Suppose the circuit in Exercise 7.18 has a resistance of 15 Ω.

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, ν = 50 Hz

Angular frequency of signal, ω = 2πν= 2π × (50) = 100π rad/s

The elements are connected in series to each other. Hence, impedance of the circuit is given as:

$Z=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$

$=\sqrt{(15)^{2}+\left(100 \pi\left(80 \times 10^{-3}\right)-\frac{1}{\left(100 \pi \times 60 \times 10^{-6}\right)}\right)^{2}}$

$=\sqrt{(15)^{2}+(25.12-53.08)^{2}}=31.728 \Omega$

Current flowing in the circuit, $I=\frac{V}{Z}=\frac{230}{31.728}=7.25 \mathrm{~A}$

Average power transferred to resistance is given as:

$P_{R}=I^{2} R$

= (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.