JEE Main 2025 · 22 January · evening · Single correct · +4 / −1
A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is I₀. If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be
- (a) I₀/2
- (b) I₀/√2
- (c) 2 I₀
- (d) I₀
Answer and solution
Answer (a) I₀/2
Solution
Physics sir · 3 steps
At resonance X_L and X_C cancel, so only R limits the current: double R, halve the current.
- 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 = E₀/Z = E₀/R
- Double the resistance. Changing R does not change the resonant frequency ω₀ = 1/√LC, so the circuit is still at resonance. The new amplitude is I₀' = E₀/2R = I₀/2.
- So. The current amplitude at resonance becomes I₀/2, which is option (a).
The idea to keep. At resonance the circuit behaves like a pure resistor: Z = R and I₀ = E₀/R. L and C only decide where resonance happens, R decides how tall the peak is.
Answer from the NTA final answer key. Solution by PrepWiser.