A 2,mu F capacitor is charged as shown in the | vedclass

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(D) Initially,the $2\,\mu F$ capacitor is charged to a potential $V$. The initial energy stored is $U_i = \frac{1}{2} C_1 V^2$,where $C_1 = 2\,\mu F$.

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(D) Initially,the $2\,\mu F$ capacitor is charged to a potential $V$. The initial energy stored is $U_i = \frac{1}{2} C_1 V^2$,where $C_1 = 2\,\mu F$.When the switch $S$ is moved to position $2$,the capacitor $C_1$ is connected in parallel with an uncharged capacitor $C_2 = 8\,\mu F$.The charge redistributes until both capacitors reach a common potential $V' = \frac{C_1 V}{C_1 + C_2} = \frac{2V}{2+8} = 0.2V$.The final energy stored in the system is $U_f = \frac{1}{2} (C_1 + C_2) (V')^2 = \frac{1}{2} (10\,\mu F) (0.2V)^2 = 0.2 C_1 V^2$.The energy dissipated as heat is $\Delta U = U_i - U_f = \frac{1}{2} C_1 V^2 - 0.2 C_1 V^2 = 0.3 C_1 V^2$.The percentage of energy dissipated is $\frac{\Delta U}{U_i} 	imes 100 = \frac{0.3 C_1 V^2}{0.5 C_1 V^2} 	imes 100 = 60\%$. Wait,using the standard formula for energy loss in charge sharing: $\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2$.Here $V_1 = V$ and $V_2 = 0$,so $\Delta U = \frac{1}{2} \frac{2 	imes 8}{2 + 8} (V - 0)^2 = \frac{1}{2} 	imes 1.6 	imes V^2 = 0.8 	imes \frac{1}{2} C_1 V^2$.Thus,the percentage of energy dissipated is $\frac{\Delta U}{U_i} 	imes 100 = \frac{C_2}{C_1 + C_2} 	imes 100 = \frac{8}{2+8} 	imes 100 = 80\%$.

$A$ $2\,\mu F$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position $2$ is.....$\%$

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