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Two capacitors of capacities $1 \mu \mathrm{F}$ and $\mathrm{C} \mu \mathrm{F}$ are connected in series and the combination is charged to a potential difference of $120 \mathrm{~V}$. If the charge on the combination is $80 \mu \mathrm{C}$, the energy stored in the capacitor of capacity $\mathrm{C}$ in $\mu \mathrm{J}$ is
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The correct answer is:
1600
Capacitance $1 \mu \mathrm{F}$ and $\mathrm{C} \mu \mathrm{F}$ are connected in series,
$$
\therefore \mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}}{1+\mathrm{C}}
$$
Given, $\mathrm{V}=120 \mathrm{~V}$ and $\mathrm{q}=80 \mu \mathrm{C}$
$$
\begin{array}{l}
\because \mathrm{q}=\mathrm{C}_{\mathrm{eq}} \mathrm{V} \\
80=\frac{\mathrm{C}}{\mathrm{C}+1} \times 20 \\
\text { or } \mathrm{C}=2 \mu \mathrm{F}
\end{array}
$$
Energy stored in the capacitor of capicity C
$$
\begin{array}{l}
\mathrm{U}=\frac{1}{2} \frac{\mathrm{q}^{2}}{\mathrm{C}} \\
=\frac{1}{2} \times \frac{\left(80 \times 10^{-6}\right)^{2}}{2 \times 10^{-6}} \\
=\frac{1}{2} \times \frac{80 \times 10^{-6} \times 80 \times 10^{-6}}{2 \times 10^{-6}} \\
\mathrm{U}=1600 \mu \mathrm{J}
\end{array}
$$
$$
\therefore \mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}}{1+\mathrm{C}}
$$
Given, $\mathrm{V}=120 \mathrm{~V}$ and $\mathrm{q}=80 \mu \mathrm{C}$
$$
\begin{array}{l}
\because \mathrm{q}=\mathrm{C}_{\mathrm{eq}} \mathrm{V} \\
80=\frac{\mathrm{C}}{\mathrm{C}+1} \times 20 \\
\text { or } \mathrm{C}=2 \mu \mathrm{F}
\end{array}
$$
Energy stored in the capacitor of capicity C
$$
\begin{array}{l}
\mathrm{U}=\frac{1}{2} \frac{\mathrm{q}^{2}}{\mathrm{C}} \\
=\frac{1}{2} \times \frac{\left(80 \times 10^{-6}\right)^{2}}{2 \times 10^{-6}} \\
=\frac{1}{2} \times \frac{80 \times 10^{-6} \times 80 \times 10^{-6}}{2 \times 10^{-6}} \\
\mathrm{U}=1600 \mu \mathrm{J}
\end{array}
$$
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