JEE Main 2025 Jan 22 Shift 2, Physics Q4: LR, CR, LCR Circuits

JEE Main2025Jan 22, Shift 2Physics

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

A
$I_{0}$
B
$\frac{I _{0}}{2}$
C
$\frac{I _{0}}{2}$
D
$2 I_{0}$

Solution

At resonance in a series LCR circuit, $X_{L} = X_{C}$ , so the impedance reduces purely to $R$ . The current amplitude is therefore:

$I_{0} = \frac{ε _{m}}{R}$ .

When $R$ is doubled, while the source emf amplitude $ε_{m}$ is unchanged:

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

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JEE Main 2025 Jan 22 Shift 2, Physics Q4: LR, CR, LCR Circuits

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