Chapter 26 - Resolução Sears - Vol 3

26.1. IDENTIFY: The newly-formed wire is a combination of series and parallel resistors. SET UP: Each of the three linear segments has resistance R3. The circle is two R6 resistors in parallel. EXECUTE: The resistance of the circle is R12 since it consists of two R6 resistors in parallel; resistance is two R3 resistors in series with an R12 resistor, giving R equiv = R3 + R3 + R12 = 3R4. EVALUATE: The equivalent resistance of the original wire has been reduced because the circle's resistance is less than it was as a linear wire. 26.2. IDENTIFY: It may appear that the meter measures X directly. But note that X is in parallel with three other resistors, so the meter measures the equivalent parallel resistance between ab. SET UP: We use the formula for resistors in parallel. EXECUTE: 1/(0.20 Ω) = 1/X + (1/5.00 Ω) + (1/5.00 Ω) + (1/10.0 Ω), so X = 7.5 Ω. EVALUATE: X is greater than the equivalent parallel resistance of 2.00 Ω. 26.3. (a) IDENTIFY: Suppose we have two resistors in parallel, with R1 < R2. SET UP: The equivalent resistance is 1/R equiv = 1/R1 + 1/R2. Therefore 1/R equiv < 1/R1 and R equiv < R1. EXECUTE: It is always true that 1/R equiv = 1/R1 + 1/R2. Therefore 1/R1 < 1/R equiv. EVALUATE: The equivalent resistance is always less than that of the smallest resistor. (b) IDENTIFY: For resistors in parallel the voltages are the same and equal to the voltage across the equivalent resistance. SET UP: V = IR equiv = I1/R1 + I2/R2. EXECUTE: (a) R equiv = (1/(1/(20 Ω) + 1/(12 Ω))) = 12.3 Ω. (b) I2 = 240 V/12.3 Ω = 19.5 A. (c) I3 = 240 V/20 Ω = 12 A. EVALUATE: More current flows through the resistor that has the smaller R. 26.5. IDENTIFY: The equivalent resistance will vary for the different connections because the series-parallel combinations vary, and hence the current will vary. SET UP: First calculate the equivalent resistance using the series-parallel formulas, then use Ohm's law (V = IR). EXECUTE: (a) I1 = 240V - (1/(5.00 Ω) + (1/(10.0 Ω))) gives R = 10.0 Ω, I = 35.0 V/(10.0 Ω) = 3.50 A. (b) I1 = (1/(0.20 Ω) + (1/(35.0 Ω))) gives R = 7.78 Ω, I = (35.0 V)/(7.78 Ω) = 4.50 A. (c) 1/R = (1/(20 Ω)) + (1/(25.0 Ω)) gives R = 11.11 Ω, so I = (35.0 V)/(11.11 Ω) = 3.15 A. 26-1 26-2 Chapter 26 26.6. EXECUTE: If from part (b), the resistance of the triangle union is 7.78 Ω. Adding the 3.00 Ω internal resistance of the battery gives an equivalent resistance for the circuit of 10.78 Ω. Therefore the current is I = (35.0 V)/(10.78 Ω) = 3.25 A. EVALUATE: It makes a difference how the triangle is connected to the battery. IDENTIFY: The potential drop is the same across the resistors in parallel, and the current into the parallel combination is the same as the current through the 45.0 Ω resistor. (a) SET UP: Apply Ohm's law in the parallel branch to find the current through the 45.0 Ω resistor. Then apply Ohm's law to the 45.0 Ω resistor to find the potential drop across it. EXECUTE: The potential drop across the 25.0 Ω resistor is V = (25.0 Ω)(2.25 A) = 31.25 V. The potential drop across each of the parallel branches is 31.25 V. For the 15.0 Ω resistor: I3 = (31.25 V)/(15.0 Ω) = 2.08 A. The resistance of the 10.0 + 15.0 combination is 25.0 Ω, so the current through it must be the same as the current through the upper 25.0 Ω resistor: I3 = 1.25 A. The sum of currents in the parallel branch will be the current through the 45.0 Ω resistor. I2 = 1.25 A + 2.08 A + 1.25 A = 4.58 A. Apply Ohm's law to the 45.0 Ω resistor: V = (4.58 A)(45.0 Ω) = 206 V. (b) SET UP: First find the equivalent resistance of the circuit and then apply Ohm's law to it. EXECUTE: The resistance of the parallel branch is 1/R = (1/25.0 Ω) + (1/15.0 Ω) + (1/25.0 Ω), so R = 6.82 Ω = 398 V. EVALUATE: The error of the battery is the sum of the potential drops across each of the branches and series resistors. 26.7. EXECUTE: The (45.0 Ω and 15.0 Ω resistors are in series-parallel position possible. Then find the equivalent series resistance of the circuit. EXECUTE: 1/(7.50 Ω + 10.0 Ω)) and R = 11.25 Ω = The total equivalent resistance is 8.00 Ω. I = 1.36 Ω = 32.25 Ω. Ohm's law gives I = 2.50 V/(32.50 Ω) = 0.769 Ω. EVALUATE: The circuit appears equivalent since we realize that the 45.0 and 15.0 Ω resistors are in parallel. For resistors in parallel, the voltages are the same and the currents add. (a) SET UP: The circuit is sketched in Figure 26.8a. 26-3 Direct-Current Circuits EXECUTE: Using P = V^2/R, P1 = V1^2/R1, P2 = (28.0 V)^2/(1.60 Ω) = 490 W, P2 = V2^2/R2 = (28.0 V)^2/(2.40 Ω) = 327 W, and P3 = V3^2/R3 = 490 W = 163 W. EVALUATE: The total power dissipated is P = P1 + P2 + P3 = 980 W. This is the same as the power delivered by the battery. (h) P = P/V = I^2R. The resistors in parallel each have the same voltage, so the power P is largest for the one with the least resistance. 26.9. IDENTIFY: For a series circuit, the current is the same in each resistor and the sum of voltages for each resistor equals the battery voltage. The equivalent resistance is R_eq = R1 + R2 + R3 = P = R^2. SET UP: Let R1 = 1.60 Ω, R2 = 2.40 Ω, R3 = 4.80 Ω. EXECUTE: (a) R_eq = 1.60 Ω + 2.40 Ω + 4.80 Ω = 8.80 Ω (b) I2 = 28.0 V/3.18 A. (c) I3 = 1.3 A; the same as for each resistor. (d) I3 = (3.18 A)(6.0 Ω) = 5.09 V. I_r = (3.18 A)(20 Ω) = 7.63 V. I_r = (3.18 A)(4.80 Ω) = 12.53 V. Note that I_r = I3 + I4 + I5 = 28.0 V. (a) I_r = R_eq. I_r = (3.18 A)(30 Ω) = 162 W. I_r = R_r = (3.18 A)(20 Ω) = 24.3 W. I_r = (3.18 A)(18 Ω) = 48.5 W. Since P = I^2R and the current is the same for each resistor, the resistor with the greatest power. EVALUATE: When resistors are connected in parallel, the resistor with the smallest R dissipates the greatest power. SET UP: P = V/I. EXECUTE: (a) P = /R = P = V = /R = (35.0 V)(15.0) = 274 V (b) P = P/(20 V)(8.00 Ω) = 1.6 W. (c) 26-4\nChapter 26\nEVALUATE: Note that i1 + i2 = i3 + i4.