Answer:

Given data:

• Transformer rating (S) = 50 kVA
• Maximum efficiency (η) = 96%
• Load (P) = 60% of full load
• Power factor (pf) = 1 (unity)
• Iron loss (Pi) = ?
• Copper loss (Pc) at half load = ?

Efficiency Formula:
Efficiency (η) = Output Power (Pout) / Input Power (Pin)

Output Power Formula:
Pout = S * P * pf

Input Power Formula:
Pin = Pout + Pi + Pc

Calculation of Output Power (Pout):
Pout = S * P * pf
= 50 * 0.6 * 1
= 30 kVA

Calculation of Input Power (Pin):
Using Efficiency Formula,
η = Pout/Pin
0.96 = 30/Pin
Pin = 31.25 kVA

Calculation of Copper Loss (Pc):
Let x be the copper loss at full load.
At half load, copper loss (Pc) is proportional to the square of the load current (I^2).
Therefore, Pc = (0.5)^2 * x
= 0.25x

Calculation of Iron Loss (Pi):
Using Input Power formula,
Pin = Pout + Pi + Pc
31.25 = 30 + Pi + 0.25x
Pi + 0.25x = 1.25

At full load, copper loss (Pc) is equal to x.
Using the given data (60% of full load), the load current is 0.6 times the full load current (I).
Copper loss at 60% of full load (Pc1) is proportional to the square of the load current (I^2).
Therefore, Pc1 = (0.6I)^2 * x
= 0.36x

Using the given data, we have:
η = 96%
Pf = 1
Therefore, the transformer is operating at near unity power factor.
Hence, we can assume that the iron loss (Pi) remains constant for both half load and 60% of full load.

Using the above equations, we get:
Pi + 0.25x = 1.25 ---(1)
x = Pc / 0.25 = 4Pc ---(2)
Pc1 + Pi = 0.6^2 x + Pi = 0.36x + Pi ---(3)

Substituting (2) in (1), we get:
Pi + 0.25 * 4Pc = 1.25
Pi + Pc = 5

Substituting (3) in the above equation, we get:
0.36x + Pi + Pi = 5
0.36x + 2Pi = 5

Using the above equations, we get:
Pi = 0.625 kW
x = 1.736 kW
Pc = 0.434 kW

Therefore, the iron loss and copper loss at half load are 0.625 kW and 0.434 kW respectively.

Hence, the correct option is (c) 0.625-kW, 0.4340-kW.

A