As given that, inductance $\mathrm{L}=0.01 \mathrm{H}$
resistance $\mathrm{R}=1 \Omega$,
voltage $(\mathrm{V})=200 \mathrm{~V}$
and frequency $(\mathrm{f})=50 \mathrm{~Hz}$
Impedance of the circuit
$$
\begin{aligned}
\mathrm{Z} &=\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2} \\
&=\sqrt{\mathrm{R}^2+(2 \pi \mathrm{fL})^2} \\
&=\sqrt{1^2+(2 \times 3.14 \times 50 \times 0.01)^2} \\
\text { or } & Z=\sqrt{10.86}=3.3 \Omega
\end{aligned}
$$


For phase angle $(\phi)$
$$
\begin{gathered}
\tan \phi=\frac{Z}{R}=\frac{X L}{R}=\frac{\omega \mathrm{L}}{\mathrm{R}}=\frac{2 \pi \mathrm{fL}}{\mathrm{R}} \\
=\frac{2 \times 3.14 \times 50 \times 0.01}{1}=3.14 \\
\phi=\tan ^{-1}(3.14) \approx 72^{\circ} \\
\text { Phase difference } \phi=\frac{72 \times \pi}{180} \text { radian } \\
=1.20 \text { radian. }
\end{gathered}
$$
Time lag between alternating voltage and current, $\phi=\omega t$
So, $\Delta \mathrm{t}=\frac{\phi}{\omega}=\frac{72 \pi}{180 \times 2 \pi \times 50}=\frac{1}{250} \mathrm{sec}$.