Question

Two capacitors of capacity $6 \, \mu F$ and $12 \, \mu F$ in series are connected by potential of $150 \, V$ . The potential of the capacitor of capacity $12 \, \mu F$ will be

• A. $25V$
• B. $50V$
• C. $100 \, V$
• D. $150 \, V$

Answer 1

Answer: (B)

Given, $C_{1}=6\mu F,C_{2}=12\mu F,V=150volt$ .
Total capacity, $\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{6}+\frac{1}{12}$
$\frac{1}{C}=\frac{3}{12}\Rightarrow C=4\mu F$
Potential of $12\mu F$ capacitor
$ \, V=\frac{q}{C}$
$ \, V= \, \frac{4 \times 150}{12} \, $
$ \, V=50V$