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

40 Ω

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