circuit voltage The mesh current equations are 260i1 16i2 180i3 400 16i1 48i2 32i3 316i1 i3 0 180i1 32i2 260i3 200 Solving i1 3 A i2 175 A i3 5 A i i1 i3 2 A Also VTh voc 180i 360 V Now find the shortcircuit current 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 495 Note with the short circuit from a to b that i is zero hence 316i is also zero The mesh currents are 80i1 16i2 0i3 400 16i1 48i2 32i3 0 0i1 32i2 80i3 200 Solving i1 5 A i2 0 A i3 25 A Then isc i1 i3 75 A RTh 360 75 48 Ω For maximum power transfer Ro RTh 48 Ω b pmax 1802 48 675 W c The problem reduces to the analysis of the following circuit In constructing the circuit we have used the fact that i is 1 A and hence 316i is 316 V Using the node voltage method to find v1 and v2 yields v1 400 64 v1 180 16 v2 180 32 v2 200 48 0 v2 v1 316 Solving v1 336 V v2 20 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 498 CHAPTER 4 Techniques of Circuit Analysis It follows that v a 44813 19213 V and v v a 163103 5 8v a 12013 V 9231 V v v v 770 13 120 13 50 V b p v2 10 250 W P 494 Voltage source acting alone io1 10 45 5 510 10 45 5 02 A vo1 20 20 6010 25 V Current source acting alone v2 5 2 v2 v3 5 0 v3 10 v3 v2 5 v3 45 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 499 Solving v2 725 V vo2 v3 45 V io2 v3 45 01 A i20 6020 20 2 15 A vo2 20i20 2015 30 V vo vo1 vo2 25 30 325 V io io1 io2 02 01 03 A P 495 6 A source 30 Ω5 Ω60 Ω 4 Ω io1 20 20 56 48 A 10 A source io2 4 2510 16 A 75 V source io3 4 2515 24 A io io1 io2 io3 48 16 24 4 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 4100 CHAPTER 4 Techniques of Circuit Analysis P 496 240 V source acting alone vo1 205 5 7 205240 60 V 84 V source acting alone vo2 2012 1 4 201284 504 V 16 A current source acting alone v1 v2 5 v1 7 16 0 v2 v1 5 v2 20 v2 v3 4 0 v3 v2 4 v3 1 16 0 Solving v2 184 V vo3 Therefore vo vo1 vo2 vo3 60 504 184 28 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 4102 CHAPTER 4 Techniques of Circuit Analysis P 498 90V source acting alone 2000i1 i2 25vb 90 2000i1 7000i2 4000i3 0 4000i2 6000i3 25vb 0 vb 1000i2 Solving i1 37895 mA i3 30789 mA i i1 i3 7105 mA 40V source acting alone 2000i1 i2 25vb 0 2000i1 7000i2 4000i3 0 4000i2 6000i3 25vb 40 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 4103 vb 1000i2 Solving i1 2105 mA i3 15789 mA i i1 i3 17895 mA Hence i i i 7105 17895 25 mA P 499 a In studying the circuit in Fig P499 we note it contains six meshes and six essential nodes Further study shows that by replacing the parallel resistors with their equivalent values the circuit reduces to four meshes and four essential nodes as shown in the following diagram The nodevoltage approach will require solving three node voltage equations along with equations involving vx vy and ix The meshcurrent approach will require writing one mesh equation and one supermesh equation plus five constraint equations involving the five sources Thus at the outset we know the supermesh equation can be reduced to a single unknown current Since we are interested in the power developed by the 50 V source we will retain the mesh current ib and eliminate the mesh currents ia ic and id The supermesh is denoted by the dashed line in the following figure b Summing the voltages around the supermesh yields 125vx 1003ia 2003ib 50 50ib 250ic id 50ic 0 The remaining mesh equation is 50ix 350id 250ic 0 The constraint equations are vy 25 ib ic 09 ic ia vx 2003ib vy 50ib ix ic id 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 4104 CHAPTER 4 Techniques of Circuit Analysis Solving ia 03 A ib 06 A ic 06 A id 04 A Finally p50V 50ib 30 W The 50 V source delivers 30 W of power P 4100 At v1 v1 120 10 v1 v2 20 v1 v3 10 0 At v2 v2 120 20 v2 120 20 v2 v1 20 v1 v3 20 v2 v4 20 0 At v3 v3 v1 10 v3 v2 20 v3 v4 20 v3 10 v3 v5 10 0 At v4 v4 v2 20 v4 v3 20 v4 v5 10 0 At v5 v5 v4 10 v5 v3 10 v5 5 0 A calculator solution yields v1 80 A v2 80 A v3 40 A v4 40 A v5 20 A i1 v2 v1 20 0 A i1 v3 v4 20 0 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 4108 CHAPTER 4 Techniques of Circuit Analysis P 4106 From the solution to Problem 4104 we have dv1 dIg2 255075 30125 3750 125 VA and dv2 dIg2 507530 30125 3750 15 VA By hypothesis Ig2 17 16 1 A v1 1251 125 V Thus v1 25 125 375 V Also v2 151 15 V Thus v2 90 15 105 V The PSpice solution is v1 375 V and v2 105 V These values are in agreement with our predicted values P 4107 From the solutions to Problems 4104 4106 we have dv1 dIg1 175 12 VA dv1 dIg2 125 VA dv2 dIg1 125 VA dv2 dIg2 15 VA By hypothesis Ig1 11 12 1 A Ig2 17 16 1 A Therefore v1 175 12 125 270833 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 4109 v2 125 15 275 V Hence v1 25 270833 520833 V v2 90 275 1175 V The PSpice solution is v1 520830 V and v2 1175 V These values are in agreement with our predicted values P 4108 By hypothesis R1 275 25 25 Ω R2 45 5 05 Ω R3 55 50 5 Ω R4 675 75 75 Ω So v1 0583325 541705 0455 0275 49168 V v1 25 49168 299168 V v2 0525 6505 0545 02475 11 V v2 90 11 889 V The PSpice solution is v1 296710 V and v2 885260 V Note our predicted values are within a fraction of a volt of the actual values 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 5 The Operational Amplifier Assessment Problems AP 51 a This is an inverting amplifier so vo RfRivs 8016vs so vo 5vs vs V 04 20 35 06 16 24 vo V 20 100 150 30 80 100 Two of the values 35 V and 24 V cause the op amp to saturate b Use the negative power supply value to determine the largest input voltage 15 5vs vs 3 V Use the positive power supply value to determine the smallest input voltage 10 5vs vs 2 V Therefore 2 vs 3 V AP 52 From Assessment Problem 51 vo RfRivs Rx16000vs Rx160000640 064Rx16000 4 105Rx Use the negative power supply value to determine one limit on the value of Rx 4 105Rx 15 so Rx 154 105 375 kΩ 51 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 52 CHAPTER 5 The Operational Amplifier Since we cannot have negative resistor values the lower limit for Rx is 0 Now use the positive power supply value to determine the upper limit on the value of Rx 4 105Rx 10 so Rx 104 105 250 kΩ Therefore 0 Rx 250 kΩ AP 53 a This is an inverting summing amplifier so vo RfRava RfRbvb 2505va 25025vb 50va 10vb Substituting the values for va and vb vo 5001 10025 5 25 75 V b Substitute the value for vb into the equation for vo from part a and use the negative power supply value vo 50va 10025 50va 25 10 V Therefore 50va 75 so va 015 V c Substitute the value for va into the equation for vo from part a and use the negative power supply value vo 50010 10vb 5 10vb 10 V Therefore 10vb 5 so vb 05 V d The effect of reversing polarity is to change the sign on the vb term in each equation from negative to positive Repeat part a vo 50va 10vb 5 25 25 V Repeat part b vo 50va 25 10 V 50va 125 va 025 V Repeat part c using the value of the positive power supply vo 5 10vb 15 V 10vb 20 vb 20 V AP 54 a Write a node voltage equation at vn remember that for an ideal op amp the current into the op amp at the inputs is zero vn 4500 vn vo 63000 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 56 CHAPTER 5 The Operational Amplifier Problems P 51 a The five terminals of the op amp are identified as follows b The input resistance of an ideal op amp is infinite which constrains the value of the input currents to 0 Thus in 0 A c The openloop voltage gain of an ideal op amp is infinite which constrains the difference between the voltage at the two input terminals to 0 Thus vp vn 0 d Write a node voltage equation at vn vn 2 4000 vn vo 12000 0 But vp 0 and vn vp 0 Thus 2 4000 vo 12000 0 so vo 6 V P 52 a Let the value of the voltage source be vs vn vs 4000 vn vo 12000 0 But vn vp 0 Therefore vo 12000 4000 vs 3vs When vs 6 V vo 36 18 V saturates at vo 15 V When vs 35 V vo 335 105 V When vs 125 V vo 3125 375 V When vs 24 V vo 324 72 V When vs 45 V vo 345 135 V When vs 54 V vo 354 162 V saturates at vo 15 V b 3vs 15 so vs 15 3 5 V 3vs 15 so vs 15 3 5 V The range of source voltages that avoids saturation is 5 V vs 5 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 57 P 53 vp 5000 5000 100006 2 V vn vn 5 3000 vn vo 6000 0 22 5 2 vo 0 vo 16 V iL vo 8000 16 8000 2000 106 iL 2 mA P 54 vb va 5000 vb vo 40000 0 therefore vo 9vb 8va a va 15 V vb 0 V vo 12 V b va 05 V vb 0 V vo 4 V c va 1 V vb 25 V vo 145 V d va 25 V vb 1 V vo 11 V e va 25 V vb 0 V vo 20 V saturates at 16 V f If vb 2 V vo 18 8va 16 025 V va 425 V P 55 vo 05 10310000 5 V io vo 5000 5 5000 1 mA P 56 a ia 240 103 8000 30 µA v1 40 103i2 3 V b 0 va 60000 ia so va 60000ia 18 V c va 60000 va 40000 va vo 30000 0 vo 225va 405 V d io vo 20000 va vo 30000 2775 µA 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 513 Now create the 5 resistor values needed from the realistic resistor values in Appendix H Note that Rf 120 kΩ Ra 15 kΩ and Rc 12 kΩ are already values from Appendix H Create Rb 30 kΩ by combining two 15 kΩ resistors in series Create Rd 20 kΩ by combining two 10 kΩ resistors in series Of course there are many other acceptable possibilities The final circuit is shown here P 518 a The circuit shown is a noninverting amplifier b We assume the op amp to be ideal so vn vp 2 V Write a KCL equation at vn 2 25000 2 vo 150000 0 Solving vo 14 V P 519 a This circuit is an example of the noninverting amplifier b Use voltage division to calculate vp vp 8000 8000 32000vs vs 5 Write a KCL equation at vn vp vs5 vs5 7000 vs5 vo 56000 0 Solving vo 8vs5 vs5 18vs c 18vs 12 so vs 667 V 18vs 15 so vs 833 V Thus 833 V vs 667 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 515 P 522 a From the equation for the noninverting amplifier Rs Rf Rs 25 so Rs Rf 25Rs and therefore Rf 15Rs Choose Rs 22 kΩ which is a component in Appendix H Then Rf 1522 33 kΩ which is also a resistor value in Appendix H The resulting noninverting amplifier circuit is shown here b vo 25vg 16 so vg 64 V vo 25vg 16 so vg 64 V Therefore 64 V vg 64 V P 523 a This circuit is an example of a noninverting summing amplifier b Write a KCL equation at vp and solve for vp in terms of vs vp 5 16000 vp vs 24000 0 3vp 15 2vp 2vs 0 so vp 2vs5 3 Now write a KCL equation at vn and solve for vo vn 24000 vn vo 96000 0 so vo 5vn Since we assume the op amp is ideal vn vp Thus vo 52vs5 3 2vs 15 c 2vs 15 10 so vs 25 V 2vs 15 10 so vs 125 V Thus 125 V vs 25 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 519 ia va 08va Ra 02 va Ra Rin va ia 5Ra 22 kΩ Ra 44 kΩ Rb 176 kΩ Create Ra 44 kΩ by combining two 22 kΩ resistors in series Create Rb 176 kΩ by combining a 12 kΩ resistor and a 56 kΩ resistor in series P 530 a vp 20000 vp vc 30000 vp vd 20000 0 8vp 2vc 3vd 8vn vn va 20000 vn vb 18000 vn vo 180000 0 vo 20vn 9va 10vb 2014vc 38vd 9va 10vb 20075 15 91 102 16 V b vo 5vc 30 9 20 5vc 1 20 5vc 1 vb 42 V and vb 38 V 42 V vb 38 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 522 CHAPTER 5 The Operational Amplifier If we use 1500 Rx 2994 Ω If we use 1500 Rx 3006 Ω 2994 Ω Rx 3006 Ω P 535 It follows directly from the circuit that vo 7515vg 5vg From the plot of vg we have vg 0 t 0 vg t 0 t 2 vg 4 t 2 t 6 vg t 8 6 t 10 vg 12 t 10 t 14 vg t 16 14 t 18 etc Therefore vo 5t 0 t 2 vo 5t 20 2 t 6 vo 40 5t 6 t 10 vo 5t 60 10 t 14 vo 80 5t 14 t 18 etc These expressions for vo are valid as long as the op amp is not saturated Since the peak values of vo are 9 the output is clipped at 9 The plot is shown below P 536 vp 54 72vg 075vg 3 cosπ4t V vn 20000 vn vo 60000 0 4vn vo vn vp 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 526 CHAPTER 5 The Operational Amplifier First it enables the source to control 16 times as much power delivered to the load resistor When a small amount of power controls a larger amount of power we refer to it as power amplification Second it allows the full source voltage to appear across the load resistor no matter what the source resistance This is the voltage follower function of the operational amplifier Third it allows the load resistor voltage and thus its current to be set without drawing any current from the input voltage source This is the current amplification function of the circuit P 541 a Let vo1 output voltage of the amplifier on the left Let vo2 output voltage of the amplifier on the right Then vo1 47 10 1 47 V vo2 220 33 015 10 V ia vo2 vo1 1000 57 mA b ia 0 when vo1 vo2 so from a vo2 1 V Thus 47 10 vL 1 vL 10 47 21277 mV P 542 a Assume the opamp is operating within its linear range then iL 8 4000 2 mA For RL 4 kΩ vo 4 42 16 V Now since vo 20 V our assumption of linear operation is correct therefore iL 2 mA b 20 24 RL RL 6 kΩ c As long as the opamp is operating in its linear region iL is independent of RL From b we found the opamp is operating in its linear region as long as RL 6 kΩ Therefore when RL 6 kΩ the opamp is saturated We can estimate the value of iL by assuming ip in iL Then iL 204000 16000 1 mA To justify neglecting the current into the opamp assume the drop across the 50 kΩ resistor is negligible since the input resistance to the opamp is at least 500 kΩ Then ip in 8 4500 103 8 µA But 8 µA 1 mA hence our assumption is reasonable 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 528 CHAPTER 5 The Operational Amplifier Write two node voltage equations one at the left node the other at the right node vn vg 5000 vn vo 100000 vn 500000 0 vo 3 105vn 5000 vo vn 100000 vo 500 0 Simplify and place in standard form 106vn 5vo 100vg 6 106 1vn 221vo 0 Let vg 1 V and solve the two simultaneous equations vo 199844 V vn 7361 µV Thus the voltage gain is vovg 199844 b From the solution in part a vn 7361 µV c ig vg vn 5000 vg 7361 106vg 5000 Rg vg ig 5000 1 7361 106 500368 Ω d For an ideal op amp the voltage gain is the ratio between the feedback resistor and the input resistor vo vg 100000 5000 20 For an ideal op amp the difference between the voltages at the input terminals is zero and the input resistance of the op amp is infinite Therefore vn vp 0 V Rg 5000 Ω P 545 a vn 088 1600 vn 500000 vn vTh 24000 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 530 CHAPTER 5 The Operational Amplifier vn 088 1600 vn 500000 vn 1318 24000 0 vn 942 µV ig 088 942 106 1600 54941 µA Rg 088 ig 160171 Ω P 546 a vTh 24000 1600 088 132 V RTh 0 since opamp is ideal b Ro RTh 0 Ω c Rg 16 kΩ since vn 0 P 547 a vn vg 2000 vn vo 10000 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 534 CHAPTER 5 The Operational Amplifier P 552 1 R48 104 10495 104 100 R 9500 48 19791667 Ω change in R 19719667 104 100 198 P 553 a It follows directly from the solution to Problem 550 that vo R2 2RRf R1R Rf RRfRfvin RR1R Rf RRf Now R1 R R Substituting into the expression gives vo R RfRfRvin RR RR Rf RRf Now let R R and get vo R RfRfRvin R2R 2Rf b It follows directly from the solution to Problem 550 that approx value true value RR RR Rf RRf R2R 2Rf Error R RR Rf RRf RR 2Rf RR 2Rf RR Rf RR 2Rf error RR Rf RR 2Rf 100 c R R 9810 Ω R 10000 9810 190 Ω vo 48 10447 10419015 10895 104 6768 V d error 19048 104100 10495 104 096 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 6 Inductance Capacitance and Mutual Inductance Assessment Problems AP 61 a ig 8e300t 8e1200tA v Ldig dt 96e300t 384e1200tV t 0 v0 96 384 288 V b v 0 when 384e1200t 96e300t or t ln 4900 154 ms c p vi 384e1500t 768e600t 3072e2400t W d dp dt 0 when e1800t 125e900t 16 0 Let x e900t and solve the quadratic x2 125x 16 0 x 144766 t ln 145 900 41105 µs x 110523 t ln 1105 900 267 ms p is maximum at t 41105 µs e pmax 384e15041105 768e06041105 3072e24041105 3272 W f W is max when i is max i is max when didt is zero When didt 0 v 0 therefore t 154 ms g imax 8e03154 e12154 378 A wmax 124 1033782 286 mJ 61 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 64 CHAPTER 6 Inductance Capacitance and Mutual Inductance 8di2 dt 3168e4t 40e5t 20i2 020 1980e4t 20e5t 5ig 98 98e4t 8dig dt 6272e4t Test 18560e4t 240e5t 10 290e4t 300e5t 3168e4t 40e5t 020 1980e4t 20e5t 98 98e4t 6272e4t 98 300 240 40 20e5t 18560 290 3168 1980e4t 98 5292e4t 98 0e5t 23708 290e4t 98 5292e4t 98 5292e4t 98 5292e4t OK Also 8di1 dt 37120e4t 480e5t 20i1 8 232e4t 240e5t 16di2 dt 6336e4t 80e5t 800i2 8 792e4t 800e5t 16dig dt 12544e4t Test 37120e4t 480e5t 8 232e4t 240e5t 6336e4t 80e5t 8 792e4t 800e5t 12544e4t 8 8 800 480 240 80e5t 37120 232 6336 792e4t 12544e4t 800 800e5t 66656 792e4t 12544e4t 12544e4t 12544e4t OK 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 65 Problems P 61 a v Ldi dt 150 10625e500t 500te500t 375e500t1 500t mV b i5 ms 250005e25 1026 mA v5 ms 000375e251 25 46173 µV p5 ms vi 1026 10346173 106 474 µW c delivering 474 µW d i5 ms 1026 mA from part b w 1 2Li2 1 2150 1060010262 79 nJ e The energy is a maximum where the current is a maximum diL dt 0 when 1 500t 0 or t 2 ms imax 250002e1 1839 mA wmax 1 2150 1060018392 2538 nJ P 62 a i 0 t 0 i 4t A 0 t 25 ms i 02 4t A 25 t 50 ms i 0 50 ms t b v Ldi dt 500 1034 2 V 0 t 25 ms v 500 1034 2 V 25 t 50 ms v 0 t 0 v 2 V 0 t 25 ms v 2 V 25 t 50 ms v 0 50 ms t p vi 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 612 CHAPTER 6 Inductance Capacitance and Mutual Inductance b p vi 20 cos 80t25 sin 80t 50 cos 80t sin 80t p 25 sin 160t W w 1 2Li2 1 20125 sin 80t2 3125 sin2 80t mJ w 15625 15625 cos 160t mJ 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 614 CHAPTER 6 Inductance Capacitance and Mutual Inductance Rm 201000 20 kΩ vm16 s 096 10320 103 192 V P 613 a i C dv dt 5 106500t2500e2500t 500e2500t 25 103e2500t1 2500t A b v100 µ 500100 106e025 3894 mV i100 µ 25 103e0251 025 146 mA p100 µ vi 3894 103146 103 5686 µW c p 0 so the capacitor is absorbing power d v100 µ 3894 mV w 1 2Cv2 1 25 1063894 1032 379 nJ e The energy is maximum when the voltage is maximum dv dt 0 when 1 2500t 0 or t 04 ms vmax 50004 1032e1 7358 mV pmax 1 2Cv2 max 1353 nJ P 614 a v 0 t 0 v 10t A 0 t 2 s v 40 10t A 2 t 6 s v 10t 80 A 6 t 8 s v 0 8 s t b i C dv dt i 0 t 0 i 2 mA 0 t 2 s i 2 mA 2 t 6 s i 2 mA 6 t 8 s i 0 8 s t 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 616 CHAPTER 6 Inductance Capacitance and Mutual Inductance From the plot of power above it is clear that power is being absorbed for 0 t 2 s and for 4 s t 6 s because p 0 Likewise power is being delivered for 2 s t 4 s and 6 s t 8 s because p 0 P 615 a w0 1 2Cv02 1 25 106602 9 mJ b v A1 A2te1500t v0 A1 60 V dv dt 1500e1500tA1 A2t e1500tA2 1500A2t 1500A1 A2e1500t dv0 dt A2 1500A1 i C dv dt i0 C dv0 dt dv0 dt i0 C 100 103 5 106 20 103 20 103 A2 150060 Thus A2 20 103 90 103 110 103 V s c v 60 110 103te1500t i C dv dt 5 106 d dt60 110 103te1500t i 5 106110000e1500t 150060 110 000te1500t 01 825te1500t A t 0 P 616 iC Cdvdt 0 t 2 s iC 100 10915t2 15 106t2 A 2 t 4 s iC 100 10915t 42 15 106t 42 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 617 P 617 a i C dv dt 0 t 0 b i C dv dt 120 106 d dt30 5e500t6 cos 2000t sin 2000t 120 1065500e500t6 cos 2000t sin 2000t 52000e500t6 sin 2000t cos 2000t 