Question

In the given network, if potential difference between $p$ and $q$ is $2 V$ and $C_{2}=3 C_{1}$, then find the potential difference (in volt) between $a$ and $b$.

Answer 1

$Q_{p q}=2 C_{2}=6 C_{1}=Q_{b p}$
$\therefore V_{b p}=\frac{6 C_{1}}{C_{1}}=6\, V$
$\therefore V_{b q}=6+2=8\, V$
Total charge flown through
$a b =C_{2} V_{b q}+C_{1} V_{b p}$
$=3 C_{1} \times 8+C_{1} \times 6=30 C_{1}$
$\therefore V_{a b}=\frac{Q_{\text {total }}}{C_{a b}}=\frac{30 C_{1}}{C_{1}}$
$=30\, V$