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(B) The initial energy stored in the capacitor is $U_0 = \frac{1}{2} C V_0^2$.When $25\%$ of the energy is transferred to the inductor,the remaining e

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$\frac{\sqrt{3}}{2} V_0$

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(B) The initial energy stored in the capacitor is $U_0 = \frac{1}{2} C V_0^2$.When $25\%$ of the energy is transferred to the inductor,the remaining energy in the capacitor is $U = U_0 - 0.25 U_0 = 0.75 U_0 = \frac{3}{4} U_0$.Since the energy stored in a capacitor is given by $U = \frac{1}{2} C V^2$,we have $\frac{1}{2} C V^2 = \frac{3}{4} (\frac{1}{2} C V_0^2)$.Simplifying this,we get $V^2 = \frac{3}{4} V_0^2$.Taking the square root of both sides,we find $V = \frac{\sqrt{3}}{2} V_0$.

$A$ capacitor of capacitance $C$ is charged to a potential difference $V_0$. Now,this capacitor is connected to an ideal inductor. When $25\%$ of the energy of the capacitor is transferred to the inductor,what will be the potential difference across the capacitor?

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