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 magnetic and electric energy at $t=1.2\text{ms}$. $(\sin {137.6}^{\circ }=0.67\text{and}\cos {137.6}^{\circ }=-0.74)$

• A. $0.3\text{J},0.4\text{J}$
• B. $0.8\text{J},0.4\text{J}$
• C. $0.2\text{J},0.3\text{J}$
• D. $0.5\text{J},0.6\text{J}$

Answer 1

The correct option is D $0.5\text{J},0.6\text{J}$

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 capacitor, $V_0=300\text{V}$

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

So, angular frequency of oscillation

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

Further charge on capacitor at any time $t$,

$q=q_0\cos \omega t$

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

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

$=25\times {10}^{-6}\times 300\times \cos \left({137.6}^{\circ }\right)$

$\approx -5.5\times {10}^{-3}\text{C}...\left(1\right)$

Circuit current at this instant,

$i=-q_0\omega \sin \omega t$

$=-C{V}_0\omega \sin \omega t$

$=-25\times {10}^{-6}\times 300\times 2000\times \sin (2000\times 1.2\times {10}^{-3}\times \frac{{180}^{\circ }}{3.14})$

$=-25\times {10}^{-6}\times 300\times 2000\times \sin \left({137.6}^{\circ }\right)$

$\approx -10\text{A}...\left(2\right)$

Magnetic energy,
$U_L=\frac{1}{2}L{i}^2$

$=\frac{1}{2}\times 10\times {10}^{-3}\times (-10{)}^2=0.5\text{J}$

Electric energy,

$U_C=\frac{1}{2}\frac{q^2}{C}$

$=\frac{1}{2}\times \frac{(-5.5\times {10}^{-3}{)}^2}{25\times {10}^{-6}}=0.6\text{J}$

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