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.

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, 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:

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.