JEE Physics 2025 Question with Solution

Given: A series LCR circuit is at resonance. Initial current amplitude is $I_0$ and resistance becomes twice its initial value.

Find: The new current amplitude at resonance.

At resonance, the impedance of a series LCR circuit is equal to the resistance.

$$I = \frac{E}{R}$$

So initially,

$$I_0 = \frac{E}{R}$$

When the resistance becomes $2R$, the new current amplitude is

$$I' = \frac{E}{2R}$$

Using $I_0 = \frac{E}{R}$,

$$I' = \frac{I_0}{2}$$

Therefore, the current amplitude becomes $\frac{I_0}{2}$. The correct option is A.

Given: Current amplitude in a series LCR circuit is to be found after doubling the resistance.

Find: New resonance current in terms of $I_0$.

The current amplitude is

$$I = \frac{E}{Z}$$

where impedance is

$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$

At resonance,

$$X_L = X_C$$

Hence,

$$Z = R$$

So the initial current amplitude is

$$I_0 = \frac{E}{R}$$

If the resistance becomes $2R$, then at resonance the new impedance is

$$Z' = 2R$$

Therefore,

$$I' = \frac{E}{Z'} = \frac{E}{2R}$$

Comparing with $I_0 = \frac{E}{R}$,

$$I' = \frac{I_0}{2}$$

Therefore, the new current amplitude is $\frac{I_0}{2}$.