An inductor of 50 mH, a capacitor of $50 \mu\mathrm{F}$, and a resistor of $40 \Omega$ are connected

Question

An inductor of 50 mH, a capacitor of $50 \mu\mathrm{F}$, and a resistor of $40 \Omega$ are connected in series across an AC source of emf $V = 100 \sin(1000t)$ Volts. The average power loss in the circuit is

Professor PC Thomas Classes · Answer

(c) : From the voltage equation, $\omega = 1000$ rad/s and peak voltage $V_0 = 100$ V. Inductive reactance $X_L = \omega L = 1000 	imes 50 	imes 10^{-3} = 50 \Omega$. Capacitive reactance $X_C = 1/(\omega C) = 1/(1000 	imes 50 	imes 10^{-6}) = 20 \Omega$. The impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + (50 - 20)^2} = 50 \Omega$. The rms current is $I_{rms} = V_{rms} / Z = (100/\sqrt{2}) / 50 = \sqrt{2}$ A. The average power dissipated is strictly across the resistor: $P = I_{rms}^2 R = (\sqrt{2})^2 	imes 40 = 2 	imes 40 = 80$ W.