Two capacitors C 1 = 6.1 ?F, C 2 = 14.7 ?F are charged individually to V 1 = 16.

Question

Two capacitors C1 = 6.1 ?F, C2 = 14.7 ?F are charged individually to V1 = 16.8 V, V2 = 3.3 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.

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

By how much (absolute value) is the total stored energy reduced when the two capacitors are connected?

Explanation / Answer

charge on 1 = cx v = 6.1 x 10^-6 x 16.8 = 102.48 x 10^-6 C

charge on 2 = cx v = 14.7 x 10^-6 x 3.3 = 48.51 x 10^-6 C

on connecting both in parallel Ceq = C1 + C2

total charge = 102.48 + 48.51 ] x 10^-6 C = 151 x 10^-6 C

both will have equal voltage

[C1 + C2 ]x V = 151 x 10^-6 C

[6.1 + 14.7] x 10^-6 x V = 151 x 10^-6

V final = 7.26 V

charge on C1 finally = 6.1 x 7.26 x 10^-6 = 4.43 x 10^-5 C

charge flow on C1 = 4.43 x 10^-5 C - 102.48 x 10^-6 C = -5.82 x 10^-5 C [charges flow from this to C2]

initial enegy = .5 x c1V1^2 + .5 x C2V2^2

final enegy = .5 x Ceq x Vfinal ^2

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