MHT CET · Physics · Capacitance
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
- A \(\frac{\mathrm{Q}-\mathrm{Q}^{\prime}}{\sqrt{\mathrm{V}^{\prime}}}\)
- B \(\frac{\mathrm{V}^{\prime}}{\mathrm{Q}-\mathrm{Q}^{\prime}}\)
- C \(\frac{\mathrm{Q}+\mathrm{Q}^{\prime}}{\mathrm{V}^{\prime}}\)
- D \(\frac{\mathrm{Q}-\mathrm{Q}^{\prime}}{\mathrm{V}^{\prime}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{Q}-\mathrm{Q}^{\prime}}{\mathrm{V}^{\prime}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Q}=\mathrm{CV}\)
\(\mathrm{Q}^{\prime}=\mathrm{C}\left(\mathrm{V}-\mathrm{V}^{\prime}\right)=\mathrm{CV}=\mathrm{CV}^{\prime}\)
\(\mathrm{Q}^{\prime}=\mathrm{Q}-\mathrm{CV}^{\prime}\)
\(\therefore \mathrm{CV}^{\prime}=\mathrm{Q}-\mathrm{Q}^{\prime}\)
\(\mathrm{C}=\frac{Q-Q^{\prime}}{V^{\prime}}\)
\(\mathrm{Q}^{\prime}=\mathrm{C}\left(\mathrm{V}-\mathrm{V}^{\prime}\right)=\mathrm{CV}=\mathrm{CV}^{\prime}\)
\(\mathrm{Q}^{\prime}=\mathrm{Q}-\mathrm{CV}^{\prime}\)
\(\therefore \mathrm{CV}^{\prime}=\mathrm{Q}-\mathrm{Q}^{\prime}\)
\(\mathrm{C}=\frac{Q-Q^{\prime}}{V^{\prime}}\)
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