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:

Solution

Physics sir · 4 steps

At resonance the reactances cancel, so R alone limits the current: halve R and the current doubles.

1. Impedance at resonance. In a series LCR circuit, Z = √(R² + (X_L - X_C)²). At resonance X_L = X_C, so the bracket is zero and Z = R.

   I₀ = V₀/Z = V₀/R

2. Resonance still holds. The resonant frequency ω₀ = 1/√LC does not contain R. With L and C unchanged, the circuit is still at resonance after R changes.
3. Halve R. The new amplitude is I₀' = V₀/(R/2) = 2V₀/R = 2I₀.
4. So. The current amplitude at resonance doubles, which is option (d).

The idea to keep. At resonance X_L and X_C cancel, so the peak current is V₀/R. Resistance sets how tall the resonance peak is, not where it sits.

Answer from the NTA final answer key. Solution by PrepWiser.