If P₁ and P₂, be the iron and copper losses of a transformer on full-load, what is the ratio of P₁ and P₂ such that maximum efficiency occurs at 75% full-load ? 

Copper loss:

Copper loss = i2R

Iron loss:

Iron losses = Eddy current loss + hysteresis loss

Eddy current loss = Ket2f2B2

Hysteresis loss = KhfBx

These are constant losses.

Efficiency:

The efficiency of the Transformer \((\eta ) = \frac{{XS\cos \phi }}{{XS\cos \phi + {P_i} + {X^2}{P_e}}}\)

Where, X = Fraction of load

S = Apparent power in kVA

Wi = Iron losses

Wcu = Copper losses

The transformer will give the maximum efficiency when their copper loss is equal to the iron loss.

Efficiency is maximum at some fraction x of full load.

\(x = \sqrt {\frac{{{W_i}}}{{{W_{cu}}}}}\)

Where Wi = iron losses

Wcu = copper losses

kVA at maximum efficiency is given by,

\(kVA\;at\;{\eta _{max}} = full\;load\;kVA \times \sqrt {\frac{{{W_i}}}{{{W_{cu}}}}}\)

Calculation:

Given that,

P₁ = iron losses

P₂ = copper losses

x = fraction of load at maximum efficiency = .75,

\(x = \sqrt {\frac{{{P_1}}}{{{P_{2}}}}}\)

(.75)² = P₁/P₂

⇒P1/P₂ = 0.5625