06e500tcos 2000t 125 sin 2000t A t 0 c no v0 60 V v0 30 56 60 V d yes i0 0 A i0 06 A e v 30 V w 1 2Cv2 1 2120 106302 54 mJ P 618 a v20 µs 125 10920 1062 5 V end of first interval v20 µs 10620 106 125400 103 10 5 V start of second interval v40 µs 10640 106 1251600 103 10 10 V end of second interval b p10µs 625 10121053 625 mW v10 µs 125 V i10µs 50 mA p10 µs vi 12550 m 625 mW checks p30 µs 43750 mW v30 µs 875 V i30 µs 005 A p30 µs vi 875005 625 mW checks 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 620 CHAPTER 6 Inductance Capacitance and Mutual Inductance d 40 µs t v40 µs 64 12 52 V 40 µs t P 622 a 1530 10 mH 10 10 20 mH 2020 10 mH 1224 8 mH 10 8 18 mH 189 6 mH Lab 6 8 14 mH b 12 18 30 µH 3020 12 µH 12 38 50 µH 307550 15 µH 15 15 30 µH 3060 20 µH Lab 20 25 45 µH P 623 a Combine two 10 mH inductors in parallel to get a 5 mH equivalent inductor Then combine this parallel pair in series with three 1 mH inductors 10 m10 m 1 m 1 m 1 m 8 mH b Combine two 10 µH inductors in parallel to get a 5 µH inductor Then combine this parallel pair in series with four more 10 µH inductors 10 µ10 µ 10 µ 10 µ 10 µ 10 µ 45 µH 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 624 CHAPTER 6 Inductance Capacitance and Mutual Inductance 1 C5 1 20 1 20 1 10 1 5 C5 5 nF Equivalent capacitance is 5 nF with an initial voltage drop of 15 V b 1 36 1 18 1 12 1 6 Ceq 6 µF 24 6 30 µF 25 5 30 µF 1 30 1 30 1 30 3 30 Ceq 10 µF Equivalent capacitance is 10 µF with an initial voltage drop of 25 V P 628 a Combine a 470 pF capacitor and a 10 pF capacitor in parallel to get a 480 pF capacitor 470 p in parallel with 10 p 470 p 10 p 480 pF 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 629 50 103 dio dt 0 0 so v20 0 v10 25io0 v20 252 0 50 V P 636 a Rearrange by organizing the equations by di1dt i1 di2dt i2 and transfer the ig terms to the right hand side of the equations We get 4di1 dt 25i1 8di2 dt 20i2 5ig 8dig dt 8di1 dt 20i1 16di2 dt 80i2 16dig dt b From the given solutions we have di1 dt 320e5t 272e4t di2 dt 260e5t 204e4t Thus 4di1 dt 1280e5t 1088e4t 25i1 100 1600e5t 1700e4t 8di2 dt 2080e5t 1632e4t 20i2 20 1040e5t 1020e4t 5ig 80 80e5t 8dig dt 640e5t Thus 1280e5t 1088e4t 100 1600e5t 1700e4t 2080e5t 1632e4t 20 1040e5t 1020e4t 80 80e5t 640e5t 80 1088 1700 1632 1020e4t 1600 1280 2080 1040e5t 80 720e5t 80 2720 2720e4t 2640 3360e5t 80 720e5t OK 8di1 dt 2560e5t 2176e4t 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 630 CHAPTER 6 Inductance Capacitance and Mutual Inductance 20i1 80 1280e5t 1360e4t 16di2 dt 4160e5t 3264e4t 80i2 80 4160e5t 4080e4t 16dig dt 1280e5t 2560e5t 2176e4t 80 1280e5t 1360e4t 4160e5t 3264e4t 80 4160e5t 4080e4t 1280e5t 80 80 2560 1280 4160 4160e5t 1360 2176 3264 4080e4t 1280e5t 0 1280e5t 0e4t 1280e5t OK P 637 a Yes using KVL around the lower right loop vo v20Ω v60Ω 20i2 i1 60i2 b vo 201 52e5t 51e4t 4 64e5t 68e4t 601 52e5t 51e4t 203 116e5t 119e4t 60 3120e5t 3060e4t vo 5440e5t 5440e4t V c vo L2 d dtig i2 M di1 dt 16 d dt15 36e5t 51e4t 8 d dt4 64e5t 68e4t 2880e5t 3264e4t 2560e5t 2176e4t vo 5440e5t 5440e4t V P 638 a vg 5ig i1 20i2 i1 60i2 516 16e5t 4 64e5t 68e4t 201 52e5t 51e4t 4 64e5t 68e4t 601 52e5t 51e4t 60 5780e4t 5840e5t V b vg0 60 5780 5840 0 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 636 CHAPTER 6 Inductance Capacitance and Mutual Inductance b The mutualcapacitance and the touch capacitance are effectively connected in series Therefore the mutual capacitance between the xgrid and ygrid electrodes closest to the touch point is C mxy CmxyCt Cmxy Ct 3015 30 15 10 pF c In the selfcapacitance design touching the screen increases the capacitance being measured at the point of touch For example in part a the measured capacitance before the touch is 30 pF and after the touch is 45 pF In the mutualcapacitance design touching the screen decreases the capacitance being measured at the point of touch For example in part b the measured capacitance before the touch is 30 pF and after the touch is 10 pF P 653 a The four touch points identified are the two actual touch points and two ghost touch points Their coordinates in inches from the upper left corner of the screen are 21 43 32 25 21 25 and 32 43 These four coordinates identify a rectangle within the screen shown below b The touch points identified at time t1 are those listed in part a The touch points recognized at time t2 are 18 49 39 18 18 18 and 39 49 The first two coordinates are the actual touch points and the last two coordinates are the associated ghost points Again the four coordinates identify a rectangle at time t2 as shown here 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 637 Note that the rectangle at time t2 is larger than the rectangle at time t1 so the software would recognize the two fingers are moving toward the edges of the screen This pinch gesture thus specifies a zoomin for the screen c The touch points identified at time t1 are those listed in part a The touch points recognized at time t2 are 28 39 30 28 28 28 and 30 39 The first two coordinates are the actual touch points and the last two coordinates are the associated ghost points Again the four coordinates identify a rectangle at time t2 as shown here 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 638 CHAPTER 6 Inductance Capacitance and Mutual Inductance Here the rectangle at time t2 is smaller than the rectangle at time t1 so the software would recognize the two fingers are moving toward the middle of the screen This pinch gesture thus specifies a zoomout for the screen 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 7 Response of FirstOrder RL and RC Circuits Assessment Problems AP 71 a The circuit for t 0 is shown below Note that the inductor behaves like a short circuit effectively eliminating the 2 Ω resistor from the circuit First combine the 30 Ω and 6 Ω resistors in parallel 306 5 Ω Use voltage division to find the voltage drop across the parallel resistors v 5 5 3120 75 V Now find the current using Ohms law i0 v 6 75 6 125 A b w0 1 2Li20 1 28 1031252 625 mJ c To find the time constant we need to find the equivalent resistance seen by the inductor for t 0 When the switch opens only the 2 Ω resistor remains connected to the inductor Thus τ L R 8 103 2 4 ms d it i0etτ 125et0004 125e250t A t 0 e i5 ms 125e2500005 125e125 358 A So w 5 ms 1 2Li25 ms 1 28 1033582 513 mJ 71 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 74 CHAPTER 7 Response of FirstOrder RL and RC Circuits Find the current in the loop and use it to find the initial voltage drops across the two RC circuits i 15 75000 02 mA v50 4 V v10 8 V There are two time constants in the circuit one for each RC subcircuit τ5 is the time constant for the 5 µF 20 kΩ subcircuit and τ1 is the time constant for the 1 µF 40 kΩ subcircuit τ5 20 1035 106 100 ms τ1 40 1031 106 40 ms Therefore v5t v50etτ5 4et01 4e10t V t 0 v1t v10etτ1 8et004 8e25t V t 0 Finally vot v1t v5t 8e25t 4e10t V t 0 b Find the value of the voltage at 60 ms for each subcircuit and use the voltage to find the energy at 60 ms v160 ms 8e25006 179 V v560 ms 4e10006 220 V w160 ms 1 2Cv2 160 ms 1 21 1061792 159 µJ w560 ms 1 2Cv2 560 ms 1 25 1062202 1205 µJ w60 ms 159 1205 1364 µJ Find the initial energy from the initial voltage w0 w10 w20 1 21 10682 1 25 10642 72 µJ Now calculate the energy dissipated at 60 ms and compare it to the initial energy wdiss w0 w60 ms 72 1364 5836 µJ dissipated 5836 10672 106100 8105 AP 75 a Use the circuit at t 0 shown below to calculate the initial current in the inductor i0 242 12 A i0 Note that i0 i0 because the current in an inductor is continuous b Use the circuit at t 0 shown below to calculate the voltage drop across the inductor at 0 Note that this is the same as the voltage drop across the 10 Ω resistor which has current from two sources 8 A from the current source and 12 A from the initial current through the inductor 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 75 v0 108 12 200 V c To calculate the time constant we need the equivalent resistance seen by the inductor for t 0 Only the 10 Ω resistor is connected to the inductor for t 0 Thus τ LR 200 10310 20 ms d To find it we need to find the final value of the current in the inductor When the switch has been in position a for a long time the circuit reduces to the one below Note that the inductor behaves as a short circuit and all of the current from the 8 A source flows through the short circuit Thus if 8 A Now it if i0 ifetτ 8 12 8et002 8 20e50t A t 0 e To find vt use the relationship between voltage and current for an inductor vt Ldit dt 200 1035020e50t 200e50t V t 0 AP 76 a From Example 76 vot 60 90e100t V Write a KCL equation at the top node and use it to find the relationship between vo and vA vA vo 8000 vA 160000 vA 75 40000 0 20vA 20vo vA 4vA 300 0 25vA 20vo 300 vA 08vo 12 Use the above equation for vA in terms of vo to find the expression for vA vAt 0860 90e100t 12 60 72e100t V t 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 710 CHAPTER 7 Response of FirstOrder RL and RC Circuits Problems P 71 a io0 20 16 12 4 8 20 40 05 A io 0 A b io 05etτ τ L R 80 103 12 8 4 ms io 05e250t A t 0 c 05e250t 01 e250t 5 t 644 ms P 72 a For t 0 ig 50 20 7550 50 50 1 A io0 50 75 501 04 A io0 For t 0 iot io0etτ A t 0 τ L R 002 3 6015 133 ms 1 τ 750 iot 04e750t A t 0 b vL Ldio dt 00275004e750t 6e750t V vo 6015 3 6015vL 12 156e750t 48e750t V t 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 715 e The current in a resistor can change instantaneously The switching operation forces i20 to equal 12 mA and i20 4 mA P 79 a For t 0 the circuit is io0 0 since the switch is open iL0 25 250 01 100 mA vL0 0 since the inductor behaves like a short circuit b For t 0 the circuit is iL0 iL0 100 mA ig 25 50 05 500 mA io0 ig iL0 500 100 400 mA 200iL0 vL0 0 vL0 200iL0 20 V c As t the circuit is iL 0 vL 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 718 CHAPTER 7 Response of FirstOrder RL and RC Circuits c w0 1 2LI2 o 1 20010012 05 µJ wt 1 2001001e1000t2 05 106e2000t So 05 106e2000t 1 2w0 025 106 e2000t 05 then e2000t 2 t ln 2 2000 34657 µs for a 10 mH inductor P 714 t 0 iL0 iL0 12 A t 0 Find Thevenin resistance seen by inductor iT 25vT vT iT RTh 1 25 04 Ω τ L R 20 103 04 50 ms 1τ 20 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 719 io 12e20t A t 0 vo Ldio dt 20 103240e20t 48e20t V t 0 P 715 a t 0 iL0 72 24 6 24 A t 0 i 100 160iT 5 8iT vT 20i iT 10060 160 125iT 375iT vT iT RTh 125 375 25 Ω τ L R 250 103 25 1 τ 100 iL 24e100t A t 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 722 CHAPTER 7 Response of FirstOrder RL and RC Circuits t 0 120 iab 18 12 iab 90 A t 0 b At t iab 2402 120 A t c i10 18 τ1 2 103 10 02 ms i20 12 τ2 6 103 15 04 ms i1t 18e5000t A t 0 i2t 12e2500t A t 0 iab 120 18e5000t 12e2500t A t 0 120 18e5000t 12e2500t 114 6 18e5000t 12e2500t Let x e2500t so 6 18x2 12x Solving x 1 3 e2500t 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 723 e2500t 3 and t ln 3 2500 43944 µs P 721 a For t 0 v0 20000001 200 V b w0 1 2Cv02 1 2400 1092002 8 mJ c For t 0 Req 10000 5000075000 40 kΩ τ ReqC 40000400 109 16 ms d vt v0etτ 200e625t V t 0 P 722 For t 0 Vo 200006000020 103 300 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 724 CHAPTER 7 Response of FirstOrder RL and RC Circuits For t 0 Req 10000 2000060000 25 kΩ τ ReqC 2500040 109 1 ms vt Voetτ 300e1000t V t 0 P 723 a For t 0 Vo 10000 15000120 80 V For t 0 Req 2500040000 10000 1667 kΩ τ ReqC 1666667160 109 267 ms vt Voetτ 80e375t V t 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 728 CHAPTER 7 Response of FirstOrder RL and RC Circuits vc 02e312000t V t 0 i1 vc 2 01e312000t A t 0 f i2 vc 8 25e312000t mA t 0 P 728 t 0 t 0 vT 5io 15io 20io 20iT RTh vT iT 20 Ω τ RC 40 µs 1 τ 25000 vo 15e25000t V t 0 io vo 20 075e25000t A t 0 P 729 a R v i 8 kΩ 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 730 CHAPTER 7 Response of FirstOrder RL and RC Circuits v 5 103iT vT 25 103iT 20 103α5 103iT RTh 25000 100 106α τ RThC 40 103 RTh08 106 RTh 50 kΩ 25000 100 106α α 25000 100 106 25 104 AV b vo0 5 1033600 18 V t 0 t 0 vo 18e25t V t 0 v 5000 v vo 20000 25 104v 0 4v v vo 5v 0 v vo 10 18e25t V t 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 734 CHAPTER 7 Response of FirstOrder RL and RC Circuits t 0 iL 32 48 12 8 08 A τ L R 5 103 12 8 250 µs 1 τ 4000 iL iL iL0 iLetτ 08 16 08e4000t 08 24e4000t A t 0 vo 8iL 48 808 24e4000t 48 416 192e4000t V t 0 b vL LdiL dt 5 103400024e4000t 48e4000t V t 0 vL0 48 V From part a vo0 0 V Check at t 0 the circuit is vo0 48 8 Ω16 A 608 V vL0 vo0 1216 32 vL0 192 32 608 48 V P 737 a t 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 735 KVL equation at the top node vo 240 60 vo 20 vo 5 0 Multiply by 60 and solve 240 3 1 12vo vo 15 V io0 vo 5 155 3 A t 0 Use voltage division to find the Thevenin voltage VTh vo 20 20 5225 180 V Remove the voltage source and make series and parallel combinations of resistors to find the equivalent resistance RTh 5 205 5 4 9 Ω The simplified circuit is τ L R 10 103 9 111 ms 1 τ 900 io 190 9 20 A io io io0 ioetτ 20 3 20e900t 20 17e900t A t 0 b vo 5io Ldio dt 520 17e900t 00190017e900t 100 85e900t 153e900t vo 100 68e900t V t 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 737 v Vs IoReRLt Vs R 4 Io Vs R 4 Vs IoR 80 R L 40 Io 4 Vs R 8 A Now since Vs 4R we have 4R 8R 80 R 20 Ω Vs 80 V L R 40 05 H b i 4 4e40t i2 16 32e40t 16e80t w 1 2Li2 1 20516 32e40t 16e80t 4 8e40t 4e80t 4 8e40t 4e80t 9 or e80t 2e40t 125 0 Let x e40t x2 2x 125 0 Solving x 05 x 25 But x 0 for all t Thus e40t 05 e40t 2 t 25 ln 2 1733 ms P 740 a Note that there are many different possible solutions to this problem R L τ Choose a 1 mH inductor from Appendix H Then R 0001 8 106 125 Ω Construct the resistance needed by combining 100 Ω 10 Ω and 15 Ω resistors in series 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 740 CHAPTER 7 Response of FirstOrder RL and RC Circuits t 0 Find Thevenin equivalent with respect to a b Use a test source to find the Thevenin equivalent resistance 1 01vφ vT vx 20 0 vx vT 20 vx 10 vx 55 0 vφ 40 55vx Solving vT 74 V so RTh vT 1 A 74 Ω Find the open circuit voltage with respect to a b 01vφ vTh vx 20 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 741 vx vTh 20 vx 140 10 vx 150 55 0 vφ 40 55vx 150 Solving vTh 96 V io 96124 0774 A τ 40 103 124 03226 ms 1τ 3100 io 0774 6 0774e3100t 0774 5226e3100t A t 0 P 745 t 0 calculate vo0 va 15 va vo0 5 20 103 va 075vo0 75 103 15 103 vo0 va 5 vo0 8 9i 50 103 0 13vo0 8va 360i 2600 103 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 742 CHAPTER 7 Response of FirstOrder RL and RC Circuits i vo0 8 9i 50 103 i vo0 80 5 103 360i 45vo0 1800 103 8va 6vo0 600 103 13vo0 6vo0 600 103 45vo0 1800 103 2600 103 25vo0 200 103 vo0 80 mV vo 0 Find the Thevenin resistance seen by the 4 mH inductor iT vT 20 vT 8 9i i vT 8 9i 10i vT 8 i vT 80 iT vT 20 10vT 80 9vT 80 iT vT 1 20 1 80 5 80 1 16 S RTh 16Ω τ 4 103 16 025 ms 1τ 4000 vo 0 80 0e4000t 80e4000t mV t 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 746 CHAPTER 7 Response of FirstOrder RL and RC Circuits f i C dvc dt 25 1091000270e1000t 675e1000t mA t 0 P 752 a for t 0 vc0 4000015 6 V For t 0 vc 10 V Req 20 Ω so τ ReqC 25025 106 625 ms vct vc vc0 vcetτ 10 6 10e160t 10 4e160t V b For t 0 vc0 10 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 747 For t 0 vc 4000015 6 V Req 100 400 500 Ω so τ ReqC 50025 106 125 ms vct vc vc0 vcetτ 6 10 6e80t 6 4e80t V P 753 a Use voltage division to find the initial value of the voltage vc0 v9k 9 k 9 k 3 k120 90 V b Use Ohms law to find the final value of voltage vc v40k 15 10340 103 60 V c Find the Thevenin equivalent with respect to the terminals of the capacitor VTh 60 V RTh 10 k 40 k 50 kΩ τ RThC 1 ms 1000 µs d vc vc vc0 vcetτ 60 90 60e1000t 60 150e1000t V t 0 We want vc 60 150e1000t 0 Therefore t ln15060 1000 9163 µs P 754 a For t 0 vo0 10000 1500075 50 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 749 RTh 50 kΩ50 kΩ 25 kΩ C 16 nF τ 2500016 109 04 ms 1 τ 2500 vot 50 150e2500t V t 0 ic C dvo dt 6e2500t mA t 0 i50k vo 50000 1 3e2500t mA t 0 io ic i50k 1 3e2500t mA t 0 P 756 For t 0 Simplify the circuit 8010000 8 mA 10 kΩ40 kΩ24 kΩ 6 kΩ 8 mA 3 mA 5 mA 5 mA 6 kΩ 30 V Thus for t 0 vo0 vo0 30 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 750 CHAPTER 7 Response of FirstOrder RL and RC Circuits t 0 Simplify the circuit 8 mA 2 mA 10 mA 10 k40 k24 k 6 kΩ 10 mA6 kΩ 60 V Thus for t 0 vo 10 1036 103 60 V τ RC 10 k005 µ 05 ms 1 τ 2000 vo vo vo0 voetτ 60 30 60e2000t 60 90e2000t V t 0 P 757 Use voltage division to find the initial voltage vo0 60 40 6050 30 V Use Ohms law to find the final value of voltage vo 5 mA20 kΩ 100 V τ RC 20 103250 109 5 ms 1 τ 200 vo vo vo0 voetτ 100 30 100e200t 100 130e200t V t 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 752 CHAPTER 7 Response of FirstOrder RL and RC Circuits c vt 25 75e4t 50 so e4t 1 3 t ln 3 4 27465 ms P 760 For t 0 VTh 2516000ib 400 103ib ib 33000 80000120 106 495 µA VTh 400 103495 106 198 V RTh 16 kΩ vo 198 V vo0 0 τ 16 000025 106 4 ms 1τ 250 vo 198 198e250t V t 0 wt 1 2025 106v2 o w1 e250t2 J 1 e250t2 036w w 036 1 e250t 06 e250t 04 t 367 ms 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 757 b io0 i10 i20 0 consistent with initial conditions vo0 120 V consistent with io0 0 vo 3di1 dt 120e3t V t 0 or vo 15di2 dt 120e3t V t 0 The voltage solution is consistent with the current solutions λ1 3i1 40 40e3t Wbturns λ2 15i2 40 40e3t Wbturns λ1 λ2 as it must since vo dλ1 dt dλ2 dt λ1 λ2 40 Wbturns λ1 3i1 3403 40 Wbturns λ2 15i2 1583 40 Wbturns i1 and i2 are consistent with λ1 and λ2 P 768 a From Example 710 Leq L1L2 M2 L1 L2 2M 0125 00625 075 05 50 mH τ L R 1 5000 1 τ 5000 iot 40 40e5000t mA t 0 b vo 10 250io 10 250004 004e5000t 10e5000t V t 0 c vo 05di1 dt 025di2 dt 10e5000t V io i1 i2 dio dt di1 dt di2 dt 200e5000t As di2 dt 200e5000t di1 dt 10e5000t 05di1 dt 50e5000t 025di1 dt 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 759 P 770 a Leq 002 004 20015 003 30 mH τ L R 003 4500 667 µs 1 τ 150000 i 002 002e150000t A t 0 b v1t 002di dt 0015di dt 0005di dt 00053000e150000t 15e150000t V t 0 c v2t 004di dt 0015di dt 0025di dt 00253000e150000t 75e150000t V t 0 d i0 0 which agrees with initial conditions 90 4500i1 v1 v2 4500002 002e150000t 15e150000t 75e150000t 90 V Therefore Kirchhoffs voltage law is satisfied for all values of t 0 Thus the answers make sense in terms of known circuit behavior P 771 a From Example 710 Leq L1L2 M2 L1 L2 2M 50 25 15 10 1 H τ L R 1 20 1 τ 20 iot 4 4e20t A t 0 b vo 80 20io 80 80 80e20t 80e20t V t 0 c vo 5di1 dt 5di2 dt 80e20t V io i1 i2 dio dt di1 dt di2 dt 80e20t As di2 dt 80e20t di1 dt 80e20t 5di1 dt 400e20t 5di1 dt 10di1 dt 480e20t di1 48e20t dt 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 761 0 t 10 ms i 10e100t A i10 ms 10e1 