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

Answer 1

Answer: (A)

Potential difference across $5 F$ capacitor is $120 / 5=24\, V$.
Hence, potential difference across all three capacitors connected in parallel is $V_{1}=24 \,V$.
$V_{1}=V\left(\frac{C_{2}}{C_{1}+C_{2}}\right)$
or $V=V_{1}\left(\frac{C_{1}+C_{2}}{C_{2}}\right)$
$=24\left[\frac{12+4}{4}\right]=96\, V$
$\therefore V_{a b}=V=96 \,V$

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