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A capacitor of unknown capacitance $\mathrm{C}$ is connected across a battery of $\mathrm{V}$ volt. The
charge stored in it becomes $Q$ coulomb. When potential across the capacitor is
reduced by $\mathrm{V}^{\prime}$ volt, the charge stored in it becomes Q'coulomb. The capacitance C
is
Options:
charge stored in it becomes $Q$ coulomb. When potential across the capacitor is
reduced by $\mathrm{V}^{\prime}$ volt, the charge stored in it becomes Q'coulomb. The capacitance C
is
Solution:
1422 Upvotes
Verified Answer
The correct answer is:
$\frac{\mathrm{Q}-\mathrm{Q}^{\prime}}{\mathrm{V}^{\prime}}$
$\mathrm{Q}=\mathrm{CV}$
$\mathrm{Q}^{\prime}=\mathrm{C}\left(\mathrm{V}-\mathrm{V}^{\prime}\right)=\mathrm{CV}=\mathrm{CV}^{\prime}$
$\mathrm{Q}^{\prime}=\mathrm{Q}-\mathrm{CV}^{\prime}$
$\therefore \mathrm{CV}^{\prime}=\mathrm{Q}-\mathrm{Q}^{\prime}$
$\mathrm{C}=\frac{Q-Q^{\prime}}{V^{\prime}}$
$\mathrm{Q}^{\prime}=\mathrm{C}\left(\mathrm{V}-\mathrm{V}^{\prime}\right)=\mathrm{CV}=\mathrm{CV}^{\prime}$
$\mathrm{Q}^{\prime}=\mathrm{Q}-\mathrm{CV}^{\prime}$
$\therefore \mathrm{CV}^{\prime}=\mathrm{Q}-\mathrm{Q}^{\prime}$
$\mathrm{C}=\frac{Q-Q^{\prime}}{V^{\prime}}$
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