Question

For the shown $L-C$ circuit, the maximum value of charge on capacitor is $200\mu \text{C}$. When the charge become $100\mu \text{C}$ what is the value of $\left|\frac{di}{dt}\right|$, where $i$ is the instantaneous current and $t$ is the time ? [Note that the capacitor is first charged and then connected with the inductor ]

• A. ${10}^2\text{A/s}$
• B. ${10}^3\text{A/s}$
• C. ${10}^4\text{A/s}$
• D. ${10}^5\text{A/s}$

Answer 1

The correct option is C ${10}^4\text{A/s}$

Given:-

Maximum value of charge on capacitor,

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

Instantaneous value of charge on capacitor,

$q=100\mu \text{C}=100\times {10}^{-6}\text{C}$

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

$\omega =\frac{1}{\sqrt{LC}}$
$=\frac{1}{\sqrt{2\times {10}^{-3}\times 5\times {10}^{-6}}}$

$={10}^4\text{rad/s}$

Charge on capacitor at any time $t$.

$q=q_0\cos \omega t...\left(1\right)$

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

$\Rightarrow \frac{di}{dt}=-q_0{\omega }^2\cos \omega t...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,

$\frac{di}{dt}=-q\times {\omega }^2$

$=-100\times {10}^{-6}\times ({10}^4{)}^2=-{10}^4\text{A/s}$

$\therefore \left|\frac{di}{dt}\right|={10}^4\text{A/s}$

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