As shown in the figure, series combination of ideal capacitor, pure resistor and a practical inductor are connected in series with an ac source 200 V, 60 Hz. Find power consumed in inductor.

Options

A. 320 W
B. 176 W
C. 144 W
D. 0 W

Answer

C

Explanation

`|Z|=sqrt((R_1+R_2)^2+(X_L-X_C)^2)`
`=sqrt((44+36)^2+(90-30)^2)`
`=sqrt(6400+3600)`
`=sqrt10000=100 Omega`
Current in the circuit `I=V/"|Z|"=200/100`=2A
Power consumed in inductor ,
`P=I^2R_2=(2)^2 xx36`
`therefore` P=144 W