An ideal capacitor of capacitance $0.2 \mu \mathrm{F}$ is charged to a potential difference of $10…
An ideal capacitor of capacitance $0.2 \mu \mathrm{F}$ is charged to a potential difference of $10 \mathrm{~V}$. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance $0.5 \mathrm{mH}$. The current at a time when the potential difference across the capacitor is $5 \mathrm{~V}$, is:
$0.17 \mathrm{~A}$
$0.15 \mathrm{~A}$
$0.34 \mathrm{~A}$
$0.25 \mathrm{~A}$
Solution
Given: Capacitance, $C=0.2 \mu \mathrm{F}=0.2 \times 10^{-6}$ $\mathrm{F}$
Inductance $\mathrm{L}=0.5 \mathrm{mH}=0.5 \times 10^{-3} \mathrm{H}$
Current $\mathrm{I}$ = ?
Using energy conservation
$
\frac{1}{2} C V^2=\frac{1}{2} C V_1^2+\frac{1}{2} L I^2
$
$\frac{1}{2} \times 0.2 \times 10^{-6} \times 10^2+0$
$=\frac{1}{2} \times 0.2 \times 10^{-6} \times 5^2+\frac{1}{2} \times 0.5 \times 10^{-3} I^2$
$\therefore \quad I=\sqrt{3} \times 10^{-1} \mathrm{~A}=0.17 \mathrm{~A}$