A capacitor of 2μF is charged as shown in the diagram. When the switch S is turned to position 2 , the percentage of its stored energy dissipated is ?
(a) 0%
(b) 20%
(c) 75%
(d) 80%
Answer : (d) 80%
\( q_{\text{before}} = q_{\text{after}} \)
\( q_1 + q_2 = q_1' + q_2' \)
\( C_1 V + C_2 V_2 = (C_1 + C_2)V_f \)
\( (2\mu) V + 8\mu (0) = (10\mu) V_f \)
\( 0.2V = V_f \)
\( U_i = \frac{1}{2} C V^2 = \frac{1}{2} (2\mu) V^2 \)
\( U_f = \frac{1}{2}(2\mu)\left( \frac{V}{5} \right)^2 + \frac{1}{2}(8\mu)\left( \frac{V}{5} \right)^2 \)
%Energy loss = \( \frac{U_i - U_f}{U_i} \times 100 \)
\( = \frac{\frac{1}{2}(2\mu)V^2 - \left( \right)}{\frac{1}{2}(2\mu)V^2} \times 100 \)
\( = \left[ 1 - \left( \frac{1}{5^2} + \frac{4}{5^2} \right) \right] \times 100 \)
\( = 80\% \)
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