\nR1 R3 E\nR2 R4\nFigure 26.11 (b)\n26.12. IDENTIFY: Replace the series combinations of resistors by their equivalents. In the resulting parallel network the battery voltage is the voltage across each resistor.\nSet UP: The circuit is sketched in Figure 26.12a.\ni1 48.0 \nR1 = 1.00 R3 = 3.00 \nT12 \n5.00 \nR5 = 7.00 R4 = 5.00 \nFigure 26.12a\nThe circuit is equivalent to the circuit sketched in Figure 26.12b.\n\nE = 48.0 V\ni2 = 4.00 R2 = 12.0 \nR2 and R4 are in parallel and are equivalent to\nR2 and R4 in series have an equivalent resistance of R_s = R4 + R2 = 4.00 \n\nR2 and R4 are in series have an equivalent resistance of R_s = R1 + R3 = 12.0 \n\n\nThe voltage across each branch of the parallel combination is E, so E - i3R4 = 0.\ni3 = R2 = 4.00 \n\nR_s = 4.00 12.0 \n(4.00 12.0) = 3.00 \n\nE - i3R_s = 0\nThe current is 120 A through the 1.00 and 3.00 resistors, and it is 4.0 A through the 7.00 and 5.00 resistors.\nEVALUATE: The current through the battery is I = i1 + i2 + i3 = 12.0 A + 4.0 A + 16.0 A, and this is equal to E/R_s = 48.0 V/3.00 = 16.0 A.\n26.13. IDENTIFY: In both circuits, with and without R4, replace series and parallel combinations of resistors by their equivalents. Calculate the currents and voltages in the equivalent circuit and infer from this the currents and voltages in the original circuit. Use P = I2R to calculate the power dissipated in each bulb. (a) Set UP: The circuit is sketched in Figure 26.13a.\n \n\nR1\n \nFigure 26.13a\nEXECUTE: R1, R3, and R4 are in parallel, so their equivalent resistance R_eq is given by R_eq = 1/(1/R1 + 1/R3 + 1/R4).\n\nn2 1.50 R3 = 4.50 \n\nR_eq = 3.00 \n\nE = 9.00 V\n\nR_eq = 45.0 + 47.0 \n\nR_eq = \n 26.13b\nThe equivalent circuit is drawn in Figure 26.13b.\nE - I(R1 + R3) = 0\nI = E / (R1 + R3)\n\nI = 9.00 V / (4.50 + 1.50) = 1.50 A\n \n\nR4 = 4.50 = 4.50 V / (1.50 A * 4.50) = 6.75 V\nTherefore V1 = I1R1 = (1.50 A)(4.50 ) = 6.75 V\nI4 = 1.50 A, V4 = I4R4 = (1.50 A)(4.50 ) = 20.25 V.\n\nV1 = I1R2 = (9.00V)(2.25) = 20.25 V\n\nV2 = 2.25 V.\n\nV3 = 0.500 A, I1 = V4 / R4\n \nBalt not.is_EVALUE: Note that i1 + i2 + i3 = 1.50 A, which is I_eq for resistors in parallel.\n(b) Set UP: P = I2R\nEXECUTE: P1 = I1^2(R1)\n\nP1 = (1.50 A)2(4.50) = 110.1 W.\n\nP2 = P2 = P_r\n\nI4l = 1.50 A)(4.50 ) = 1.25 W, which rounds to 1.12 W, R4 glows brightest.\nEVALUATE: Note that P1 + P2 + P3 = 3.37 W. This equals P3 = I R in the equivalent circuit, and corresponds to\n \nIeq = 3.00 (4.50 = 48.0V\n\nor I. = 1.333 A\n\n= 0V = 4.50 \nR_eq = 22.5 \n E - I(R1 + R3) = 0\nI = eR1 + R4 + R4\n = 1.00 A \nR_eq = 4.50 + 4.50 \n(b) If R1 burns out, the current through R3 will increase. The circuit will be slower since the two R4 will help to maintain greater brightness.\n \n\n for R1 makes the battery 0.67. When R4 is removed, the equivalent resistance of the total provides the required values, returns current\n through R2. The voltage drop across R1 is less so the continue to increase. The:. E = GR R R + R + R1 + R2\n\n(b) If R1 burns out, the total power increases. The circuit will increase the output of the two resistors.\n\n(a) Set UP: The circuit is sketched in Figure 26.17b.\nFigure 26.17b\nFor resistors in parallel the voltage across each resistor is the same.\n EXECUTE: I = V / R = 120 V / 400 \\u03A9 = 0.300 A, I = V / R = 120 V / 800 \\u03A9 = 0.150 A. EVALUATE: Note that each current is larger than the current when the resistors are connected in series. (a) EXECUTE: I = I_H + I_L = (0.300 A)^2(400 \\u03A9) = 36.0 W. P_T = P_S + P_R = 54.0 W. EVALUATE: Note that the total current drawn from the line is I_T = I_H + I_L = 0.450 A. The power input from the line P_L = V * I = (120 V)(0.450 A) - 54.0 W, which equals the total power dissipated by the bulbs. (b) The bulb that is dissipating the most power glows most brightly. For the series connection the currents are the same and by P = I^2R the bulb with the larger R has the larger P; the 800 \\u03A9 bulb glows more brightly. For the parallel connection the voltages are the same and by P = V^2/R the bulb with the smaller R has the larger P; the 400 \\u03A9 bulb glows more brightly. (b) The total power output P_total equals P = V_R / V, so P_total is larger for the parallel connection where the current drawn from the line is larger (because the equivalent resistance is smaller). 26.18. IDENTIFY: Use P = V^2/R with V = 120 V and the wattage for each bulb to calculate the resistance of each bulb. When connected in series the voltage across each bulb will not be V and the power for each bulb will be different. SET UP: For resistors in series the currents are the same and R_eq = R_1 + R_2. EXECUTE: (a) R_1 = P_1 / I^2 = (120 V)^2 / 240 W = 72.0 \\u03A9; R_2 = P_2 / I^2 = (120 V)^2 / 200 W = 72.2 \\u03A9. Therefore, I_total = I_series = V / R_eq = (240 V) / (72.0 + 72.2) \\u03A9 = 0.769 A. 26.19. The series R_f represents the largest resistance dissipates the greatest amount of power. IDENTIFY: SET UP: Replace series and parallel combinations of resistors by their equivalents until the circuit is reduced to a single loop. Use the loop equation to find the current through the 20.0 \\u03A9 resistor. Set P = I^2R for the 200 \\u03A9 resistor to find the time rate at which heat goes into the water and set Q = mc\\u00D7T. EXECUTE: Replace the network by the equivalent resistor, as shown in Figure 26.19. 30.0 V - I(20.0 \\u03A9 + 5.0 \\u03A9 + 5.0 \\u03A9) = 0, I = 1.00 A. For the 20.0 \\u03A9 resistor thermal energy is generated at the rate P = I^2R = 20.0 W. Q = P t and Q = mc\\u00D7T gives (10.0 kg)(4.19x10^3 J/kg \\u00B7 \\u00B0C) = 1.01x10^5 J. EVALUATE: The battery is supplying heat at the rate P = E/t = 3.00 W. In the series circuit, more energy is dissipated in the larger resistor (20.0 \\u03A9) than in the smaller ones (5.00 \\u03A9). 