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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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:

G1 Physics · Accepted Answer

At resonance: X(L) = X(C) Therefore impedance: Z = R Current amplitude: I = V/R If resistance is halved: R' = R/2 Then: I' = V/(R/2) = 2V/R = 2I Hence the current amplitude doubles.