25 kVA transformer: load voltage and primary current

Refer all impedance to the secondary, find the voltage drop at half load, and subtract it from the no-load secondary voltage.

Assumptions: the primary is held at 11 kV, the magnetising current is neglected, and the load power factor is unity (not stated in the problem).

Step 1: Turns ratio and equivalent impedance

$$a = \frac{11000}{400} = 27.5, \qquad a^2 = 756.25$$ $$\begin{aligned} R_{02} &= R_2 + \frac{R_1}{a^2} = 0.02 + \frac{0.4}{756.25} = 0.02053\ \Omega \\ X_{02} &= X_2 + \frac{X_1}{a^2} = 1 + \frac{2}{756.25} = 1.00264\ \Omega \end{aligned}$$

Step 2: Half-load current

$$I_2 = \frac{0.5 \times 25000}{400} = 31.25\ \text{A}$$

Step 3: Load terminal voltage

No-load secondary voltage $E_2 = 400$ V. At unity power factor the phasor relation is $E_2^2 = (V_2 + I_2R_{02})^2 + (I_2X_{02})^2$. So

$$\begin{aligned} V_2 &= \sqrt{E_2^2 - (I_2X_{02})^2} - I_2R_{02} \\ &= \sqrt{400^2 - (31.33)^2} - 0.642 \\ &= 398.13\ \text{V} \end{aligned}$$

Step 4: Primary current

$$I_1 = \frac{I_2}{a} = \frac{31.25}{27.5} = 1.136\ \text{A}$$

Answer: load terminal voltage $\approx 398.13\ \text{V}$ (drop of about 1.87 V) and primary current $\approx 1.136\ \text{A}$ at half load.

(If the load power factor were 0.8 lagging instead, the same method gives $V_2 \approx 379.9$ V; the primary current is still about 1.136 A.)