26.20. IDENTIFY: P = I^2R determines R_1, R_2, and R_3. The 10.0 \\u03A9 resistor are all in parallel so have the same voltage. Apply the junction rule to find the current through R_3. SET UP: P = I^2R for a resistor and P = E/t for an emf. The emf inputs electrical energy into the circuit and electrical energy is removed in the resistors. EXECUTE: (a) P_1 = 2 R_1 = 20 W = (2 A)^2R_3 and R_3 = 5.00 \\u03A9, R_1 and R_2 are in parallel, so (10.0 \\u03A9) = (5.0 \\u03A9). I_0 = 1.50 A, So I_1 = 3.50 A - I_0 = 0.50 A. R_1 and R_2 are in parallel, so (0.50)(20.0 \\u03A9), R_2 = 20.0 \\u03A9. (b) E = V = (2.00 A)(5.0 \\u03A9) = 10.0 V. (c) From part (a), I_2 = 0.500 A, I_w = 1.00 A. (d) P_T = 200 W (given). P_2 = I^2R_2 = (0.50 A)^2(20 \\u03A9) = 500 W. P_T = P_1 + P_2 + P_3 = 10.0 W. The total rate at which the resistors remove electrical energy is P_rain = 20 W + 5 W + 35.0 W. The total rate at which the battery inputs electrical energy is P_supply = I = (3.50 A)(10.0 V) = 350 W. P_rain = P_supply, which agrees with conservation of energy. EVALUATE: The three resistors are in parallel, so the voltage for each is the battery voltage, 10.0 V. The currents in the three resistors add to give the current in the battery. 26.21. IDENTIFY: Apply Kirchhoff's point rule to point a to find the current through R_a. Apply Kirchhoff's loop rule to loops (1) and (2) shown in Figure 26.21a to calculate R and E. Travel around each loop in the direction shown. (a) SET UP: Figure 26.21a. EXECUTE: Apply Kirchhoff's point rule to point a: \\u2211 I = 2.00 A + 4.00 A - 6.00 A = 0, I = 2.00 A (in the direction shown in the diagram). -18.0 V - (2.00 A)(R) + 28.0 V = 0 R = 28.0 \\u03A9. (c) Apply Kirchhoff's loop rule to loop (2): -(6.00 A)(3.00 \\u03A9) - (2.00 A)(6.00 \\u03A9) + E = 0. 18.0 V + 24.0 V = 42.0 V. EVALUATE: You can check that the loop rule is satisfied for loop (3), as a check of our work: (1) - (2.00 A)(6.00 \\u03A9) - (2.00 A)(36.0) - (2.00 A) = 0. 28.0 V - 42.0 V + 24.0 V + 10.0 V = 0. 52.0 V = 42.0 V + 10.0 V. 52.0 V = 52.0 V, so the loop rule is satisfied for this loop. (d) IDENTIFY: If the circuit is broken at point j, there can be no current in the 6.00 \\u03A9 resistor. There is now only a single current path and we can apply the loop rule to this path. SET UP: The circuit is sketched in Figure 26.21b. EXECUTE: +28.0 V - (3.00 \\u03A9) I - (5.00 \\u03A9) I = 0 I = 28.0 V / 8.00 \\u03A9 = 3.50 A. EVALUATE: Breaking the circuit at x removes the 42.0 V emf from the circuit and the current through the 3.00 \\u03A9 resistor is reduced. IDENTIFY: Apply the loop rule and junction rule. SET UP: The circuit diagram is given in Figure 26.22. The junction rule has been used to find the magnitude and direction of the current in the middle branch of the circuit. There are no remaining unknown currents. EXECUTE: The loop rule applied to loop (1) gives: 2.00 V - (1.00 A)(10.0 \\u03A9) - (1.00 A)(10.0 \\u03A9) = E_1 - (1.00 A)(6.00 \\u03A9) = 0. 2.00 V - 1.00 A(10.0 \\u03A9) + 7.0 V + 4.00 V - 6.00 V = 1.80 V. The loop rule applied to loop (2) gives: +60.0 V - (1.00 A)(6.00 \\u03A9) - (2.00 A)(6.00 \\u03A9) - E_2 - (2.00 A)(10.0 \\u03A9) = 0. E_2 = 2.00 V - 20.0 V - 4.00 V - 6.00 V = 10.0 V. Going from b to a along the upper branch of the circuit. We can also calculate V_b - V_a by going from b to a along the upper branch of the circuit. V_b - V_a = (20.0 \\u03A9)(6.00 \\u03A9) + 20.0 V - 2.00 V - 4.00 V = V_a - V_b - 6.00 V = 10.0 V. (b) EXECUTE: Apply the loop rule to loop (1): \\( \\epsilon_1 - (3.00 \\Omega)(4.00 \\Omega) - (1.00)(3.00 \\Omega) = 0 \\) \\n \\epsilon_1 = 12.0 V \\text{ (8.00 A)(3.00 \\Omega)} - 36.0 V \\n Apply the loop rule to loop (2): \\( \\epsilon_2 - (5.00)(6.00 \\Omega) - (1.00)(3.00 \\Omega) = 0 \\) \\n \\epsilon_2 = 300 \\text{ V (8.00 A)(3.00 \\Omega)} - 54.0 V \\n (c) Apply the loop rule to loop (3): \\( -(2.00 \\Omega) \\epsilon_3 = \\epsilon_1 - \\epsilon_2 + \\epsilon_5 = 0 \\) \\n R = \\epsilon_1 - \\epsilon_2 + 36.0 - 9.00 = 2.00 A. loop (2) +5.00 V - (1.00 \\Omega)(1) \\text{100 \\Omega } - (4.00 \\Omega) = 0 \\n \\text{5.00 V + (1.00)(2) A - (150.0 A)} = 0 \\n 1.00 A - 2.300 A = 1. = 0 \\n Use this in the second equation: 1.00 A - 2.00 A\\text{=}0 \\n\\text{5.00V} - (5.00 A)(100)\\Omega \\text {= 0.} \\n Eliminate the unknowns for this loop: \\text{ . . .} 26.30. IDENTIFY: The circuit is as shown in Figure 26.30. Since all the external resistors are equal, the current must be symmetrical in this circuit. Since the total current enters through R, this is the same. \\n\\text{(a)} To construct an ammeter, add a shunt resistor in parallel with the galvanometer coil. To construct a voltmeter, add a resistor in series with the galvanometer coil. \\n \\text{(b)} As worked out above, \\text{RV = 1Ω} \\n \\text{(c)}} \\text{VA = 0, since no current flows through R. } Direct-Current Circuits 26-13\n\n(b) For a 500-nV voltmeter, the resistances are in series and the current is the same through each: \( V_s = I_s (R_k + R_r) \) and \( R_k = \\frac{500\\times10^{-9}}{25.0\\times10^{-6}} = 975.2 \\Omega. \n\nEVALUATE: The equivalent resistance of the voltmeter is \\( R_s = R_k + R_r = 1000 \\Omega. \\) The equivalent resistance of the ammeter is given by \( \\frac{1}{I_s} = \\frac{1}{R_s} + \\frac{1}{R_r} \\) and \\( R_m = 0.625 \\Omega. \\) The voltmeter is a high-resistance device and the ammeter is a low-resistance device.\n\nIDENTIFY: The galvanometer is represented in the circuit as a resistance \( R_k. \\) Use the junction rule to relate the current through the galvanometer and the current through the shunt resistor. The voltage drop across each parallel path is the same; use this to write an equation for the resistance \( R_k. \n\nSet Up: The circuit is sketched in Figure 26.32.