For a series LCR circuit, I vs ω curve is shown :
(a) To the left of ωᵣ, the circuit is mainly capacitive.
(b) To the left of ωᵣ, the circuit is mainly inductive.
(c) At ωᵣ, impedance of the circuit is equal to the resistance of the circuit.
(d) At ωᵣ, impedance of the circuit is 0.
Choose the most appropriate answer from the options given below :

Solution

Physics sir · 3 steps

Compare X_L and X_C on each side of the peak: the bigger reactance decides the circuit's nature.

1. Two reactances. In a series LCR circuit, X_L = ω L rises with ω and X_C = 1/(ω C) falls with ω. At the peak ωᵣ the two are equal.

   Z = √(R² + (X_L - X_C)²)

   Region	Reactances	Nature of circuit
   ω < ωᵣ (left of peak)	X_C > X_L	mainly capacitive, statement (a) true
   ω = ωᵣ (peak)	X_L = X_C, so Z = R	purely resistive, statement (c) true
   ω > ωᵣ (right of peak)	X_L > X_C	mainly inductive

2. Z is not zero. At resonance only the reactive part cancels, and R is still there. So Z = R and the current is Iₘₐₓ = V/R, a finite peak as the graph shows. Statement (d) is false.
3. So. Statements (a) and (c) are correct, which is option (c).

The idea to keep. Below resonance X_C wins and the circuit is capacitive. Above it X_L wins and it is inductive. At resonance the reactances cancel and Z = R, its minimum value, never zero.

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