Question

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.

Answer 1

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 × 10⁻³ H

Capacitance of capacitor, C = 60 μF = 60 × 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 = 2πν= 2π × (50) = 100π rad/s

The elements are connected in series to each other. Hence, impedance of the circuit is given as:

Current flowing in the circuit,

Average power transferred to resistance is given as:

P_R= I²R

= (7.25)² × 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.