In an a. c. series LCR circuit, the applied voltage is $V = (100\sqrt{2})\sin (ωt + π/4)$ volt. Given that, $R = 30 Ω, X_L = 50 Ω$ and $X_C= 10$. The value of power dissipated in the circuit is

120 W

The correct answer is Option (4) → 120 W

Given:

Voltage: $V = 100\sqrt{2} \sin(\omega t + \pi/4)$ V

Resistance: $R = 30 \ \Omega$

Inductive reactance: $X_L = 50 \ \Omega$

Capacitive reactance: $X_C = 10 \ \Omega$

Impedance: $Z = \sqrt{R^2 + (X_L - X_C)^2}$

$Z = \sqrt{30^2 + (50 - 10)^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \ \Omega$

RMS voltage: $V_\text{rms} = \frac{100\sqrt{2}}{\sqrt{2}} = 100$ V

Power dissipated: $P = \frac{V_\text{rms}^2 R}{Z^2}$

$P = \frac{100^2 \times 30}{50^2} = \frac{10000 \times 30}{2500} = 120$ W

Answer: $P = 120$ W