JEE Main 2025 — Alternating Current Question with Solution

At resonance in a series LCR circuit, the reactive components (inductive and capacitive) cancel each other out, leaving only the resistance. Therefore, the amplitude of the current is given by:

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

Now, if the resistance is doubled, the new resistance becomes $2R$. The current amplitude at resonance then becomes:

$$I=\frac{E}{2R}=\frac{1}{2}\left(\frac{E}{R}\right)=\frac{I_0}{2}$$

Thus, the amplitude of current at resonance when the resistance is doubled is $\frac{I_0}{2}$.

The correct option is A.