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A 20 kVA, 2000/200 V, 1-f transformer has name plate leakage impedance of 8%. Voltagerequired to be applied on the high voltage side to circulate full load current with the low voltagewinding 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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A 20 kVA, 2000/200 V, 1-f transformer has name plate leakage impedance...
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A 20 kVA, 2000/200 V, 1-f transformer has name plate leakage impedance...
Solution:
Given, Transformer rating = 20 kVA
High voltage side rating = 2000 V
Low voltage side rating = 200 V
Leakage impedance = 8% of transformer rating
Let us calculate the leakage impedance of the transformer.
Leakage impedance = 8/100 * 20 kVA = 1.6 kΩ
Let us calculate the full load current on the low voltage side.
Full load current (I2) = Transformer rating / Low voltage side rating
= 20,000 / 200
= 100 A
Let us assume that the high voltage side voltage required to circulate full load current with the low voltage winding short-circuited is V.
Impedance of the transformer (Z) = V / I2
Let us calculate the equivalent circuit of the transformer.
Refer to the following equivalent circuit of the transformer.
[Transformer equivalent circuit](
Impedance of the transformer (Z) = R + jX
Where,
R = Resistance of the transformer
X = Reactance of the transformer
Reactance of the transformer (X) = Leakage impedance (8% of transformer rating)
X = 1.6 kΩ
Resistance of the transformer (R) = (High voltage side voltage / Full load current) - Leakage resistance
R = (V / 100) - 1.6 kΩ
Impedance of the transformer (Z) = (V / 100) - j1.6 kΩ
Magnitude of the impedance of the transformer (Z) = √(R² + X²)
Magnitude of the impedance of the transformer (Z) = √[((V / 100) - 1.6 kΩ)² + (1.6 kΩ)²]
Magnitude of the impedance of the transformer (Z) = √[(V² / 10000) - (3.2 V / 50) + 2.56 + 2.56]
Magnitude of the impedance of the transformer (Z) = √[(V² / 10000) - (3.2 V / 50) + 5.12]
Magnitude of the impedance of the transformer (Z) = √[(V² / 10000) - (64 V / 1000) + 5.12]
Magnitude of the impedance of the transformer (Z) = √[(V² / 10000) - (V / 15.625) + 5.12]
Let us calculate the voltage required to circulate full load current with the low voltage winding short-circuited.
Magnitude of the impedance of the transformer (Z) = High voltage side voltage / Full load current
√[(V² / 10000) - (V / 15.625) + 5.12] = 2000 / 100
√[(V² / 10000) - (V / 15.625) + 5.12] = 20
(V² / 10000) - (V / 15.625) + 5.12 =
Given, Transformer rating = 20 kVA
High voltage side rating = 2000 V
Low voltage side rating = 200 V
Leakage impedance = 8% of transformer rating
Let us calculate the leakage impedance of the transformer.
Leakage impedance = 8/100 * 20 kVA = 1.6 kΩ
Let us calculate the full load current on the low voltage side.
Full load current (I2) = Transformer rating / Low voltage side rating
= 20,000 / 200
= 100 A
Let us assume that the high voltage side voltage required to circulate full load current with the low voltage winding short-circuited is V.
Impedance of the transformer (Z) = V / I2
Let us calculate the equivalent circuit of the transformer.
Refer to the following equivalent circuit of the transformer.
[Transformer equivalent circuit](
https://www.edurev.in/studytube/transformer-equivalent-circuit/dc2cfd04-9124-4e4e-9a53-7c4d1eaf6f9d_t) https://www.edurev.in/studytube/transformer-equivalent-circuit/dc2cfd04-9124-4e4e-9a53-7c4d1eaf6f9d_t) |  |
Impedance of the transformer (Z) = R + jX
Where,
R = Resistance of the transformer
X = Reactance of the transformer
Reactance of the transformer (X) = Leakage impedance (8% of transformer rating)
X = 1.6 kΩ
Resistance of the transformer (R) = (High voltage side voltage / Full load current) - Leakage resistance
R = (V / 100) - 1.6 kΩ
Impedance of the transformer (Z) = (V / 100) - j1.6 kΩ
Magnitude of the impedance of the transformer (Z) = √(R² + X²)
Magnitude of the impedance of the transformer (Z) = √[((V / 100) - 1.6 kΩ)² + (1.6 kΩ)²]
Magnitude of the impedance of the transformer (Z) = √[(V² / 10000) - (3.2 V / 50) + 2.56 + 2.56]
Magnitude of the impedance of the transformer (Z) = √[(V² / 10000) - (3.2 V / 50) + 5.12]
Magnitude of the impedance of the transformer (Z) = √[(V² / 10000) - (64 V / 1000) + 5.12]
Magnitude of the impedance of the transformer (Z) = √[(V² / 10000) - (V / 15.625) + 5.12]
Let us calculate the voltage required to circulate full load current with the low voltage winding short-circuited.
Magnitude of the impedance of the transformer (Z) = High voltage side voltage / Full load current
√[(V² / 10000) - (V / 15.625) + 5.12] = 2000 / 100
√[(V² / 10000) - (V / 15.625) + 5.12] = 20
(V² / 10000) - (V / 15.625) + 5.12 =
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A 20 kVA, 2000/200 V, 1-f transformer has name plate leakage impedance of 8%. Voltagerequired to be applied on the high voltage side to circulate full load current with the low voltagewinding 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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