gen.wileyplus.com/edugen/student/mainfr.uni Chapter 25, Problem 038 Your answer

Question

gen.wileyplus.com/edugen/student/mainfr.uni Chapter 25, Problem 038 Your answer is partially correct. Try again. In the figure a potential difference of V 120 V is applied across a capacitor arrangement with capacitances C1-9.05 6.80 uf, and C3 = 12.2 F, what are (a) charge q3, (b) potential difference v3, and (c) stored energy U3 for capacitor 3, (d) qi, (e) Vi, and () U1 for capacitor 1, and (g) 92. (h) V2, and (i) U2 for capacitor2 is applied across a capacitor arrangement with capacitances C-9.05 uF, C2- ergy Us for capacitor 3, (d)

Explanation / Answer

capacitor C1 and C2 are in parallel so equivalent capacitance of C1 and C2 = C1 + C2 = 9.05 + 6.8 = 15.85 uF

now C3 is in series with combination of C1 and C2 so

equivalent capacitance of circuit = 15.85*12.2/(15.85 + 12.2) = 6.89 uF

charge through battery = Q = Ceq*V = 6.89*120 = 827 uC

charge on capacitor C3 = 827 uC

potential difference on C3 = V3 = Q/C3 = 827/12.2 = 67.78 volt

energy stored = U3 = (1/2)C3*V3^2 = (1/2)*12.2*10^(-6)*(67.78)^2 = 0.028 J

charge on capacitor C2 = 827*6.8/(6.8 + 9.05) = 354.8 uC

potential difference on C2 = V2 = Q2/C2 = 354.8/6.8 = 52.17 volt

energy stored = U2 = (1/2)C2*V2^2 = (1/2)*6.8*10^(-6)*(52.17)^2 = 0.00925 J = 9.25*10^(-3) J

charge on capacitor C1 = 827*9.05/(6.8 + 9.05) = 472.2 uC

potential difference on C1 = V1 = Q1/C1 = 472.2/9.05 = 52.17 volt

energy stored = U1 = (1/2)C1*V1^2 = (1/2)*9.05*10^(-6)*(52.17)^2 = 0.0123 J = 12.3*10^(-3) J

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