An LCR series circuit consisting of resistance 45 Ω, inductive reactance 4 Ω and capacitive reactance 4 is connected to an alternating source of emf 90 V. The readings of the ammeter connected to the circuit would be

2 A

The correct answer is Option (1) → 2 A

Given:

Resistance, $ R = 45\ \Omega $

Inductive reactance, $ X_L = 4\ \Omega $

Capacitive reactance, $ X_C = 4\ \Omega $

Supply voltage, $ V = 90\ \text{V} $

Net reactance:

$ X = X_L - X_C = 4 - 4 = 0 $

Impedance of the circuit:

$ Z = \sqrt{R^2 + X^2} = \sqrt{45^2 + 0^2} = 45\ \Omega $

Current in the circuit:

$ I = \frac{V}{Z} = \frac{90}{45} = 2\ \text{A} $

Therefore, the reading of the ammeter is $ 2\ \text{A} $.