Question

Suppose the circuit in Exercise 18 has a resistance of `15 Omega`. Obtain the average power transferred to each element of the circuit, and the total power absorbed.

Answer 1

Given, the rms value of voltage
`V_("rms")=230 V`
Resistance `R = 15 Omega`
Frequency `f = 50 Hz`
Average power across inductor and capacitor is zero as the phase difference between current and voltage is `90^(@)`.
Total power absorbed = power absorbed in resistor, `P_("av")`
`= V_("rms").I_("rms")`
So, `Z=sqrt(R^(2)+(X_(L)-X_(C ))^(2))`
`= sqrt((15)^(2)+(2xx3.14xx50xx80xx10^(-13)-(1)/(2xx3.14xx50xx60xx10^(-6))))`
`= sqrt(1002.85)=31.67 Omega`
`I_("rms")=(V_("rms"))/(Z)=(230)/(31.67)=7.26 A`
Total power absorbed `P_("av")`
`= V_("rms").I_("rms")`
`= I_("rms").R.I_("rms")`
`= (7.26)^(2)xx15=790.6W`
Total power absorbed = 790.6W