JEE Main 2024 — Alternating Current Question with Solution

To solve this problem, we need to understand the relationship between the current amplitude in a series LCR circuit at resonance and the resistance $R$. At resonance, the impedance $Z$ of the series LCR circuit is equal to the resistance $R$, and thus:

$$Z=R$$

The amplitude of the current $I_0$ at resonance is given by Ohm's law:

$$I_0=\frac{V_0}{R}$$

where $V_0$ is the amplitude of the voltage supplied.

Now, if the resistance $R$ is halved while keeping the voltage amplitude $V_0$, capacitance $C$, and inductance $L$ the same, the new resistance becomes:

$$R_{new}=\frac{R}{2}$$

The new current amplitude $I_0^{\prime }$ at resonance is given by the modified Ohm's law:

$$I_0^{\prime }=\frac{V_0}{R_{new}}=\frac{V_0}{\frac{R}{2}}=\frac{2V_0}{R}=2\times I_0$$

Therefore, the current amplitude at resonance will be doubled. Hence, the correct answer is:

Option D: double