3-phase 500 kVA transformer primary current

For an ideal transformer, losses and voltage drops are zero, so input VA equals output VA. The primary is delta connected, so the line current is $\sqrt{3}$ times the phase current.

Output at full load: $S = 500$ kVA, power factor 0.8 lagging.

Primary line current

$$S = \sqrt{3}\,V_L I_L \quad\Rightarrow\quad I_{L1} = \frac{500\times 10^3}{\sqrt{3}\times 11000} = 26.24\ \text{A}$$

Primary phase current (delta)

$$I_{ph1} = \frac{I_{L1}}{\sqrt{3}} = \frac{26.24}{\sqrt{3}} = 15.15\ \text{A}$$

(Check: $I_{ph1} = \dfrac{500000}{3\times 11000} = 15.15$ A.)

Power drawn

$$P = S\cos\phi = 500 \times 0.8 = 400\ \text{kW}, \qquad Q = 500\times 0.6 = 300\ \text{kVAR}$$

For an ideal transformer the input power factor equals the load power factor (0.8 lagging), so the current lags the voltage by $\cos^{-1}0.8 = 36.87^\circ$.

Answer: the transformer draws a line current of $26.24\ \text{A}$ from the 11 kV supply (phase current $15.15\ \text{A}$ in each delta winding) at 0.8 lagging power factor.