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 $12\, V$ 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,
$Q=C V=3 \times 12=36\, \mu C$
Voltage across $6\,\mu F$ capacitor $ = \frac{36\,\mu C}{6\,\mu F} =6\,V$
$\therefore $ Voltage across each of $2\,\mu F$ and $4\,\mu F$ capacitor
$= 12\,V - 6\,V = 6\,V$