Suppose the circuit in Exercise 18 has a resistance of 15Ω. Obtain the average power transferred to each element of the circuit, and the total power absorbed.
Given: Average power transferred to the resistor = 788.44 W
Average power transferred to the capacitor = 0 W
Power absorbed by the circuit = 788.44 W
Inductance L = 80 mH
In henry, Inductance L = 80 × 10-3H
Capacitance C = 60 μF = 60 × 10-6F
Resistance R = 15 Ω
Voltage V = 230 V
Frequency v = 50 Hz
Angular frequency is calculated as follows:
ω = 2πv
ω = 2 × 3.14 × 50(Hz) = 100π rad/s
The impedance can be calculated as follows:
On calculating, we get
Z = 31.728 Ω
Current can be calculated as follows: I = V/Z
I = 230 (A)/31.728 (Ω) = 7.25 A
The average power transferred to resistor can be calculated as follows:
Pr = I2R
Substituting values, we get
Pr = (7.25)2(A) × 15(Ω)
Pr = 788.44 W
The average power transferred to capacitor is equal to the average power transferred to resistor = 0
Total power absorbed = PR + Pc + PL = 788.4W