\n\nWe want that \\( I_a = 20.0 A \\) in the external circuit to produce \\( 1 = 0.0224 A \\) through the galvanometer coil.\n\nApplying the junction rule to point A gives \\( I_a - I_g - I_s = 0\\)\n\\[ I_a = I_g + I_s \] \\( I_g = A - 0.0224 A = 19.998 A. \\)\nThe potential difference \\( V_b \\) between points B and must be the same for both paths between these two points:\n\\( (R_k + R_r) I_g = R_k I_R \)\n\\( I_g \\approx 19.98 A \\)\n\\( R_k = 9.36 \\Omega\\)\n\\( R = 0.0224 A\\)\n\\( E = 9.36 = 22.30 - 9.36 = 12.9 \\) 26.34. IDENTIFY: The resistance of the galvanometer can alter the resistance in a circuit.\n\nSet Up: \\( R_k \\) and the meter are in series, so \\( R_{tot} = R_m + R_r + R_k, \\) where \\( R_m = 65.0 \\Omega \\) is the resistance of the current through the ammeter, and hence the current it measures. \n\nEXTRACT: (a) When the wires are shorted, the full-scale deflection current is obtained: \\( E = \\frac{1.52 V}{(2.50\\times10^{-3} A)(65.0 \\Omega + R_k)} \\) and \\( R = 543.2 \\Omega. \\)\n\n(b) If the resistance \\( R_{m} = 200.0 \Omega : \\) \\( I_{m} = \\frac{I}{R_{m}} = \\frac{1.52 V}{65.0 + 543.2 \\Omega} = 1.88 mA. \\) \\( R_{m} = 65.0 + 543.2 \\Omega \\) \\( R_d = R_m \\);\n\n(c) % error = \\( (2.72V - 1.72 V) (2.72\\times10^{-3} = 21.4 % )\\)\n\n26.35. IDENTIFY: An uncharged capacitor is placed into a circuit. Apply the loop rule at each step.\n\nEXECUTE: (a) At the instant the circuit is completed, there is no voltage over the capacitor, since it has no charge across it.\n\n(d) The current through the resistor is \\( i = \\frac{E}{R_{m}} = \\frac{125 V}{7500 \Omega} = 0.0167 A. \\) 26.41. IDENTIFY: The capacitors, which are in parallel, will discharge exponentially through the resistors.\n\nSet Up: Since \\( V \\) is proportional to \\( q, \\) must obey the same exponential equation as \\( q. \\)\n\nEXECUTE: (a) Solve for time when the potential across each capacitor is 10.0 V: \\( W_{m} = RC \\) \\( {V}(t) = -680.0 \\times 0.35 \\mu F)(60.0 V)[-e^{-t/(0.045)}] = 4.21 ms.\n\n(b) I = \\( \\frac{q}{R} \\). Using the above values, with \( V_{o} = 45.0 V \\), gives \\( I = 0.2125 A.\n\nEVALUATE: Since the current and the potential both obey the same exponential equation, they are both reduced by the same factor of 0.222 (i.e., 0.2221 in 42.1 ms). \n\n26.42. IDENTIFY: In \\( R = RC \\) use the equivalent capacitance of the two capacitors.\n\nSet Up: For capacitors in series, \\( \\frac{1}{C_{tot} = \\frac{1}{C_1} + \\frac{1}{C_2}}. \\) For capacitors in parallel, \\( C_{\\,tot} = C_1 + C_2; \\) . Originally,\n\n{ C = C_1 = 0.870 s.\n\nWith the two capacitors in parallel the new total capacitance is simply 20C. Thus the time constant is \\( R(2C) = 2(0.870s) = 1.74 s.\\) \n\nEVALUATE: The time constant is proportional to \\( C_2. \\) For capacitors in series the capacitance is decreased and for capacitors in parallel the capacitance is increased. 26-16 Chapter 26\n\n26.43. IDENTIFY and SET UP: Apply the loop rule. The voltage across the resistor depends on the current through it and the voltage across the capacitor depends on the charge on its plates.\nEXECUTE: \u2018E = 120 V, V_a = IR = (800 A)(80.0 \u00b5F) = 72 V, V_c = 48 V\nC = CV = (4.00 x 10^(-6) F)(48 V) = 192 \u00b5C.\n\nEVALUATE: The initial charge is zero and the final charge is CE = 480 \u00b5C. Since current is flowing at the instant considered in the problem the capacitor is still being charged and its charge has not reached its final value.\nIDENTIFY: Charge is increasing while the current is decreasing. Both obey exponential equations, but they are not the same equation.\nSET UP: The charge obeys the equation Q = Q_m(1 - e^{(-t/RC)}), but the equation for the current is I = I_m e^{(-t/RC)}. When the charge has reached Q_m it is maximum value, we have Q_m = Q_m(1 - e^{(-t/RC)}) which says that the exponential term has to be e^{(-t/RC)} = 1. The current at this time is I = I_m e^{(-t/34) = (314/ (100 V)(20 \u00b5F)) = 0.625 A.\nEVALUATE: Note that the current will be b^t, 20, t of its maximum value when the charge is Q_m.\n\n26.45. IDENTIFY: The stored energy is proportional to the square of the charge on the capacitor, so it will obey an exponential output, but not the same equation as the charge.\nSET UP: The energy stored in the capacitor is E = \frac{1}{2}C V^2. \nEXECUTE: U = \frac{1}{2}C(1 - e^{(-t/RC)})^2C = U_0. \nWhen the capacitor has lost 80% of its stored energy, the energy is 20% of the initial energy, which is U/5 = U_m e^{(-t/RC)} \to \ \frac{U_m}{C} = 5.0 (50.0 2)(0.2) and (5)(0.52)(52) =929 \mu s. \nAt this time, the current is I = \frac{5}{(RC)^{m}.}\nEVALUATE: When the energy reduced by 80%, neither the current nor the charge are reduced by that percent. is proportional to the charge of the charge. \nSET UP: The charge obeys the equation Q = Q_m(1 - e^{(-t/RC)}). \nEXECUTE: (a) The charge is reduced by half: \frac{1}{2}Q_m = Q_m(1 - e^{(-t/RC)}) gives \n(b) The energy is reduced by half: U = \frac{1}{2}U_m \to U = (1.456 ms^2 = 0.728 ms\n\n26.47. EXECUTE: (a) Q^2 = C = (5.90x10^(-6) F)(28.0 V) = 1.65 \u00b5C.\n(b) q = Q(1 - e^{(-t/RC)}), so e^{-t/RC} = 1 - \frac{q}{Q} and R = -t/\ln(1 - q/Q). After \nT = 3x10^3 s R = -3x10^{8} s.\n\n26.48. IDENTIFY: For a charging capacitor, q = Q(1 - e^{-\frac{t}{RC}})\nEXECUTE: The charge Q on the capacitor is given as a function of time by Eq(26.12): \nQ_C = CC_e^{(180.0 V) = 2.70x10^{0} C.\nRC = (980 \L(20.0 5)^{1} = 0.0147 s. \nThus, at t = 0.00 s, the circuit is the charging circuit shown in Figure 26.49a.\nAt t /0 = 0.0.