Question

A 10×10^-6F capacitor is charged to a pd of 100V. A 2×10^-6F capacitor is connected in parallel with it, disconnected and discharged. This process of connection, disconnection and discharge is carried out four times. Calculate :

1. The fraction of the original charge which remains on the larger capacitor after the first connection.
2. The fraction of the original charge that remains after the fourth connection.
3. The final pd across the 10×10^-6 capacitor

Answer 1

$Part\>(i)\\ Q=CV$

In parallel arrangement, V is constant

$\implies\>Q\propto\>C$

Total capacitance in parallel $=10+2$

$=12\mu\>F$

Fraction of original charge on larger capacitor

$=\frac{10}{12}=\frac{5}{6}$

Part(ii)

(ii) 2nd connection:

Total charge $=(10×100)+\frac{1}{6}×1000$

$=\frac{7}{6}×1000$

(iii) 3^rd Connection

Total charge $=1000+\frac{1}{6}(\frac{7}{6}×1000)$

$=\frac{43}{36}×1000$

4^th Connection

Total charge $=1000+\frac{1}{6}×(\frac{43}{36}×1000)$

$=\frac{259}{216}×1000$

Fraction of the original charge that remains after the fourth connection

On small capacitor $=\frac{1}{6}×\frac{259}{216}$

$=\frac{259}{1296}$

On large capacitor $=\frac{5}{6}×\frac{259}{216}$

$=\frac{1295}{1296}$

Part(iii)

Final p.d $=\frac{Q}{C}=\frac{\frac{1295}{1296}×100}{10}$

$=99.92V$