JEE MAIN - Physics (2025 - 22nd January Evening Shift - No. 12)
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
$\frac{\mathrm{I}_0}{2}$
$\frac{\mathrm{I}_0}{\sqrt{2}}$
$2 \mathrm{I}_0$
$\mathrm{I_0}$
Explanation
Initially, $\mathrm{I}_0=\frac{\varepsilon_{\mathrm{m}}}{\mathrm{R}}$
Finally, $I_0^1=\frac{\varepsilon_m}{2 R}=\frac{I_0}{2}$
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