Given, inductive reactance, \( \omega L=\frac{1}{\sqrt{3}} \Omega \); resistance, \( R=1 \Omega \); ac voltage, \( V=200 \mathrm{~V}, 50 \mathrm{~Hz} \);
Now, \( \tan \phi=\frac{\omega L}{R} \)
\[
\begin{array}{l}
\Rightarrow \tan \phi=\frac{1}{\sqrt{3}} \\
\Rightarrow \phi=30^{\circ} \text { or } \phi=\frac{\Pi}{6} \\
\Rightarrow \omega t=\frac{\Pi}{6} \\
\Rightarrow t=\frac{\Pi}{6} \times \frac{1}{\omega}
\end{array}
\]
Since \( \omega=2 \Pi f=2 \Pi \times 50 \)
\( t=\frac{\Pi}{6} \times \frac{1}{2 \Pi \times 50}=\frac{1}{600} s \)
Therefore, time lag between maximum voltage and current is \( 1 / 600 \mathrm{~s} \).