\n\n26.50. IDENTIFY: P = V \times I \to R\nEXECUTE: (a) P = I^2R = (240 V)^{2} / 4100 W = 17.1 A. So we need at least 14-gauge wire (good up to 18 A). 12 gauge is also ok (good up to 25 A).\n IDENTIFY: Another solution is two resistors in parallel in series with two more in parallel. SET UP: The network can be simplified as shown in Figure 26.54b. EXECUTE: I = 1 + 1 = 1 + 1 = 1 / 200 || 2 / 400 Ω R1 = R2 = 200 Ω; R3 = R4 = 400 Ω. EVALUATE: This combination has the required 400 Ω equivalent resistance It will be shown in part (b) that a total of 2.4 W can be dissipated without exceeding the power rating of each resistor. (b) IDENTIFY and SET UP: Find the applied voltage V0 such that a total of 2.4 W is dissipated and then for this V0, find the power dissipated by each resistor. EXECUTE: For a combination with equivalent resistance R_eq = 400 Ω to dissipate 2.4 W the voltage V0 applied to the network must be given by P = V^2 / R_eq. So V0 = √(2.4 W)(400 Ω) = 31.0 V and the current through across each parallel branch and (3.10 V) = 31.0 V across each 400 Ω resistor. The power dissipated by each individual resistor is P = V^2 / R = (15.5 V)^2 / 400 Ω = 0.60 W, which is less than the maximum allowed value of 1.20 W. For the second combination this means a voltage of IR = (0.0775 A)(400 Ω) = 15.5 V across each parallel combination and hence across each separate resistor. The power dissipated by each resistor is P = (0.155 V)² / 400 Ω = 0.60 W, which is less than the maximum allowed value of 1.20 W. 26.56. IDENTIFY and SET UP: Let R = 1.00 Ω, the resistance of one wire. Each half of the wire has R' = R/2 = 0.50 Ω. The combined resistors are the same as a resistor network. Due to the rules for equivalent resistance for resistors in series and parallel find the resistance of the network, as shown in Figure 26.56. EXECUTE: The equivalent resistance is R_eq = R + R' / 2 + R' + R' = R_eq = (0.500 Ω) = 1.25 Ω. EVALUATE: If the two wires were connected end-to-end, the total resistance would be 2.00 Ω. If they were connected side-by-side, the total resistance would be 0.500 Ω. Our answer is between these two limiting values. 26.57. IDENTIFY: The terminal voltage of the battery depends on the current through it and therefore on the equivalent resistance connected to it. The power delivered to each bulb is P = I R, where I is the current through SET UP: The terminal voltage of the source is E = E - I r. EXECUTE: (a) The equivalent resistance of the two bulbs is 1.0 Ω. This equivalent resistance is in series with the internal resistance of the source, so the current through the battery is I = E / R_eq = 8.0 V / 1.0 Ω = 4.4 A and the current through each bulb is 2.2 A. The voltage applied to each bulb is E - I r = 8.0 V - (0.44 A)(0.80 Ω) = 4.4 V. 26.60. Identify: Heat, which is generated in the resistor, melts the ice. SET UP: Find the rate at which heat is generated in the 20.0 Ω resistor using P = V^2 / R. Then use the heat of fusion of ice to find the rate at which the ice melts. The rate at which heat enters the ice, Qdot, is the power P in the resistor, so P = L_f dm/dt. Therefore the rate of melting of the ice is dm/dt = P / L_f. EXECUTE: The total current in the circuit is I_total = 1.286 A. The potential difference across the parallel branch is V = I R_eq = (6.429)(20.0 Ω) = 6.429 V. The rate at which the ice melts is dm/dt = P_L / L_f (2.066 W)(3.34 × 10^5 J/kg) = 6.19 × 10^3 kg/s = 6.19 × 10^3 g/s. 26-22 Chapter 26\nApply the loop rule to loop (2):\n-(1.00 (Ω) + 9.00 V - (I1 + I2)(8.00 Ω)) = 0\nI1(9.00 Ω) + I2(8.00 Ω) = eq.(2)\n\nApply the loop rule to loop (3):\n-(1.00 (Ω) + 9.00 V + I1(1.00 Ω) - I2(1.00 Ω)) = 12.0 V\n-1.00(Ω)I2 + I1(1.00 Ω) = 3.00 V\neq(3)\n\nEq.(1) gives I1 = 2.00 A + 3/2 I2; eq.(2) gives I1 = 1.00 A + I2;\nUsing these results in eq.(3) gives -(1.00 - 3/2 I2)(1.00 Ω) + (2.00 A + 3/2 I2)(1.00 Ω) = 3.00 V\n\nThen I1 = 2.00 A + 1/3 I2; I2 = 2.14 A and I1 = 1.00 A - 5/8 I2 = 0.848 A.\nEVALUATE: We could check that the loop rule is satisfied for a loop that goes through the 5.00 Ω, 8.00 Ω and 10.00 Ω resistors. Going around the loop clockwise: -(I2 - I1)(5.00 Ω) + (I1 + I2)(8.00 Ω) + I1(1.00 Ω) = -0.85 V + 8.15 V + 17.1 V, which does equal zero, apart from rounding.\n\n26.62. DENSITY: Apply the junction rule and the loop rule to the circuit.\nSet Up: Because of the polarity of each part of the circuit, the current in the 2.00 Ω resistor must be in the direction shown in Figure 26.62a. Let I1 be the current in the 20 A battery.\nExecute: The loop rule applied to loop (1) gives: -240 V - (1.80 A)(7.00 Ω) - (1.00 Ω)(2.00 A) = -0.03 Ω = I1. I1 = 1.380 A. The junction rule says that the current in the middle branch is 2.00 A.\nAs shown in Figure 26.62b. The loop rule applied to loop (2) gives: -(1.80 A)(7.00 Ω) + (2.00 A)(2.00 Ω) - 4.0 = E and E = 8.6 V.\nEVALUATE: We can check our results by applying the loop rule to loop (3) in Figure 26.62b: +240 V - E - (2.00 A)(2.00 Ω) - (3.80 A)(3.00 Ω) = E - 240 V - 114.4 V = 8.6 V, which agrees with our result from loop (2). 26.63. IDENTIFY and SET UP: The circuit is sketched in Figure 26.63.\nTwo unknown currents I1 (through the 2.00 Ω resistor) and I2 (through the 5.00 Ω resistor) are labeled on the circuit diagram. The current through the 4.00 Ω resistor has been written as I2 - I1 using the junction rule.\nApply the loop rule to loops (1) and (2) to get two equations for the unknown currents, I1 and I2. Loop (3) can then be used to check the results.\n\nExecute: Loop (1): +200 V - I1(2.00 Ω) - 14.0 V - (I2 - I1)(4.00 Ω) = 0\n6.00 / -4.00 = 6.00 A\n3.00 / -2.00 = 3.00 A\nLoop (2): +36.0 V - I2(5.00 Ω) - (I1 - I2)(4.00 Ω) = 0\n4.00 / -9.00 / 36.0 A = 36.0 A\nThen I1 = 1.00 A + 3/1 I2 = 5.21 A.\n\nIn summary then\nCurrent through the 2.00 Ω resistor: I1 = 5.21 A.