368 A 10 ms t 20 ms Req 520 25 4 Ω 1 τ R L 4 50 103 80 i 368e80t001 A 20 ms t i20 ms 368e80002001 165 A i 165e100t002 A vo Ldi dt L 50 mH di dt 165100e100t002 165e100t002 vo 50 103165e100t002 826e100t002 V t 20 ms vo25 ms 826e1000025002 5013 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 762 CHAPTER 7 Response of FirstOrder RL and RC Circuits P 773 From the solution to Problem 772 the initial energy is w0 1 250 mH10 A2 25 J 004w0 01 J 1 250 103i2 L 01 so iL 2 A Again from the solution to Problem 773 t must be between 10 ms and 20 ms since i10 ms 368 A and i20 ms 165 A For 10 ms t 20 ms i 368e80t001 2 e80t001 368 2 so t 001 00076 t 176 ms P 774 t 0 iL0 75 mA iL0 0 t 25 ms τ 0010 iLt 0075et 0075e0 75 mA 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 763 25 ms t τ 001 500 20 µs 1τ 50000 iLt 75e50000t0025 mA t 25 ms P 775 a t 0 Using Ohms law ig 800 40 6040 125 A Using current division i0 60 60 40125 75 A i0 b 0 t 1 ms i i0etτ 75etτ 1 τ R L 40 12060 80 103 1000 i 75e1000t i200µs 75e103200106 75e02 614 A c i1 ms 75e1 27591 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 766 CHAPTER 7 Response of FirstOrder RL and RC Circuits vo 20 20e250t V 0 t 25 ms 25 ms t t i 70 V 25 kΩ 28 mA vo 28 1032000 50 6 V vo00025 20 20e0625 929 V vo 6 929 6et 00025τ RTh 2000500 400 Ω τ 4008 106 32 ms 1τ 3125 vo 6 329e3125t 00025 25 ms t P 778 Note that for t 0 vo 1015vc where vc is the voltage across the 25 nF capacitor Thus we will find vc first t 0 vc0 30 4010 75 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 767 0 t 02 ms τ ReC Re 1500030000 10 kΩ τ 10 10325 109 025 ms 1 τ 4000 vc 75e4000t V t 0 vc02 ms 75e08 337 V 02 ms t 08 ms τ 15 10325 109 375 µs 1 τ 266667 vc 337e266667t200106 V 08 ms t τ 025 ms 1 τ 4000 vc08 ms 337e266667800200106 337e16 068 V vc 068e4000t08103 V vc1 ms 068e4000108103 068e08 0306 V vo 10150306 0204 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 769 0 t 5 s τ 1τ 0 vo 5e0 5 V 5 s t τ 10001 10 s 1τ 01 vo 5e01t 5 V Summary vo 5 V 0 t 5 s vo 5e01t 5 V 5 s t P 781 a io0 0 io 50 mA 1 τ R L 3000 75 103 40000 io 50 50e40000t mA 0 t 25 µs vo 0075dio dt 150e40000t V 0 t 25 µs 25 µs t io25µs 50 50e1 316 mA io 0 io 316e40000t25106 mA vo 0075dio dt 9482e40000t25µs t 0 vo 0 0 t 25 µs vo 150e40000t V 25 µs t vo 9482e40000t25µs V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 770 CHAPTER 7 Response of FirstOrder RL and RC Circuits b vo25µs 150e1 5518 V vo25µs 9482 V c io25µs io25µs 316 mA P 782 a 0 t 25 ms vo0 80 V vo 0 τ L R 2 ms 1τ 500 vot 80e500t V 0 t 25 ms vo25 ms 80e125 2292 V io25 ms 80 2292 20 285 A vo25 ms 20285 5708 V vo 0 τ 2 ms 1τ 500 vo 5708e500t 00025 V t 25 ms b c vo5 ms 1635 V io 1635 20 81768 mA P 783 a t 0 vo 0 0 t 25 µs τ 400050 109 02 ms 1τ 5000 vo 10 10e5000t V 0 t 25 µs 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 771 vo25 µs 101 e0125 1175 V 25 µs t 50 µs vo 10 11175e5000t25106 V 25 µs t 50 µs vo50 µs 10 11175e0125 0138 V t 50 µs vo 0138e5000t50106 V t 50 µs b c t 0 vo 0 0 t 25 µs τ 80050 109 40 µs 1τ 25000 vo 10 10e25000t V 0 t 25 µs vo25 µs 10 10e0625 465 V 25 µs t 50 µs vo 10 1465e25000t25106 V 25 µs t 50 µs vo50 µs 10 1465e0625 216 V t 50 µs vo 216e25000t50106 V t 50 µs 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 772 CHAPTER 7 Response of FirstOrder RL and RC Circuits P 784 a 0 t 1 ms vc0 0 vc 50 V RC 400 103001 106 4 ms 1RC 250 vc 50 50e250t vo 50 50 50e250t 50e250t V 0 t 1 ms 1 ms t vc1 ms 50 50e025 1106 V vc 0 V τ 4 ms 1τ 250 vc 1106e250t 0001 V vo vc 1106e250t 0001 V t 1 ms b 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 773 P 785 vT 2000iT 4000iT 2 103vφ 6000iT 8vφ 6000iT 82000iT vT iT 10000 τ 10 10000 1 ms 1τ 1000 i 25e1000t mA 25e1000t 103 5 t ln 200 1000 53 ms P 786 a Using Ohms law vT 4000iσ Using current division iσ 12000 12000 4000iT βiσ 075iT 075βiσ 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 774 CHAPTER 7 Response of FirstOrder RL and RC Circuits Solve for iσ iσ1 075β 075iT iσ 075iT 1 075β vT 4000iσ 3000iT 1 075β Find β such that RTh 4 kΩ RTh vT iT 3000 1 075β 4000 1 075β 075 β 233 b Find VTh Write a KCL equation at the top node VTh 30 4000 VTh 12000 233iσ 0 The constraint equation is iσ VTh 30 4000 0 Solving VTh 40 V Write a KVL equation around the loop 40 4000i 008di dt Rearranging di dt 500 50000i 25000i 001 Separate the variables and integrate to find i di i 001 50000 dt 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 776 CHAPTER 7 Response of FirstOrder RL and RC Circuits iT vT 5 2 8 10000 1 10000 vT iT 10000 1 10 kΩ Open circuit voltage calculation The node voltage equations voc 2000 voc v1 1000 4i 0 v1 voc 1000 v1 4000 5 103 0 The constraint equation i v1 4000 Solving voc 80 V v1 60 V vc0 0 vc 80 V τ RC 1000016 106 16 ms 1 τ 625 vc vc vc0 vcetτ 80 80e625t 14400 Solve for the time of the maximum voltage rating e625t 181 625t ln 181 t 8309 ms 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 783 b ib2 VCC R 25993 µA 5 t 0 µs ib2 0 0 t RC ln 2 ib2 VCC R VCC RL etRC ln 2RLC 25993 300e02106t4106 µA RC ln 2 t P 799 a While T2 has been ON C2 is charged to VCC positive on the left terminal At the instant T1 turns ON the capacitor C2 is connected across b2 e2 thus vbe2 VCC This negative voltage snaps T2 OFF Now the polarity of the voltage on C2 starts to reverse that is the righthand terminal of C2 starts to charge toward VCC At the same time C1 is charging toward VCC positive on the right At the instant the charge on C2 reaches zero vbe2 is zero T2 turns ON This makes vbe1 VCC and T1 snaps OFF Now the capacitors C1 and C2 start to charge with the polarities to turn T1 ON and T2 OFF This switching action repeats itself over and over as long as the circuit is energized At the instant T1 turns ON the voltage controlling the state of T2 is governed by the following circuit 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 784 CHAPTER 7 Response of FirstOrder RL and RC Circuits It follows that vbe2 VCC 2VCCetR2C2 b While T2 is OFF and T1 is ON the output voltage vce2 is the same as the voltage across C1 thus It follows that vce2 VCC VCCetRLC1 c T2 will be OFF until vbe2 reaches zero As soon as vbe2 is zero ib2 will become positive and turn T2 ON vbe2 0 when VCC 2VCCetR2C2 0 or when t R2C2 ln 2 d When t R2C2 ln 2 we have vce2 VCC VCCeR2C2 ln 2RLC1 VCC VCCe10 ln 2 VCC e Before T1 turns ON ib1 is zero At the instant T1 turns ON we have ib1 VCC R1 VCC RL etRLC1 f At the instant T2 turns back ON t R2C2 ln 2 therefore ib1 VCC R1 VCC RL e10 ln 2 VCC R1 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 785 g h P 7100 a tOFF2 R2C2 ln 2 18 1032 109 ln 2 25 µs b tON2 R1C1 ln 2 25 µs c tOFF1 R1C1 ln 2 25 µs d tON1 R2C2 ln 2 25 µs e ib1 9 3 9 18 35 mA f ib1 9 18 9 3e6 ln 2 05469 mA g vce2 9 9e6 ln 2 886 V P 7101 a tOFF2 R2C2 ln 2 18 10328 109 ln 2 35 µs b tON2 R1C1 ln 2 374 µs c tOFF1 R1C1 ln 2 374 µs d tON1 R2C2 ln 2 35 µs e ib1 35 mA f ib1 9 18 3e56 ln 2 0562 mA g vce2 9 9e56 ln 2 881 V Note in this circuit T2 is OFF 35 µs and ON 374 µs of every cycle whereas T1 is ON 35 µs and OFF 374 µs every cycle P 7102 If R1 R2 50RL 100 kΩ then C1 48 106 100 103 ln 2 69249 pF C2 36 106 100 103 ln 2 51937 pF 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 816 CHAPTER 8 Natural and Step Responses of RLC Circuits P 819 a α 1 2RC 800 rads ω2 o 1 LC 106 ωd 106 8002 600 rads v B1e800t cos 600t B2e800t sin 600t v0 B1 30 iR0 30 5000 6 mA iC0 12 mA dv dt 0 0012 125 109 96000 Vs 96000 αB1 ωdB2 80030 600B2 B2 120 v 30e800t cos 600t 120e800t sin 600t V t 0 b dv dt 6000e800t13 sin 600t 16 cos 600t dv dt 0 when 16 cos 600t 13 sin 600t or tan 600t 16 13 600t1 08885 t1 148 ms 600t2 08885 π t2 672 ms 600t3 08885 2π t3 1195 ms c t3 t1 1047 ms Td 2π ωd 2π 600 1047 ms d t2 t1 524 ms Td 2 1048 2 524 ms e vt1 30e1184cos 08885 4 sin 08885 227 V vt2 30e5376cos 4032 4 sin 4032 0334 V vt3 30e956cos 717 4 sin 717 522 mV 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 830 CHAPTER 8 Natural and Step Responses of RLC Circuits b i0 0 Ldi0 dt vc0 1 23125 109v2 c0 9 106 v2 c0 576 vc0 24 V di0 dt 24 05 48 As c it A1e4000t A2e16000t i0 A1 A2 0 di0 dt 4000A1 16000A2 48 Solving A1 4 mA A2 4 mA it 4e4000t 4e16000t mA t 0 d dit dt 16e4000t 64e16000t di dt 0 when 64e16000t 16e4000t or e12000t 4 t ln 4 12000 11552 µs e imax 4e04621 4e18484 189 mA f vLt 05di dt 8e1000t 32e4000t V t 0 P 842 a ω2 o 1 LC 109 125032 25 106 α R 2L ωo 5000 rads R 50002L 1250 Ω b i0 iL0 6 mA vL0 15 00061250 75 V di dt0 75 0125 60 As 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 839 α2 ω2 o overdamped s12 50 502 402 20 80 rads vo Vf A 1e20t A 2e80t Vf 150 V vo0 150 A 1 A 2 200 so A 1 A 2 50 vo dt0 20A 1 80A 2 I0 C 0 Solving A 1 6667 A 2 1667 vot 150 6667e20t 1667e80t V t 0 P 854 t 0 iL0 150 30 5 A vC0 18iL0 90 V t 0 α R 2L 10 201 50 rads ω2 o 1 LC 1 012 103 5000 ωo α2 underdamped s12 50 502 5000 50 j50 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 845 iC0 4 100 mA C dvo0 dt dvo0 dt 004 103 106 40 Vs dvo dt 10A 1e10t 20A 2e20t dvo dt 0 10A 1 20A 2 40 Therefore A 1 2A 2 4 and A 1 A 2 1 Thus A 1 2 and A 2 3 vo 5 2e10t 3e20t V c Same as Example 814 dvo1 dt 20vo1 25 d From Example 814 vo1 125 V v10 2 V given Therefore vo1 125 2 125e20t 125 075e20t V P 862 a 2C dva dt va vg R va R 0 1 Therefore dva dt va RC vg 2RC 0 va R C d0 vb dt 0 2 Therefore dvb dt va RC 0 va RC dvb dt 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 846 CHAPTER 8 Natural and Step Responses of RLC Circuits 2vb R C dvb dt C dvb vo dt 0 3 Therefore dvb dt vb RC 1 2 dvo dt From 2 we have dva dt RC d2vb dt2 and va RC dvb dt When these are substituted into 1 we get 4 RC d2vb dt2 dvb dt vg 2RC Now differentiate 3 to get 5 d2vb dt2 1 RC dvb dt 1 2 d2vo dt2 But from 4 we have 6 d2vb dt2 1 RC dvb dt vg 2R2C2 Now substitute 6 into 5 d2vo dt2 vg R2C2 b When R1C1 R2C2 RC d2vo dt2 vg R2C2 The two equations are the same except for a reversal in algebraic sign c Two integrations of the input signal with one operational amplifier P 863 a d2vo dt2 1 R1C1R2C2 vg 1 R1C1R2C2 106 1004000502 106 106 250 d2vo dt2 250vg 0 t 05 vg 80 mV d2vo dt2 20 Let gt dvo dt then dg dt 20 or dg 20 dt 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 849 s 1s 2 0 s1 1 s2 2 vo Vf A 1et A 2e2t Vf 20 2 10 V vo 10 A 1et A 2e2t vo0 0 10 A 1 A 2 dvo dt 0 0 A 1 2A 2 A 1 20 A 2 10 V vot 10 20et 10e2t V 0 t 05 s dvo1 dt 2vo1 16 vo1 08 08e2t V 0 t 05 s vo05 10 20e05 10e1 155 V vo105 08 08e1 051 V At t 05 s iC 0 051 400 103 126 µA C dvo dt 126 µA dvo dt 126 02 632 Vs 05 s t d2vo dt2 3dvo dt 2 10 vo 5 vo 5 A 1et05 A 2e2t05 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 9 Sinusoidal Steady State Analysis Assessment Problems AP 91 a V 17040 V b 10 sin1000t 20 10 cos1000t 70 I 1070 A c I 53687 105313 4 j3 6 j8 10 j5 11182657 A d sin20000πt 30 cos20000πt 60 Thus V 30045 10060 21213 j21213 50 j8660 16213 j29873 339906151 mV AP 92 a v 186 cosωt 54 V b I 2045 50 30 1414 j1414 433 j25 2916 j3914 488112668 Therefore i 4881 cosωt 12668 mA c V 20 j80 3015 20 j80 2898 j776 898 j7224 72799708 v 7279 cosωt 9708 V AP 93 a ωL 10420 103 200 Ω b ZL jωL j200 Ω 91 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 92 CHAPTER 9 Sinusoidal Steady State Analysis c VL IZL 103020090 103 2120 V d vL 2 cos10000t 120 V AP 94 a XC 1 ωC 1 40005 106 50 Ω b ZC jXC j50 Ω c I V ZC 3025 5090 06115 A d i 06 cos4000t 115 A AP 95 I1 10025 9063 j4226 I2 100145 8192 j5736 I3 10095 872 j9962 I4 I1 I2 I3 0 j0 A therefore i4 0 A AP 96 a I 12560 Zθz 125 Z 60 θZ But 60 θZ 105 θZ 45 Z 90 j160 jXC XC 70 Ω XC 1 ωC 70 C 1 705000 286 µF b I Vs Z 12560 90 j90 0982105A I 0982 A AP 97 a ω 2000 rads ωL 10 Ω 1 ωC 20 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 94 CHAPTER 9 Sinusoidal Steady State Analysis The circuit is further simplified by combining the parallel branches 20 j405 j15 12 j16 Ω Therefore I 1360 14 12 j16 4 42807 A AP 910 V1 2405313 144 j192 V V2 9690 j96 V jωL j400015 103 j60 Ω 1 jωC j 6 106 400025 j60 Ω Perform a source transformation V1 j60 144 j192 j60 32 j24 A V2 20 j 96 20 j48 A Combine the parallel impedances Y 1 j60 1 30 1 j60 1 20 j5 j60 1 12 Z 1 Y 12 Ω Vo 1232 j24 384 j288 V 483687 V vo 48 cos4000t 3687 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 95 AP 911 Use the lower node as the reference node Let V1 node voltage across the 20 Ω resistor and VTh node voltage across the capacitor Writing the node voltage equations gives us V1 20 245 V1 10Ix j10 0 and VTh j10 10 j1010Ix We also have Ix V1 20 Solving these equations for VTh gives VTh 1045V To find the Thevenin impedance we remove the independent current source and apply a test voltage source at the terminals a b Thus It follows from the circuit that 10Ix 20 j10Ix Therefore Ix 0 and IT VT j10 VT 10 ZTh VT IT therefore ZTh 5 j5 Ω AP 912 The phasor domain circuit is as shown in the following diagram The node voltage equation is 10 V 5 V j209 V j5 V 10090 20 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 97 AP 915 I1 Vs Z1 2s2Z2 25 1030 1500 j6000 2524 j144 4 j3 53687 A V1 Vs Z1I1 250000 4 j31500 j6000 37000 j28500 V2 1 25V1 1480 j1140 18681514239 V I2 V2 Z2 18681514239 4 j144 12521687 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 98 CHAPTER 9 Sinusoidal Steady State Analysis Problems P 91 a ω 2πf 800 rads f ω 2π 12732 Hz b T 1f 785 ms c Im 125 mA d i0 125 cos3687 100 mA e φ 3687 φ 36872π 360 06435 rad f i 0 when 800t 3687 90 Now resolve the units 800 radst 5313 573rad 0927 rad t 116 ms g didt 0125800 sin800t 3687 didt 0 when 800t 3687 180 or 800t 14313 573rad 2498 rad Therefore t 312 ms P 92 a Right as φ becomes more negative b Left P 93 a 25 V b 2πf 400π f 200 Hz c ω 400π 125664 rads 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 913 P 915 a ZL j50050 103 j25 Ω ZC j 50032 106 j625 Ω b I 25 60 25 j25 j625 5547 369 mA c i 5547 cos500t 369 mA P 916 a jωL j2500001 j25 Ω 1 jωC j 1 250010 106 j40 Ω Ig 125 30 A b Vo 125 30Ze Ze 1 Ye Ye 1 25 1 j25 1 30 j40 Ye 52 j24 mS Ze 1 0052 j0024 17462478 Ω Vo 125 30 17462478 21826 522 V c vo 21826 cos2500t 522 V P 917 a Y 1 3 j4 1 16 j12 1 j4 012 j016 004 j003 j025 160 j120 2003687 mS 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 914 CHAPTER 9 Sinusoidal Steady State Analysis b G 160 mS c B 120 mS d I 80 A V I Y 8 023687 403687 V IC V ZC 403687 490 105313 A iC 10 cosωt 5313 A Im 10 A P 918 a Z1 R1 jωL1 Z2 R2jωL2 R2 jωL2 ω2L2 2R2 jωL2R2 2 R2 2 ω2L2 2 Z1 Z2 when R1 ω2L2 2R2 R2 2 ω2L2 2 and L1 R2 2L2 R2 2 ω2L2 2 b R1 4000212525000 50002 400021252 2500 Ω L1 50002125 50002 400021252 625 mH P 919 a Y2 1 R2 j ωL2 Y1 1 R1 jωL1 R1 jωL1 R2 1 ω2L2 1 Therefore Y2 Y1 when R2 R2 1 ω2L2 1 R1 and L2 R2 1 ω2L2 1 ω2L1 b R2 80002 1000242 8000 10 kΩ L2 80002 1000242 100024 20 H P 920 a Z1 R1 j 1 ωC1 Z2 R2jωC2 R2 1jωC2 R2 1 jωR2C2 R2 jωR2 2C2 1 ω2R2 2C2 2 Z1 Z2 when R1 R2 1 ω2R2 2C2 2 and 1 ωC1 ωR2 2C2 1 ω2R2 2C2 2 or C1 1 ω2R2 2C2 2 ω2R2 2C2 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 915 b R1 1000 1 40 10321000250 1042 200 Ω C1 1 40 10321000250 1092 40 10321000250 109 625 nF P 921 a Y2 1 R2 jωC2 Y1 1 R1 1jωC1 jωC1 1 jωR1C1 ω2R1C2 1 jωC1 1 ω2R2 1C2 1 Therefore Y1 Y2 when R2 1 ω2R2 1C2 1 ω2R1C2 1 and C2 C1 1 ω2R2 1C2 1 b R2 1 50 10321000240 1092 50 1032100040 1092 1250 Ω C2 40 109 1 50 10321000240 1092 8 nF P 922 Zab 5 j8 10 j20 8 j1640 j80 5 j8 8 j4 12 j16 25 j20 Ω 32023866 Ω P 923 First find the admittance of the parallel branches Yp 1 6 j2 1 4 j12 1 5 1 j10 0375 j0125 S Zp 1 Yp 1 0375 j0125 24 j08 Ω Zab j128 24 j08 136 16 j12 Ω Yab 1 Zab 1 16 j12 004 j003 S 40 j30 mS 503687 mS P 924 a 1 jωC RjωL 1 jωC jωRL jωL R jωL R ω2RLC jωCjωL R 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 916 CHAPTER 9 Sinusoidal Steady State Analysis R ω2RLC jωLω2LC jωRC ω2LC jωRCω2LC jωRC The denominator in the expression above is purely real set the imaginary part of the numerator in the above expression equal to zero and solve for ω ω3L2C ωR2C ω3R2C2L 0 ω2L2 R2 ω2R2LC 0 ω2 R2 R2LC L2 2002 20020420 106 042 250000 ω 500 rads b Zab500 j100 200j200 200 j200 100 Ω P 925 a R 300 Ω 120 Ω 180 Ω ωL 1 ωC 400 so 10000L 1 10000C 400 Choose L 10 mH Then 1 10000C 100 400 so C 1 10000500 02 µF We can achieve the desired capacitance by combining two 01 µF capacitors in parallel The final circuit is shown here b 001ω 1 ω02 106 so ω2 1 00102 106 5 108 ω 223607 rads P 926 a Using the notation and results from Problem 919 RL 40 j20 so R1 40 L1 20 5000 4 mH R2 402 5000200042 40 50 Ω L2 402 5000200042 500020004 20 mH 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 917 R2jωL2 50j100 40 j20 Ω checks The circuit using combinations of components from Appendix H is shown here b Using the notation and results from Problem 921 RC 40 j20 so R1 40 C1 10 µF R2 1 5000240210 µ2 500024010 µ2 50 Ω C2 10 µ 1 5000240210 µ2 2 µF R2jωC2 50j100 40 j20 Ω checks The circuit using combinations of components from Appendix H is shown here P 927 a 40 j20jωC 50j100jωC To cancel out the j100 Ω impedance the capacitive impedance must be j100 Ω j 5000C j100 so C 1 1005000 2 µF Check RjωLjωC 50j100j100 50 Ω Create the equivalent of a 2 µF capacitor from components in Appendix H by combining two 1 µF capacitors in parallel b 40 j20jωL 50j100jωL To cancel out the j100 Ω impedance the inductive impedance must be j100 Ω j5000L j100 so L 100 5000 20 mH 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 918 CHAPTER 9 Sinusoidal Steady State Analysis Check RjωLjωC 50j100j100 50 Ω Create the equivalent of a 20 mH inductor from components in Appendix H by combining two 10 mH inductors in series P 928 Z 4000 j200005 j 1 2000100 109 3000 j1000 j5000 3000 j4000 Ω Io V Z 800 3000 j4000 96 j128 165313 mA iot 16 cos2000t 5313 mA P 929 1 jωC 1 j5 1068000 j25 Ω jωL j80003125 103 j25 Ω Vg 60 90 V Ze j25 50 j25 10 j5 Ω Ig 60 90 10 j5 j24 j48 A Vo 50 j25Ig 10 j20j24 j48 120 V vo 120 cos 8000t V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 919 P 930 Vs 25 90 V 1 jωC j20 Ω jωL j10 Ω Zeq 5 j1010 20 j20 10 j10 Ω Io Vo Zeq 25 90 10 j10 125 j125 177 135 A io 177 cos4000t 135 A P 931 a 1 jωC j250 Ω jωL j200 Ω Ze j200100 500 j250 200 j200 Ω Ig 00250 Vg IgZe 0025200 j200 5 j5 V Vo 100 j200 200 j2005 j5 5 j25 5592657 V vo 559 cos50000t 2657 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 920 CHAPTER 9 Sinusoidal Steady State Analysis b ω 2πf 50000 f 25000 π T 1 f π 25000 40π µs 2657 360 40π µs 927 µs vo leads ig by 927 µs P 932 V1 j5j2 10 V 25 10 4 j3I1 0 I1 15 4 j3 24 j18 A Ib I1 j5 24 j18 j5 24 j32 A VZ j5I2 4 j3I1 j524 j32 4 j324 j18 1 j12 V 25 1 j3I3 1 j12 0 I3 62 j66 A IZ I3 I2 62 j66 24 j32 38 j34 A Z VZ IZ 1 j12 38 j34 142 j188 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 921 P 933 Va 100 j50 20 Va j5 Va 140 j30 12 j16 0 Solving Va 40 j30 V IZ 30 j20 140 j30 j10 40 j30 140 j30 12 j16 0 Solving IZ 30 j10 A Z 100 j50 140 j30 30 j10 2 j2 Ω P 934 ZL j1000010 103 j100 Ω ZC j 100002 106 j50 Ω Construct the phasor