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Engenharia de Alimentos ·
Eletromagnetismo
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Lab 4 - FSC\n\n1) C_1 = 10 μF C_2 = 10.5 = 15 μF\n\nG_1 + G_2 = 1/15 + 1/4 = 1/19\n\nG_t = 19/15 60 60\n\nC_s = C_0 + C_1 = 3.36 μF\n\n2) C = s.e.A => C = ε_0 * e \u00b7 A \u00b7 (d/2) => C_total = 2C\n\nd = (d/2)\nC_total = 2 (60) = 12 μF\n\nΔq = VΔC = V(C_C)\nΔq = (10)(12 - d) - d/2\n\nTotal charge = 180 μC\n\n3) I = sen paralele\nC_T = 3. E.A => G_1 + G_2 = E_0.A\n\nΔC_T = E_0 A / d\n\n4) C = 102 μF\nC + Q = 108 μF => Q = 5659.2\nV = 52.4\n\nQ = 5659.2 - 154.078\n35^\n\n158.2 = 10 + C_1 => C_2 = 50 μF 5) 2100. C_s = F\n\n55 kV \u00b7 R = 53,000 / kW.A\n\nU = 1 CV^2\n1.5 \u00b7 10^6 = 50000 * 2,100 = 15881250 J = 3412498.75 kWh\n\nx 3 (3, 14 kWh)\n\nA: A = 42 cm²\n a) C = ε_0 A = 8.85 \u00b7 10^-12 * 428.36 * 28.6 * 10^-12\n d = 130 mm\n\nV = 6.625 V\n\nC_q = Q = 38.6 * 10^-12 * 0.625 V = 1.7 nC\n\nc) U = 1 CV^2\n= (2 \(1) * 10^6) / 2 = 5.58 μJ\n\nd) E = 0.126 m\n\nε_0 = Q = 5.049 \u00b7 10^-6 = 12324.7 2803\n\nV = 150 V\n\n= 0.83 * 10^-6 = (128 \u00b7 10^3) / 2 6) C_T = 26.6 \u00b7 10^-6\n\nL = ... (to be continued)\n\n7) D = 0.126 m\n\nV = 150 V\n\nC = E·A\n\n8) U = (q^2) / (2C)\n\nU = ... (to be continued)\n\nb) U = (q^2)/(C) + (E_0 A^2) - d\n\nc) W = U - U_0 =\n= q^2 (1/C) - (1/C_0) + ...\n\nd) Como e trabalhar como angular e da e da T 1) C1=k1ε0A\n ---------------\n d\n C2=k2ε0A\n ---------------\n d\n C3=\n \n C3=k1ε0A\n -------------------\n d\n 2\n d\n\n C2=k2ε0A\n -------------------\n d\n d\n\n------------------------------\n2) C1+C2\n --------------------\n 1 + 1\n C3 C1 C2 \n C3=2ε0A\n --------------------\n d\n (k1-k2) \n (k1-k2)k1
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Lab 4 - FSC\n\n1) C_1 = 10 μF C_2 = 10.5 = 15 μF\n\nG_1 + G_2 = 1/15 + 1/4 = 1/19\n\nG_t = 19/15 60 60\n\nC_s = C_0 + C_1 = 3.36 μF\n\n2) C = s.e.A => C = ε_0 * e \u00b7 A \u00b7 (d/2) => C_total = 2C\n\nd = (d/2)\nC_total = 2 (60) = 12 μF\n\nΔq = VΔC = V(C_C)\nΔq = (10)(12 - d) - d/2\n\nTotal charge = 180 μC\n\n3) I = sen paralele\nC_T = 3. E.A => G_1 + G_2 = E_0.A\n\nΔC_T = E_0 A / d\n\n4) C = 102 μF\nC + Q = 108 μF => Q = 5659.2\nV = 52.4\n\nQ = 5659.2 - 154.078\n35^\n\n158.2 = 10 + C_1 => C_2 = 50 μF 5) 2100. C_s = F\n\n55 kV \u00b7 R = 53,000 / kW.A\n\nU = 1 CV^2\n1.5 \u00b7 10^6 = 50000 * 2,100 = 15881250 J = 3412498.75 kWh\n\nx 3 (3, 14 kWh)\n\nA: A = 42 cm²\n a) C = ε_0 A = 8.85 \u00b7 10^-12 * 428.36 * 28.6 * 10^-12\n d = 130 mm\n\nV = 6.625 V\n\nC_q = Q = 38.6 * 10^-12 * 0.625 V = 1.7 nC\n\nc) U = 1 CV^2\n= (2 \(1) * 10^6) / 2 = 5.58 μJ\n\nd) E = 0.126 m\n\nε_0 = Q = 5.049 \u00b7 10^-6 = 12324.7 2803\n\nV = 150 V\n\n= 0.83 * 10^-6 = (128 \u00b7 10^3) / 2 6) C_T = 26.6 \u00b7 10^-6\n\nL = ... (to be continued)\n\n7) D = 0.126 m\n\nV = 150 V\n\nC = E·A\n\n8) U = (q^2) / (2C)\n\nU = ... (to be continued)\n\nb) U = (q^2)/(C) + (E_0 A^2) - d\n\nc) W = U - U_0 =\n= q^2 (1/C) - (1/C_0) + ...\n\nd) Como e trabalhar como angular e da e da T 1) C1=k1ε0A\n ---------------\n d\n C2=k2ε0A\n ---------------\n d\n C3=\n \n C3=k1ε0A\n -------------------\n d\n 2\n d\n\n C2=k2ε0A\n -------------------\n d\n d\n\n------------------------------\n2) C1+C2\n --------------------\n 1 + 1\n C3 C1 C2 \n C3=2ε0A\n --------------------\n d\n (k1-k2) \n (k1-k2)k1