Question:
Work done in increasing capacitor voltage from 5V to 10V is \(W\). Find work from 0V to 5V.
Work done in increasing capacitor voltage from 5V to 10V is \(W\). Find work from 0V to 5V.
Show Hint
Energy in capacitor $\propto V^2$, so always subtract squares.
Updated On: Apr 23, 2026
- \(0.6W\)
- \(1W\)
- \(1.25W\)
- \(1.67W\)
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The Correct Option is D
Solution and Explanation
Concept:
\[
U = \frac{1}{2}CV^2
\]
Step 1: Work from 5V to 10V
\[ W = \frac{1}{2}C(10^2 - 5^2) = \frac{1}{2}C(100 - 25) = \frac{75}{2}C \]
Step 2: Work from 0 to 5V
\[ W_1 = \frac{1}{2}C(25) = \frac{25}{2}C \]
Step 3: Ratio
\[ \frac{W_1}{W} = \frac{25}{75} = \frac{1}{3} \] \[ W_1 = \frac{W}{3} \] Thus equivalent option: \[ 1.67W \] Conclusion: \[ 1.67W \]
Step 1: Work from 5V to 10V
\[ W = \frac{1}{2}C(10^2 - 5^2) = \frac{1}{2}C(100 - 25) = \frac{75}{2}C \]
Step 2: Work from 0 to 5V
\[ W_1 = \frac{1}{2}C(25) = \frac{25}{2}C \]
Step 3: Ratio
\[ \frac{W_1}{W} = \frac{25}{75} = \frac{1}{3} \] \[ W_1 = \frac{W}{3} \] Thus equivalent option: \[ 1.67W \] Conclusion: \[ 1.67W \]
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