domain equivalent circuit 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 923 b ia 100 cos1500t 5313 mA ic 559 cos1500t 2657 mA ig 14577 cos1500t 2217 mA P 937 a In order for vg and ig to be in phase the impedance to the right of the 500 Ω resistor must be purely real Zeq jωLR 1ωC jωLR 1jωC jωL R 1jωC jωLjωRC 1 jωRC ω2LC 1 ω2RLC jωL1 ω2LC jωRC 1 ω2LC jωRC1 ω2LC jωRC The denominator of the above expression is purely real Now set the imaginary part of the numerator in that expression to zero and solve for ω ωL1 ω2LC ω3R2lC2 0 So ω2 1 LC R2C2 1 02106 20021062 6250000 ω 2500 rads and f 3979 Hz b Zeq 500 j500200 j400 1500 Ω Ig 900 1500 600 mA igt 60 cos 2500t mA P 938 a For ig and vg to be in phase the impedance to the right of the 480 Ω resistor must be purely real jωL 1jωC200 1jωC 200 ωL 200 1 200jωC jωL 2001 200jωC 1 2002ω2C2 jωL1 2002ω2C2 2001 200jωC 1 2002ω2C2 In the above expression the denominator is purely real So set the imaginary part of the numerator to zero and solve for ω ωL1 2002ω2c2 2002ωC 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 924 CHAPTER 9 Sinusoidal Steady State Analysis ω2 2002C L 2002C 20023125 106 01 20023125 1062 64000 ω 800 rads b When ω 800 rads Zg 480j80 200 j400 120 Ω Vg ZgIg 120006 72 V vo 72 cos 800t V P 939 a The voltage and current are in phase when the impedance to the left of the 1 kω resistor is purely real Zeq 1 jωC R jωL R jωL 1 jωRC ω2LC R jωL1 ω2LC jωRC 1 ω2LC2 ω2R2C2 The denominator in the above expression is purely real so set the imaginary part of the expressions numerator to zero and solve for ω ωR2C ωL ω3L2C 0 ω2 L R2C L2C 001 2402625 109 0012625 1092 1024 106 ω 32000 rads b Zt 1000 j500240 j320 166667 Ω Io Vo ZT 150 166667 90 mA io 9 cos 32000t mA P 940 a ZC j 1000106 j1000 Ω Z1 2500jωL 2500jωL 2500 jωL j2500ωL2500 jωL 25002 ω2L2 ZT 500 ZC Z1 500 j1000 2500ω2L2 25002jωL 25002 ω2L2 ZT is resistive when 25002ωL 25002 ω2L2 1000 or 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 925 L21000ω2 L25002ω 100025002 0 For ω 1000 L2 625L 625 0 Solving L1 5 H and L2 125 H b When L 5 H ZT 500 j1000 2500j5000 2500 Ω Ig 400 2500 160 mA ig 16 cos 1000t mA When L 125 H ZT 500 j1000 2500j1250 1000 Ω Ig 400 1000 400 mA ig 40 cos 1000t mA P 941 a Zp R jωC R 1jωC R 1 jωRC 10000 1 j500010000C 10000 1 j50 106C 100001 j50 106C 1 25 1014C2 10000 1 25 1014C2 j 5 1011C 1 25 1014C2 jωL j500008 j4000 4000 5 1011C 1 25 1014C2 1014C2 125 106C 1 0 C2 5 108C 4 1016 0 Solving C1 40 nF C2 10 nF 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 926 CHAPTER 9 Sinusoidal Steady State Analysis b Re 10000 1 25 1014C2 When C 40 nF Re 2000 Ω Ig 800 2000 400 mA ig 40 cos 5000t mA When C 10 nF Re 8000 Ω Ig 800 8000 100 mA ig 10 cos 5000t mA P 942 Simplify the top triangle using series and parallel combinations 1 j11 j1 1 Ω Convert the lower left delta to a wye Z1 j11 1 j1 j1 j1 Ω Z2 j11 1 j1 j1 j1 Ω Z3 j1j1 1 j1 j1 1 Ω Convert the lower right delta to a wye Z4 j11 1 j1 j1 j1 Ω Z5 j1j1 1 j1 j1 1 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 927 Z6 j11 1 j1 j1 j1 Ω The resulting circuit is shown below Simplify the middle portion of the circuit by making series and parallel combinations 1 j1 j11 1 12 23 Ω Zab j1 23 j1 23 Ω P 943 a jωL j400400 103 j160 Ω 1 jωC j 4003125 106 j80 Ω Using voltage division Vab j80j160 320 j80j160j50 20 j10 V VTh Vab 20 j10 V b Remove the voltage source and combine impedances in parallel to find ZTh Zab ZTh Zab 320 j80j160 64 j128 Ω c 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 928 CHAPTER 9 Sinusoidal Steady State Analysis P 944 Step 1 to Step 2 18090 j30 60 A Step 2 to Step 3 j3015 12 j6 Ω 6012 j6 72 j36 V Step 3 to Step 4 ZN 12 j6 j30 12 j24 Ω IN 72 j36 12 j24 j3 A P 945 Step 1 to Step 2 0120250 300 V Step 2 to Step 3 250 j400 j150 250 j250 Ω 300 250 j250 60 j60 mA Step 3 to Step 4 250 j250500 200 j100 Ω 200 j100006 j006 18 j6 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 929 P 946 Open circuit voltage j20Ia 40Ia 20Ia 04 j02 0 Solving Ia 2004 j02 60 j20 01 j01 A Voc 40Ia j1604 j02 08 j104 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 930 CHAPTER 9 Sinusoidal Steady State Analysis Short circuit current j20Ia 40Ia Isc 20Ia 04 j02 0 40Isc Ia j16Isc 04 j02 0 Solving Isc 03 j05 A ZTh VTh Isc 08 j104 03 j05 16 j8 Ω P 947 VTh 015 j015ZTh 30 j30 j200 ZTh j200VTh 40 j40 so j200VTh 40 j40ZTh j200 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 931 Place the two equations in standard form VTh 015 j015ZTh 30 j30 j200VTh 40 j40CTh 40 j40j200 Solving VTh j240 V ZTh 600 j800 Ω P 948 Short circuit current Vx j10I1 Isc 40 j40 5Vx j10Isc I1 0 Vx 10I1 Solving IN Isc 6 j4 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 932 CHAPTER 9 Sinusoidal Steady State Analysis Open circuit voltage I 40 j40 10 j10 4 A Vx 10I 40 V Voc 5Vx j10I 200 j40 V ZN 200 j40 6 j4 20 j20 Ω P 949 Open circuit voltage V1 250 20 j10 003Vo V1 50 j100 0 Vo j100 50 j100V1 V1 20 j10 j3V1 50 j100 V1 50 j100 250 20 j10 V1 500 j250 V Vo 300 j400 V VTh 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 933 Short circuit current Isc 2500 70 j10 35 j05 A ZTh VTh Isc 300 j400 35 j05 100 j100 Ω The Thevenin equivalent circuit P 950 Open circuit voltage V2 10 88Iφ V2 1 5V2 j50 0 Iφ 5 V25 200 Solving V2 66 j88 11012687 V VTh 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 936 CHAPTER 9 Sinusoidal Steady State Analysis VT j25IT 25 j60 30 j60IT VT IT Zab 20 j15 25 3687 Ω P 953 1 ωC1 109 5000025 8 kΩ 1 ωC2 109 500005 4 kΩ VT 2400 j8000IT 40IT90 ZTh VT IT 6000 j8000 Ω P 954 V1 240 j10 V1 50 V1 30 j10 0 Solving for V1 yields V1 19863 2444 V Vo 30 30 j10V1 18843 4288 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 938 CHAPTER 9 Sinusoidal Steady State Analysis Vg 2090 V 5 V1 V2 j8 V1 j20 j4 0 V2 V1 j8 V2 j4 V2 j20 12 0 Solving V2 8 j4 V Io V2 j4 1 j2 2246343 A io 224 cos2500t 6343 A P 957 jωL j40050 103 j20 Ω 1 jωC j 40050 106 j50 Ω Vg1 255313 15 j20 V Vg2 18033369 15 j10 V Vo 15 j20 j50 Vo 150 Vo 15 j10 j20 0 Solving Vo 150 vot 15 cos 400t V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 939 P 958 Write a KCL equation at the top node Vo j8 Vo 24I j4 Vo 5 10 j10 0 The constraint equation is I Vo j8 Solving Vo j80 8090 V P 959 Vo 50 Vo j25 20Io 0 2 j4Vo 2000Io Vo 200 j400Io Io V1 Vo10 j25 V1 20 j65Io 0006 j0013 V1 50 Io 04 j13Io Io 06 j13Io Io 06 j1310 103 06 j13 100 mA Vo 200 j400Io 2 j4 44711657 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 940 CHAPTER 9 Sinusoidal Steady State Analysis P 960 12 j12Ia 12Ig 5j8 0 12Ia 12 j4Ig j20 5j4 0 Solving Ig 4 j2 447 2657 A P 961 The circuit with the mesh currents identified is shown below The mesh current equations are 15 j20 j50I1 150I1 I2 0 15 j10 150I2 I1 j20I2 0 In standard form I1150 j50 I2150 15 j20 I1150 I2150 j20 15 j10 Solving on a calculator yields I1 04 A I2 05 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 941 Thus Vo 150I1 I2 15 V and vot 15 cos 400t V P 962 100 1 j1I1 1I2 j1I3 50 1I1 1 j1I2 j1I3 1 j1I1 j1I2 I3 Solving I1 11 j10 A I2 11 j5 A I3 6 A Ia I3 1 5 A Ib I1 I3 5 j10 A Ic I2 I3 5 j5 A Id I1 I2 j5 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 942 CHAPTER 9 Sinusoidal Steady State Analysis P 963 Va j18 V Vb 12 V jωL j400025 103 j100Ω j ωC j 4000625 106 j400 Ω j18 j300Ia j100Ib 12 j100Ia 400 j100Ib Solving Ia 675 j75 mA Ib 225 j225 mA Vo j100Ia Ib 3 j9 9497157 A vot 949 cos4000t 7157 A P 964 jωL1 j50004 103 j20 Ω jωL2 j5000110 103 j550 Ω 1 jωC j 50004 106 j50 Ω 750 j500I 100I j550Ia 0 10 j20Ia 100I j550Ia I 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 943 Solving Ia j25 A Vo 10Ia j25 2590 vo 25 cos5000t 90 25 sin 5000t V P 965 1 jωc j 1000003125 109 j3200 Ω jωL j10000080 103 j8000 Ω Let Z1 1200 j3200 Ω Z2 2400 j8000 Ω Vo Z2 Z1 Z2 Vg 2400 j8000 3600 j4800120 1568 j576 1670452017 vo 167045 cos100000t 2017 V P 966 1 jωC j 250100 106 j40 Ω jωL j25008 j200 Ω Let Z1 20 j40 Ω Z2 100 j200 Ω Ig 600 mA Io IgZ1 Z1 Z2 00620 j40 120 j160 6 j12 mA 1342 11657 mA io 1342 cos250t 11657 mA P 967 a Superposition must be used because the frequencies of the two sources are different 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 945 I o V o 400 10 j30 mA 3162 7157 mA For ω 4000 rads V o j400 V o j100 V o 20 400 0 V oj j4 1 20 so V o 20 1 j3 2 j6 V I o V j100 2 j6 j100 60 j20 mA 6325 1843 mA Thus iot 3126 cos16000t 7157 6325 cos4000t 1843 mA P 969 Vg 200 V 1 jωC j400 Ω Let Va voltage across the capacitor positive at upper terminal Then Va 200 400 Va j400 Va 400 0 Va 8 j4 V 0 Va 400 0 Vo 200 0 Vo Va 2 Vo 4 j2 44715343 V vo 447 cos5000t 15343 V P 970 a Va 200 400 jωCoVa Va 400 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 946 CHAPTER 9 Sinusoidal Steady State Analysis Va 20 2 j400ωCo Vo Va 2 Vo 10 2 j2 106Co 10180 2 j2 106Co denominator angle 45 so 2 106Co 2 C 1 µF b Vo 10180 2 j2 354135 V vo 354 cos5000t 135 V P 971 a Vg 250 V Vp 20 100Vg 50 Vn Vp 50 V 5 80000 5 Vo Zp 0 Zp j8000040000 32000 j16000 Ω Vo 5Zp 80000 5 7 j 707 813 vo 707 cos50000t 813 V b Vp 02Vm0 Vn Vp 02Vm0 02Vm 80000 02Vm Vo 32000 j16000 0 Vo 02Vm 32000 j16000 80000 Vm02 Vm028 j004 Vm028 j004 10 Vm 3536 V P 972 1 jωC1 j10 kΩ 1 jωC2 j100 kΩ 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 954 CHAPTER 9 Sinusoidal Steady State Analysis V2 1 V3 20 1I2 20I3 V1 50 V2 1 50I1 1I2 Zab V1 I1 Substituting Zab V1 I1 50V2 I250 502V2 I2 502V320 20I3 502V3 202I3 625ZL 625200 45 1250 45 Ω P 983 The phasor domain equivalent circuit is Vo Vm 2 IRx I Vm Rx jXC As Rx varies from 0 to the amplitude of vo remains constant and its phase angle increases from 0 to 180 as shown in the following phasor diagram P 984 a I 240 24 240 j32 10 j75 A Vs 2400 01 j0810 j75 247 j725 24711168 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 955 b Use the capacitor to eliminate the j component of I therefore Ic j75 A Zc 240 j75 j32 Ω Vs 240 01 j0810 241 j8 24113190 V c Let Ic denote the magnitude of the current in the capacitor branch Then I 10 j75 jIc 10 jIc 75 A Vs 240α 240 01 j0810 jIc 75 247 08Ic j725 01Ic It follows that 240 cos α 247 08Ic and 240 sin α 725 01Ic Now square each term and then add to generate the quadratic equation I2 c 60577Ic 532548 0 Ic 30288 29396 Therefore Ic 892 A smallest value and Zc 240j892 j2690 Ω Therefore the capacitive reactance is 2690 Ω P 985 a Iℓ 240 8 240 j6 30 j40 A Vℓ 01 j0830 j40 35 j20 40312974 V Vs 2400 Vℓ 275 j20 27573416 V b 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 956 CHAPTER 9 Sinusoidal Steady State Analysis c Iℓ 30 j40 240 j5 30 j8 A Vℓ 01 j0830 j8 34 j248 25039781 Vs 2400 Vℓ 2366 j248 2379598 P 986 a I1 120 24 240 84 j63 2329 j1371 2702305 A I2 120 12 120 24 50 A I3 120 12 240 84 j63 2829 j1371 31442587 A I4 120 24 50 A I5 120 12 100 A I6 240 84 j63 1829 j1371 22863687 A b When fuse A is interrupted I1 0 I3 15 A I5 10 A I2 10 5 15 A I4 5 A I6 5 A c The clock and television set were fed from the uninterrupted side of the circuit that is the 12 Ω load includes the clock and the TV set d No the motor current drops to 5 A well below its normal running value of 2286 A e After fuse A opens the current in fuse B is only 15 A P 987 a The circuit is redrawn with mesh currents identified 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 957 The mesh current equations are 1200 23Ia 2Ib 20Ic 1200 2Ia 43Ib 40Ic 0 20Ia 40Ib 70Ic Solving Ia 240 A Ib 21960 A Ic 19400 A The branch currents are I1 Ia 240 A I2 Ia Ib 2040 A I3 Ib 21960 A I4 Ic 19400 A I5 Ia Ic 460 A I6 Ib Ic 2550 A b Let N1 be the number of turns on the primary winding because the secondary winding is centertapped let 2N2 be the total turns on the secondary From Fig 958 13200 N1 240 2N2 or N2 N1 1 110 The ampere turn balance requires N1Ip N2I1 N2I3 Therefore Ip N2 N1 I1 I3 1 11024 2196 0420 A P 988 a 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 958 CHAPTER 9 Sinusoidal Steady State Analysis The three mesh current equations are 1200 23Ia 2Ib 20Ic 1200 2Ia 23Ib 20Ic 0 20Ia 20Ib 50Ic Solving Ia 240 A Ib 240 A Ic 1920 A I2 Ia Ib 0 A b Ip N2 N1 I1 I3 N2 N1 Ia Ib 1 11024 24 04360 A c Yes when the two 120 V loads are equal there is no current in the neutral line so no power is lost to this line Since you pay for power the cost is lower when the loads are equal P 989 a 125 R 005 j005I1 003 j003I2 RI3 125 003 j003I1 R 005 j005I2 RI3 Subtracting the above two equations gives 0 R 008 j008I1 R 008 j008I2 I1 I2 so In I1 I2 0 A b V1 RI1 I3 V2 RI2 I3 Since I1 I2 from part a V1 V2 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 959 c 250 44004 j004Ia 440Ib 0 440Ia 448Ib Solving Ia 31656207 j0160343 A Ib 31090917 j0157479 A I1 Ia Ib 056529 j0002864 A V1 40I1 22612 j011456 22612 0290282 V V2 400I1 226116 j11456 2261189 0290282 V d 125 4005 j005I1 003 j003I2 40I3 125 003 j003I1 40005 j005I2 400I3 0 40I1 400I2 448I3 Solving I1 3419 j0182 A I2 31396 j0164 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 960 CHAPTER 9 Sinusoidal Steady State Analysis I3 31085 j0163 A V1 40I1 I3 1242 035 V V2 400I2 I3 1244 018 V e Because an open neutral can result in severely unbalanced voltages across the 125 V loads P 990 a Let N1 primary winding turns and 2N2 secondary winding turns Then 14000 N1 250 2N2 N2 N1 1 112 a In part c Ip 2aIa Ip 2N2Ia N1 1 56Ia 1 5631656 j016 Ip 5653 j29 mA In part d IpN1 I1N2 I2N2 Ip N2 N1 I1 I2 1 1123419 j0182 31396 j0164 1 11265586 j0346 Ip 5856 j31 mA b Yes because the neutral conductor carries nonzero current whenever the load is not balanced 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 10 Sinusoidal Steady State Power Calculations Assessment Problems AP 101 a V 100 45 V I 2015 A Therefore P 1 210020 cos45 15 500 W A B Q 1000 sin 60 86603 VAR B A b V 100 45 I 20165 P 1000 cos210 86603 W B A Q 1000 sin210 500 VAR A B c V 100 45 I 20 105 P 1000 cos60 500 W A B Q 1000 sin60 86603 VAR A B d V 1000 I 20120 P 1000 cos120 500 W B A Q 1000 sin120 86603 VAR B A AP 102 pf cosθv θi cos15 75 cos60 05 leading rf sinθv θi sin60 0866 101 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 103 Parallel circuit derivation P 2502 R therefore R 2502 40000 15625 Ω Q 2502 XC therefore XC 2502 30000 2083 Ω AP 106 S1 1500006 j1500008 9000 j12000 VA S2 600008 j600006 4800 j3600 VA ST S1 S2 13800 j8400 VA ST 200I therefore I 69 j42 I 69 j42 A Vs 200 jI 200 j69 42 242 j69 251641591 V rms AP 107 a The phasor domain equivalent circuit and the Thevenin equivalent are shown below Phasor domain equivalent circuit Thevenin equivalent VTh 3 j800 20 j40 48 j24 5367 2657 V ZTh 4 j18 j800 20 j40 20 j10 22362657 Ω For maximum power transfer ZL 20 j10 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1015 250Vm cos θ V 2 m 5000 250Vm sin θ 2500 2502V 2 m V 2 m 50002 25002 62500V 2 m V 4 m 10000V 2 m 3125 106 or V 4 m 52500V 2 m 3125 106 0 Solving V 2 m 26250 2564786 Vm 22781 V and Vm 2454 V If Vm 22781 V sin θ 2500 22781250 0044 θ 252 If Vm 2454 V sin θ 2500 2454250 04075 θ 2405 b 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1017 P 1021 ST 40800 j30600 VA S1 20000096 j028 19200 j5600 VA S2 ST S1 21600 j36200 421544859176 VA rf sin59176 08587 pf cos59176 05124 lagging P 1022 a S1 10000 j4000 VA S2 VL2 Z 2 10002 60 j80 6 j8 kVA S1 S2 16 j4 kVA 1000I L 16000 j4000 IL 16 j4 Arms Vg VL IL05 j005 1000 16 j405 j005 10082 j12 10082 00682 Vrms b T 1 f 1 50 20 ms 00682 360 t 20 ms t 379 µs c VL leads Vg by 00682 or 379 µs P 1023 a I1 6000 j3000 150 40 j20 A rms 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1020 CHAPTER 10 Sinusoidal Steady State Power Calculations P 1026 250I 1 6000 j8000 I 1 24 j32 I1 24 j32 Arms 250I 2 9000 j3000 I 2 36 j12 I2 36 j12 Arms I3 2500 25 10 j0 A I4 2500 j5 0 j50 A Ig I1 I2 I3 I4 70 j70 A Vg 250 70 j70j01 243 j7 2431165 V rms P 1027 a From Problem 978 Zab 20 j2725 so I1 75 10 j1275 20 2725 75 30 j40 09 j12 A I2 jωM Z22 I1 j54 160 j12009 j12 405 j0 mA VL 60 j200405 j0 243 j81 VL 2561 V b Pgideal 7509 675 W Pgpractical 675 I1210 45 W PL I2260 98415 W delivered 98415 45 100 2187 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1022 CHAPTER 10 Sinusoidal Steady State Power Calculations Z 1 Y 3744 4203 Ω pf cos4203 074 leading rf sin4203 067 P 1031 a I 2700 36 j48 27 j36 45 5313 Arms P 4526 1215 W b YL 1 30 j40 12 j16 mS XC 1 16 103 625 Ω c ZL 1 12 103 8333 Ω d I 2700 8933 j8 301 512 A P 30126 5437 W e 5437 1215100 4475 Thus the power loss after the capacitor is added is 4475 of the power loss before the capacitor is added P 1032 IL 120000 j90000 4800 25 j1875 A rms IC 4800 jXC j 4800 XC jIC Iℓ 25 j1875 jIC 25 jIC 1875 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1023 Vs 4800 10 j525 jIC 1875 514375 5IC j625 10IC Vs2 514375 5IC2 625 10IC2 48002 125I2 C 526875IC 3422070313 0 IC 341284 Arms and IC 802165 Arms Select the smaller value of IC to minimize the magnitude of Iℓ XC 4800 802165 59838 C 1 59838120π 4443 µF P 1033 a SL 20000085 j053 17000 j1053565 VA 125I L 17000 j1053565 I L 136 j8429 Arms IL 136 j8429 Arms Vs 125 136 j8429001 j008 13310 j1004 13348431 Vrms Vs 13348 Vrms b Pℓ Iℓ2001 1602001 256 W c 1252 XC 1053565 XC 148306 Ω 1 ωC 148306 C 1 148306120π 178859 µF d Iℓ 136 j0 Arms Vs 125 136001 j008 12636 j1088 12683492 Vrms Vs 12683 Vrms e Pℓ 1362001 18496 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1024 CHAPTER 10 Sinusoidal Steady State Power Calculations P 1034 a So original load 1600 j 1600 08 06 1600 j1200 kVA Sf final load 1920 j 1920 096 028 1920 j560 kVA Qadded 560 1200 640 kVAR b deliver c Sa added load 320 j640 71554 6343 kVA pf cos6343 0447 leading d I L 1600 j1200 103 2400 66667 j500 A IL 66667 j500 83333 3687 Arms IL 83333 Arms e I L 1920 j560 103 2400 800 j23333 IL 800 j23333 83333 1626 Arms IL 83333 Arms P 1035 a Pbefore Pafter 833332025 173611 kW b Vsbefore 2400 66667 j500025 j01 261667 j5833 261732 128 Vrms Vsbefore 261732 Vrms Vsafter 2400 800 j23333025 j01 262333 j2167 262342047 Vrms Vsafter 262342 Vrms P 1036 a 50 j10I1 I2 j10I3 I2 10I1 I3 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1028 CHAPTER 10 Sinusoidal Steady State Power Calculations P 1040 a 25a2 1 4a2 2 500 I25 a1I P25 a2 1I225 I4 a2I P4 a2 2I24 P4 4P25 a2 2I24 100a2 1I2 100a2 1 4a2 2 25a2 1 100a2 1 500 a1 2 254 4a2 2 500 a2 10 b I 20000 500 500 20 Arms I25 a1I 4 A P25Ω 1625 400 W c I4 a2I 102 20 Arms V4 204 800 Vrms P 1041 a ZTh j4000 4000j4000 4000 j4000 2000 j2000 Ω ZL Z Th 2000 j2000 Ω b VTh 1004000 4000 j4000 5 j5 5 245 V I 5 245 4000 125 245 mA Irms 125 mA Pload 00012522000 3125 mW 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1029 c The closest resistor values from Appendix H are 18 kΩ and 22 kΩ Find the capacitor value 1 8000C 2000 so C 625 nF The closest capacitor value is 47 nF Choose R 18 kΩ I 545 2000 j2000 1800 j265957 07462 j106 mArms 135485 mArms Pload 0001321800 303 mW instead of 3125 mW Therefore the load impedance is ZL 1800 j265957 Ω P 1042 a From Problem 975 ZTh 85 j85 Ω and VTh 850 