An inductor of reactance 100 , a capacitor of reactance 50 , and a resistor of resistance 50 are connected in series with an AC source of 10 V, 50 Hz. Average power dissipated by the circuit is W.
Answer
1
Solution
Impedance: Z = R2 + (XL - XC)2 = 502 + (100 - 50)2 = 50 2
RMS current (Vrms = 10 V): Irms = VrmsZ = 1050 2 = 15 2 A
Average power: P = Irms2 R
P = 150 50 = 1 W
Answer: 1