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:

An LCR circuit at resonance has its resistance halved. Find the new current amplitude.

At resonance in an LCR circuit, the inductive reactance equals the capacitive reactance: $$X_L = X_C$$. These cancel out, so the impedance equals just the resistance:

$$Z = R \quad \text{(at resonance)}$$

$$I = \frac{V}{Z} = \frac{V}{R}$$

The resonance condition depends on $$L$$ and $$C$$ (not $$R$$), so the circuit remains at resonance. With $$R' = R/2$$:

$$I' = \frac{V}{R'} = \frac{V}{R/2} = \frac{2V}{R} = 2I$$

The current amplitude doubles.

The correct answer is Option (4): double.