j850 V Thus for maximum power transfer ZL Z Th 85 j85 Ω I2 850 j850 170 5 j5 A 4250 5 j5I1 j205 j5 I1 325 j100 5 j5 425 j225 A Sgdel 425425 j225 180625 j95625 VA Pg 180625 W b Ploss I125 I2245 115625 2250 138125 W loss in transformer 138125 180625100 7647 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1030 CHAPTER 10 Sinusoidal Steady State Power Calculations P 1043 a 156 j42 300 ZTh 156 j42 200 j500 0 ZTh 300 156 j42 018 j024 400 j300 Ω ZL 400 j300 Ω b I 3000 8000 03750 Arms P 03752400 5625 W c Let R 180 Ω 220 Ω 400 Ω 2π50L 300 so L 300 100π 955 mH Use 9 seriesconnected 10 mH inductors to get 90 mH Use 2 parallelconnected 10 mH inductors to get 5 mH Use 2 parallelconnected 1 mH inductors to get 05 mH Combine the 90 mH the 5 mH and the 05 mH in series to get 955 mH P 1044 a Open circuit voltage V1 5Iφ 5100 5Iφ 25 j10 25 j10Iφ 100 5Iφ Iφ 100 30 j10 3 j A VTh j3 1 j35Iφ 15 V Short circuit current V2 5Iφ 100 5Iφ 25 j10 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1031 Iφ 3 j1 A Isc 5Iφ 1 15 j5 A ZTh 15 15 j5 09 j03 Ω ZL Z Th 09 j03 Ω IL 03 18 833 Arms P IL209 625 W b VL 09 j03833 75 j25 Vrms I1 VL j3 0833 j25 Arms I2 I1 IL 75 j25 Arms 5Iφ I2 VL Iφ 3 j1 A Ids Iφ I2 45 j15 A Sg 1003 j1 300 j100 VA Sds 53 j145 j15 75 j0 VA Pdev 300 75 375 W developed 625 375 100 1667 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1033 b VTh j200 500 j300 j2003000 60 j60 Vrms I 60 j60 640 j320 375 j1125 mArms 118597157 mArms P 0118592400 5625 W c Pick the 390 Ω resistor from Appendix H for the closest match I 60 j60 630 j320 1200847193 mArms P 01200842390 5624 W P 1047 a Open circuit voltage Vφ 100 5 Vφ j5 01Vφ 0 Vφ 40 j80 Vrms VTh Vφ 01Vφj5 Vφ1 j05 80 j60 Vrms Short circuit current Isc 01Vφ Vφ j5 01 j02Vφ 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1034 CHAPTER 10 Sinusoidal Steady State Power Calculations Vφ 100 5 Vφ j5 Vφ j5 0 Vφ 100 Vrms Isc 01 j02100 10 j20 Arms ZTh VTh Isc 80 j60 10 j20 4 j2 Ω Ro ZTh 447 Ω b I 80 j60 4 20 j2 736 j882 A rms P 11492 20 59017 W c I 80 j60 8 10 j75 A rms P 102 7524 625 W d Vφ 100 5 Vφ j5 Vo 25 j50 j5 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1037 P 1051 a 5120 30I1 j100I1 j40I1 I2 j64I1 I2 j40I1 0 160I2 j64I2 I1 j40I1 Solving I1 8 j20 Arms I2 130 Arms Vo 13160 2080 Vrms b P 132160 27040 W c Sg 51208 j20 40960 j102400 VA Pg 40960 W delivered 27040 40960100 66 P 1052 a Open circuit voltage 5120 30I1 j100I1 j40I1 j64I1 j40I1 I1 5120 30 j244 254 j2067 Arms VTh j64I1 j40I1 j104I1 21498 j26432 V Short circuit current 5120 30I1 j100I1 j40I1 Isc j64I1 Isc j40I1 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1039 P 1053 Open circuit voltage I1 300 15 j30 04 j08 A VTh j30I1 j18I1 j12I1 96 j48 10732657 Short circuit current 300 15I1 j30I1 Isc j18Isc 0 j15Isc j18Isc I1 j30Isc I1 j18Isc Solving Isc 195 43025 A ZTh 96 j48 195 43025 192 j516 Ω I2 96 j48 384 27952657 A 300 15I1 j30I1iI2 j18I2 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1040 CHAPTER 10 Sinusoidal Steady State Power Calculations I1 30 j12I2 15 j30 30 j1227952657 15 j30 1 A Zg 300 1 30 j0 300 Ω P 1054 a Open circuit VTh 150 120 j160j100 60 j45 V Short circuit 120 j160I1 j100Isc 150 j100I1 30 j160Isc 0 Solving Isc 05 j25 A ZTh 60 j45 05 j25 64 j112 Ω ZL Z Th 64 j112 Ω IL VTh ZTh ZL 60 j45 128 05863687 Arms PL IL264 2198 W b I1 Z22I2 jωM 94 j48 j100 05863687 062 2608 A rms Ptransformer 150062 cos2608 062240 6815 W delivered 2198 6815100 3225 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1041 P 1055 a jωL1 jωL2 j400625 103 j250 Ω jωM j4003125 103 j125 Ω 400 125 j250Ig j125IL 0 j125Ig 375 j250IL Solving Ig 08 j12 A IL 04 A Thus ig 144 cos400t 5631 A iL 04 cos 400t A b k M L1L2 03125 0625 05 c When t 125π ms 400t 400125π 103 05π rad 90 ig125π ms 144 cos90 5631 12 A iL125π ms 04 cos90 0 A w 1 2L1i2 1 1 2L2i2 2 Mi1i2 1 20625122 0 0 450 mJ When t 25π ms 400t 40025π 103 π 180 ig25π ms 144 cos180 5631 08 A iL25π ms 04 cos180 04 A w 1 20625082 1 20625042 031250804 150 mJ d From a IL 04 A P 1 2042375 30 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1042 CHAPTER 10 Sinusoidal Steady State Power Calculations e Open circuit VTh 400 125 j250j125 160 j80 V Short circuit 400 125 j250I1 j125Isc 0 j125I1 j250Isc Solving Isc 04923 j07385 ZTh VTh Isc 160 j80 04923 j07385 25 j200 Ω RL 20156 Ω f I 160 j80 22656 j200 0592 1487 A P 1 20592220156 353 W g ZL Z Th 25 j200 Ω h I 160 j80 50 3582657 P 1 2358225 160 W P 1056 a 54 I1 j2I1 I2 j3I2 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1043 0 7I2 j2I2 I1 j3I2 j8I2 j3I1 I2 Solving I1 12 j21 A rms I2 3 A rms Vo 7I2 210 Vrms b P I227 63 W c Pg 5412 648 W delivered 63 648100 972 P 1057 a 54 I1 j2I1 I2 j4kI2 0 7I2 j2I2 I1 j4kI2 j8I2 j4kI1 I2 Place the equations in standard form 54 1 j2I1 j4k 2I2 0 j4k 2I1 7 j10 8kI2 I1 54 I2j4k 2 1 j2 Substituting I2 j544k 2 7 j10 8k1 j2 4k 2 For Vo 0 I2 0 so if 4k 2 0 then k 05 b When I2 0 I1 54 1 j2 108 j216 Arms Pg 54108 5832 W Check Ploss I121 5832 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1045 b I1 24 100 2400 mA N1I1 N2I2 I2 6I1 1440 A IL I1 I2 1680 A VL 2 j16IL 336 j2688 2719713 Vrms P 1060 a Replace the circuit to the left of the primary winding with a Thevenin equivalent VTh 1520j10 60 j120 V ZTh 2 20j10 6 j8 Ω Transfer the secondary impedance to the primary side Zp 1 25100 jXC 4 j XC 25 Ω Now maximize I by setting XC25 8 Ω C 1 20020 103 025 µF 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1046 CHAPTER 10 Sinusoidal Steady State Power Calculations b I 60 j120 10 6 j12 A P I24 720 W c Ro 25 6 Ω Ro 150 Ω d I 60 j120 12 5 j10 A P I26 750 W P 1061 Va 1 Vo 50 1Ia 50Io Va Ia Vo50 50Io VoIo 502 5000 502 2 Ω Vb 25 Va 1 25Ib 1Ia Vb25 25Ib VbIb 252 Va Ia Vb Ib 252Va Ia 2522 1250 Ω Thus Ib 145200 1250 100 mA rms since the ideal transformers are lossless P5kΩ P1250Ω and the power delivered to the 1250 Ω resistor is 0121250 or 125 W P 1062 a Vb Ib 2525000 a2 200 Ω therefore a2 15625 a 125 b Ib 145 400 3625 mA P 036252200 2628125 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1048 CHAPTER 10 Sinusoidal Steady State Power Calculations b VTh 2550 40 j30j200 10205313 V IL 10205313 3060 j1020 0316347 Arms Since the transformer is ideal P6800 P1700 P I21700 170 W c 2550 40 j30I1 j200026 j018 I1 413 j180 Arms Pgen 255413 1053 W Pdiss 1053 170 883 W dissipated 883 1053100 8385 P 1065 a Open circuit voltage 400 4I1 I3 12I3 VTh I1 4 I3 I1 4I3 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1049 Solving VTh 400 V Short circuit current 400 4I1 4I3 I1 V1 4V1 16I14 4I1 V1 I1 400 6I1 4I3 Also 400 4I1 I3 12I3 Solving I1 6 A I3 1 A Isc I14 I3 25 A RTh VTh Isc 40 25 16 Ω I 400 32 1250 Arms P 125216 25 W b 40 4I1 I3 12I3 20 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1051 Short circuit current 500 80Isc I1 360Isc 05I1 2V1 40I1 2 360Isc 05I1 500 80I1 Isc 20I1 V1 Solving Isc 147 A RTh VTh Isc 150 147 102 Ω P 752 102 5515 W b 500 80I1 75102 75 360I2 75102 575 6000 102 27000 102 80I1 180I2 I1 3456 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1053 P 1070 jωL1 j2π60025 j9425 Ω I 120 005 j9425 127 8697 Arms P R1I2 0051272 811 mW Note that while the current supplied by the voltage source is virtually identical to that calculated in Problem 1069 the much smaller value of transformer resistance results in a much smaller value of real power consumed by the transformer P 1071 An ideal transformer has no resistance so consumes no real power This is one of the important characteristics of ideal transformers 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 11 Balanced ThreePhase Circuits Assessment Problems AP 111 Make a sketch We know VAN and wish to find VBC To do this write a KVL equation to find VAB and use the known phase angle relationship between VAB and VBC to find VBC VAB VAN VNB VAN VBN Since VAN VBN and VCN form a balanced set and VAN 240 30V and the phase sequence is positive VBN VANVAN 120 240 30 120 240 150 V Then VAB VAN VBN 240 30 240 150 415460 V Since VAB VBC and VCA form a balanced set with a positive phase sequence we can find VBC from VAB VBC VABVAB 120 415690 120 41569 120 V 111 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 112 CHAPTER 11 Balanced ThreePhase Circuits Thus VBC 41569 120 V AP 112 Make a sketch We know VCN and wish to find VAB To do this write a KVL equation to find VBC and use the known phase angle relationship between VAB and VBC to find VAB VBC VBN VNC VBN VCN Since VAN VBN and VCN form a balanced set and VCN 450 25 V and the phase sequence is negative VBN VCNVCN 120 450 23 120 450 145 V Then VBC VBN VCN 450 145 450 25 77942 175 V Since VAB VBC and VCA form a balanced set with a negative phase sequence we can find VAB from VBC VAB VBCVBC 120 77942 295 V But we normally want phase angle values between 180 and 180 We add 360 to the phase angle computed above Thus VAB 7794265 V AP 113 Sketch the aphase circuit 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 113 a We can find the line current using Ohms law since the aphase line current is the current in the aphase load Then we can use the fact that IaA IbB and IcC form a balanced set to find the remaining line currents Note that since we were not given any phase angles in the problem statement we can assume that the phase voltage given VAN has a phase angle of 0 24000 IaA16 j12 so IaA 24000 16 j12 96 j72 120 3687 A With an acb phase sequence IbB IaA 120 and IcC IaA 120 so IaA 120 3687 A IbB 1208313 A IcC 120 15687 A b The line voltages at the source are Vab Vbc and Vca They form a balanced set To find Vab use the aphase circuit to find VAN and use the relationship between phase voltages and line voltages for a yconnection see Fig 119b From the aphase circuit use KVL Van VaA VAN 01 j08IaA 24000 01 j0896 j72 24000 24672 j696 246818162 V From Fig 119b Vab Van 3 30 427502 2838 V With an acb phase sequence Vbc Vab 120 and Vca Vab 120 so Vab 427502 2838 V Vbc 4275029162 V Vca 427502 14838 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 116 CHAPTER 11 Balanced ThreePhase Circuits Problems P 111 a First convert the cosine waveforms to phasors Va 13763 Vb 137 57 Vc 137183 Subtract the phase angle of the aphase from all phase angles V a 63 63 0 V b 57 63 120 V c 183 63 120 Compare the result to Eqs 111 and 112 Therefore abc b First convert the cosine waveforms to phasors making sure that all waveforms are represented as cosines Va 820 36 Vb 82084 Vc 820 156 Subtract the phase angle of the aphase from all phase angles V a 36 36 0 V b 84 36 120 V c 156 36 120 Compare the result to Eqs 111 and 112 Therefore acb P 112 a Va 48 45 V Vb 48 165 V Vc 4875 V Balanced positive phase sequence b Va 18860 V Vb 1880 V 188180 V Vc 188 60 V Balanced negative phase sequence 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 117 c Va 4260 V Vb 462120 V Vc 426 120 V Unbalanced due to unequal amplitudes d Va 1121 20 V Vb 1121 140 V Vc 1121100 V Balanced positive phase sequence e Va 540 90 V Vb 540 120 V Vc 540120 V Unbalanced due to unequal phase separation f Va 14480 V Vb 144 160 V Vc 144 40 V Balanced negative phase sequence P 113 Va Vm0 Vm j0 Vb Vm 120 Vm05 j0866 Vc Vm120 Vm05 j0866 Va Vb Vc Vm1 j0 05 j0866 05 j0866 Vm0 0 P 114 I 18860 188180 188 60 3RW jXW 0 P 115 I 4260 462120 426 120 3RW jXW 36120 3RW jXW P 116 a The voltage sources form a balanced set the source impedances are equal and the line impedances are equal But the load impedances are not equal Therefore the circuit is unbalanced Also IaA 110 32 j24 2753687 A rms 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 118 CHAPTER 11 Balanced ThreePhase Circuits IbB 110 120 6 j8 11 17313 A rms IcC 110120 40 j30 228313 A rms The magnitudes are unequal and the phase angles are not 120 apart so the currents are not balanced and thus the circuit is not balanced b Io IaA IbB IcC 11796758 A rms P 117 a IaA 2770 80 j60 277 3687 A rms IbB 277 120 80 j60 277 15687 A rms IcC 277120 80 j60 2778313 A rms Io IaA IbB IcC 0 b VAN 78 j54IaA 26279 217 V rms c VAB VAN VBN VBN 77 j56IbB 26373 12084 V rms VAB 26279 217 26373 12084 452892855 V rms d Unbalanced see conditions for a balanced circuit in the text P 118 Zga Zla ZLa 60 j80 Ω Zgb Zlb ZLb 90 j120 Ω Zgc Zlc ZLc 30 j40 Ω VN 320 60 j80 VN 320 120 90 j120 VN 320120 30 j40 VN 20 0 Solving for VN yields VN 49477514 V rms Io VN 20 2477514 A rms 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 119 P 119 VAN 285 45 V VBN 285 165 V VCN 28575 V VAB VAN VBN 49883 15 V VBC VBN VCN 49883 135 V VCA VCN VAN 49883105 V vAB 49883 cosωt 15 V vBC 49883 cosωt 135 V vCA 49883 cosωt 105 V P 1110 a Van 1 3 30Vab 110 90 V rms The aphase circuit is b IaA 110 90 40 j30 22 12687 A rms c VAN 37 j28IaA 10208 8975 V rms VAB 330VAN 17681 5975 A rms P 1111 a IaA 6600 3240 j70 15241626 A rms IaA IL 1524 A rms 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1110 CHAPTER 11 Balanced ThreePhase Circuits b Van 15241626240 j66 380124091 Vab 3380124 658394 V rms P 1112 Make a sketch of the aphase a Find the aphase line current from the aphase circuit IaA 1250 01 j08 199 j142 1250 20 j15 4 j3 5 3687 A rms Find the other line currents using the acb phase sequence IbB 5 3687 120 58313 A rms IcC 5 3687 120 5 15687 A rms b The phase voltage at the source is Van 1250 V Use Fig 119b to find the line voltage Van from the phase voltage Vab Van 3 30 21651 30 V rms Find the other line voltages using the acb phase sequence Vbc 21651 30 120 2165190 V rms Vca 21651 30 120 21651 150 V rms c The phase voltage at the load in the aphase is VAN Calculate its value using IaA and the load impedance VAN IaAZL 4 j3199 j142 1222 j29 12223 136 V rms Find the phase voltage at the load for the b and cphases using the acb sequence VBN 12223 136 120 1222311864 V rms VCN 12223 136 120 12223 12136 V rms 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1111 d The line voltage at the load in the aphase is VAB Find this line voltage from the phase voltage at the load in the aphase VAN using Fig 119b VAB VAN 3 30 21172 3136 V rms Find the line voltage at the load for the b and cphases using the acb sequence VBC 21172 3136 120 211728864 V rms VCA 21172 3136 120 21172 15136 V rms P 1113 a IAB 7200 216 j288 205313 A rms IBC 2017313 A rms ICA 20 6687 A rms b IaA 330IAB 34648313 A rms IbB 3464 15687 A rms IcC 3464 3687 A rms c Van 7200 3 30 3 j534648313 40853262 V rms Vab 3 30Van 707543262 V rms Vbc 70754312262 V rms Vca 707543 11738 V rms P 1114 a Van Vbn 1120 15015 V rms Zy Z3 43 j57 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1113 P 1117 a IaA 300 60 j45 300 30 j30 86 1759 A rms IaA 86 A rms b IAB 300 330 90 j90 408 15 A rms IAB 408 A rms c IAN 3000 60 j45 43687 A rms IAN 4 A rms d Van 86 17592 j2 3000 32179199 V rms Vab 332179 55737 V rms P 1118 a b IaA 13800 32375 j1349 2917 296 A rms IaA 2917 A rms c VAN 2352 j11392917 296 762293 376 V rms VAB 3VAN 1320331 V rms 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1114 CHAPTER 11 Balanced ThreePhase Circuits d Van 2372 j13192917 296 761693 052 V rms Vab 3Van 1371252 V rms e IAB IaA 3 168413 A rms f Iab IAB 168413 A rms P 1119 a IAB 132000 100 j75 10563687 A rms IBC 105615687 A rms ICA 1056 8313 A rms b IaA 3 30IAB 18296687 A rms IbB 1829 17313 A rms IcC 1829 5313 A rms c Iba IAB 10563687 A rms Icb IBC 105615687 A rms Iac ICA 1056 8313 A rms P 1120 a IAB 4800 24 j07 1921626 A rms IBC 480120 8 j6 488313 A rms ICA 480 120 20 24 120 A rms b IaA IAB ICA 2102079 IbB IBC IAB 17868 17804 IcC ICA IBC 707 10453 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1115 P 1121 a Since the phase sequence is abc positive we have Van 49883 30 V rms Vbn 49883 150 V rms Vcn 4988390 V rms ZY 1 3Z 15 j1 Ωφ b Vab 49883 30 49883 150 49883 30 8640 V rms Since the phase sequence is positive it follows that Vbc 864 120 V rms Vca 864120 V rms 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1116 CHAPTER 11 Balanced ThreePhase Circuits c Iba 864 45 j3 15975 3369 A rms Iac 159758631 A rms IaA Iba Iac 27670 6369 A rms Since we have a balanced threephase circuit and a positive phase sequence we have IbB 2767017631 A rms IcC 27670 5631 A rms 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1117 d IaA 4988330 15 j1 27670 6369 A rms Since we have a balanced threephase circuit and a positive phase sequence we have IbB 2767017631 A rms IcC 276705631 A rms P 1122 a b IaA 49883 30 1200 j1600 24942 8313 mA rms VAN 1192 j1584024942 8313 49445 3009 V rms VAB 349445 85641 V rms 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1118 CHAPTER 11 Balanced ThreePhase Circuits c Iab 024942 3 144 mA rms d Van 11985 j1599024942 8313 49842 2998 V rms Vab 349842 86329 V rms P 1123 a IaA 13650 30 j40 273 5313 A rms ICA IaA 3 150 15769687 A rms b Sgφ 1365I aA 2235875 j2981156 VA Pdevelopedphase 22359 kW Pabsorbedphase IaA2285 21241 kW delivered 21241 22359100 95 P 1124 The complex power of the source per phase is Ss 20000 cos1 06 200005313 12000 j16000 kVA This complex power per phase must equal the sum of the perphase impedances of the two loads Ss S1 S2 so 12000 j16000 10000 S2 S2 2000 j16000 VA Also S2 Vrms2 Z 2 Vrms Vload 3 120 V rms Thus Z 2 Vrms2 S2 1202 2000 j16000 011 j089 Ω Z2 011 j089 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1119 P 1125 The aphase of the circuit is shown below I1 12020 8 j6 12 1687A rms I 2 60036 12020 516A rms I I1 I2 12 1687 5 16 17 1661 A rms Sa VI 12020171661 20403661 VA ST 3Sa 61203661 VA P 1126 a I aA 128 j96103 1600 80 j60 IaA 80 j60 A rms Van 1600 80 j6002 j08 1664 j52 V rms IC 1664 j52 j25 208 j6656 A rms Ina IaA IC 7792 j656 782481 A rms b Sgφ 1664 j527792 j656 130000 j6864 VA SgT 3Sgφ 390000 j20592 VA Therefore the source is delivering 390 kW and absorbing 20592 kvars 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1121 P 1129 a S1 10200087 j102000493 8874 j502913 VA S2 4200 j19136 VA 3VLIL sin θ3 7250 sin θ3 7250 3220368 0517 Therefore cos θ3 0856 Therefore P3 7250 0517 0856 120039 W S3 120039 j7250 VA ST S1 S2 S3 25078 j14192 kVA STφ 1 3ST 83593 j47307 VA 220 3 I aA 83593 j47307 I aA 6581 j3724 A IaA 6581 j3724 7562 2951 A rms b pf cos0 2951 087 lagging P 1130 From the solution to Problem 1118 we have SAB 4800192 1626 884737 j258045 VA SBC 48012048 8313 1843198 j1382403 VA SCA 480 12024120 11520 j0 VA P 1131 Let pa pb and pc represent the instantaneous power of phases a b and c respectively Then assuming a positive phase sequence we have pa vaniaA Vm cos ωtIm cosωt θφ pb vbnibB Vm cosωt 120Im cosωt θφ 120 pc vcnicC Vm cosωt 120Im cosωt θφ 120 The total instantaneous power is pT pa pb pc so pT VmImcos ωt cosωt θφ cosωt 120 cosωt θφ 120 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1122 CHAPTER 11 Balanced ThreePhase Circuits cosωt 120 cosωt θφ 120 Now simplify using trigonometric identities In simplifying collect the coefficients of cosωt θφ and sinωt θφ We get pT VmImcos ωt1 2 cos2 120 cosωt θφ 2 sin ωt sin2 120 sinωt θφ 