JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2024
- Hard
- MCQ
\(LCR\) 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 what?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) double
Step-by-step Solution
Detailed explanation
In resonance \(Z=R\)
\(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}\)
\(\mathrm{R} \rightarrow \text { halved }\)
\(\Rightarrow \mathrm{I} \rightarrow 2 \mathrm{I}\)
I becomes doubled.
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