25 kVA transformer load voltage, half load

Given: 25 kVA, 11 kV/400 V. Primary impedance Z1 = 0.4 + j2 Ω; secondary Z2 = 0.02 + j1 Ω. Half load.

Assumptions (not stated in the question): the primary is supplied with the rated 11 kV; the load power factor is 0.8 lagging; the no-load current is neglected (not given).

Step 1: refer to the secondary

K = 400/11 000 = 0.03636, K² = 1.322×10⁻³.

R02 = R2 + K² R1 = 0.02 + 1.322×10⁻³ × 0.4 = 0.02053 Ω
X02 = X2 + K² X1 = 1.0 + 1.322×10⁻³ × 2 = 1.00264 Ω
Z02 = 0.02053 + j1.00264 Ω

Equivalent circuit (secondary side):

 V1' = 400 V ──[0.0205 Ω]──[j1.0026 Ω]──┬── load V2
                                        └── I2

Step 2: load current

Half-load: I2 = 0.5 × 25 000/400 = 31.25 A, at 0.8 lag (I2 = 31.25∠−36.87° A, with V2 as the reference).

Step 3: voltage drop and V2

I2 Z02 = 31.25∠−36.87° × (0.02053 + j1.00264)
       = 19.31 + j24.68 V
V1' = V2 + I2 Z02 = 400 (magnitude)
(V2 + 19.31)² + 24.68² = 400²
V2 + 19.31 = √(160 000 − 609) = 399.24
V2 = 379.9 V

(The approximate formula, drop ≈ I2(R02 cos φ + X02 sin φ) = 31.25 × 0.618 = 19.3 V, gives V2 ≈ 380.7 V. The exact result is 379.9 V.)

Step 4: primary current

I1 = K × I2 = 0.03636 × 31.25 = 1.136 A

(For a unity pf load the terminal voltage would be about 398.1 V.)

Answer: load terminal voltage ≈ 380 V (secondary, at 0.8 pf lag); primary current ≈ 1.14 A.