PART A: Two capacitors C1 = 3.4 µF, C2 = 18.0 µF are charged individually to V1

Question

PART A:

Two capacitors C1 = 3.4 µF, C2 = 18.0 µF are charged individually to V1 = 18.5 V , V2 = 7.5 V . The two capacitors are then connected together in parallel with the positive plates together and the negative plates together. Calculate the final potential difference across the plates of the capacitors once they are connected.

PART B:

Calculate the amount of charge that flows from one capacitor to the other when the capacitors are connected together.

PART C:

By how much is the total stored energy reduced when the two capacitors are connected?

Explanation / Answer

PART A) Q1 = C1*V1 = 3.4*10^-6*18.5 = 62.9*10^-6 C

Q2 = C2*V2 = 18.0*10^-6*7.5 = 135*10^-6 C

After connecting parallel let the common voltage is V volts

total charge will remain constant

=> V*(C1+C2) = Q1+Q2 = (62.9 +135)*10^-6

=> V = 197.9*10^-6 /((3.4+18)*10^-6) = 9.24 v

PART B) Q1new = C1*V = 3.4*10^-6*9.24 = 31.41*10^-6 C

Q2new = C2*V = 18*10^-6*9.24 = 166.32*10^-6 C

amount of charge transfer = Q1 - Q1new = 31.49*10^-6 C

PART C) new energy stored En= 0.5*(C1+C2)*V^2 = 0.5*(21.4*10^-6)*9.24 = 98.86*10^-6 J

previously energy stored Ep= 0.5*C1*V1^2 + 0.5*C2*V2^2 =0.5*3.4*10^-6*(18.5)^2 + 0.5*18*10^-6*(7.5)^2

=> Ep = 581.82*10^-6 + 506.25*10^-6 = 1088.07*10^-6 J

SO energy stored reduced = (1088.07 - 98.86)*10^-6 = 989.21*10^-6 J

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