A capacitor of 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%

20%

80%

answer is C.

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Detailed Solution

Initial energy stored $= \frac{1}{2} (2 μF) \times V^{2}$
Energy dissipated on connection across 8µF
$= \frac{1}{2} \frac{c_{1} c_{2}}{c_{1} + c_{2}} V^{2} = \frac{1}{2} \times \frac{2 μF \times 8 μF}{10 μF} V^{2}$
$∴ % loss of energy = \frac{1 .6}{2} \times 100 = 80 %$

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