Question C.2: The transformer of Example C.1 gave the following results on......

The active component of the current I_{e} = 3.33/13.8 = 0.241 A per phase. Therefore,
g_{m} = \frac{ 10 \ × \ 10^3}{3 \ × \ (13.8 \ × \ 10^3)^2} = 0.017 × 10^{-3} mhos
The magnetizing current is
I_{m} = \sqrt{I_{0}^{2} – I^{2}_{e}} = \sqrt{1.44^2 – 0.241^2} = 1.42 A
The power factor angle of the no-load current is 9.63°, and b_{m} from Equation C.11

b_{m}= \frac{Q_{0}}{V^{2}_{1}}= \sqrt{\frac{S^{2}_{0} \ – \ P^{2}_{0}}{V_{1}^{2}}}     (C.11)

is -0.103 × 10^{-3} mhos per phase.