\nCurrent through the 5.00 Ω resistor: I2 = 6.32 A.\nCurrent through the 4.00 Ω resistor: I2 - I1 = 6.32 A - 5.21 A = 1.11 A.\nEVALUATE: Use loop (3) to check: -20.0 V - I2(2.00 Ω) + 36.0 V - I1(5.00 Ω) = 0 \n10 V / (2.00 Ω) + (6.32 A)(5.00 Ω) = 4.20 V. 26.65. (a) IDENTIFY: Break the circuit between points a and b means no current in the branch that contains the 3.00 Ω resistor and the 10.0 V battery. The circuit therefore has a single current path. Find the current, so that potential drops across the resistors can be calculated. Calculate Vab by traveling from a to b, keeping track of the potential changes along the path taken.\n\nEXECUTE: Apply the loop rule to loop (1): +12.0 V - I1(1.00 Ω + 2.00 Ω + 2.00 Ω + 1.00 Ω) - 8.0 V - Y/[(2.00 Ω + 1.00 Ω)] = 0. I1 = 12.0 V - 8.0 V - 0.4444 A = 0.422 V (point a as it has higher potential).\nTo find I1 start at point a and travel to b, adding up the potential rises and drops. Travel on path (2) shown on the diagram. The 1.00 Ω and 3.00 Ω resistors in the middle branch have no current through them and hence no voltage across them. Therefore, I1 = 10.0 V + 12.0 V - I1(1.00 Ω + 1.00 Ω) = V; thus Vb - Va = 2.0 V - (0.4444 A)(5.00 Ω) = -42.2 V (point a is at higher potential)+\n\n(b) IDENTIFY and SET UP: With points a and b connected by a wire there are three current branches, as shown in Figure 26.65b. The junction rule has been used to write the third current (in the 8.0 V battery) in terms of the other currents. Apply the loop rule to loops (1) and (2) to obtain two equations for the two unknowns I1 and I2. 26-25 Direct-Current Circuits 26.66. IDENTIFY: In Figure 26.67, points a and c are at the same potential and points d and b are at the same potential, so we can calculate V_ab by calculating V_cd. We know the current through the resistor that is between points c and d. We thus can calculate the terminal voltage of the 24.0 V battery without calculating the current through it. SET UP: I_a = 0.076 A EXECUTE: V_a + I_a(10.0 Ω) + 12.0 V = V_e V_c - V_a = 12.7 V V_b = V_e - V_a - V_e = 12.7 V EVALUATE: The voltage across each parallel branch must be the same. The current through the 24.0 V battery must be (24.0 V - 12.7 V)/(10.0 Ω) = 1.13 A in the direction b to a. 26-26 Chapter 26 26.68. IDENTIFY: The current through the 40.0 Ω resistor equals the current through the emf, and the current through each of the other resistors is less than or equal to this current. So, set P_a = 1.00 W and use this to solve for the current I through the emf. P_a = 1.00 W, then P for each of the other resistors is less than 1.00 W. SET UP: Use the equivalent resistance for series and parallel combinations to simplify the circuit. EXECUTE: I = P_a gives I^2(40) = 1 W, and I = 0.158 A. Now use series / parallel reduction to simplify the circuit. The upper parallel branch is 6.33 Ω and the lower one is 25.2 Ω. The series sum is now 126.0 Ω's law gives E = (126 Ω)(0.158 A) = 19.9 V. EVALUATE: The power input from the emf is E_I = 3.14 W, so nearly one-third of the total power is dissipated in the 40.0 Ω resistor. 26.69. IDENTIFY and SET UP: Simplify the circuit by replacing the parallel networks of resistors by their equivalents. In this simplified circuit apply the loop and junction rules to find the current in each branch. EXECUTE: The 200.0 and 30.0 Ω resistors are in parallel and have equivalent resistance 12.0 Ω. The two resistors R2 are in parallel and have equivalent resistance R2. The circuit is equivalent to the circuit sketched in Figure 26.69. Figure 26.69 (a) Calculate V_c by traveling along the branch that contains the 200 V battery, since we know the current in that branch. V_c = (500 A)(120 Ω) - (500 A)(180 Ω) - 200 V = V_c V_c = V_c = 200 V + 90.0 V + 60.0 V = 170.0 V V_c - V_a = 160 V V = 170.0 V so X = 186.0 V, with the upper terminal + (b) I_1 = (16.0 V)(80 Ω) = 2.00 A The junction rule applied to point a gives I_1 + I_1 = 5.00 A, so I_3 = 3.00 A. The current through the 200.0 V battery is in the direction from the - to the + terminal, as shown in the diagram. (c) 200.0 V - I_1(R/2) = 170.0 V (3.00 A)(R/2) = 300.0 V so R = 20.0 Ω EVALUATE: We can check the loop rule by going clockwise around the outer circuit loop. This gives +200 V - (5.00 A)(180 Ω + 12.0 Ω) + (3.00 A)(10 Ω) - 200.0 V = 200.0 V + 150.0 V - 200.0 V, which does equal zero. 26-27 Direct-Current Circuits 26.71. IDENTIFY and SET UP: For part (a) use that the full emf is across each resistor. In part (b), calculate the power dissipated by the equivalent resistance, and in this expression express R_2 and R_1 in terms of P_2 and E. 26.72. When the resistors are connected in parallel to the emf, the voltage across each resistor is E and the power dissipated by each resistor is the same as if only the one resistor were connected. P_eq = R_1 + R_2 (b) When the resistors are connected in series the equivalent resistance is R_eq = R_1 + R_2 EVALUATE: The result in part (b) can be written as 1/P_eq = 1/P_1 + 1/P_2. Our results are that for parallel the powers add and that for series they are the reciprocals of the power. This is opposite the result for combining resistance. Since P_eq = E^2/R_1 it tells us that P_1 is proportional to 1/R, this makes sense. 