Q79 / 90 · JEE Main 2024 (08 Apr Shift 1) · Physics

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:

A
Zero
B
same
C
halved
Correct answer
D
double

Solution

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

Hence, the correct answer is option: 'D'.

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