Question

A capacitor of capacitance $25\mu \text{F}$ is charged to $300\text{V}$. It is then connected across a $10\text{mH}$ inductor. The resistance in the circuit is negligible. Find the potential difference across capacitor at $1.2\text{ms}$ after the inductor and capacitor are connected. $(\sin {137.6}^{\circ }=0.67\text{and}\cos {137.6}^{\circ }=-0.74)$

• A. $210\text{V}$
• B. $215\text{V}$
• C. $220\text{V}$
• D. $225\text{V}$

Answer 1

The correct option is C $220\text{V}$

Given:-
Capacitance, $C=25\mu \text{F}=25\times {10}^{-6}\text{F}$

Inductance, $L=10\text{mH}=10\times {10}^{-3}\text{H}$

Initial potential difference of charged capacitor, $V_0=300\text{V}$

Time, $t=1.2\text{ms}=1.2\times {10}^{-3}\text{s}$


So, angular frequency of oscillation.

$\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{10\times {10}^{-3}\times 25\times {10}^{-6}}}=2000\text{rad/s}$

Charge on capacitor at any time $t$,

$q=q_0\cos \omega t$

$=C{V}_0\cos \omega t$

$=25\times {10}^{-6}\times 300\times \cos (2000\times 1.2\times {10}^{-2}\times \frac{{180}^{\circ }}{3.14})[\because 1\text{rad}=\frac{{180}^{\circ }}{\pi }]$

$=25\times {10}^{-6}\times 300\times \cos {137.6}^{\circ }$

$=-5.5\times {10}^{-3}\text{C}$

Hence, potential difference,

$V=\frac{|q|}{C}=\frac{5.5\times {10}^{-3}}{25\times {10}^{-6}}=220\text{V}$

Hence, option $\left(C\right)$ is correct.