JEE Physics 2026 Question with Solution

Given: Maximum current is $50 \, \text{mA}$, source voltage is $V = 5 \sin(100t)$, so the peak voltage is $5 \, \text{V}$ and angular frequency is $\omega = 100 \, \text{rad/s}$. From the figure, $L = 2 \, \text{H}$ and $R = 100 \, \Omega$.

Find: The capacitance $C$.

The maximum current is

$$I_{\max} = 50 \, \text{mA} = 0.05 \, \text{A}$$

For a series LCR circuit,

$$I_{\max} = \frac{V_{\text{peak}}}{Z}$$

So the impedance is

$$Z = \frac{V_{\text{peak}}}{I_{\max}} = \frac{5}{0.05} = 100 \, \Omega$$

Since $Z = 100 \, \Omega$ and $R = 100 \, \Omega$, the impedance equals the resistance. Therefore the circuit is in resonance, so the reactances cancel:

$$X_L = X_C$$

That is,

$$\omega L = \frac{1}{\omega C}$$

Hence,

$$C = \frac{1}{\omega^2 L}$$

Substituting $\omega = 100 \, \text{rad/s}$ and $L = 2 \, \text{H}$,

$$C = \frac{1}{(100)^2 \times 2} = \frac{1}{20000} \; \text{F}$$ $$C = 5 \times 10^{-5} \; \text{F} = 50 \times 10^{-6} \; \text{F}$$

Therefore, the capacitance is $50 \, \mu \text{F}$.

Given: At maximum current, the measured current is $0.05 \, \text{A}$ and the applied voltage is $V = 5 \sin(100t)$. The figure gives $R = 100 \, \Omega$ and $L = 2 \, \text{H}$.

Find: The value of $C$.

First calculate the impedance from peak values:

$$Z = \frac{5}{0.05} = 100 \, \Omega$$

Now compare with the resistance from the figure:

$$R = 100 \, \Omega$$

Thus,

$$Z = R$$

In a series LCR circuit, this happens only at resonance, where

$$X_L - X_C = 0$$

So,

$$X_L = X_C$$

Using

$$X_L = \omega L, \qquad X_C = \frac{1}{\omega C}$$

we get

$$\omega L = \frac{1}{\omega C}$$

Rearranging,

$$C = \frac{1}{\omega^2 L}$$

Now substitute the known values:

$$C = \frac{1}{(100)^2 \times 2} = \frac{1}{20000} \; \text{F}$$

Convert to microfarads:

$$\frac{1}{20000} \; \text{F} = 5 \times 10^{-5} \; \text{F} = 50 \, \mu \text{F}$$

Therefore, the correct numerical answer is 50.