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
Answer & explanation
Correct answer: option 1
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} $.