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Capacity of a capacitor is $48 \mu \mathrm{F}$. When it is charged from $0.1 \mathrm{C}$ to $0.5 \mathrm{C}$, change in the energy stored is
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The correct answer is:
$2500 \mathrm{~J}$
Change in energy $\Delta U=\frac{1}{2}\left[\frac{q_{1}^{2}-q_{2}^{2}}{C}\right]$
$$
\begin{array}{l}
=\frac{1}{2}\left[\frac{(0.5)^{2}-(0.1)^{2}}{48 \times 10^{-6}}\right] \\
=\frac{1}{2}\left[\frac{0.25-0.01}{48 \times 10^{-6}}\right] \\
=\frac{1}{2}\left[\frac{0.24}{48 \times 10^{-6}}\right] \\
=\frac{1}{2}\left[\frac{10^{4}}{2}\right] \\
=2500 \mathrm{~J}
\end{array}
$$
$$
\begin{array}{l}
=\frac{1}{2}\left[\frac{(0.5)^{2}-(0.1)^{2}}{48 \times 10^{-6}}\right] \\
=\frac{1}{2}\left[\frac{0.25-0.01}{48 \times 10^{-6}}\right] \\
=\frac{1}{2}\left[\frac{0.24}{48 \times 10^{-6}}\right] \\
=\frac{1}{2}\left[\frac{10^{4}}{2}\right] \\
=2500 \mathrm{~J}
\end{array}
$$
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