Question

A capacitor of $30\mu F$ charged to $100V$ is connected in parallel to capacitor of $20\mu F$ charged to $50V$. The common potential is -

• A. $75V$
• B. $150V$
• C. $50V$
• D. $80V$

Answer 1

The correct option is D $80V$

$C_1=30\mu F$
$Q_1=30\times {10}^{-6}\times 100$
$Q_1=3\times {10}^{-3}$
$C_2=20\times {10}^{-6}F$
$Q_2=20\times {10}^{-6}\times 50$
$Q_2=1\times {10}^{-3}$
Total charge $=Q_1+Q_2=4\times {10}^{3}C$
Now let common potential be $V$
$\Rightarrow x=C_{1}V$ and $(Q-x)=C_{2}V$
$\Rightarrow \frac{Q-x}{x}=\frac{C_2}{C_1}$
$\Rightarrow (4\times {10}^{-3}-x)30=x.20$
$\Rightarrow 12\times {10}^{-3}-3x=2x$
$\Rightarrow x=\frac{12}{5}\times {10}^{-3}$
Now $V=\frac{x}{C_1}$

$V=\frac{12\times {10}^{-3}}{5\times 30\times {10}^{-5}}$

$V=80V$