JEE Physics 2023 Question with Solution

Given: Capacitance $C = 62.5 \times 10^{-9} \, \text{F}$ and frequency $f = 2000 \, \text{Hz}$.

Find: The inductance $L$ for which current amplitude is maximum.

For maximum current amplitude in an LCR series circuit, resonance occurs. Hence,

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

Rearranging for $L$,

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

Substituting the given values,

$$L = \frac{1}{4\pi^2 \times 2000^2 \times 62.5 \times 10^{-9}}$$

Thus,

$$L = 0.1 \, \text{H}$$

So,

$$L = 100 \, \text{mH}$$

Therefore, the value of inductance is $100 \, \text{mH}$.

Given: An LCR series circuit with $C = 62.5 \, \text{nF}$ and source frequency $f = 2.0 \, \text{kHz}$.

Find: The inductance required for maximum current.

At resonance, the inductive reactance and capacitive reactance cancel each other, so the current becomes maximum. The resonance frequency is

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

Solving for $L$,

$$L = \frac{1}{(2\pi f)^2 C}$$

Now substitute

$$f = 2000 \, \text{Hz}, \quad C = 62.5 \times 10^{-9} \, \text{F}$$

Therefore,

$$L = \frac{1}{(2 \pi \cdot 2000)^2 \cdot 62.5 \times 10^{-9}}$$

This gives

$$L = 0.1 \, \text{H} = 100 \, \text{mH}$$

Therefore, the correct numerical answer is 100.