A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant…

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
  1. $2 \mathrm{I}_0$
  2. $I_0$
  3. $\frac{\mathrm{I}_0}{2}$
  4. $\frac{\mathrm{I}_0}{\sqrt{2}}$

Solution

Initially, $\mathrm{I}_0=\frac{\varepsilon_{\mathrm{m}}}{\mathrm{R}}$
Finally, $\mathrm{I}_0^1=\frac{\varepsilon_{\mathrm{m}}}{2 \mathrm{R}}=\frac{\mathrm{I}_0}{2}$

Asked in: JEE Main 2025 (22 Jan Shift 2)

Practice more AC Circuits questions on Aicharya