230/2300 V transformer with load impedance

The approximate equivalent circuit referred to the primary is used: the shunt branch (R₀ ∥ jX₀) is placed across the supply, and the series branch carries the load current.

Equivalent circuit (referred to primary)

      I1       Re=0.1   Xe=j0.4    I2'
 o----->-------[===]----[~~~]--+---->----+
 |                             |         |
 V1=230V   I0 --->   +------+  |       Zl'=4+j6
 |                   |      |  |         |
 |                  Ro=500  jXo=200      |
 |                   |      |  |         |
 o-------------------+------+--+---------+

Turns ratio a = 2300/230 = 10.

Step 1: refer load to primary

Zl' = Zl / a² = (400 + j600) / 100 = 4 + j6 Ω

Step 2: load current referred to primary

Z_total = (0.1 + j0.4) + (4 + j6) = 4.1 + j6.4 Ω, |Z| = 7.601 Ω

I₂' = 230 / (4.1 + j6.4) = 30.26 ∠−57.36° A = (16.32 − j25.48) A

Step 3: no-load (exciting) current

• Core-loss component I_c = 230 / 500 = 0.46 A
• Magnetising component I_m = 230 / 200 = 1.15 A (lagging)
• I₀ = 0.46 − j1.15 = 1.239 A

(i) Primary current and input power factor

I₁ = I₂' + I₀ = (16.32 + 0.46) − j(25.48 + 1.15) = 16.78 − j26.63 A

|I₁| = 31.48 A at −57.78°, so pf = cos 57.78° = 0.533 lagging

(ii) Secondary terminal voltage

Voltage drop in the series impedance = I₂' × (0.1 + j0.4) = 30.26 × 0.412 = 12.47 V

V₂' = 230 − I₂'(0.1 + j0.4) = 218.2 V (check: I₂' × |Zl'| = 30.26 × 7.211 = 218.2 V)

V₂ = a × V₂' = 10 × 218.2 = 2182 V

Answer: I₁ = 31.5 A, input pf = 0.533 lagging, secondary terminal voltage = 2182 V.