Alternating Current

JEE Main level, single correct, medium

66 of 190

Question 66 of 190

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

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Answer: (D) double

In resonance Z=RZ=R
I=VRR→ halved ⇒I→2I\begin{aligned} & \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}} \\ & \mathrm{R} \rightarrow \text { halved } \\ & \Rightarrow \mathrm{I} \rightarrow 2 \mathrm{I}\end{aligned}
I becomes doubled.

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