QJEEM24Phf390: JEE Main 2024 Physics - Class 12 JEEM24Phf390 Solution: A LCR circuit is at resonance for a capacitor C, inductance L…

To solve this problem, we need to understand the relationship between the current amplitude in a series LCR circuit at resonance and the resistance $R$ . At resonance, the impedance $Z$ of the series LCR circuit is equal to the resistance $R$ , and thus:

$Z = R$

The amplitude of the current $I_0$ at resonance is given by Ohm's law:

$I_0 = \frac{V_0}{R}$

where $V_0$ is the amplitude of the voltage supplied.

Now, if the resistance $R$ is halved while keeping the voltage amplitude $V_0$ , capacitance $C$ , and inductance $L$ the same, the new resistance becomes:

$R_{new} = \frac{R}{2}$

The new current amplitude $I_0'$ at resonance is given by the modified Ohm's law:

$I_0' = \frac{V_0}{R_{new}} = \frac{V_0}{\frac{R}{2}} = \frac{2V_0}{R} = 2 \times I_0$

Therefore, the current amplitude at resonance will be doubled. Hence, the correct answer is:

Option D: double