A LCR circuit is at resonance for a capacitor $C$, inductance $L$ and resistance $R$. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now:

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Aicharya · Accepted Answer

In resonance $Z=R$ $\begin{aligned} & \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}} \ & \mathrm{R} ightarrow 	ext { halved } \ & \Rightarrow \mathrm{I} ightarrow 2 \mathrm{I}\end{aligned}$ I becomes doubled.

A LCR circuit is at resonance for a capacitor $C$, inductance $L$ and resistance $R$. Now the value of…

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