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 $120 \,V$. 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

Answer: (B)

Capacitance $1 \, \mu F$ and $C \, \mu F$ are connected in series,
$\therefore C_{eq} = \frac{C}{1+C}$
Given, $V = 120 \, V $ and $q = 80 \,\mu C $
$ \because q =C_{eq}V $
$ 80= \frac{C}{C+1} \times20$
or $ C = 2 \,\mu F$
Energy stored in the capacitor of capicity $C$
$ U = \frac{1}{2} \frac{q^{2}}{C} $
$ = \frac{1}{2}\times\frac{\left(80\times10^{-6}\right)^{2}}{2 \times10^{-6}}$
$ = \frac{1}{2} \times\frac{80\times 10^{-6}\times 80\times 10^{-6}}{2\times 10^{-6}}$
$ U = 1600 \,\mu J$