Question

Two capacitors C1 = 4.5 μF, C2 = 19.4 μF are charged individually to V1 = 19.7 V, V2 = 7.7 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?

Answer 1

final voltage across them will be same say V.

Applying charge conservation,

(charge Q = C V)

C1 V2 + C2 V2 = C1 V + C2 V

88.65 + 149.38 = (4.5 + 19.4)V

V = 9.96 Volt .............Ans

on C1,

Q1i = 4.5 x 19.7 = 88.65 uC  

Q1f = 4.5 x 9.96 = 44.82 uC

charge flow = Q1i - Q1f = 43.8 uC ......Ans

(U = C V^2/ 2)

Ui = C1 V1^2 / 2 + C2 V2^2 /2 = 1.45 x 10^-3 J  

Uf = (C1 + C2)V^2 /2 = 1.19 x 10^-3 J  

Energy reduced = Ui - Uf = 2.6 x 10^-4 J .....Ans