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
Answer & explanation
Correct answer: option 4
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