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

Answer: (B)

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 capicity $C$
$U=\frac{1}{2}\frac{q^2}{C}$
$=\frac{1}{2}\times \frac{{\left(80\times 10^{-6}\right)}^2}{2\times 10^{-6}}$
$=\frac{1}{2}\times \frac{80\times 10^{-6}\times 80\times 10^{-6}}{2\times 10^{-6}}$
$U=1600\mu J$