Question 3.1: A transformer on no-load has a core-loss of 50 W,draws a cur...

Power factor,
\cos \theta _{0} =\frac{50}{2\times 230}=0.108 lagging;
\theta _{0} =83.76°
Magnetizing current,
I_{m} =I_{0} \sin\theta _{0}=2\sin \left(\cos ^{-1} 0.108 \right) =1.988 A
Since \theta _{0}\approx 90°, there is hardly any difference between the magnitudes of the exciting current and its magnetizing component.
Core-loss current, I_{i} = I_{0}\cos \theta _{0}=2\times 0.108=0.216 A
In the no-load circuit model of Fig.3.7 core loss is given by
G_{i} V_{1} ^{2} =P_{i}
or G_{i}=\frac{P_{i}}{V_{1} ^{2}} =\frac{50}{\left(230\right) ^{2} } =0.945\times 10^{-3} \mho
Also    I_{m} =B_{m}V_{1}
or      B_{m}=\frac{I_{m} }{V_{1}} =\frac{1.988}{230} =8.64\times 10^{-3} \mho