A capacitor of capacitance C is charged to potential difference V_0 . Now this capacitor is connecte

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2. Electric Potential and Capacitance

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A capacitor of capacitance $C$ is charged to potential difference $V_0$. Now this capacitor is connected to an ideal inductor. When $25\%$ of energy of capacitor is transferred to inductor then at that time what will be potential difference across capacitor

A

$\frac {V_0}{2}$

B

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

C

$\frac {V_0}{4}$

D

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

Solution

Initial energy in the capacitor is

$\mathrm{U}_{0}=\frac{\mathrm{q}_{0}^{2}}{2 \mathrm{C}}$

after connecting the inductor energy

$\mathrm{U}=\frac{3 \mathrm{U}_{0}}{4}$

So, $\mathrm{q}=\frac{\sqrt{3}}{2} \mathrm{q}_{0}$

and $\mathrm{V} \propto \mathrm{q}$

So, $\mathrm{V}=\frac{\sqrt{3}}{2} \mathrm{V}_{0}$

Standard 12

Physics

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