Question

For the shown $L-C$ circuit, the capacitor is first charged to $200\mu \text{C}$ and then connected with inductor. When instantaneous current is equal to one- half of its maximum value, then the value of instantaneous charge on capacitor is

• A. $1.7\times {10}^{-4}\text{C}$
• B. $1.5\times {10}^{-4}\text{C}$
• C. $1.3\times {10}^{-4}\text{C}$
• D. $1.9\times {10}^{-4}\text{C}$

Answer 1

The correct option is A $1.7\times {10}^{-4}\text{C}$

Maximum value of charge on capacitor,

$q_0=200\mu \text{C}=200\times {10}^{-6}\text{C}$

Capacitance, $C=5\mu \text{F}=5\times {10}^{-6}\text{F}$

Inductance, $L=2\text{mH}=2\times {10}^{-3}\text{H}$

Now, angular frequency of oscillation

$\omega =\frac{1}{\sqrt{LC}}$

$=\frac{1}{\sqrt{2\times {10}^{-3}\times 5\times {10}^{-6}}}={10}^4\text{rad/sec}$

Charge on capacitor at any time $t,$

$q=q_0\cos \omega t(\because \text{at}t=0,q=q_0)$

$\Rightarrow i=\frac{dq}{dt}=-q_0\omega \sin \omega t$

So, maximum value of current

$i_0=q_0\omega$

$=200\times {10}^{-6}\times {10}^4=2\text{A}$

Now, instantaneous value of current

$i_{ins}=\frac{i_0}{2}=1\text{A}$

From energy conservation,

$\frac{1}{2}L{i}_0^2=\frac{1}{2}L{i}_{ins}^2+\frac{1}{2}\frac{q^2}{C}$

$\Rightarrow q=\sqrt{LC(i_0^2-i_{ins}^2)}$

Substituting the values,

$q=\sqrt{2\times {10}^{-3}\times 5\times {10}^{-6}\times (2^2-1^2)}$

$\approx 1.7\times {10}^{-4}\text{C}$

Hence, $\left(\text{A}\right)$ is the correct answer.