Question

Five capacitors are connected as shown in the figure. Initially, $S$ is opened and all capacitors are uncharged. When $S$ is closed, steady state is obtained. If the potential difference between the points $M$ and $N$ is found to be $V_M-V_N=4P$ volt. The value of $P$ is _____.

Answer 1

Answer: 3

Let $q$ be the charge on all the capacitor

Apply KVL
$31-\frac{q}{4}-\frac{q}{2}-\frac{q}{4}-7-\frac{q}{6}-\frac{q}{12}=0$
$24=\left[\frac{3+6+3+2+10}{12}\right]q$
$q=12\mu C$
Now $V_N+\frac{q}{6}+7+\frac{q}{4}=V_M$
$V_M-V_N=12V$