Question

Two capacitors of capacities $1\mu F$ and $C\mu F$ are connected in series and the combination is charged to a potential difference of $120V$. If the charge on the combination is $80\mu C$, the energy stored in the capacitor of capacity $C$ in $\mu J$ is :

• A. $1800$
• B. $1600$
• C. $14400$
• D. $7200$

Answer 1

The correct option is B $1600$

Capacitance $1\mu F$ and $C\mu F$ are connected in series,
$\therefore C_{eq}=\frac{C}{1+C}$
Given, $V=120V$ and $q=80\mu C$
$\because q=C_{eq}V$
$80=\frac{C}{C+1}\times 20$
or $C=2\mu F$
Energy stored in the capacitor of capacity $C$
$U=\frac{1}{2}\frac{q^2}{C}$
$=\frac{1}{2}\times \frac{{(80\times {10}^{-6})}^2}{2\times {10}^{-6}}$
$=\frac{1}{2}\times \frac{80\times {10}^{-6}\times 80\times {10}^{-6}}{2\times {10}^{-6}}$
$U=1600\mu J$