Question

Several capacitors are connected as shown in figure. If the charge on the $5\mu F$ capacitor is $120\mu C$, the potential between points $A$ and $D$ is

• A. $16V$
• B. $32V$
• C. $64V$
• D. none of these

Answer 1

Answer: (A)

Potential difference across $5F$ capacitor is $120/5=24V$.
Hence, potential difference across all three capacitors connected in parallel is $V_1=24V$.
$V_1=V\left(\frac{C_2}{C_1+C_2}\right)$
or $V=V_1\left(\frac{C_1+C_2}{C_2}\right)$
$=24\left[\frac{12+4}{4}\right]=96V$
$\therefore V_{ab}=V=96V$

$V_{AD}=V\left(\frac{6}{30+6}\right)=96\left[\frac{6}{30+6}\right]=16V$