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

Average power transferred to the resistor=7 88.44W
Average power transferred to the capacitor=0W
Total power absorbed by the circuit=788.44W
Inductance of inductor, L=80mH =`80xx 10^(-3)H`
Capacitnace of capacitor, `C=60muF=60xx10^(-6)F`
Resistance of resistor , R=`15 Omega`
Potential of voltage supply, V=230V
Frequency of signal, v=50Hz
Angular frequency of signal `omega=2nv=2nxx(50)=100n` rad/s
The elements are connected in serie sto each other, Hence, impedance of the circuit is
`Z=sqrt(R^(2)+(omegaL-(1)/(omegaC))^(2))`
`=sqrt((15)^(2)+(100pi(80xx10^(-3))-(1)/((100pixx60xx10^(-6)))^(2))`
`sqrt((15)^(2)+(25.12-53.08)^(2))=31.728Omega`
`I=(V)/(Z)=(230)/(31.728)=7.25A`
Current folowing in the circuit,
Average power transferred to resistance is given as:
`P_(R)=I^(2)R`
`=(7.25)^(2)xx15=788.44W`
Average power transferred to capacitor, `P_(c)=` Average powertransferred to inductor, `P_(L)` =0
Total power absorbed by the circuit: Total power absorbed by the circuit
`=P_(R)+P_(C)+P_(L)`
`=788.44+0+0=788.44W`
Hence, the total power absorbed by the circuit is 7 88.44W.