When `100` volt `DC` source is applied across a coil, a current of `1 A` flows through it. When `100 V AC` source of `50 Hz` is applied to the same coil, only` 0.5 A` current flows. Calculate the inductance of the coil

Options

A. 17.28 W
B. 8.64 W
C. 10 W
D. 15 W

Answer

A

Explanation

Now when the capacitor is connected to the above circuit in series,
`X_(C)=1/(omega C) = (1)/(50 xx 2500 xx 10^(-6)) =(10^3)/(125) = 8 Omega`
so, `Z=sqrt(R^(2)+(X_(L)-X_(C ))^(2)) = sqrt(3^(2)+(4-8)^(2))= 5 Omega`
and hence, `I_(rms) = V/Z 12/5 = 2.4 A`
so, `P_(av) = V_(rms) I_(rms) cos phi = (I_(rms)xxZ) xx (I_(rms))xx((R )/(Z))`
i.e., `P(av)=I_(rms)^(2)R=(2.4)^(2) xx 3 = 17.28 W`.