JEE Mains · Physics · STD 12 -7. Alternating current
\(LCR\) 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?
- A Zero
- B double
- C same
- D halved
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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