Question

Two capacitors of $2\mu F$ and $4\mu F$ are connected in parallel. A third capacitor of $6\mu F$ is connected in series. The combination is connected across a $12V$ battery. The voltage across $2\mu F$ capacitor is

• A. 2 V
• B. 8 V
• C. 6 V
• D. 1 V

Answer 1

Answer: (C)

$C_P=2+4=6\mu F$
$\frac{1}{C}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}$
or $C=3\mu F$
Total charge,
$Q=CV=3\times 12=36\mu C$
Voltage across $6\mu F$ capacitor $=\frac{36\mu C}{6\mu F}=6V$
$\therefore$ Voltage across each of $2\mu F$ and $4\mu F$ capacitor
$=12V-6V=6V$