An AC circuit having an inductor and a resistor in series draws a power of 560 W from an AC source marked 210 V-60 Hz. The power factor of the circuit is 0.8, the impedance of the circuit and the inductance of the inductor is

The average power over a complete cycle is P=Erms×Irms×cosϕ ∴  Irms=PErms×cosϕ=560210×0.8=103 A  The impedance of the circuit is  Z=ErmsIrms=210 V(10/3)A=63Ω Power factor=cosϕ=RZ ⇒R=0.8×63=50.4Ω  Now, the impedance of an L-R circuit is  Z=R2+(ωL)2 ∴(ωL)2=Z2-R2=(63)2-(50.4)2=1428.84Ω2 ⇒ωL=1428.84=37.8Ω ∴  L=37.82πf=37.82×3.14×60=0.1H