Given Data:

Power rating of transformer, S = 50 KVA

Primary voltage, $V_1=2200\ Volts$

Secondary voltage, $V_2=440\ Volts$

Supply Frequency, $f= 50\ Hz$

No. of turns in primary winding, $N_1= 200$

Lets, consider a Two winding transformer shown in figure below-

(1) flux in core:

Considering Ideal transformer, $V_1=E_1=4.44\phi_m fN_1....(1) $

Where, $\phi_m \ is \ the \ flux\ in\ core$

Hence, flux in core, $\phi_m = \dfrac{V_1}{4.44fN_1}=\dfrac{2200}{4.44\times 50\times 200}=0.05 \ Web$

(2) Secondary turns

As we know that the transformation ratio of transformer is given by, $k=\dfrac{V_2}{V_1}=\dfrac{N_2}{N_1}.....(2)$

Where, $N_2 \ is \ the\ no.\ of \ turns\ in\ secondary$

Hence, no. of turns in secondary, $N_2=\dfrac{V_2}{V_1}\times N_1=\dfrac{440}{2200}\times200=40\ turns$

3) Rated primary current :

We know that for an ideal transformer, apparent power, $S=V_1I_1=V_2I_2....(3)$

Hence rated primary current of transformer, $I_1=\dfrac{S}{V_1}=\dfrac{50\times 10^3}{2200}=22.73A$

4) Rated secondary current:

Rated secondary current of transformer, $I_2=\dfrac{S}{V_2}=\dfrac{50\times 10^3}{440}=113.64A$