26.72. IDENTIFY and SET UP: Just after the switch is closed the charge on the capacitor is zero, so the voltage across the capacitor is zero and the capacitor can be replaced by a wire in analyzing the circuit. After a long time the current to the capacitor is zero, so the current through R_h is zero. After a long time the capacitor can be replaced by a break in the circuit. EXECUTE: (a) Ignoring the capacitor for the moment gives 1/(6.00 Ω) + 1/(3.00 Ω) + 1/(10.00 Ω) = R_eq = 2.00 Ω. In the absence of the capacitor, the total current in the circuit (the current through the 8.00 Ω resistor would be i = E/(8.00 Ω + 2.00 Ω) = 2.40 A, so that which 2/3 or 2.80 A would go through the 3.00 Ω resistor and 1/3, or 1.40 A would go through the 6.00 Ω resistor. Since the current through the capacitor is given by I = e^(−t/RC) at the instant t = 0 the circuit behaves as though the capacitor were not present, so the currents through the various resistors are as calculated above. (b) Once the capacitor is fully charged, the current through that part of the circuit (the 8.00 and 2.00 Ω resistors) are now in series, and the current through them is i = E/(R = (42.0 V)/(8.00 Ω + 6.00 Ω)) = 3.00 A. The voltage across both the 6.00 Ω resistor and the capacitor is V = iR = i/R = 18.0 V. EVALUATE: The equivalent resistance of R_c and R_h in parallel is less than R_h, so initially the current through R_h is larger than its value after a long time has elapsed. 26-28 Chapter 26\nEXECUTE: -I1 = (6.00 Ω + 3.00 Ω) + 36.0 V = 0\n360 Ω\nI1 = 6.00 Ω + 3.00 Ω\nI1 = 4.00 A\n-I3 + (3.00 Ω) + 36.0 V = 0\n-I3 = (300 Ω + 6.00 Ω) + 36.0 V = 0\nI2 = 4.00 A\nTo calculate Vb - Va = Vb - V, start at point b and travel to point a, adding up all the potential rises and drops along the way. We can do this by going from b up through the 3.00 Ω resistor:\nVb - Va = (4.00 A)(3.00 Ω) = 12.0 V => Vb = -12.0 V\nVb = -12.0 V (point a is 12.0 V lower in potential than point b)\nEVALUATE: Alternatively, we can go from point b down through the 6.00 Ω resistor:\nVb - Va = (4.00 A)(6.00 Ω) + (4.00 A)(3.00 Ω) = -240 V + 12.0 V = -12.0 V\n(b) IDENTIFY: Two term are multiple current paths, as shown in Figure 26.73. Use junction rule to write the current in each branch in terms of three unknown currents I1, I2, and I3. Apply the loop rule to three loops to get three equations for the three unknowns. The target variable is I5, the current through the switch Rth is calculated from I = Ires, where I is the total current that passes through the network.\nSET UP:\n1/I = 36V\nThe three unknown currents I1, I2, and I3 are labeled on Figure 26.73b.\nEXECUTE: Apply the loop rule to loops (1), (2), and (3).\nloop (1): -I1(6.00 Ω + 3.00 Ω) + I2(3.00 Ω + 3.00 Ω) = 0\ni1 = 2/I1\nloop (2): -I2 - I4 -I3 + (I - I4) + I1(6.00 Ω) - I3(3.00 Ω) = 0\n67 - 12I2 - 31I4 = 0\neq(1) to replace I2:\n4I2 - 21I2 - 4I4 = 0\neq(2)\nloop(3): (This loop is completed through the battery [not shown], in the direction from the - to the + terminal.):\nI1 + I2 + 3.60 V + I4 + It = 12.0 A\nEXECUTE: 1 = 8.5. A + 2.42 A + 5.13 A = 8.55 A\nI = 8.55 A + 4.21 Ω\n (c) From the results in part (a) the current through the battery is I = I1 + I2 + 3.42 A + 5.13 A = 8.55 A. The equivalent circuit is a single resistor that produces the same current through the 36.0 V battery, as shown in Figure 26.76.\nEXECUTE: Evaluate: With the switch open (part a), point b is at higher potential than point a, so when the switch is closed the current flows in the direction from b to a. With the switch closed the circuit can be simplified using series and parallel combinations but there is still an equivalent resistance that represents the network and the voltage across each resistor remains the same.\nSET UP: With the switch closed, C1 = 180 V. When S is closed, current I flows through the 6.00 Ω and 3.00 Ω resistors.\nEXECUTE: (i) Initially the capacitor's charge were, C1 = CV = (6.00x10^6)(180 V) = 1.80x10^-10 C After the switch is closed:\nQ1 = C1 = (6.00x10^6)(180 V) = 1.80x10^-10 C After the switch is closed:\nQ2 = C2 = (6.00x10^6)(180 V) - 12.0 V = 180x10^-10 C Both capacitors lose 1.80x10^-10 C. EVALUATE: The voltage across each capacitor decreases when the switch is closed because there is then current through each resistor and therefore a potential drop across each resistor.\nIDENTIFY: The current through the advantages for full-scale deflection is 0.020 A For each connection, there are two parallel branches and the voltage across is the same. SET UP: The sum of the two currents in the parallel branches for each connection equals the current into the meter (or reading).\nRth + R1 + R10.00 A - 0.0020 A) = (48.0Ω)(0.020 A) and R1 + Rth = 12.0 Ω.\nEXECUTE: (ii) (R + R1)(0.20 A) = (48.0Ω)(0.020 A) and R1 + R1 - 0.0020 Ω R2 = 0.980 Ω.\nEVALUATE: From (i) and (ii), Rth = 10.8 Ω. From (iii), Rth = 0.212 Ω. \n EXECUTE: 1 = 15.0 V\nFor V = 15.0 V, R = R1 + R2 and the total meter resistance Rth is Rth = RTH + R. V = 15.0 V - 1.50x10^6 ohm.\nI = 0.0020 A\nFor V = 15.0 V, R = R1 + R2 + Rth, and the total meter resistance is Rth = R + Rth + 15.0 V.\nRth = (1.00 MΩ, 1.00 kΩ) + 68 V gives (68 V)(3.00 Ω) R = (110 V)(30 kΩ and R = 18.5 kΩ.\nI = 0.20 kΩ (a) The voltage between points a and d is E, so I = e/(N + M) and I2 = e/(P + X). Using these expressions in [I/N = I/P gives = e/(N + M) = e/(P + X).\n(b) K = MP/N = (8500.0)/(33.48 x 18.97 = 1897 Ω\nEVALUATE: The measurement of X does not require that we know the value of the emf.\nIDENTIFY: Add resistors in series and parallel with the second galvanometer, so that the equivalent resistance is 65.0 Ω and to start for a current of 15.0 mA into the device the current through the galvanometer is 3.60 μA.