Question

A 100 kVA, 2400 V/240 V, 50 Hz single phase transformer has an exciting current of 0.64 A and core loss 700 W when its high voltage side is energized at rated voltage and frequency. If load current is 40 A at 0.8 p.f. lagging on the LV side, then magnitude of the primary current will be

Answer 1

$I_c=\frac{P_{oc}}{V_{oc}}=\frac{700}{2400}=0.292A$

$I_{\varphi }=\sqrt{I_e^2-I_c^2}=0.57A$

$\therefore I_e=0.64\angle {27.12}^{\circ }A$

$I_1=I_e+I_2^1$

$=0.64\angle {27.12}^{\circ }+4\angle {53.13}^{\circ }$

$=4.58\angle {49.62}^{\circ }A$