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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.

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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...
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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:

PRI2R

= (7.25)2 × 15 = 788.44 W

Average power transferred to capacitor, PC = Average power transferred to inductor, PL = 0

Total power absorbed by the circuit:

P+ PC + PL

= 788.44 + 0 + 0 = 788.44 W

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

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