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A capacitor of $2 μ F$ is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is

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75%

80%

20%

answer is B.

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Detailed Solution

Loss of energy $V_{1} = V and V_{2} = 0$

$ΔU = \frac{1}{2} \frac{C_{1} C_{2}}{C_{1} + C_{2}} {(V_{1} - V_{2})}^{2}$

% loss of energy

$\frac{ΔU}{U} \times 100$

$= \frac{\frac{1}{2} \frac{C_{1} C_{2}}{C_{1} + C_{2}} \cdot V_{1}^{2}}{\frac{1}{2} C_{1} V_{1}^{2}} \times 100$

$= \frac{C_{2}}{C_{1} + C_{2}} \times 100 = 80 %$

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