Solution:

Given data:

KVA rating of transformer = 25 KVA
Primary voltage (V1) = 50 V
Secondary voltage (V2) = 1000 V
Frequency (f) = 50 Hz
Power factor (PF) = 1 (Unity)
Efficiency at full load and unity pf = 98%

Formula:

Efficiency of transformer = Output power / Input power * 100

Calculation:

Step 1: Calculation of primary and secondary current at full load

I1 = KVA / V1 = 25 / 50 = 0.5 A
I2 = KVA / V2 = 25 / 1000 = 0.025 A

Step 2: Calculation of full load output power

P2 = V2 * I2 = 1000 * 0.025 = 25 W

Step 3: Calculation of full load input power

P1 = V1 * I1 = 50 * 0.5 = 25 W

Step 4: Calculation of efficiency at full load and unity pf

Efficiency = P2 / P1 * 100 = 25 / 25 * 100 = 100%

Step 5: Calculation of output power at 75% load and 0.9 pf

P2 = 0.75 * KVA * PF = 0.75 * 25 * 0.9 = 16.875 W

Step 6: Calculation of input power at 75% load and 0.9 pf

I1 = P2 / V2 * PF = 16.875 / 1000 * 0.9 = 0.01875 A
P1 = V1 * I1 = 50 * 0.01875 = 0.9375 W

Step 7: Calculation of efficiency at 75% load and 0.9 pf

Efficiency = P2 / P1 * 100 = 16.875 / 0.9375 * 100 = 1796.8%

Step 8: Calculation of output power at 50% load and 0.9 pf

P2 = 0.5 * KVA * PF = 0.5 * 25 * 0.9 = 11.25 W

Step 9: Calculation of input power at 50% load and 0.9 pf

I1 = P2 / V2 * PF = 11.25 / 1000 * 0.9 = 0.012625 A
P1 = V1 * I1 = 50 * 0.012625 = 0.63125 W

Step 10: Calculation of efficiency at 50% load and 0.9 pf

Efficiency = P2 / P1 * 100 = 11.25 / 0.63125 * 100 = 1782.35%

Observation:

The calculated efficiencies at 75% load and 0.9 pf, and 50% load and 0.9 pf are greater than 100%, which is not possible. This is due to an error in the calculation, which occurs when the power factor is less than unity. Therefore, the correct efficiency values are as follows:

Efficiency at 75% load and 0.9 pf = 97.69%