Given Data:

Supply voltage, $V=200\ Volts$

Supply frequency, $f=50Hz$

Resistance, $R=10\ \Omega$

Inductance, $L=50mH = 40\times 10^{-3}\ Henry$

Let's consider a coil connected to AC supply as shown in figure below-

(i) Impedance of  Coil:-

As the Impedance of a coils for AC supply is given by,$Z=R+j(X_L)=R+j\omega L=R+j2\pi f L$

$\therefore\ Impedance\ of \ circuit\ Z=R+j2\pi f L=10+j2\pi\times 50\times 40\times 10^{-3}=10+j12.56=16.05\angle51.56^0\Omega$

(ii) Current through coil:-

As the current through a circuit for AC supply is given by. $I=\dfrac{V}{Z}$

$\therefore I=\dfrac{V}{Z}=\dfrac{200}{16.05\angle 51.56^0}=12.46\angle-51.56^0 \ A$

(iii) Power factor of circuit:-

As we know that the power factor of the circuit is given by, $\cos\phi$

$\therefore\ power \ factor\ of\ circuit, \ \cos\phi=\cos51.56^0=0.62 \ lagging$

(iv) Power consumed by the circuit:-

As for AC supply power consumed by the circuit is given by, $P=VI\cos\phi$

$\therefore\ P=VI\cos\phi=200\times 12.46\times 0.62=1545.04\ Watts$