NEET-XII-Physics
07: Alternating Current
- #19Suppose the circuit in Exercise 7.18 has a resistance of 15 ``\Omega``. Obtain the average power transferred to each element of the circuit, and the total power absorbed.
Ans : 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 ``\Omega``
Potential of voltage supply, V = 230 V
Frequency of signal, ν = 50 Hz
Angular frequency of signal, ω = 2``\pi``ν= 2``\pi`` × (50) = 100``\pi`` 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:
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.