A 2µF capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is

Engineering

Physics

RC Circuit

Question

20%

75%

80%

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$ΔU = \frac{1}{2} \times \frac{2 \times 8}{2 + 8} {[V - 0]}^{2}$

$= \frac{1}{2} \times \frac{16}{10} V^{2} = \frac{8 V^{2}}{10}$

$U_{i}^{'} = \frac{1}{2} \times 2 \times V^{2}$

= V²

% dissipated = $\frac{ΔU}{U_{i}} = \frac{8}{10} \times 100$

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