Ncert Exercises & Solutions

7.19 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.

Here,

Inductance, 𝐿 = 80 mH

Capacitance, 𝐶 = 60 μF

rms voltage 𝑉 = 230 V

Frequency, 𝑓 = 50 Hz

$ω = 2 π f = 2 π \times 50 H z = 100 π r a d s^{- 1}$

Resistance, 𝑅𝑅 = 15 Ω

The impedance of the circuit –

$\begin{array}{l} Z = \sqrt{(X_{L} - X_{C})^{2} + R^{2}} \\ = \sqrt{(100 π \times 80 \times 10^{- 3} - \frac{1}{100 π \times 60 \times 10^{- 6})^{2}} + 15^{2} = 31.728 Ω} \end{array}$

Current following in the circuit-

$I_{r m s} = \frac{V}{Z} = \frac{230 V}{31.72 Ω} = 7.25 A$

The average power transferred to the resistor,

$P_{R} = I_{r m s}^{2} R = {7.25}^{2} \times 15 = 791 W$

The average power transferred to the resistor and capacitor, $P_{C} = P_{I} = 0$

As the average power transferred to the inductor and the capacitor is zero in one
complete cycle. Hence, the total transferred is 791 W.