15VmImcos ωt cosωt θφ sin ωt sinωt θφ 15VmIm cos θφ P 1132 Iline 1600 240 3 11547 A rms Zy V I 240 3 11547 12 Zy 12 50 Ω Z 3Zy 36 50 2314 j2758 Ωφ P 1133 Assume a connected load series Sφ 1 3150 103096 j028 48000 j14000 VA Z φ 6002 48000 j14000 6912 j2016 Ωφ Zφ 6912 j2016 Ω Now assume a Yconnected load series ZY φ 1 3Zφ 2304 j0672 Ωφ 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1123 Now assume a connected load parallel Pφ 6002 R Rφ 6002 48000 75 Ω Qφ 6002 X Xφ 6002 14000 25714 Ω Now assume a Yconnected load parallel RY φ 1 3Rφ 25 Ω XY φ 1 3Xφ 8571 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1124 CHAPTER 11 Balanced ThreePhase Circuits P 1134 a POUT 746 100 74600 W PIN 74600097 7690722 W 3VLIL cos θ 7690722 IL 7690722 3208088 24258 A rms b Q 3VLIL sin φ 3208242580475 4151190 VAR P 1135 4000I 1 210 j280103 I 1 210 4 j 280 4 525 j70 A rms I1 525 j70 A rms I2 40000 1536 j448 240 j70 A rms IaA I1 I2 2925 j0 A rms Van 4000 j0 292501 j08 403604332 V rms Vab 3Van 699062 V rms 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1126 CHAPTER 11 Balanced ThreePhase Circuits VAN 240 3 004 j0037076 j7076 13361 j071 13361030 V rms VAB 313361 23142 V rms b SLφ 13361 j0717076 j7076 9404 j95045 VA SL 3SLφ 28212 j28513 VA Check Sg 4160007071 j07071 29415 j29415 VA Pℓ 3IaA2004 1202 W Pg PL Pℓ 28212 1202 29414 W checks Qℓ 3IaA2003 901 VAR Qg QL Qℓ 28513 901 29414 VAR checks P 1138 a Sg 1 3540096 j028 1728 j504 kVA S1 1 3284 j2088 128 j696 kVA S2 Sg S1 160 j120 kVA I L2 160 j120103 1600 100 j75 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1128 CHAPTER 11 Balanced ThreePhase Circuits Vab 3310644 53805 V rms b I1 120 j160 A from part a S2 0 j 1 31125 103 j375000 VAR I 2 j375000 2500 j150 A rms I2 j150 A rms IaA 120 j160 j150 120 j10 A rms Van 2500 120 j101 j3 2650 j350 267301752 V rms Vab 3267301 46298 V rms c IaA 200 A rms Plossφ 20021 40 kW Pgφ 300000 40000 340 kW η 300 340100 882 d IaA 120416 A rms Pℓφ 12041621 14500 W η 300000 314500100 954 e ZcapY 25002 j375000 j1667 Ω 1 ωC 1667 C 1 1667120π 159155 µF 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1129 P 1140 a From Assessment Problem 119 IaA 1018 j1357 A rms Therefore Icap j1357 A rms Therefore ZCY 2450 3 j1357 j1042 Ω Therefore CY 1 10422π60 2545 µF ZC j10423 j3126 Ω Therefore C 2545 3 8484 µF b CY 2545 µF c IaA 1018 A rms P 1141 Wm1 VABIaA cosVAB IaA 1995824 cos6568 19726 W Wm2 VCBIcC cosVCB IcC 1995824 cos568 47664 W CHECK W1 W2 6739 242393 6739 W P 1142 tan φ 3W2 W1 W1 W2 075 φ 3687 2400 3IL cos 6687 4082309 IL 25 A Z 2400 25 96 Ω Z 963687 Ω P 1143 IaA VAN Zφ ILθφ A Zφ Zθφ VBC VL 90 V Wm VL IL cos90 θφ VL IL cosθφ 90 VL IL sin θφ therefore 3Wm 3VL IL sin θφ Qtotal 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1130 CHAPTER 11 Balanced ThreePhase Circuits P 1144 a Z 16 j12 20 3687 Ω VAN 6800 V IaA 343687 A VBC VBN VCN 680 3 90 V Wm 680 334 cos90 3687 2402707 W 3Wm 416161 W b Qφ 34212 13872 VAR QT 3Qφ 41616 VAR 3Wm P 1145 a W2 W1 VLILcosθ 30 cosθ 30 VLILcos θ cos 30 sin θ sin 30 cos θ cos 30 sin θ sin 30 2VLIL sin θ sin 30 VLIL sin θ therefore 3W2 W1 3VLIL sin θ QT b Zφ 8 j6 Ω QT 3247625 97975 2592 VAR QT 31226 2592 VAR Zφ 8 j6 Ω QT 397975 247625 2592 VAR QT 31226 2592 VAR Zφ 51 j 3 Ω QT 32160 0 374123 VAR QT 31225 3 374123 VAR Zφ 1075 Ω QT 364553 176363 417280 VAR QT 312210 sin 75 417280 VAR 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1131 P 1146 Zφ Zθ VAN IaA θ VAN IaA θ1 VAB IaA For a positive phase sequence VAB VAN 30 Thus θ1 VAN 30 IaA θ 30 Similarly Zφ Zθ VCN IcC θ VCN IcC θ2 VCB IcC For a positive phase sequence VCB VBA 120 VAB 60 IcC IaA 120 Thus θ2 VAB 60 IaA 120 θ1 60 θ 30 60 θ 30 P 1147 a Zφ 100 j75 125 3687 Ω Sφ 132002 1253687 1115136 j836352 VA 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1132 CHAPTER 11 Balanced ThreePhase Circuits b 13200 3 30I aA Sφ so IaA 18296687 Wm1 132001829 cos0 6687 94840192 W Wm2 132001829 cos60 5313 239700608 W Check PT 31115136 W Wm1 Wm2 P 1148 From the solution to Prob 1120 we have IaA 2102079 A and IbB 17868 17804 A a W1 Vac IaA cosθac θaA 480210 cos60 2079 781032 W b W2 Vbc IbB cosθbc θbB 48017868 cos120 17804 403177 W c W1 W2 118421 W PAB 192224 884736 W PBC 4828 18432 W PCA 24220 11520 W PAB PBC PCA 1184257 therefore W1 W2 Ptotal roundoff differences P 1149 a I aA 144096 j028103 7200 20 1626 A VBN 7200 120 V VCN 7200120 V VBC VBN VCN 7200 3 90 V IbB 20 10374 A Wm1 7200 320 cos90 10374 24227814 W b Current coil in line aA measure IaA Voltage coil across AC measure VAC c IaA 201676 A VCA VAN VCN 7200 3 30 V Wm2 7200 320 cos30 1626 17244186 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1133 d Wm1 Wm2 41472kW PT 432000096 41472 kW Wm1 Wm2 P 1150 a W1 VBAIbB cos θ Negative phase sequence VBA 240 3150 V IaA 2400 1333 30 1830 A IbB 18150 A W1 18240 3 cos 0 748246 W W2 VCAIcC cos θ VCA 240 3 150 V IcC 18 90 A W2 18240 3 cos60 374123 W b Pφ 182403 cos30 374123 W PT 3Pφ 1122369 W W1 W2 748246 374123 1122369 W W1 W2 PT checks P 1151 a Z 1 3Z 448 j1536 167374 Ω IaA 6000 167374 375 7374 A IbB 375 19374 A VAC 600 3 30 V VBC 600 3 90 V W1 600 3375 cos30 7374 2815615 W W2 600 3375 cos90 19374 925615 W b W1 W2 18900 W PT 3375213443 18900 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1134 CHAPTER 11 Balanced ThreePhase Circuits c 3W1 W2 64800 VAR QT 3375246083 64800 VAR P 1152 a Negative phase sequence VAB 240 3 30 V VBC 240 390 V VCA 240 3 150 V IAB 240 3 30 2030 2078 60 A IBC 240 390 600 69390 A ICA 240 3 150 40 30 1039 120 A IaA IAB IAC 18 30 A IcC ICB ICA ICA IBC 1675 10806 Wm1 240 318 cos30 30 748246 W Wm2 240 31675 cos90 10807 662123 W b Wm1 Wm2 1410369 W PA 12 3220 cos 30 748246 W PB 4 3260 2880 W PC 6 3240 cos30 374123 W PA PB PC 1410369 Wm1 Wm2 P 1153 a 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1135 b c d P 1154 a Q V2 XC XC 138002 12 106 15870 Ω 1 ωC 15870 C 1 2π6015870 1671 µF b XC 13800 32 12 106 1 315870 C 31671 5014 µF P 1155 a The capacitor from Appendix H whose value is closest to 5014 µF is 47 µF XC 1 ωC 1 2π6047 106 564 Ω Q V 2 3XC 138002 3564 11247756 VAR 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1136 CHAPTER 11 Balanced ThreePhase Circuits b I aA 1200000 j75224 13800 3 1506 j94 A Van 13800 3 0 06 j481506 j94 81348506 Vab 381348 140899 V This voltage falls within the allowable range of 13 kV to 146 kV P 1156 a The capacitor from Appendix H whose value is closest to 1671 µF is 22 µF XC 1 ωC 1 2π6022 106 12057 Ω Q V 2 XC 138002 12057 1579497 VARφ b I aA 1200000 j379497 13800 3 502 j159 A Van 13800 3 0 06 j48502 j159 78978176 Vab 378978 136794 V This voltage falls within the allowable range of 13 kV to 146 kV P 1157 If the capacitors remain connected when the substation drops its load the expression for the line current becomes 13800 3 I aA j12 106 or I aA j15061 A Hence IaA j15061 A Now Van 13800 3 0 06 j48j15061 724449 j9037 724505071 V The magnitude of the linetoline voltage at the generating plant is Vab 3724505 1254880 V This is a problem because the voltage is below the acceptable minimum of 13 kV Thus when the load at the substation drops off the capacitors must be switched off 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1137 P 1158 Before the capacitors are added the total line loss is PL 315061 j15061206 8166 kW After the capacitors are added the total line loss is PL 315061206 4083 kW Note that adding the capacitors to control the voltage level also reduces the amount of power loss in the lines which in this example is cut in half P 1159 a 13800 3 I aA 60 103 j160 103 j1200 103 I aA 60 3 j1040 3 138 753 j13053 A IaA 753 j13053 A Van 13800 3 0 06 j48753 j13053 734541 j11446 73463089 V Vab 373463 1272416 V b Yes the magnitude of the linetoline voltage at the power plant is less than the allowable minimum of 13 kV P 1160 a 13800 3 I aA 60 j160 103 I aA 60 3 j160 3 138 753 j2008 A IaA 753 j2008 A Van 13800 3 0 06 j48753 j2008 806834 j2410 806838017 V Vab 3806838 1397477 V b Yes 13 kV 1397477 146 kV c Ploss 3753 j13053206 3077 kW d Ploss 3753 j2008206 083 kW e Yes the voltage at the generating plant is at an acceptable level and the line loss is greatly reduced 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 126 CHAPTER 12 Introduction to the Laplace Transform Problems P 121 a b ft 5tut ut 10 100 5tut 10 ut 30 50ut 30 ut 40 25t 150ut 40 ut 60 P 122 P 123 a 3t 15ut 5 ut 3t 15ut ut 5 3t 5ut 5 30ut 3t 5ut 5 b 5t 20ut 4 ut 2 5tut 2 ut 2 5t 20ut 2 ut 4 5t 4ut 4 10t 2ut 2 10t 2ut 2 5t 4ut 4 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 129 Fs 5e4s 2e2s 2e2s e4s s2 P 1214 a ft 8t 80ut 10 ut 5 8tut 5 ut 5 8t 80ut 5 ut 10 8t 10ut 10 16t 5ut 5 16t 5ut 5 8t 10ut 10 Fs 8e10s 2e5s 2e5s e10s s2 b ft 8ut 10 ut 5 8ut 5 ut 5 8ut 5 ut 10 8ut 10 16ut 5 16ut 5 8ut 10 Lft 8e10s 2e5s 2e5s e10s s 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1217 b V1 sL V1 V2 R Ig V2 V1 R sCV2 0 or R sLV1s sLV2s RLsIgs V1s RCs 1V2s 0 Solving V2s sIgs Cs2 RLs 1LC P 1232 a 625 150i1 625di1 dt 25 d dti2 i1 125 d dti1 i2 25di1 dt 0 125 d dti2 i1 25di1 dt 100i2 Simplifying the above equations gives 625 150i1 25di1 dt 125di2 dt 0 100i2 125di1 dt 125di2 dt b 625 s 25s 150I1s 125sI2s 0 125sI1s 125s 100I2s c Solving the equations in b I1s 50s 8 ss 4s 24 I2s 50 s 4s 24 P 1233 From Problem 1226 V s 100s2 s2 16s 100s2 400 s2 16s 100 s 8 j6s 8 j6 s2 400 s j20s j20 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1220 CHAPTER 12 Introduction to the Laplace Transform P 1237 a 1 LC 1 165 106 125 103 1 RC 1 20005 106 100 Vos 5600 s2 100s 125 103 s12 50 j350 rads Vos 5600 s 50 j350s 50 j350 K1 s 50 j350 K 1 s 50 j350 K1 5600 j700 8 90 vot 16e50t cos350t 90ut V 16e50t sin 350tut V b Ios 56625 ss 50 j350s 50 j350 K1 s K2 s 50 j350 K 2 s 50 j350 K1 3500 125 103 28 mA K2 3500 50 j350j700 141417187 mA iot 28 2828e50t cos350t 17187ut mA P 1238 1 C 64 106 1 LC 1600 106 R L 100000 Ig 015 s V2s 96 105 s2 105s 1600 106 s1 20000 s2 80000 V2s 96 105 s 20000s 80000 160 s 20000 160 s 80000 v2t 160e20000t 160e80000tut V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1228 CHAPTER 12 Introduction to the Laplace Transform P 1249 sVos sVdcRC s2 1RCs 1LC lim s0 sVos 0 vo 0 lim s sVos 0 vo0 0 sIos VdcRC s2 1RCs 1LC lim s0 sIos VdcRLC 1LC Vdc R io Vdc R lim s sIos 0 io0 0 P 1250 a sFs 6s2 60s s 5s 8 lim s0 sFs 0 f 0 lim s sFs 6 f0 6 b sFs 20s2 141s 315 s2 10s 21 lim s0 sFs 15 f 15 lim s sFs 20 f0 20 c sFs 15s3 112s2 228s s 2s 4s 6 lim s0 sFs 0 f 0 lim s sFs 15 f0 15 d sFs 2s3 33s2 93s 54 s 1s2 5s 6 lim s0 sFs 9 f 9 lim s sFs 2 f0 2 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1229 P 1251 a sFs 280s s2 14s 245 lim s0 sFs 0 f 0 lim s sFs 0 f0 0 b sFs s2 52s 445 s2 10s 89 lim s0 sFs 5 f 5 lim s sFs 1 f0 1 c sFs 14s3 56s2 152s s 6s2 4s 20 lim s0 sFs 0 f 0 lim s sFs 14 f0 14 d sFs 8ss 12 s2 10s 34s2 8s 20 lim s0 sFs 0 f 0 lim s sFs 0 f0 0 P 1252 a sFs 320 ss 8 Fs has a secondorder pole at the origin so we cannot use the final value theorem here lim s sFs 0 f0 0 b sFs 80s 3 s 22 lim s0 sFs 60 f 60 lim s sFs 0 f0 0 c sFs 60ss 5 s 12s2 6s 25 lim s0 sFs 0 f 0 lim s sFs 0 f0 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1230 CHAPTER 12 Introduction to the Laplace Transform d sFs 25s 42 ss 52 Fs has a secondorder pole at the origin so we cannot use the final value theorem here lim s sFs 0 f0 0 P 1253 a sFs 135 s 33 lim s0 sFs 5 f 5 lim s sFs 0 f0 0 b sFs 10ss 22 s2 2s 22 lim s0 sFs 0 f 0 lim s sFs 0 f0 0 c This Fs function is an improper rational function and thus the corresponding ft function contains impulses δt Neither the initial value theorem nor the final value theorem may be applied to this Fs function d This Fs function is an improper rational function and thus the corresponding ft function contains impulses δt Neither the initial value theorem nor the final value theorem may be applied to this Fs function P 1254 a ZL j120π001 j377 Ω ZC j 120π100 106 j26526 Ω The phasortransformed circuit is IL 1 15 j377 j26526 36695661 mA iLsst 3669 cos120πt 5661 mA 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1232 CHAPTER 12 Introduction to the Laplace Transform Therefore iLt K0176e750t cos66144t 13959 1163te100t 0134e100tut mA Plot the expression above with K 1 The maximum value of the inductor current is 0068K mA Therefore K 40 0068 588 So the inductor current rating will not be exceeded if the input to the RLC circuit is 588te100t V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 13 The Laplace Transform in Circuit Analysis Assessment Problems AP 131 a Y 1 R 1 sL sC Cs2 1RCs 1LC s 1 RC 106 5000025 80000 1 LC 25 108 Therefore Y 25 109s2 80000s 25 108 s b z12 40000 16 108 25 108 40000 j30000 rads z1 40000 j30000 rads z2 40000 j30000 rads p1 0 rads AP 132 a Z 2000 1 Y 2000 4 107s s2 80000s 25 108 2000s2 105s 25 108 s2 80000s 25 108 2000s 500002 s2 80000s 25 108 b z1 z2 50000 rads p1 40000 j30000 rads p2 40000 j30000 rads 131 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 136 CHAPTER 13 The Laplace Transform in Circuit Analysis AP 138 a The sdomain circuit with the voltage source acting alone is V 20s 2 V 125s V s 20 0 V 200 s 2s 8 1003 s 2 1003 s 8 v 100 3 e2t e8tut V b With the current source acting alone V 2 V 125s V s 20 5 s V 100 s 2s 8 503 s 2 503 s 8 v 50 3 e2t e8tut V c v v v 50e2t 50e8tut V AP 139 a Vo s 2 Vos 10 Ig therefore Vo Ig Hs 10s 2 s2 2s 10 b z1 2 rads p1 1 j3 rads p2 1 j3 rads AP 1310 a Vo 10s 2 s2 2s 10 1 s Ko s K1 s 1 j3 K 1 s 1 j3 Ko 2 K1 53 12687 K 1 5312687 vo 2 103et cos3t 12687ut V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 137 b Vo 10s 2 s2 2s 10 1 K2 s 1 j3 K 2 s 1 j3 K2 527 1843 K 2 5271843 vo 1054et cos3t 1843ut V AP 1311 a Hs Lht Lvot vot 10000 cos θe70t cos 240t 10000 sin θe70t sin 240t 9600e70t cos 240t 2800e70t sin 240t Therefore Hs 9600s 70 s 702 2402 2800240 s 702 2402 9600s s2 140s 62500 b Vos Hs 1 s 9600 s2 140s 62500 K1 s 70 j240 K 1 s 70 j240 K1 9600 j480 j20 20 90 Therefore vot 40e70t cos240t 90ut V 40e70t sin 240tut V AP 1312 From Assessment Problem 139 Hs 10s 2 s2 2s 10 Therefore Hj4 102 j4 10 16 j8 447 6343 Thus vo 10447 cos4t 6343 447 cos4t 6343 V AP 1313 a Let R1 10 kΩ R2 50 kΩ C 400 pF R2C 2 105 then V1 V2 VgR2 R2 1sC 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 138 CHAPTER 13 The Laplace Transform in Circuit Analysis Also V1 Vg R1 V1 Vo R1 0 therefore Vo 2V1 Vg Now solving for VoVg we get Hs R2Cs 1 R2Cs 1 It follows that Hj50000 j 1 j 1 j1 190 Therefore vo 10 cos50000t 90 V b Replacing R2 by Rx gives us Hs RxCs 1 RxCs 1 Therefore Hj50000 j20 106Rx 1 j20 106Rx 1 Rx j50000 Rx j50000 Thus 50000 Rx tan 60 17321 Rx 2886751 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1311 Zab s 2 2ss 2 s2 s 2 2ss 22 s2 s 2 s 2 2ss 2 s2 s 2 2ss 22 s 2s2 s 2 2ss 2 2ss 2 s2 3s 2 Zeros at 0 and 2 rads poles at 1 rads and 2 rads P 139 Vo 90s8 105s 140 0005s 8 105s 144 108 ss2 28000s 16 107 144 108 ss 8000s 20000 K1 s K2 s 8000 K3 s 20000 K1 144 108 16 107 90 K2 144 108 800012000 150 K3 144 108 1200020000 60 Vo 90 s 150 s 8000 60 s 20000 vot 90 150e8000t 60e20000tut V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1319 P 1317 a Vo sLIgsC R sL 1sC IgC s2 RLs 1LC Ig C 006 0001 60 R L 140 1 LC 4000 Vo 60 s2 140s 4000 b sVo 60s s2 140s 4000 lim s0 sVo 0 vo 0 lim s sVo 0 vo0 0 c Vo 60 s 40s 100 1 s 40 1 s 100 vo e40t e100tut V P 1318 IC Ig s Vo sL 006 s 240 ss2 140s 4000 006 s 006 s 01 s 40 004 s 100 iCt 100e40t 40e100tut mA Check iC0 60 mA ok iC 0 ok 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1320 CHAPTER 13 The Laplace Transform in Circuit Analysis P 1319 a 200i1 04sI1 I2 200 s 200I2 105 s I2 04sI2 I1 0 Solving the second equation for I1 I1 s2 500s 25 104 s2 I2 Substituting into the first equation and solving for I2 04s 200s2 500s 25 104 s2 04s 200 s I2 05s s2 500s 125000 I1 s2 500s 25 104 s2 05s s2 500s 125000 05s2 500s 25 104 ss2 500s 125000 Io I1 I2 05s2 500s 25 104 ss2 500s 125000 05s s2 500s 125000 250s 500 ss2 500s 125000 K1 s K2 s 250 j250 K 2 s 250 j250 K1 1 K2 05 180 05 iot 1 1e250t cos 250tut A b Vo 04sIo 100s 500 s2 500s 125000 K1 s 250 j250 K 1 s 250 j250 K1 7071 45 vot 14142e250t cos250t 45ut V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1321 c At t 0 the circuit is vo0 100 V 14142 cos45 Io0 0 Both values agree with our solutions for vo and io At t the circuit is vo 0 io 1 A Both values agree with our solutions for vo and io P 1320 a V1 325s 250 V1 500 V1 V2 250 0 V2 0625s V2 V1 250 V2 325ss 125 104 0 Thus 5V1 2V2 650 s 5000sV1 s2 5000s 2 106V2 325s 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1326 CHAPTER 13 The Laplace Transform in Circuit Analysis P 1323 20 103 s Vo 50 104s 15 103Vφ Vo s Vφ 104s 50 104sVo 104Vo 50s 104 20 103 s Vos 50s 104 150Vo 50s 104 Vo s Vo s 200 s2 200s 104 K1 s 1002 K2 s 100 K1 100 K2 1 Vo 100 s 1002 1 s 100 vot 100te100t e100tut V P 1324 vC0 vC0 0 005 s Vo 1000 Vo 100 Vos 5 106 03Vo 100 250 103 s 5000 50000 s 15000Vo Vo 250 103 ss 40000 K1 s K2 s 40000 625 s 625 s 40000 vot 625 625e40000tut V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1329 c At t 0 the circuit is At t the circuit is Vφ 600 10 Vφ 140 Vφ 4 0 Vφ 168 V Vo checks d Vo N2 21000s 4200 ss2 8s 25 K1 s K2 s 4 j3 K 2 s 4 j3 K1 4200 25 168 K2 210004 j3 4200 4 j3j6 84 j3612 3612989133 vot 168 722595e4t cos3t 9133ut V Check vo0 0 V vo 168 V P 1327 a 20Iφ 25sIo Iφ 25Io I1 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1331 At t the circuit is Io 100 Achecks d Io 20s2 96s 200 ss 1s 2 K1 s K2 s 1 K3 s 2 K1 200 12 100 K2 20 96 200 11 84 K3 80 192 200 21 36 Io 100 s 84 s 1 36 s 2 iot 100 84et 36e2tut A io 100 Achecks io0 100 84 36 20 Achecks 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1333 P 1329 a For t 0 V2 10 10 40450 90 V b V1 25450s 125000s 25 125 103s 9 106 s2 20 000s 108 9 106 s 100002 v1t 9 106te10000tut V c V2 90 s 25000s450s 125000s 125 103s 25 90s 20000 s2 20000s 108 900000 s 100002 90 s 10000 v2t 9 105te10000t 90e10000tut V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1334 CHAPTER 13 The Laplace Transform in Circuit Analysis P 1330 a b Zeq 50000 107 3s 20 1012s2 12 106s 50000 107 3s 20 1012 12 106s 100000s 107 2s I1 20s Zeq 04 103 s 100 V1 107 3s I1 40003 ss 100 V2 107 6s 04 104 s 100 20003 ss 100 c i1t 04e100tut mA V1 403 s 403 s 100 v1t 4031 1e100tut V V2 203 s 203 s 100 v2t 2031 1e100tut V d i10 04 mA i10 20 50 103 044 mAchecks v10 0 v20 0checks v1 403 V v2 203 Vchecks v1 v2 20 Vchecks 03 106v1 4 µC 06 106v2 4 µCchecks 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1341 b Reversing the dot on the 250 mH coil will reverse the sign of M thus the circuit becomes The two simulanteous equations are 150 s 100 01875sI1 0125sI2 0 0125sI1 025s 200I2 When these equations are compared to those derived in Problem 1336 we see the only difference is the algebraic sign of the 0125s term Thus reversing the dot will have no effect on I1 and will reverse the sign of I2 Hence i2t 