Suppose 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.

Explanation

Impedance , Z = `sqrt(R^(2) + (omega L - (1)/(omega C))^(2))`
` = sqrt( 15^(2) + (2pi xx 50 xx 80 xx 10^(-3) - (1)/(2pi xx 50 xx 60 xx 10^(-6)))^(2) )`
= 31.69 `Omega`
` I_(rms) = (E_(rms))/(Z) = (230)/(31.69) 7.26` A
Average power transferred to inductor = Average
power transferred to capacitor = 0
Average power transferred to resistor = `I_(rms)^(2)` R
` = (7.87)^(2) xx 15 = 791 ` W
TOtal power absorbed = 791 W