Correct Answer - Option 1 : 0.92 lagging
In two wattmeter method,
\(\begin{array}{l}
\tan \phi = \frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}\\
\Rightarrow \tan \phi = \frac{{\sqrt 3 \left( {\frac{{{W_1}}}{{{W_2}}} - 1} \right)}}{{\left( {\frac{{{W_1}}}{{{W_2}}} + 1} \right)}}\\
\Rightarrow \tan \phi = \frac{{\sqrt 3 \left( {\frac{5}{3} - 1} \right)}}{{\left( {\frac{5}{3} + 1} \right)}} = \frac{{\sqrt 3 }}{4}\\
\Rightarrow \phi = 23.41^\circ
\end{array}\)
Power factor, \(\cos \phi = 0.917\)
Given that the load is inductive and hence power factor is lagging.