05e400t 05e1600tut A P 1338 a W 1 2L1i2 1 1 2L2i2 2 Mi1i2 W 0163002 0362002 024200300 432 kJ b 240i1 032di1 dt 024di2 dt 0 540i2 072di2 dt 024di1 dt 0 Laplace transform the equations to get 240I1 032sI1 300 024sI2 200 0 540I2 072sI2 200 024sI1 300 0 In standard form 032s 240I1 024sI2 144 024sI1 072s 540I2 216 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1347 b Vo K1 s K2 s 400 K3 s 600 K1 32 104320 400600 42667 K2 32 10480 400200 320 K3 32 104280 600200 74667 vot 42667 320e400t 74667e600tut V P 1344 a Vo Zf Zi Vg Zf 200016 106s 2000 16 106s 16 106 s 800 Zi 800 106 3125s 800s 400 s Vg 5000 s2 Vo 107 ss 400s 800 b Vo K1 s K2 s 400 K3 s 800 K1 107 400800 3125 K2 107 400400 625 K3 107 800400 3125 vot 3125 625e400t 3125e800tut V c 3125 625e400ts 3125e800ts 10 625e400ts 3125e800ts 2125 Let x e400ts Then 625x 3125x2 2125 or x2 2x 068 0 Solving x 04343 e400ts 04343 ts 21 ms 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1354 CHAPTER 13 The Laplace Transform in Circuit Analysis 1000Vo2s 5000 s2Vo 5000sVg Vos2 2000s 5 106 5000sVg Vo Vg 5000s s2 2000s 5 106 s12 1000 106 5 106 1000 j2000 Vo Vg 5000s s 1000 j2000s 1000 j2000 b z1 0 p1 1000 j2000 p2 1000 j2000 P 1356 a Vo 50 Vo s Vos 2000 Ig Vo 2000s s2 40s 2000 Ig and Io 5 104sVo Hs Io Ig s2 s2 40s 2000 b Ig 0025s s2 40000 so Io s20025s s 20 j40s 20 j40s2 40000 Io 0025s3 s 20 j40s 20 j40s j200s j200 c Damped sinusoid of the form Me20t cos40t θ1 d Steadystate sinusoid of the form N cos200t θ2 e Io K1 s 20 j40 K 1 s 20 j40 K2 s j200 K 2 s j200 K1 002520 j403 j8020 j16020 j240 71977 106 9794 K2 0025j2003 j40020 j16020 j240 1288 1031189 iot 144e20t cos40t 9794 2576 cos200t 1189 mA 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1355 P 1357 a 2000Io Ig 8000Io µIg Io2000 2sIo 0 Io 10001 µ s 10005 µIg Hs 10001 µ s 10005 µ b µ 5 c µ Hs Io 3 4000s 8000 20000ss 8000 0 1000s 5000 5000ss 5000 4 3000s 1000 15000ss 1000 5 4000s 20000s2 6 5000s 1000 25000ss 1000 µ 3 Io 25 s 25 s 8000 io 25 25e8000tut A µ 0 Io 1 s 1 s 5000 io 1 e5000tut A µ 4 Io 15 s 15 s 1000 io 15 15e1000tut A µ 5 Io 20000 s2 io 20000t ut A µ 6 Io 25 s 25 s 1000 io 251 e1000tut A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1386 CHAPTER 13 The Laplace Transform in Circuit Analysis At t 0 our circuit is iLt 50etτ A t 0 τ LR 04 ms iLt 50e2500t A t 0 vot 032diL dt 40000e2500t V t 0 which agrees with our solution P 1384 a The sdomain circuit is The nodevoltage equation is V sL1 V R V sL2 ρ s Therefore V ρR s RLe where Le L1L2 L1 L2 Therefore v ρReRLetut V b I1 V R V sL2 ρs RL2 ss RLe K0 s K1 s RLe K0 ρL1 L1 L2 K1 ρL2 L1 L2 Thus we have i1 ρ L1 L2 L1 L2eRLetut A c I2 V sL2 ρRL2 ss RLe K2 s K3 s RLe K2 ρL1 L1 L2 K3 ρL1 L1 L2 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1395 V0 12206685 Vrms Therefore v0 12206 2 cos120πt 685 V agreeing with the steadystate component of the result in part b d A plot of v0 generated in Excel is shown below P 1393 a At t 0 the phasor domain equivalent circuit is I1 j120 12 j10 10 90A rms I2 j12035 j1440 35 12 35 12180A rms I3 j120 8 j15 15 90A rms IL I1 I2 I3 35 12 j25 2517 9665A rms iL 2517 2 cos120πt 9665A iL0 iL0 292 2A i2 35 12 2 cos120πt 180A i20 i20 35 12 2 292 2A Vg Vo j1IL 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1396 CHAPTER 13 The Laplace Transform in Circuit Analysis Vg j120 25 j 35 12 25 j12292 12543 7850V rms vg 12543 2 cos120πt 7850V 12543 2cos 120πt cos 7850 sin 120πt sin 7850 25 2 cos 120πt 12292 2 sin 120πt Vg 25 2s 12292 2120π s2 120π2 sdomain circuit where Ll 1 120π H La 12 35π H Ra 12 Ω iL0 292 2 A i20 292 2 A The node voltage equation is 0 Vo Vg iL0Ll sLl Vo Ra Vo i20La sLa Solving for Vo yields Vo VgRaLl s RaLl LaLaLl RaiL0 i20 s RaLl LaLlLa Ra Ll 1440π RaLl La LlLa 12 1 120π 12 35π 12 35π 1 120π 1475π iL0 i20 292 2 292 2 0 Vo 1440π25 2s 12292 2120π s 1475πs2 120π2 K1 s 1475π K2 s j120π K 2 s j120π 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1397 K1 1455 2 K2 6103 2 8315 vot 1455 2e1475πt 12206 2 cos120πt 8315V Check vo0 1455 1455 2 0 b 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1398 CHAPTER 13 The Laplace Transform in Circuit Analysis 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1399 c In Problem 1392 the linetoneutral voltage spikes at 300 2 V Here the linetoneutral voltage has no spike Thus the amount of voltage disturbance depends on what part of the cycle the sinusoidal steadystate voltage is switched 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 13100 CHAPTER 13 The Laplace Transform in Circuit Analysis P 1394 a First find Vg before Rb is disconnected The phasor domain circuit is IL 120θ Ra 120θ Rb 120θ jXa 120θ RaRbXa Ra RbXa jRaRb Since Xl 1 Ω we have Vg 120θ 120θ RaRbXa RaRb jRa RbXa Ra 12 Ω Rb 8 Ω Xa 1440 35 Ω Vg 120θ 1400 1475 j300 25 12θ59 j12 12543θ 1150 vg 12543 2 cos120πt θ 1150V Let β θ 1150 Then vg 12543 2cos 120πt cos β sin 120πt sin βV Therefore Vg 12543 2s cos β 120π sin β s2 120π2 The sdomain circuit becomes where ρ1 iL0 and ρ2 i20 The sdomain node voltage equation is Vo Vg ρ1Ll sLl Vo Ra Vo ρ2La sLa 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 13102 CHAPTER 13 The Laplace Transform in Circuit Analysis Now K1 1440π12543 21475π cos β 120π sin β 14752π2 14400π2 144012543 21475 cos β 120 sin β 14752 14000 Since β θ 1150 K1 reduces to K1 12118 2 cos θ 1455 2 sin θ From the partial fraction expansion for Vo we see vot will go directly into steady state when K1 300 2 cos θ It follow that 1455 2 sin θ 17882 2 cos θ or tan θ 1229 Therefore θ 8535 b When θ 8535 β 7385 K2 1440π12543 2120π sin7385 j120π cos7385 1475π j120πj240π 720 212048 j3488 120 j1475 6103 2 7850 vo 12206 2 cos120πt 7850 V t 0 17261 cos120πt 7850 V t 0 c vo1 16971 cos120πt 8535V t 0 vo2 17261 cos120πt 7850V t 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 13103 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 13104 CHAPTER 13 The Laplace Transform in Circuit Analysis 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 14 Introduction to FrequencySelective Circuits Assessment Problems AP 141 fc 8 kHz ωc 2πfc 16π krads ωc 1 RC R 10 kΩ C 1 ωcR 1 16π 103104 199 nF AP 142 a ωc 2πfc 2π2000 4π krads L R ωc 5000 4000π 040 H b Hjω ωc ωc jω 4000π 4000π jω When ω 2πf 2π50000 100000π rads Hj100000π 4000π 4000π j100000π 1 1 j25 0048771 Hj100000π 004 c θ100000π 8771 AP 143 ωc R L 5000 35 103 143 Mrads 141 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 143 AP 149 ω2 o 1 LC so L 1 ω2 oC 1 400π202 106 3166 mH Q fo β 5 103 200 25 ωoRC R Q ωoC 25 400π02 106 995 kΩ AP 1410 ωo 8000π rads C 500 nF ω2 o 1 LC so L 1 ω2oC 317 mH Q ωo β ωoL R 1 ωoCR R 1 ωoCQ 1 8000π5005 109 1592 Ω AP 1411 ωo 2πfo 2π20000 40π krads R 100 Ω Q 5 Q ωo β ωo RL so L QR ωo 5100 40π 103 398 mH ω2 o 1 LC so C 1 ω2oL 1 40π 1032398 103 1592 nF 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 144 CHAPTER 14 Introduction to FrequencySelective Circuits Problems P 141 a ωc R L 1200 50 103 24 krads fc ωc 2π 24000 2π 381972 Hz b Hs ωc s ωc 24000 s 24000 Hjω 24000 24000 jω Hjωc 24000 24000 j24000 07071 45 Hj0125ωc 24000 24000 j3000 09923 7125 Hj8ωc 24000 24000 j192000 0124 82875 c votωc 14142 cos24000t 45 V vot0125ωc 19846 cos3000t 7125 V vot8ωc 248 cos192000t 82875 V P 142 a ωc 1 RC 1 1605 106 1250 rads fc ωc 2π 19894 Hz b Hjω ωc s ωc 1250 s 1250 Hjω 1250 1250 jω Hjωc 1250 1250 j1250 07071 45 Hj01ωc 1250 1250 j125 09950 571 Hj10ωc 1250 1250 j12500 00995 8429 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1414 CHAPTER 14 Introduction to FrequencySelective Circuits b Hj4697256 j46972566250 500002 46972562 j46972566250 1 2 45 Vo 1 2 45Vi vot 354 cos4697256t 45 V c Hj5322256 j53222566250 500002 53222562 j53222566250 1 2 45 Vo 1 2 45Vi vot 354 cos5322256t 45 V d Hj5000 j50006250 500002 50002 j50006250 00126893 Vo 00126893Vi vot 631 cos5000t 893 mV e Hj500000 j5000006250 500002 5000002 j5000006250 00126 893 Vo 00126 893Vi vot 631 cos500000t 893 mV P 1423 Hs 1 RLs s2 RLs 1LC s2 1LC s2 RLs 1LC Hjω 500002 ω2 500002 ω2 jω6250 a Hj50000 500002 500002 500002 500002 j500006250 0 Vo 0Vi vot 0 mV b Hj4697256 500002 46972562 500002 46972562 j46972566250 1 2 45 Vo 1 2 45Vi vot 354 cos4697256t 45 V c Hj5322256 500002 53222562 500002 53222562 j53222566250 1 2 45 Vo 1 2 45Vi vot 354 cos5322256t 45 V d Hj5000 500002 50002 500002 50002 j50006250 09999 072 Vo 09999 072Vi vot 49996 cos5000t 072 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1419 f P 1432 a b L 1 ω2 oC 1 50 103220 104 20 mH R ωoL Q 50 10320 103 625 160 Ω c Re 160480 120 Ω Re Ri 120 80 200 Ω Qsystem ωoL Re Ri 50 10320 103 200 5 d βsystem ωo Qsystem 50 103 5 10 krads βsystemHz 10000 2π 159155 Hz P 1433 a Vo Vi Z Z R where Z 1 Y and Y sC 1 sL 1 RL LCRLs2 sL RL RLLs 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1422 CHAPTER 14 Introduction to FrequencySelective Circuits h fc2 16000 2π 254648 Hz P 1436 a Hjω 1 LC ω2 1 LC ω2 j R L ω 80002 ω2 80002 ω2 j12000ω ωo 8000 rads Hjωo 80002 80002 80002 80002 j120008000 0 ωc1 4000 rads Hjωc1 80002 40002 80002 40002 j120004000 07071 45 ωc2 16000 rads Hjωc2 80002 160002 80002 160002 j1200016000 0707145 01ωo 800 rads Hj01ωo 80002 8002 80002 8002 j12000800 09887 862 10ωo 80000 rads Hj10ωo 80002 800002 80002 800002 j1200080000 09887862 b ω ωo 8000 rads Vo Hj8000Vi 0Vi vot 0 ω ωc1 4000 rads Vo Hj4000Vi 07071 4580 5657 45 vot 5657 cos4000t 45 V ω ωc2 16000 rads Vo Hj16000Vi 070714580 565745 vot 5657 cos16000t 45 V ω 01ωo 800 rads Vo Hj800Vi 09887 86280 791 862 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1426 CHAPTER 14 Introduction to FrequencySelective Circuits d Hj3125 250002 31252 250002 31252 j312510000 09987 291 Vo 09987 291Vi vot 9987 cos3125t 291 mV e Hj200000 250002 2000002 250002 2000002 j20000010000 09987291 Vo 09987291Vi vot 9987 cos200000t 291 mV P 1441 Hjω jωβ ω2o ω2 jωβ jω10000 250002 ω2 jω10000 a Hj25000 j2500010000 250002 250002 j2500010000 1 Vo 1Vi vot 10 cos 25000t V b Hj204951 j20495110000 250002 2049512 j20495110000 1 2 45 Vo 1 2 45Vi vot 7071 cos204951t 45 V c Hj304951 j30495110000 250002 3049512 j30495110000 1 2 45 Vo 1 2 45Vi vot 7071 cos304951t 45 V d Hj3125 j312510000 250002 31252 j312510000 00518709 Vo 00518709Vi vot 051 cos3125t 8709 V e Hj200000 j20000010000 250002 2000002 j20000010000 0051 8709 Vo 0051 8709Vi vot 051 cos200000t 8709 V P 1442 a ωo 2πfo 8π krads L 1 ω2 oC 1 8000π205 106 317 mH R Q ωoC 5 8000π05 106 39789 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 15 Active Filter Circuits Assessment Problems AP 151 Hs R2R1s s 1R1C 1 R1C 1 rads R1 1 Ω C 1 F R2 R1 1 R2 R1 1 Ω Hprototypes s s 1 AP 152 Hs 1R1C s 1R2C 20000 s 5000 1 R1C 20000 C 5 µF R1 1 200005 106 10 Ω 1 R2C 5000 R2 1 50005 106 40 Ω 151 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 154 CHAPTER 15 Active Filter Circuits Problems P 151 a K 101020 316 R2 R1 R2 1 ωcC 1 2π103750 109 21221 Ω R1 R2 K 21221 316 6716 Ω b P 152 a 1 RC 2π1000 so RC 15915 104 There are several possible approaches Here choose Rf 150 Ω Then C 15915 104 150 106 106 Choose C 1 µF This gives ωc 1 150106 667 103 rads so fc 1061 Hz To get a passband gain of 10 dB choose Ri Rf 316 150 316 4747 Ω Choose Ri 47 Ω to give K 20 log1015047 1008 dB 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 155 The resulting circuit is b error in fc 1061 1000 1000 100 61 error in passband gain 1008 10 10 100 08 P 153 a ωc 1 R2C so R2 1 ωcC 1 2π250010 109 6366 Ω K R2 R1 so R1 R2 K 6366 5 1273 Ω b Both the cutoff frequency and the passband gain are changed P 154 a 535 175 V so Vcc 175 V b Hjω 52π2500 jω 2π2500 Hj5000π 55000π 5000π j5000π 25 j25 5 2 135 Vo 175 2 135Vi so vot 1237 cos5000πt 135 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 158 CHAPTER 15 Active Filter Circuits b P 159 a 1 RC 2π8000 so RC 1989 106 There are several possible approaches Here choose C 0047 µF Then Ri 1989 106 0047 106 423 Choose Ri 390 Ω This gives ωc 1 0047 106390 5456 krads so fc 868 kHz To get a passband gain of 14 dB choose Rf 5Ri 5390 1950 Ω Choose Rf 18 kΩ to give a passband gain of 20 log101800390 133 dB The resulting circuit is b error in fc 868376 8000 8000 100 85 error in passband gain 133 14 14 100 51 P 1510 a ωc 1 R1C so R1 1 ωcC 1 2π4000250 109 159 Ω K R2 R1 so R2 KR1 8159 1273 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 159 b The passband gain changes but the cutoff frequency is unchanged P 1511 a 825 20 V so Vcc 20 V b Hjω 8jω jω 8000π Hj8000π 8j8000π 8000π j8000π 8 2 135 Vo 1414 135Vi so vot 1414 cos8000πt 135 V c Hj1000π 8j1000π 8000π j1000π 099 971 Vo 248 971Vi so vot 248 cos1000πt 971 mV d Hj64000π 8j64000π 8000π j64000π 794 1729 Vo 1985 1729Vi so vot 1985 cos64000πt 1729 V P 1512 For the RC circuit Hs Vo Vi s s 1RC R kmR C C kmkf RC RC kf 1 kf 1 RC kf Hs s s 1RC s s kf skf skf 1 For the RL circuit Hs s s RL 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1516 CHAPTER 15 Active Filter Circuits P 1522 a km 80 200 04 km kf 04 002 so kf 0404 002 8 C 20 106 048 625 µF b Zab 1 jωC RjωL 1 jωC jωRL R jωL R jωL ω2RLCω2LC jωRC ω2LC jωRCω2LC jωRC The denominator is purely real so set the imaginary part of the numerator equal to 0 and solve for ω ωR2C ω3L2C3 ωR2LC2 0 ω2 R2 R2LC L2 802 802002625 106 0022 16 106 Thus ω 4000 rads c In the original unscaled circuit the frequency at which the impedance Zab is purely real is ω2 us 2002 20020420 106 042 250000 Thus ωus 500 rads ωscaled ωus 4000 500 8 kf P 1523 a kf 10000 250 40 C 100 106 40 25 µF L 08 40 20 mH 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1522 CHAPTER 15 Active Filter Circuits f P 1530 ωo 2πfo 400π rads β 2π1000 2000π rads ωc2 ωc1 2000π ωc1ωc2 ωo 400π Solve for the cutoff frequencies ωc1ωc2 16 104π2 ωc2 16 104π2 ωc1 16 104π2 ωc1 ωc1 2000π or ω2 c1 2000πωc1 16 104π2 0 ωc1 1000π 106π2 016 106π2 ωc1 1000π1 116 24201 rads ωc2 2000π 24201 652519 rads Thus fc1 3852 Hz and fc2 103852 Hz Check β fc2 fc1 1000Hz ωc2 1 RLCL 652519 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1523 RL 1 6525195 106 3065 Ω ωc1 1 RHCH 24201 RH 1 242015 106 82643 Ω P 1531 ωo 1000 rads GAIN 6 β 4000 rads C 02 µF β ωc2 ωc1 4000 ωo ωc1ωc2 1000 Solve for the cutoff frequencies ω2 c1 4000ωc1 106 0 ωc1 2000 1000 5 23607 rads ωc2 4000 ωc1 423607 rads Check β ωc2 ωc1 4000 rads ωc1 1 RLCL RL 1 02 10623607 2118 kΩ 1 RHCH 423607 RH 1 02 106423607 118 kΩ Rf Ri 6 If Ri 1 kΩ Rf 6Ri 6 kΩ 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1529 d It is apparent from the calculations in part c that as n increases the amplitude characteristic at the cutoff frequency decreases at a much faster rate for the Butterworth filter Hence the transition region of the Butterworth filter will be much narrower than that of the cascaded sections P 1539 n 5 1 15s10 0 s10 1 s10 10 360k so s 136k k sk1 k sk1 0 10 5 1180 1 136 6 1216 2 172 7 1252 3 1108 8 1288 4 1144 9 1324 Group by conjugate pairs to form denominator polynomial s 1s cos 108 j sin 108s cos 252 j sin 252 s cos 144 j sin 144s cos 216 j sin 216 s 1s 0309 j0951s 0309 j0951 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1530 CHAPTER 15 Active Filter Circuits s 0809 j0588s 0809 j0588 which reduces to s 1s2 0618s 1s2 1618s 1 n 6 1 16s12 0 s12 1 s12 1180 360k k sk1 k sk1 0 115 6 1195 1 145 7 1225 2 175 8 1255 3 1105 9 1285 4 1135 10 1315 5 1165 11 1345 Grouping by conjugate pairs yields s 02588 j09659s 02588 j09659 s 07071 j07071s 07071 j07071 s 09659 j02588s 09659 j02588 or s2 05176s 1s2 14142s 1s2 19318s 1 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1532 CHAPTER 15 Active Filter Circuits C1 108 5000π8000 861 nF C2 092 5000π8000 735 nF b P 1542 a n 00555 log1020040 393 n 4 From Table 151 the transfer function of the first section is H1s 1 s2 0765s 1 For the prototype circuit 2 R2 0765 R2 261 Ω R1 1 R2 0383 Ω The transfer function of the second section is H2s 1 s2 1848s 1 For the prototype circuit 2 R2 1848 R2 1082 Ω R1 1 R2 09240 Ω The scaling factors are kf ω o ωo 2π40000 1 80000π C C kmkf 250 109 1 80000πkm km 1 80000π250 109 159 Therefore in the first section R 1 kmR1 61 Ω R 2 kmR2 415 Ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1534 CHAPTER 15 Active Filter Circuits For convenience let D1 s2 3825πs 25 106π2 D2 s2 9240πs 25 106π2 D3 s2 61200πs 6400 106π2 D4 s2 147840πs 6400 106π2 Then Hs 4096 1016π4s4 D1D2D3D4 d ωo 2π10000 20000π rads s j20000π s4 16 1016π4 D1 25 106π2 400 106π2 j20000π3825π 106π2375 j765 106π238272 16847 D2 25 106π2 400 106π2 j20000π9240π 106π2375 j1848 106π24180615377 D3 6400 106π2 400 106π2 j20000π61200π 106π26000 j1224 106π26123581153 D4 6400 106π2 400 106π2 j20000π147840π 106π26000 j29568 106π266892623 Hjωo 409616π8 1032 π8 102438272418066123586689 0999 P 1544 a First we will design a unity gain filter and then provide the passband gain with an inverting amplifier For the high pass section the cutoff frequency is 200 Hz The order of the Butterworth is n 00540 log1020040 286 n 3 Hhps s3 s 1s2 s 1 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1535 For the prototype firstorder section R1 R2 1 Ω C 1 F For the prototype secondorder section R1 05 Ω R2 2 Ω C 1 F The scaling factors are kf ω o ωo 2π200 400π km C Ckf 1 106400π 79577 In the scaled firstorder section R 1 R 2 kmR1 79577 Ω C 1 µF In the scaled secondorder section R 1 05km 3979 Ω R 2 2km 15915 Ω C 1 µF For the lowpass section the cutoff frequency is 2500 Hz The order of the Butterworth filter is n 00540 log10125002500 286 n 3 Hlps 1 s 1s2 s 1 For the prototype firstorder section R1 R2 1 Ω C 1 F For the prototype secondorder section R1 R2 1 Ω C1 2 F C2 05 F The lowpass scaling factors are km R R 2500 kf ω o ωo 25002π 5000π For the scaled firstorder section R 1 R 2 25 kΩ C C kfkm 1 5000π2500 2546 nF 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1540 CHAPTER 15 Active Filter Circuits P 1549 a From the statement of the problem K 10 20 dB Therefore for the prototype bandpass circuit R1 Q K 16 10 16 Ω R2 Q 2Q2 K 16 502 Ω R3 2Q 32 Ω The scaling factors are kf ω o ωo 2π6400 12800π km C Ckf 1 20 10912800π 124340 Therefore R 1 kmR1 16124330 199 kΩ R 2 kmR2 16502124340 3963 Ω R 3 kmR3 32124340 3979 kΩ b P 1550 a At very low frequencies the two capacitor branches are open and because the op amp is ideal the current in R3 is zero Therefore at low frequencies the circuit behaves as an inverting amplifier with a gain of R2R1 At very high frequencies the capacitor branches are short circuits and hence the output voltage is zero b Let the node where R1 R2 R3 and C2 join be denoted as a then Va ViG1 VasC2 Va VoG2 VaG3 0 VaG3 VosC1 0 or G1 G2 G3 sC2Va G2Vo G1Vi 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1542 CHAPTER 15 Active Filter Circuits d 1 Select C2 1 F 2 Select C1 such that C1 b2 1 4bo1 K 3 Calculate G2 R2 4 Calculate G1 R1 G1 KG2 5 Calculate G3 R3 G3 boC1G2 P 1551 a In the second order section of a third order Butterworth filter bo b1 1 Therefore C1 b2 1 4bo1 K 1 415 005 F C1 005 F limiting value b G2 1 21 4 