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If the charge on the capacitor is increased by 3 coulombs, the energy stored in it increases by $44 \%$. The original charge on the capacitor is
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Verified Answer
The correct answer is:
$15 \mathrm{C}$
Energy stored in a charged capacitor is
$$
\mathrm{U}=\frac{\mathrm{Q}^2}{2 \mathrm{C}} \quad \Rightarrow \mathrm{U} \propto \mathrm{Q}^2
$$
As per given condition, when, $\mathrm{Q}_2=\left(\mathrm{Q}_1+3\right)$,
$$
\begin{array}{ll}
\therefore & \mathrm{U}_2=44 \% \text { of } \mathrm{U}_1=\frac{144}{100} \mathrm{U}_1 \\
\therefore & \frac{\mathrm{Q}_2^2}{2 \mathrm{C}}=\frac{144}{100} \times \frac{\mathrm{Q}_1^2}{2 \mathrm{C}} \\
\therefore & \mathrm{Q}_1=\frac{10}{12} \mathrm{Q}_2 \\
\therefore & \mathrm{Q}_1=\frac{5}{6}\left(\mathrm{Q}_1+3\right) \\
\therefore & 6 \mathrm{Q}_1=5 \mathrm{Q}_1+15 \\
\therefore & \mathrm{Q}_1=15 \mathrm{C}
\end{array}
$$
$$
\mathrm{U}=\frac{\mathrm{Q}^2}{2 \mathrm{C}} \quad \Rightarrow \mathrm{U} \propto \mathrm{Q}^2
$$
As per given condition, when, $\mathrm{Q}_2=\left(\mathrm{Q}_1+3\right)$,
$$
\begin{array}{ll}
\therefore & \mathrm{U}_2=44 \% \text { of } \mathrm{U}_1=\frac{144}{100} \mathrm{U}_1 \\
\therefore & \frac{\mathrm{Q}_2^2}{2 \mathrm{C}}=\frac{144}{100} \times \frac{\mathrm{Q}_1^2}{2 \mathrm{C}} \\
\therefore & \mathrm{Q}_1=\frac{10}{12} \mathrm{Q}_2 \\
\therefore & \mathrm{Q}_1=\frac{5}{6}\left(\mathrm{Q}_1+3\right) \\
\therefore & 6 \mathrm{Q}_1=5 \mathrm{Q}_1+15 \\
\therefore & \mathrm{Q}_1=15 \mathrm{C}
\end{array}
$$
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