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Question
Suppose the circuit has a resistance of 15 Ω. Obtain the average power transferred to each element of the circuit, and the total power absorbed.
Solution
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, v = 50 Hz
Angular frequency of signal, ω = 2πv = 2π × (50) = 100π rad/s
The elements are connected in series to each other. Hence, the impedance of the circuit is given as:
Z = `sqrt("R"^2 + (ω"L" - 1/(ω"C"))^2)`
= `sqrt((15)^2 + [100π (80 xx 10^-3) - 1/((100π xx 60 xx 10^-6))]^2`
= `sqrt((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:
PR = I2R
= (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:
= PR + 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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