A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is $\mathrm{I}_{0}$. If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be

(C) $\frac{\mathrm{I}_{0}}{\sqrt{2}}$

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Alternating Current

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: (B)

At resonance in a series LCR circuit, the inductive and capacitive reactances cancel, leaving only resistance:

Z = R, I_{0} = \frac{ε _{m}}{R}

If R is doubled to 2R, the peak emf is unchanged, so the new resonant current amplitude is

I_{0}^{'} = \frac{ε _{m}}{2 R} = \frac{I _{0}}{2}

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option