01 S G3 1 01005 05 S G1 401 04 S Therefore R1 1 G1 25 Ω R2 1 G2 10 Ω R3 1 G3 2 Ω c kf ω o ωo 2π2500 5000π km C2 C 2kf 1 10 109kf 63662 C 1 005 kfkm 05 109 500 pF R 1 2563662 1592 kΩ R 2 1063662 6366 kΩ R 3 263662 1273 kΩ d R 1 R 2 636621 637 kΩ C C kfkm 1 108 10 nF 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1543 e P 1552 a By hypothesis the circuit becomes For very small frequencies the capacitors behave as open circuits and therefore vo is zero As the frequency increases the capacitive branch impedances become small compared to the resistive branches When this happens the circuit becomes an inverting amplifier with the capacitor C2 dominating the feedback path Hence the gain of the amplifier approaches 1jωC21jωC1 or C1C2 Therefore the circuit behaves like a highpass filter with a passband gain of C1C2 b Summing the currents away from the upper terminal of R2 yields VaG2 Va VisC1 Va VosC2 VasC3 0 or VaG2 sC1 C2 C3 VosC2 sC1Vi Summing the currents away from the inverting input terminal gives 0 VasC3 0 VoG1 0 or sC3Va G1Vo Va G1Vo sC3 Therefore we can write G1Vo sC3 G2 sC1 C2 C3 sC2Vo sC1Vi 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1546 CHAPTER 15 Active Filter Circuits km C Ckf 1 25 109kf 20000 π Therefore in the first scaled secondorder section R 1 R 2 296km 1884 kΩ R 3 113km 7194 kΩ C 1 003 kfkm 750 pF C 2 25 nF In the second scaled secondorder section R 1 R 2 14008km 892 kΩ R 3 238km 1515 kΩ C 1 03 kfkm 75 nF C 2 25 nF Highpass filter section n 00524 log1080004000 399 n 4 In the first prototype secondorder section b1 0765 bo 1 C2 C3 1 F C1 K 1 F R1 K 2 b1 3 0765 392 Ω R2 b1 boK 2 0765 3 0255 Ω In the second prototype secondorder section b1 1848 bo 1 C2 C3 1 F C1 K 1 F R1 K 2 b1 3 1848 1623 Ω R2 b1 boK 2 1848 3 0616 Ω In the highpass section of the filter kf ω o ωo 2π8000 16000π 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1547 km C Ckf 1 25 10916000π 2500 π In the first scaled secondorder section R 1 392km 31194 Ω R 2 0255km 2029 Ω C 1 C 2 C 3 25 nF In the second scaled secondorder section R 1 1623km 12915 Ω R 2 0616km 4902 Ω C 1 C 2 C 3 25 nF In the gain section let Ri 10 kΩ and Rf 100 kΩ b P 1554 a The prototype lowpass transfer function is Hlps 1 s2 0765s 1s2 1848s 1 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1551 R 2 0765km 864 kΩ In the second section R 1 0541km 61 kΩ R 2 1848km 2086 kΩ b P 1557 a Interchanging the Rs and Cs yields the following circuit At low frequencies the capacitors appear as open circuits and hence the output voltage is zero As the frequency increases the capacitor branches approach short circuits and va vi vo Thus the circuit is a unitygain highpass filter b The sdomain equations are Va VisC1 Va VoG1 0 Vo VasC2 VoG2 0 It follows that VaG1 sC1 G1Vo sC1Vi and Va G2 sC2Vo sC2 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1554 CHAPTER 15 Active Filter Circuits P 1560 a ωo 8000π rads kf ω o ωo 8000π km C Ckf 1 05 1068000π 250 π R kmR 796 Ω so R2 398 Ω σ 1 1 4Q 1 1 410 0975 σR 776 Ω 1 σR 2 Ω C 05 µF 2C 1 µF b c kf 8000π Hs s8000π2 1 s8000π2 1 10s8000π 1 s2 64 106π2 s2 800πs 64 106π2 P 1561 To satisfy the gain specification of 14 dB at ω 0 and α 1 requires R1 R2 R1 5 or R2 4R1 Use the specified resistor of 10 kΩ for R1 and a 50 kΩ potentiometer for R2 Since R1 R2R1 1 the value of C1 is C1 1 2π5050000 6366 nF 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1556 CHAPTER 15 Active Filter Circuits P 1564 a Combine the impedances of the capacitors in series in Fig P1564b to get 1 sCeq 1 α sC1 α sC1 1 sC1 which is identical to the impedance of the capacitor in Fig P1560a b Vx αsC1 1 αsC1 αsC1 V αV Vy αR2 1 αR2 αR2 αV Vx c Since x and y are both at the same potential they can be shorted together and the circuit in Fig 1534 can thus be drawn as shown in Fig 1553c d The feedback path between Vo and Vs containing the resistance R4 2R3 has no effect on the ratio VoVs as this feedback path is not involved in the nodal equation that defines the voltage ratio Thus the circuit in Fig P1564c can be simplified into the form of Fig 152 where the input impedance is the equivalent impedance of R1 in series with the parallel combination of 1 αsC1 and 1 αR2 and the feedback impedance is the equivalent impedance of R1 in series with the parallel combination of αsC1 and αR2 Zi R1 1α sC1 1 αR2 1 αR2 1α sC1 R1 1 αR2 R1R2C1s 1 R2C1s Zf R1 α sC1 αR2 αR2 α sC1 R1 αR2 R1R2C1s 1 R2C1s P 1565 As ω 0 Hjω 2R3 R4 2R3 R4 1 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1557 Therefore the circuit would have no effect on low frequency signals As ω Hjω 1 βR4 RoβR4 R3 1 βR4 R3βR4 Ro When β 1 Hjβ1 RoR4 R3 R3R4 Ro If R4 Ro Hjβ1 Ro R3 1 Thus when β 1 we have amplification or boost When β 0 Hjβ0 R3R4 Ro RoR4 R3 If R4 Ro Hjβ0 R3 R0 1 Thus when β 0 we have attenuation or cut Also note that when β 05 Hjωβ05 05R4 Ro05R4 R3 05R4 R305R4 Ro 1 Thus the transition from amplification to attenuation occurs at β 05 If β 05 we have amplification and if β 05 we have attenuation Also note the amplification an attenuation are symmetric about β 05 ie Hjωβ06 1 Hjωβ04 Yes the circuit can be used as a treble volume control because The circuit has no effect on low frequency signals Depending on β the circuit can either amplify β 05 or attenuate β 05 signals in the treble range The amplification boost and attenuation cut are symmetric around β 05 When β 05 the circuit has no effect on signals in the treble frequency range 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1559 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 164 CHAPTER 16 Fourier Series Hs Vo Vg βs s2 βs ω2c β 1 RC 109 10420 5000 rads ω2 c 1 LC 109103 400 25 108 Hs 5000s s2 5000s 25 108 Hjω j5000ω 25 108 ω2 j5000ω H1 j5 107 24 108 j5 107 0028881 H3 j15 107 16 108 j15 107 0098464 H5 j25 107 25 107 10 H7 j35 107 24 108 j35 107 014 8170 Vo1 Vg1H1 17508881 V Vo3 Vg3H3 2614 9536 V Vo5 Vg5H5 1680 V Vo7 Vg7H7 17329830 V vo 1750 cos10000t 8881 2614 cos30000t 9536 168 cos50000t 1732 cos70000t 9830 V b The 5th harmonic because the circuit is a passive bandpass filter with a Q of 10 and a center frequency of 50 krads AP 167 w0 2π 103 20944 3 rads jω0k j3k 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1632 CHAPTER 16 Fourier Series Yeq 1 R1 1 R2 sL sC Zeq 1Cs R2L s2 sR1R2C LR1LC R1 R2R1LC Therefore Hs 20000s 400 s2 10400s 450 104 We want the output for the third harmonic ω0 2π T 2π 05 4π 3ω0 12π Ig3 100 π2 1 9 sin 3π 5 1070 Hj12π 20000j12π 400 j12π2 10400j12π 450 104 1780403 Therefore Vo3 Hj12πIg3 17804031070 190403 V vo3 19 sin12πt 0403 V P 1632 ωo 2π T 2π 10π 106 200 krads n 3 106 02 106 15 n 5 106 02 106 25 Hs Vo Vg 1RCs s2 1RCs 1LC 1 RC 1012 250 1034 106 1 LC 1031012 104 25 1012 Hs 106s s2 106s 25 1012 Hjω jω 106 25 1012 ω2 j106ω 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1645 c Vo 250 Vo sL VosC Vo Vg 625 0 250LCs2 5sL 250Vo 4sLVg Vo Vg Hs 4250Cs s2 150C 1LC Hs 16000s s2 2 104s 4 1010 ωo 2π T 2π 10π 106 2 105 rads Hj0 0 Hj2 105k j8k 1001 k2 j10k Therefore H1 080 H1 080 H2 j16 300 j20 005328619 H2 00532 8619 H3 j24 800 j30 003008785 H2 00300 8785 H4 j32 1500 j40 002138847 H2 00213 8847 The output voltage coefficients C0 0 C1 101912248080 81512248 V C1 815 12248 V C2 430 900058619 02287 381 V C2 02287381 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1647 A1 1801 V θ1 45 A2 1273 V θ2 90 A3 6 V θ3 135 A4 0 A5 36 V θ5 45 A6 424 V θ6 90 A7 257 V θ7 135 A8 0 b Cn an jbn 2 Cn an jbn 2 C n C0 av 10 V C3 3135 V C6 21290 V C1 945 V C3 3 135 V C6 212 90 V C1 9 45 V C4 C4 0 C7 129135 V C2 63790 V C5 1845 V C7 129 135 V C2 637 90 V C5 18 45 V P 1650 a From the solution to Problem 1633 we have Ak ak jbk Im π2k2cos kπ 1 j Im πk A0 075Im 375 A A1 5 π22 j 5 π 18912248 A A2 j 5 2π 079690 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1648 CHAPTER 16 Fourier Series A3 5 9π22 j 5 3π 054210198 A A4 j 5 4π 039890 A A5 5 25π22 j 5 5π 03219726 A A6 j 5 6π 026590 A b C0 A0 375 mA C1 1 2A1θ1 094512248 A C1 0945 12248 A C2 1 2A2θ2 039890 A C2 0398 90 A C3 1 2A3θ3 027110198 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1649 C3 0271 10198 A C4 1 2A4θ4 019990 A C4 0199 90 A C5 1 2A5θ5 01619726 A C5 0161 9726 A C6 1 2A6θ6 0132590 A C6 01325 90 A P 1651 a v A1 cosωot 90 A3 cos3ωot 90 A5 cos5ωot 90 A7 cos7ωot 90 v A1 sin ωot A3 sin 3ωot A5 sin 5ωot A7 sin 7ωot b vt A1 sin ωot A3 sin 3ωot A5 sin 5ωot A7 sin 7ωot vt vt odd function 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1651 f P 1653 From Table 151 we have Hs s3 s 1s2 s 1 After scaling we get Hs s3 s 2500s2 2500s 625 104 ωo 2π T 2π 400π 106 5000 rads Hjnωo j8n3 1 j2n1 4n2 j2n It follows that Hj0 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 175 362000 7220002 4000 36 10620003 12 24 kJ W6kΩ 24 103 6 103 4 J 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1718 CHAPTER 17 The Fourier Transform vot v1t v2t 25 25sgnt 50e200tut 50ut 50e200tut vot 501 e200tut V b P 1723 a From the solution to Problem 1722 Hω 200 jω 200 Now Vgω 50 jω Then Voω HωVgω 10000 jωjω 200 K1 jω K2 jω 200 50 jω 50 jω 200 vot 25sgnt 50e200tut V b 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1720 CHAPTER 17 The Fourier Transform vot 50sgnt 100e1000tut V b Yes at the time the current source jumps from 40 to 40 mA the capacitor is charged to 50 V That is at t 0 vo0 125040 103 50 V At t the capacitor will be charged to 50 V That is vo 125040 103 50 V The time constant of the circuit is 125008 103 1 ms so 1τ 1000 The function vot is plotted below P 1726 a Vo Vg Hs 16000s 100 01s 16000s Hs 160000 s2 1000s 160000 160000 s 200s 800 Hjω 160000 jω 200jω 800 Vgω 16 jω Voω VgωHjω 256 104 jωjω 200jω 800 Voω K1 jω K2 jω 200 K3 jω 800 K1 256 104 16 104 16 K2 256 104 200600 2133 K3 256 104 800600 533 Voω 16 jω 2133 jω 200 533 jω 800 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1722 CHAPTER 17 The Fourier Transform b io0 0 c io0 0 d Io 16s 100 01s 16000s 160 s2 1000s 160000 160 s 200s 800 0267 s 200 0267 s 800 iot 0267e200t 0267e800tut A e Yes P 1728 a ig 2e100t Igω 2 jω 100 2 jω 100 400 jω 100jω 100 Vo 500 104sVo Ig Vo Ig Hs 104 s 20 Hω 104 jω 20 Voω IgωHω 4 106 jω 20jω 100jω 100 K1 jω 20 K2 jω 100 K3 jω 100 K1 4 106 12080 41667 K2 4 106 80200 250 K3 4 106 120200 16667 Voω 41667 jω 20 250 jω 100 16667 jω 100 vot 41667e20t 250e100tut 16667e100tut V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1724 CHAPTER 17 The Fourier Transform K1 220 120 033 K2 2100 120 167 K3 220 80 05 K4 2100 80 25 Ioω 0833 jω 20 167 jω 100 25 jω 100 iot 167e100tut 0833e20t 25e100tut A b io0 167 V c io0 167 V d Note since io0 167 A vo0 1000 83333 16667 V Io Vg 16667s 500 104s sVg 16667 500s 104 Vg 1000 s 100 Io 167s 3333 s 20s 100 0833 s 20 25 s 100 iot 0833e20t 25e100tut A e Yes for t 0 the solution in part a is also iot 0833e20t 25e100tut A P 1730 Vo Vg 2s 100Vo s Vos 100s 125 104 0 Vo s100s 125 104Vg 125s2 12000s 25 106 Io sVo 100s 125 104 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1727 iot 75 cos40000t 14313 mA P 1732 a Vo Vgs 106 Vo 4s Vo 800 0 Vo s2Vg s2 1250s 25 104 Vo Vg Hs s2 s 250s 1000 Hjω jω2 jω 250jω 1000 vg 45e500t Vgω 45000 jω 500jω 500 Voω HjωVgω 45000jω2 jω 250jω 500jω 1000jω 500 K1 jω 250 K2 jω 500 K3 jω 1000 K4 jω 500 K1 450002502 250750750 20 K2 450005002 2505001000 90 K3 4500010002 7505001500 80 K4 450005002 75010001500 10 vot 20e250t 90e500t 80e1000tut 10e500tut V b vo0 10 V Vo0 20 90 80 10 V vo 0 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1728 CHAPTER 17 The Fourier Transform c IL Vo 4s 025sVg s 250s 1000 Hs IL Vo 025s s 250s 1000 Hjω 025jω jω 250jω 1000 ILω 025jω45000 jω 250jω 500jω 1000jω 500 K1 jω 250 K2 jω 500 K3 jω 1000 K4 jω 500 K4 02550045000 75010001500 5 mA iLt 5e500tut iL0 5 mA K1 02525045000 250750750 20 mA K2 02550045000 2505001000 45 mA K3 025100045000 7505001500 20 mA iL0 K1 K2 K3 20 45 20 5 mA Checks ie iL0 iL0 5 mA At t 0 vC0 45 10 35 V At t 0 vC0 45 10 35 V d We can check the correctness of out solution for t 0 by using the Laplace transform Our circuit becomes Vo 800 Vo 4s Vo Vgs 106 35 106 5 103 s 0 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1729 s2 1250s 24 104Vo s2Vg 35s 5000 vgt 45e500tut V Vg 45 s 500 s 250s 1000Vo 45s2 35s 5000s 500 s 500 Vo 10s2 22500s 250 104 s 250s 500s 1000 20 s 250 90 s 500 80 s 1000 vot 20e250t 90e500t 80e1000tut V This agrees with our solution for vot for t 0 P 1733 a From the plot of vg note that vg is 50 V for an infinitely long time before t 0 Therefore vo0 50 V There cannot be an instantaneous change in the voltage across a capacitor so vo0 50 V b io0 0 A At t 0 the circuit is io0 150 50 25 200 25 8 A 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1731 Vos 16s 10 s 16sVgs Hs Vos Vgs 16 s2 10s 16 16 s 2s 8 Hjω 16 jω 2jω 8 Vojω Hjω Vgω 1152jω 4 jω4 jω2 jω8 jω K1 4 jω K2 4 jω K3 2 jω K4 8 jω K1 11524 8612 8 K2 11524 824 72 K3 11522 626 32 K4 11528 1246 32 Vojω 8 4 jω 72 4 jω 32 2 jω 32 8 jω vot 8e4tut 72e4t 32e2t 32e8tutV b vo0 8V c vo0 72 32 32 8V The voltages at 0 and 0 must be the same since the voltage cannot change instantaneously across a capacitor P 1735 a Vo Vgs 25 100s s Vgs2 s2 25s 100 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1732 CHAPTER 17 The Fourier Transform Hs Vo Vg s2 s 5s 20 Hjω jω2 jω 5jω 20 vg 25ig 450e10tut 450e10tut V Vg 450 jω 10 450 jω 10 Voω HjωVg 450jω2 jω 10jω 5jω 20 450jω2 jω 10jω 5jω 20 K1 jω 10 K2 jω 5 K3 jω 20 K4 jω 5 K5 jω 10 K6 jω 20 K1 450100 1530 100 K4 45025 515 150 K2 45025 1515 50 K5 450100 510 900 K3 450400 3015 400 K6 450400 1510 1200 Voω 100 jω 10 100 jω 5 1600 jω 20 900 jω 10 vo 100e10tut 900e10t 100e5t 1600e20tut V b vo0 100 V c vo0 900 100 1600 800 V d At t 0 the circuit is Therefore the solution predicts v10 will be 350 V Now v10 v10 because the inductor will not let the current in the 25 Ω resistor change instantaneously and the capacitor will not let the voltage across the 001 F capacitor change instantaneously 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1733 At t 0 the circuit is From the circuit at t 0 we see that vo must be 800 V which is consistent with the solution for vo obtained in part a It is informative to solve for either the current in the circuit or the voltage across the capacitor and note the solutions for io and vC are consistent with the solution for vo The solutions are io 10e10tut 20e5t 80e20t 90e10tut A vC 100e10tut 900e10t 400e5t 400e20tut V P 1736 Vos 40 s 60 s 100 100 s 300 24000s 50 ss 100s 300 Vos Hs 20 s Hs 1200s 50 s 100s 300 Hω 1200jω 50 jω 100jω 300 Voω 40 jω 1200jω 50 jω 100jω 300 48000jω 50 jωjω 100jω 300 Voω 80 jω 120 jω 100 200 jω 300 vot 160sgnt 120e100t 200e300tut V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1735 Wo0 100 rads 3 π tan12 06 π tan110 106 028 078 J Therefore the percent between 0 and 100 rads is 078 12 100 6469 P 1739 Io IgR R 1sC RCsIg RCs 1 Hs Io Ig s s 1RC RC 200025 106 0005 1 RC 1 0005 200 Hs s s 200 Hjω jω jω 200 Igω 001 jω 50 Ioω HjωIgω 001jω jω 50jω 200 Ioω ω001 ω2 502 ω2 2002 Ioω2 104ω2 ω2 502ω2 2002 K1 ω2 2500 K2 ω2 4 104 K1 1042500 37500 667 106 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 184 CHAPTER 18 TwoPort Circuits c VTh Vg a11 a21Zg 50 103 6 104 500 6 V Therefore V2 250 6 V Pmax 1225062 706 103 744 mW AP 187 a For the given bridgedtee circuit we have a 11 a 22 125 a 21 1 20 S a 12 1125 Ω The aparameters of the cascaded networks are a11 1252 1125005 2125 a12 1251125 1125125 28125 Ω a21 005125 125005 0125 S a22 a11 2125 RTh 451253125 1444 Ω b Vt 100 3125 32 V therefore V2 16 V c P 162 1444 1773 W 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1825 P 1829 I1 g11V1 g12I2 V2 g21V1 g22I2 From the first measurement g11 I1 V1 100 106 01 1 mS g21 V2 V1 200 01 2000 From the second measurement g12 I1 I2 25 106 5 103 0005 g22 V2 I2 200 5 103 40 kΩ Summary g11 1 mS g12 0005 g21 2000 g22 40 kΩ From the circuit Zg 1 kΩ Vg 45 mV ZTh g22 g12g21Zg 1 g11Zg 40000 101000 1 1 45000 VTh g21Vg 1 g11Zg 200000045 1 1 45 V i 45 90000 50 µA P 50 106245000 1125 µW 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1834 CHAPTER 18 TwoPort Circuits V1 4 s V2 2500 106 ss 12500s 50000 4 s 533 s 12500 133 s 50000 v2 4 533e12500t 133e50000tut V P 1838 a 11 z11 z21 353 40003 875 103 Ω a 12 z z21 25 1043 40003 625 Ω a 21 1 z21 1 40003 075 103 Ω a 22 z22 z21 100003 40003 25 Ω a 11 y22 y21 40 106 800 106 005 S a 12 1 y21 1 800 106 1250 S a 21 y y21 4 108 800 106 50 106 S a 22 y11 y21 200 106 800 106 025 S a11 a 11a 11 a 12a 21 875 103005 62550 106 35625 103 a12 a 11a 12 a 12a 22 875 1031250 625025 265625 a21 a 21a 11 a 22a 21 075 103005 2550 106 1625 106 a22 a 21a 12 a 22a 22 075 1031250 25025 15625 V2 ZLVg a11 a21ZgZL a12 a22Zg 15000003 35625 103 1625 1061015000 265625 1562510 375 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 Problems 1835 P 1839 The a parameters of the first two port are a 11 h h21 5 103 40 125 106 a 12 h11 h21 1000 40 25 Ω a 21 h22 h21 25 40 106 625 109 S a 22 1 h21 1 40 25 103 The a parameters of the second two port are a 11 5 4 a 12 3R 4 a 21 3 4R a 22 5 4 or a 11 125 a 12 54 kΩ a 21 1 96 mS a 22 125 The a parameters of the cascade connection are a11 125 106125 2510396 102 24 a12 125 10654 103 25125 38 Ω a21 625 109125 25 10310396 104 96 S a22 625 10954 103 25 103125 65 103 Vo Vg ZL a11 a21ZgZL a12 a22Zg a21Zg 104 96 800 102 12 a11 a21Zg 102 24 102 12 102 8 a11 a21ZgZL 102 8 72000 90 a22Zg 65 103800 52 Vo Vg 72000 90 38 52 400 vo Vo 400Vg 36 V 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1838 CHAPTER 18 TwoPort Circuits For VTh note that Voc z21 Zg z11 Vg since I2 0 P 1843 a V1 z11 z12I1 z12I1 I2 z11I1 z12I2 V2 z21 z12I1 z22 z12I2 z12I2 I1 z21I1 z22I2 b With port 2 terminated in an impedance ZL the two mesh equations are V1 z11 z12I1 z12I1 I2 0 ZLI2 z21 z12I1 z22 z12I2 z12I1 I2 Solving for I1 I1 V1z22 ZL z11ZL z22 z12z21 Therefore Zin V1 I1 z11 z12z21 z22 ZL P 1844 a I1 y11V1 y21V2 y12 y21V2 I2 y21V1 y22V2 I1 y11V1 y12V2 I2 y12V1 y22V2 y21 y12V1 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 1840 CHAPTER 18 TwoPort Circuits g22 h11 h 1000 5 103 200 kΩ From Problem 367 Ref 72 kΩ hence our circuit reduces to V2 8000V172 272 I2 V2 72000 8V1 272 vg 9 mV V1 9 103 800 V15 103 018V1 272 0 V1 9 103 4V1 80V1 34 0 V1 34 103 V2 800072 272 34 103 72 V From Problem 367 Vo V2 05 Vo 36 V This result matches the solution to Problem 1839 P 1846 a To determine b11 and b21 create an open circuit at port 1 Apply a voltage at port 2 and measure the voltage at port 1 and the current at port 2 To determine b12 and b22 create a short circuit at port 1 Apply a voltage at port 2 and measure the currents at ports 1 and 2 b The equivalent bparameters for the blackbox amplifier can be calculated as follows b11 1 h12 1 103 1000 b12 h11 h12 500 103 500 kΩ 2015 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permissions write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458