Question

The extra charge flowing through the cell on closing the key $k$ is equal to

• A. $\frac{3}{4}CV$
• B. $\frac{4}{3}CV$
• C. $4CV$
• D. $\frac{CV}{4}$

Answer 1

Answer: (D)

Given circuit is

When the key is open :
$C_{net}=\frac{C\times 3C}{C+3C}$
$C_{net}=\frac{3C}{4}$
The total charge in the network , $q=C_{net}V=\frac{3CV}{4}.....\left(1\right)$
When key $k$ was closed :
$3C$ becomes short circulated. Now the net charge on $C$ is
$q‘=CV.....\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$ ,
The extra flowing through the cell is , $\Delta q=q^{\prime }-q=CV-\frac{3}{4}CV$
$\Delta q=\frac{CV}{4}$
Hence the extra charge is $\frac{CV}{4}$