\nEVALUATE: R1 = 1.496 mΩ and R = 9.14 mΩ. And for the total resistance to be 65.0 Ω:\n65.0 Ω = R + R1 = 1/R + 1/R1 = .0914 Ω and R1 = 64.9 Ω. EVALUATE: Adding R1 in parallel lowers the equivalent resistance so R2 must be added in series to raise the equivalent resistance to 65.0 Ω. 26.81. IDENTIFY and SET UP: Without the meter, the circuit consists of the two resistors in series. When the meter is connected, its resistance is added to the circuit in parallel with the resistor it is connected across. (a) EXECUTE: I = I1 + I2 = 900.0 V / (R + R2) = (224 Ω + 589 Ω) I = R + R2. I = (0.1107 A)(224 Ω) = 248. V; V1 = I1R2 = (0.1107 A)(589 Ω) = 652 V (b) SET UP: The resistor network is sketched in Figure 26.81a. EXECUTE: The voltage drop across the 589 Ω resistor is 90.0 V - 238.2 V = 662. V; so I = V / R = 66.2 V / 589 Ω = 0.1124 A. The voltage drop across the 224 Ω resistor is 238.2 V. I = V / R = 238.2 V / 224 Ω = 1.0632 A. Then I = I1 + I2 gives I1 = 0.1124 A - 0.1062 A = 0.0062 A. R1 = V / I = 238.2 V / 3840 Ω. Figure 26.81b EXECUTE: Replace the two resistors in parallel by their equivalent, as shown in Figure 26.81c. I = 90.0 V / (224 Ω + Req) 90.0 V = 0.1225 A. The potential drop across the 224 Ω resistor then is IR = (0.1225 A)(224 Ω) = 27.4 V; so the potential drop across the 589 Ω resistor and across the voltmeter (what the voltmeter reads) is 90.0 V - 27.4 V = 62.6 V. 26.82. IDENTIFY: The rate at which the resistor dissipates electrical energy is P = V2/R, where V is the voltage across the resistor. The power output of the source is P = E/t. EXECUTE: (a) (i) P1 = V2 / R = (120 V)2 / 426.2 Ω = 3380 W. (ii) P2 = dU/dt = D*(d(U/t))C = (120 V)2 / 426.2 = 3380 W. (b) After a long time, i = 0, so P1 = 0, P2 = 0, P0 = 0. EVALUATE: Initially all the power output of the source is dissipated in the resistor. After a long time energy is stored in the capacitor but the amount stored isn't changing. 26.83. IDENTIFY: Apply the loop rule to the circuit. The initial current determines R. We can then use the time constant to calculate C. SET UP: The circuit is sketched in Figure 26.83. EXECUTE: The loop rule therefore gives Ei - IR = 0 and R = (10 V) / (6.5 x 10-5 A) = R. The time constant is given by t = RC (Eq.26.14). So, R = 1.7 x 10^1 Ω = 3.6 μF. 26.84. IDENTIFY: The energy stored in a capacitor is U = (1/2) Q2 / C. The electrical power dissipated in the resistor is P = I2R. EXECUTE: (a) U = Q2 / (2C) = (0.0081 C)2 / (24.62 × 10-6 F) = 7.101. (b) P = I2R (Q2 / RC). R = (850 Ω)(24.62 x 10-6 F) = 3616 W. (e) When U = U_Q, U1 = 1/2 Q1 / 2C, this gives P = (Q2 / RC) = (Q2 / RC) gives P = (Q2 / R)(Q2 / 2C). EVALUATE: Increasing the energy stored in the capacitor increases current through the resistor as the capacitor discharges. (f) Changing the 57-V battery for an 80-V battery just affects the calculation in part (e). It changes too: Left loop: 140 + 210; 0 = Right loop: 80 - 351 - 210 = 0. Junction rule: I1 - I2 - I3 = 0. Solving for the three currents: I1 = -0.403 A. I2 = 0.269 A. I3 = 0.672 A. The total current for the full circuit is the sum of (b), (d) and (f) above: I = I1 + I2 + I3 = 0.184 A. I4 = 0.576 A. I5 = 0.392 A. 26.90. EVALUATE: This problem presents an alternative means of solving for currents in multi-loop circuits. IDENTIFY and SET UP: When C changes after the capacitor is charged, the voltage across the capacitor changes. Current flows through the resistor until the voltage across the capacitor again equals the emf. EXECUTE: (a) Fully discharged: Q = CV = (0.10x10^(-12) F)(2000V) = 1.00x10^(-9) C (b) The initial current just after the capacitor is charged is I = (E - Vc) / R = E / R (RC) This gives i(0) = (E / RC)e^(-t/RC), where C = 1.1C. (c) We need a resistance such that the current will be greater than 1 µA for longer than 200 µs. This requires that at t = 200 µs, I = 1.01×10^(-6) A = (1000 V - 1.01×10^(-6) C) * e^(-200 µs)/(RC) = (1.0)(1.0x10^(-10)) e^(-t/(200 µs)). This says 1.0x10^(-6) A = (0.90)e^(-t/200 µs) and 18.3 R = Rmin - 1.8x10^(-10) Ω. Solving for R numerically we find 7.15x10^(-5) Ω ≈ 5.721x10^(-11) Ω. 26.91. IDENTIFY: The resistance is too small, then the capacitor discharges too quickly, and if the resistance is too large, the current is not large enough. SET UP: We can re-draw the circuit as shown in Figure 26.91. EXECUTE: R2 = R' + (1 / R1) = 2R', R' = -2R', R0 = -R', so R = R' + 2R', R > 0, so R = R' + (√R + 2R). EVALUATE: Even though there are an infinite number of resistors, the equivalent resistance of the network is finite. 26.92. IDENTIFY: Assume a voltage V applied between points a and b and consider the currents that flow along each path between a and b. SET UP: The currents are shown in Figure 26.92. EXECUTE: Let current I enter at a and exit at b. At a there are three equivalent branches, so current is I/3 in each. At the next junction point there are two equivalent branches so each gets current I/6. Then at b there are three equivalent branches with current I/3 in each. The voltage drop from a to b then is V = (I / (1/3 R) + (1 / (1/3 R) + (1 / (1/3 R)) = V. This must be the same as V = I R', so R = 1 / R. EVALUATE: The equivalent resistance is less than R, even though there are 12 resistors in the network. 26.93. IDENTIFY: The network is the same as the one in Challenge Problem 26.91, and that problem shows that the equivalent resistance of the network is R = R1^2 / R2 + 2R2. SET UP: The circuit can be redrawn as shown in Figure 26.93. EXECUTE: (a) Va = V0 = V1 + βR' = V2 / R2 / R1 and Rse = Rb = R' = R2 R1 / Rb, so β = 2R(R2 + R1) / Rb. (b) V0 = Va - (V1 / (1 + β)) = V - Vc = (V1 - V1 / (1 + β)) = Vb / (1 + β). If R = Rn then Rb = Rn + (R'^2 + R'^2 + R1√3) and R1 = R1 + (3 / (1 + √3) = 2.73. So, for the nth segment to have 1% of the original voltage, we need: 1 / (α - β) or 5001. This says n = 4, and V2 = 0.005V. (c) R = R1 + R / (√R + 2R^2), gives R = 6400Ω + (6400Ω) and R = 3.2x10^6 and β = 26400(9.2).3x10^(-9) Ω + (8.0x10^(-1)), and V = (3.2x10^(-9) Ω / (2)8x10^(-1) Ω). (3.2)x10^(-1). (d) Along a length of 2.0 m of axon, there are 200 segments each 1.0 m long. The voltage therefore attenuates by V = (I β) = V0 = (4 + (4.0x10^(-10))). So, V = Vmax = (1 / (6.2x10^(-10))) = 0.88. EVALUATE: As R increases, V decreases and the potential difference decreases from one section to the next is less.