Question -
Answer -
Average power transferred to theresistor = 788.44 W
Average power transferred to thecapacitor = 0 W
Total power absorbed by thecircuit = 788.44 W
Inductance of inductor, L =80 mH = 80 × 10−3 H
Capacitance of capacitor, C =60 μF = 60 × 10−6 F
Resistance of resistor, R =15 Ω
Potential of voltagesupply, V = 230 V
Frequency of signal, ν =50 Hz
Angular frequency ofsignal, ω = 2πν= 2π × (50) = 100π rad/s
The elements are connected inseries to each other. Hence, impedance of the circuit is given as:
Current flowing in the circuit, 
Average power transferred toresistance is given as:
PR= I2R
= (7.25)2 × 15 =788.44 W
Average power transferred tocapacitor, PC = Average power transferred toinductor, PL = 0
Total power absorbed by thecircuit:
= PR + PC +PL
= 788.44 + 0 + 0 = 788.44 W
Hence, the total power absorbedby the circuit is 788.44 W.