JEE Physics 2023 Question with Solution

Given: $L = 2 \times 10^{-6} \, \text{H}$, $f = 7 \times 10^3 \, \text{Hz}$.

Find: The value of $x$ if $C = \frac{1}{x} \, \text{F}$.

For maximum current, the circuit must be in resonance, so inductive reactance equals capacitive reactance:

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

Rearranging,

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

Substitute the given values:

$$C = \frac{1}{4\pi^2 (7 \times 10^3)^2 (2 \times 10^{-6})}$$

Using the simplification shown,

$$C = \frac{1}{4 \times (3.14)^2 \times 49 \times 10^6 \times 2 \times 10^{-6}} = \frac{1}{3872}$$

Hence, $x = 3872$.

Therefore, the required numerical value is $3872$.

Given: A series LCR circuit with inductance $L$ and variable capacitance $C$.

Find: The value of $x$ in $C = \frac{1}{x} \, \text{F}$ when current is maximum.

In a series LCR circuit, current is maximum at resonance. The resonance condition is:

$$X_L = X_C$$

That is,

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

Solving for capacitance,

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

Now put $f = 7 \times 10^3 \, \text{Hz}$ and $L = 2 \times 10^{-6} \, \text{H}$:

$$C = \frac{1}{4\pi^2 (7 \times 10^3)^2 (2 \times 10^{-6})}$$

the solution evaluates this as:

$$C = \frac{1}{3872}$$

Comparing with $C = \frac{1}{x} \, \text{F}$, we get

$$x = 3872$$

Therefore, the answer is $3872$.