Question

In the LC circuit shown below, $C=1\mu \text{F}$. When the capacitor is fully charged up to $100\text{V}$, the switch $S$ is closed at $t=0$. The current in the circuit oscillates at a frequency of ${10}^3\text{Hz}$. The value of $L$ will be approximately equal to



Take ${\pi }^2\approx 10$

• A. $0.025\text{H}$
• B. $0.015\text{H}$
• C. $0.045\text{H}$
• D. $0.085\text{H}$

Answer 1

The correct option is A $0.025\text{H}$

Given:

Frequency of oscillations,

$f={10}^3\text{Hz}$

We know that,

$f=\frac{1}{2\pi \sqrt{LC}}$

$\Rightarrow L=\frac{1}{4{\pi }^2{f}^{2}C}$


$\Rightarrow L=\frac{1}{4\times 10\times {10}^6\times {10}^{-6}}=0.025\text{H}$

Hence, option $\left(A\right)$ is the correct answer.