Question

In the circuit shown in the given figure the potential difference across $3\mu F$ capacitor is:

• A. 16V
• B. 10V
• C. 4V
• D. 6V

Answer 1

Answer: (B)

The resultant emf E which is responsible for flow of current is
$E=20-4=16V$
Given capacitors are connected in series, therefore
$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$
$C_1=3\mu F,C_2=5\mu F$
$\therefore \frac{1}{C}=\frac{1}{3}+\frac{1}{5}=\frac{8}{15}$
$\Rightarrow C=\frac{15}{8}\mu F$
Total charge stored in capacitor is
$q=CV=\frac{15}{8}\times 16=30\mu C$
Potential difference across $3\mu F$ capacitor is
$q=\frac{q}{C}=\frac{30}{3}=10V$