25 kVA ideal transformer currents, flux

Given: $S=25$ kVA, $N_1=500$, $N_2=40$, $V_1=3000$ V, $f=50$ Hz. The transformer is ideal.

Turns ratio: $K=\frac{N_2}{N_1}=\frac{40}{500}=0.08$.

(i) Full-load currents.

$$I_1=\frac{S}{V_1}=\frac{25000}{3000}=8.33\ \text{A}$$ $$I_2=\frac{I_1}{K}=\frac{8.33}{0.08}=104.17\ \text{A}$$

(ii) Secondary emf.

$$E_2=V_1\times\frac{N_2}{N_1}=3000\times0.08=240\ \text{V}$$

(Check: $25000/240=104.17$ A.)

(iii) Maximum core flux. From the emf equation $E_1=4.44fN_1\Phi_m$ (for an ideal transformer, $E_1=V_1$):

$$\Phi_m=\frac{3000}{4.44\times50\times500}=0.02703\ \text{Wb}=27.03\ \text{mWb}$$

Answer: (i) $I_1=8.33$ A, $I_2=104.17$ A; (ii) $E_2=240$ V; (iii) $\Phi_m=27.03$ mWb.