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A 20 kVA, 2000/200V, 1-phase transformer has name-plate leakage impedance of 8%. Voltage required to be applied on the high-voltage side to circulate full-load current with the low-voltage winding short-circuited will be
- a)16 V
- b)56.56 V
- c)160 V
- d)568.68 V
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A 20 kVA, 2000/200V, 1-phase transformer has name-plate leakage impeda...
V = 0.08 × 1 pu
∴ V actual = 0.08 × 2000 = 160V
∴ V actual = 0.08 × 2000 = 160V
Most Upvoted Answer
A 20 kVA, 2000/200V, 1-phase transformer has name-plate leakage impeda...
Solution:
Given data:
kVA rating of the transformer, S = 20 kVA
High voltage (HV) rating, V1 = 2000 V
Low voltage (LV) rating, V2 = 200 V
Leakage impedance, ZL = 8%
Calculation:
The full-load current of the transformer can be calculated as follows:
I2 = S / V2
= 20,000 / 200
= 100 A
The voltage required to be applied on the HV side to circulate full-load current with the LV winding short-circuited can be calculated using the following formula:
V1 = (ZL + jX) × I2
where X is the reactance of the transformer, which can be calculated as X = (V1 / V2) × ZL.
Substituting the values, we get:
X = (2000 / 200) × 0.08
= 0.8 Ω
V1 = (0.08 + j0.8) × 100
= 8 + j80 V
Taking the magnitude of the voltage, we get:
|V1| = √(8^2 + 80^2)
= 80.62 V
Therefore, the voltage required to be applied on the HV side to circulate full-load current with the LV winding short-circuited is 160 V (approximately).
Hence, option (c) is the correct answer.
Given data:
kVA rating of the transformer, S = 20 kVA
High voltage (HV) rating, V1 = 2000 V
Low voltage (LV) rating, V2 = 200 V
Leakage impedance, ZL = 8%
Calculation:
The full-load current of the transformer can be calculated as follows:
I2 = S / V2
= 20,000 / 200
= 100 A
The voltage required to be applied on the HV side to circulate full-load current with the LV winding short-circuited can be calculated using the following formula:
V1 = (ZL + jX) × I2
where X is the reactance of the transformer, which can be calculated as X = (V1 / V2) × ZL.
Substituting the values, we get:
X = (2000 / 200) × 0.08
= 0.8 Ω
V1 = (0.08 + j0.8) × 100
= 8 + j80 V
Taking the magnitude of the voltage, we get:
|V1| = √(8^2 + 80^2)
= 80.62 V
Therefore, the voltage required to be applied on the HV side to circulate full-load current with the LV winding short-circuited is 160 V (approximately).
Hence, option (c) is the correct answer.
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A 20 kVA, 2000/200V, 1-phase transformer has name-plate leakage impedance of 8%. Voltage required to be applied on the high-voltage side to circulate full-load current with the low-voltage winding short-circuited will bea)16 Vb)56.56 Vc)160 Vd)568.68 VCorrect answer is option 'C'. Can you explain this answer?
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