Question

A capacitor circuit is equipped with key switch $k$ as shown. Initially key switch is open. The excess charge flowing through the cell on closing the key $k$ is equal to

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

Answer 1

The correct option is A $\frac{CV}{4}$

The given circuit is shown in figure $C_{net}=\frac{3}{4}C$
When the key was open charge stored in the capacitors $q=\frac{3}{4}CV$
When key is closed, capacitor $3C$ becomes short circuited. Net charge on $C$ is now,
$q^{\prime }=CV$
$\therefore \Delta q=q^{\prime }-q=\frac{CV}{4}$