A one-phase, 50 Hz, 40 kVA transformer with a ratio of 2000 V/250 V has a primary resistance of 1.15 Ω and a secondary resistance of 0.0155 Ω. Calculate total copper loss on half of the full load. 

Concept:

Consider a two winding single phase transformer as shown below,

N₁ = primary winding turns

N₂ = secondary winding turns

V₁ = primary winding voltage

V₂ = secondary winding voltage

I₁ = current through the primary winding

I₂ = current through the secondary winding

Transformation ratio: It is defined as the ratio of the secondary voltage to the primary voltage. It is denoted by K.

  ----- (1)

Transformer equivalent circuit with respect to secondary can be represented a show below

Where R₀₂ = Effective resistance of the transformer as referred to the secondary side of the transformer.

R₀₂ = R₂ + R₁'    ------- (2)

R₁' = Equivalent primary resistance as referred to the secondary winding

R₁' = R₁ × K ²  ------ (3)

Similarly, effective resistance of the transformer as referred to the primary side of the transformer is given as,

R01 = R₁ + R₂' 

R₂' = Equivalent secondary resistance as referred to the primary winding

R₂' = R₂ / K 2

Calculation:

Given data

V₁ = 2000 V, V₂ = 250 V, R₁ = 1.15 Ω, R₂ = 0.0155 Ω

From equation(1)

K= V₂ / V₁

K = 250 / 2000

K = 1 / 8

Effective resistance of the transformer as referred to the secondary of the transformer.

From equations(2) & (3)

R₀₂ = 0.0155 + 1.15 / 8²

R02 = 0.0335 Ω

Power output P = 40 kVA

V₂ I₂ = 40 × 10³

I₂ = 40000 / 250

I₂ = 160 A = I_fl

I_fl is the full load current flowing through the secondary.

We required to find power loss at half full load condition

So, current at half full load is given as,

∴ Total power loss at half full load condition is given as,

P_hfl = I²_hfl × R₀₂

Phfl = 80² × 0.0335

Phfl = 214.2 W

Total power loss at half full load condition using R₀₁ will also give same value.

Here we have to calculate primary current I₁ using the formula,

Power output P = V₁ I₁

Then, half-full load current I_hfl = I₁ / 2

Power loss at half full load P_hfl = I²_hfl × R₀₁  W