$A$ circuit containing an $80 \; mH$ inductor,a resistance of $15 \; \Omega$,and a $60 \; \mu F$ capacitor in series is connected to a $230 \; V, 50 \; Hz$ supply. Obtain the average power transferred to each element of the circuit,and the total power absorbed.

(N/A) Given: Inductance $L = 80 \; mH = 80 \times 10^{-3} \; H$,Capacitance $C = 60 \; \mu F = 60 \times 10^{-6} \; F$,Resistance $R = 15 \; \Omega$,Voltage $V = 230 \; V$,Frequency $f = 50 \; Hz$.
Angular frequency $\omega = 2 \pi f = 2 \pi \times 50 = 100 \pi \; rad/s \approx 314.16 \; rad/s$.
Inductive reactance $X_L = \omega L = 100 \pi \times 80 \times 10^{-3} \approx 25.13 \; \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 \pi \times 60 \times 10^{-6}} \approx 53.05 \; \Omega$.
Impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{15^2 + (25.13 - 53.05)^2} = \sqrt{225 + (-27.92)^2} = \sqrt{225 + 779.53} = \sqrt{1004.53} \approx 31.69 \; \Omega$.
Current $I = \frac{V}{Z} = \frac{230}{31.69} \approx 7.26 \; A$.
Average power transferred to the inductor $P_L = 0 \; W$ (since phase difference is $90^\circ$).
Average power transferred to the capacitor $P_C = 0 \; W$ (since phase difference is $90^\circ$).
Average power transferred to the resistor $P_R = I^2 R = (7.26)^2 \times 15 \approx 790.5 \; W$.
Total power absorbed by the circuit $P_{total} = P_R + P_L + P_C = 790.5 + 0 + 0 = 790.5 \; W$.