A 10 kVA, 400 V/ 200 V, 1 -phase transformer with a percentage resistance of 3% and percentage reactance of 6% is supplying acurrent of 50 A to a resistive load. The value of the load voltage isa)390 Vb)194 Vc)196 Vd)192 VCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

Given data:
Transformer rating, S = 10 kVA
Primary voltage, V1 = 400 V
Secondary voltage, V2 = 200 V
Percentage resistance, R% = 3%
Percentage reactance, X% = 6%
Load current, I2 = 50 A

To find:
Load voltage, V2

Solution:
1. Calculation of impedance:
Let Z2 be the complex impedance of the load.
The equivalent circuit of the transformer and the load is shown below:

The percentage impedance of the transformer, Z%, is given by:
Z% = (V1 / V2) * 100 / S
Substituting the given values, we get:
Z% = (400 / 200) * 100 / 10,000
Z% = 2%

The complex impedance of the transformer, ZT, is given by:
ZT = (R% + jX%) * Z%
Substituting the given values, we get:
ZT = (0.03 + j0.06) * 2
ZT = 0.06 + j0.12

The complex impedance of the load, Z2, is given by:
Z2 = V2 / I2
Substituting the given values, we get:
Z2 = 200 / 50
Z2 = 4

2. Calculation of load voltage:
The voltage drop in the transformer due to its impedance is given by:
VT = ZT * I2
Substituting the values of ZT and I2, we get:
VT = (0.06 + j0.12) * 50
VT = 3 + j6

The load voltage, V2, is given by:
V2 = V1 - 2 * VT
Substituting the values of V1 and VT, we get:
V2 = 400 - 2 * (3 + j6)
V2 = 388 - j12
|V2| = √(388² + 12²)
|V2| = 388.2 V
∠V2 = tan⁻¹(-12 / 388)
∠V2 = -1.75°

Therefore, the load voltage is 194 V (at an angle of -1.75°).

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