Q 12-07-042JEE MainJEE Main 2025 (22 Jan, Shift 2)Easy
A series LCR circuit is connected to an alternating source of emf $E$. The current amplitude at resonant frequency is $I_0$. If the value of resistance $R$ becomes twice of its initial value, then amplitude of current at resonance will be
Answer: (C) $\dfrac{I_0}{2}$
At resonance $X_L = X_C$, so the impedance is just $Z = R$ and $I_0 = \dfrac{E_0}{R}$.
Doubling $R$ (resonant frequency $1/\sqrt{LC}$ is unchanged):
$$I = \frac{E_0}{2R} = \frac{I_0}{2}$$
Solution by Sreeraj P, M.Sc Physics