A series LCR circuit is connected to a 230 V, 50 rad/s source. Given L = 5 H, C = 80 μF and R=40 Ω. Impedance of the circuit is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 40 Ω
Impedance of a series LCR circuit:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
Inductive reactance:
$X_L = \omega L = 50 \times 5 = 250 \, \Omega$
Capacitive reactance:
$X_C = \frac{1}{\omega C} = \frac{1}{50 \times 80 \times 10^{-6}} = \frac{1}{0.004} = 250 \, \Omega$
Net reactance:
$X_L - X_C = 250 - 250 = 0$
Therefore impedance:
$Z = \sqrt{R^2 + 0^2} = R = 40 \, \Omega$
Impedance = 40 Ω