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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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Class 12 Physics
CBSE and UP Board
Alternating Current
Chapter-7 Exercise 7.19
Additional Exercise

NCERT Solutions for Class 12 Physics Chapter 7 Question 19

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1 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 x 10-3 H
    Capacitance of capacitor, C = 60 μF = 60 x 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 = 2 πν=  x (50) = 100π rad/s
    The elements are connected in series to each other. Hence, impedance of the circuit is given as:
    Z = √[(R)² +(ωL -1/ωC)²]

    =√[15² +((100π x 80 x 10-3 )-1/(100πx 60 x 10-6 ))²]

    = √[(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 x 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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