Q 12-07-059JEE MainJEE Main 2024 (8 Apr, Shift 1)Easy
An LCR circuit is at resonance for a capacitor $C$, inductance $L$ and resistance $R$. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now:
Answer: (D) double
At resonance $X_L = X_C$, so $Z = R$ and $I_0 = \dfrac{V_0}{R}$. The resonance frequency does not depend on $R$, so halving $R$ doubles the current amplitude.
Solution by Sreeraj P, M.Sc Physics