Suppose the circuit in Exercise 7.18 has a resistance of 15 Ω. Obtain the average power transferred to each element of the circuit, and the total power absorbed.

Average power transferred to the resistor = 788.44 W
Average power transferred to the capacitor = 0 W
Total power absorbed by the circuit = 788.44 W
Inductance of inductor, L = 80 mH = 80 x 10^-3 H
Capacitance of capacitor, C = 60 μF = 60 x 10^-6 F
Resistance of resistor, R = 15 Ω
Potential of voltage supply, V = 230 V
Frequency of signal, ν = 50 Hz
Angular frequency of signal, ω = 2 πν= 2π x (50) = 100π rad/s
The elements are connected in series to each other. Hence, impedance of the circuit is given as:
Z = √[(R)² +(ωL -1/ωC)²]

=√[15² +((100π x 80 x 10^-3 )-1/(100πx 60 x 10^-6 ))²]

= √[(15)² +(25.12 -53.08)²] = 31.728 Ω

Current flowing in the circuit, I = V/Z = 230/31.728 = 7.25 A

Average power transferred to resistance is given as:

P_R = I²R = (7.25)² x 15 = 788.44 W

Average power transferred to capacitor, P_C = Average power transferred to inductor, P_L= 0

Total power absorbed by the circuit: = P_R + P_C + P_L = 788.44 + 0 + 0 = 788.44 W

Hence, the total power absorbed by the circuit is 788.44 W.