A resistor R=300 and a capacitor C=25 F are connected in series with a 50 ~V , 50 Hz AC source. The average power dissipated in the circuit is

Step-by-step solution

aligned & Given, R=300 , C=25 F =25 10⁻⁶ ~F & V=50 ~V aligned Frequency of AC source, v= 50 Hz array cc & 2 = & =2 50 =100 rad / s array AC voltage, V=50 t =50 (100 t) rms voltage, V_ rms = V₀ 2 = 50 2 =25 2 ~V Impedance of the R C -circuit is aligned Z & = R^2+X_C^2 = R^2+ ( 1 C )^2 Z & = 300^2+ ( 1 100 25 10⁻⁶ )^2 & = 300^2+400^2 Z & =500 aligned Current through the circuit, I_ rms = V_ rms Z = 25 2 500 = 2 20 ~A Average power dissipated through the circuit is aligned & =V_ rms I_ rms & =25 2 2 20 R Z & = 50 20 3