Correct Answer - A
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`.