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A 11/6.6/0.5kV, star/star/delta, three winding transformer has a magnetizing current of 3A. At rated voltages, the secondary supplies a balanced load of 300 kVA at 0.8 pf lag and tertiary supplies a balanced load of 150 kW at 0.9 pf lead. The magnitude of primary phase current (in Amp) will be
- a)2.22
- b)22.21
Correct answer is between '2.22,22.21'. Can you explain this answer?
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Verified Answer
A 11/6.6/0.5kV, star/star/delta, three winding transformer has a magn...
Magnetising current Iμ=-j3.0
Secondary phase current 
at 0.8pf lag Secondary phase current referred to primary
at 0.8pf lag Secondary phase current referred to primaryI2=15.746∠−36.87∘
Tertiary line current 
at 0.9 leading
at 0.9 leadingTertiary phase current 
at 0.9 leading
at 0.9 leadingTertiary phase current referred to primary 
a t 0-9 lead I3 = 8.75∠25.84°
a t 0-9 lead I3 = 8.75∠25.84°Primary current =Iμ+I2+I3=−j3+15.746∠−36.87∘+8.75∠25.84∘
=22.21∠−22.86∘
So, magnitude of primary phase current= 22.21 A
Most Upvoted Answer
A 11/6.6/0.5kV, star/star/delta, three winding transformer has a magn...
To find the magnitude of the primary phase current in the given three-winding transformer, we need to calculate the secondary and tertiary phase currents first.
Given data:
- Secondary load: 300 kVA at 0.8 power factor (lagging)
- Tertiary load: 150 kW at 0.9 power factor (leading)
Calculating Secondary Phase Current:
The apparent power (S) of the secondary load can be calculated using the formula:
S = V * I * √3
Where V is the rated secondary voltage (11 kV) and I is the secondary phase current.
The apparent power (S) is given as 300 kVA, so we can rearrange the formula to solve for I:
I = S / (V * √3)
Substituting the given values:
I = 300,000 / (11,000 * √3) = 16.92 A
Calculating Tertiary Phase Current:
The apparent power (S) of the tertiary load can also be calculated using the formula:
S = V * I * √3
Where V is the rated tertiary voltage (0.5 kV) and I is the tertiary phase current.
The apparent power (S) is given as 150 kW (150,000 W), so we can rearrange the formula to solve for I:
I = S / (V * √3)
Substituting the given values:
I = 150,000 / (500 * √3) = 144.34 A
Calculating Primary Phase Current:
The primary phase current (Ip) can be calculated using the turns ratio of the transformer.
Given turns ratio: 11 kV / 6.6 kV
Since the transformer is star/star/delta connected, the primary phase current is equal to the secondary phase current divided by the square root of 3.
Ip = Is / √3
Ip = 16.92 A / √3 = 9.77 A
So, the magnitude of the primary phase current is approximately 9.77 A.
Given data:
- Secondary load: 300 kVA at 0.8 power factor (lagging)
- Tertiary load: 150 kW at 0.9 power factor (leading)
Calculating Secondary Phase Current:
The apparent power (S) of the secondary load can be calculated using the formula:
S = V * I * √3
Where V is the rated secondary voltage (11 kV) and I is the secondary phase current.
The apparent power (S) is given as 300 kVA, so we can rearrange the formula to solve for I:
I = S / (V * √3)
Substituting the given values:
I = 300,000 / (11,000 * √3) = 16.92 A
Calculating Tertiary Phase Current:
The apparent power (S) of the tertiary load can also be calculated using the formula:
S = V * I * √3
Where V is the rated tertiary voltage (0.5 kV) and I is the tertiary phase current.
The apparent power (S) is given as 150 kW (150,000 W), so we can rearrange the formula to solve for I:
I = S / (V * √3)
Substituting the given values:
I = 150,000 / (500 * √3) = 144.34 A
Calculating Primary Phase Current:
The primary phase current (Ip) can be calculated using the turns ratio of the transformer.
Given turns ratio: 11 kV / 6.6 kV
Since the transformer is star/star/delta connected, the primary phase current is equal to the secondary phase current divided by the square root of 3.
Ip = Is / √3
Ip = 16.92 A / √3 = 9.77 A
So, the magnitude of the primary phase current is approximately 9.77 A.
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A 11/6.6/0.5kV, star/star/delta, three winding transformer has a magnetizing current of 3A. At rated voltages, the secondary supplies a balanced load of 300 kVA at 0.8 pf lag and tertiary supplies a balanced load of 150 kW at 0.9 pf lead. The magnitude of primary phase current (in Amp) will bea)2.22b)22.21Correct answer is between '2.22,22.21'. Can you explain this answer?
Question Description
A 11/6.6/0.5kV, star/star/delta, three winding transformer has a magnetizing current of 3A. At rated voltages, the secondary supplies a balanced load of 300 kVA at 0.8 pf lag and tertiary supplies a balanced load of 150 kW at 0.9 pf lead. The magnitude of primary phase current (in Amp) will bea)2.22b)22.21Correct answer is between '2.22,22.21'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 11/6.6/0.5kV, star/star/delta, three winding transformer has a magnetizing current of 3A. At rated voltages, the secondary supplies a balanced load of 300 kVA at 0.8 pf lag and tertiary supplies a balanced load of 150 kW at 0.9 pf lead. The magnitude of primary phase current (in Amp) will bea)2.22b)22.21Correct answer is between '2.22,22.21'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 11/6.6/0.5kV, star/star/delta, three winding transformer has a magnetizing current of 3A. At rated voltages, the secondary supplies a balanced load of 300 kVA at 0.8 pf lag and tertiary supplies a balanced load of 150 kW at 0.9 pf lead. The magnitude of primary phase current (in Amp) will bea)2.22b)22.21Correct answer is between '2.22,22.21'. Can you explain this answer?.
A 11/6.6/0.5kV, star/star/delta, three winding transformer has a magnetizing current of 3A. At rated voltages, the secondary supplies a balanced load of 300 kVA at 0.8 pf lag and tertiary supplies a balanced load of 150 kW at 0.9 pf lead. The magnitude of primary phase current (in Amp) will bea)2.22b)22.21Correct answer is between '2.22,22.21'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 11/6.6/0.5kV, star/star/delta, three winding transformer has a magnetizing current of 3A. At rated voltages, the secondary supplies a balanced load of 300 kVA at 0.8 pf lag and tertiary supplies a balanced load of 150 kW at 0.9 pf lead. The magnitude of primary phase current (in Amp) will bea)2.22b)22.21Correct answer is between '2.22,22.21'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 11/6.6/0.5kV, star/star/delta, three winding transformer has a magnetizing current of 3A. At rated voltages, the secondary supplies a balanced load of 300 kVA at 0.8 pf lag and tertiary supplies a balanced load of 150 kW at 0.9 pf lead. The magnitude of primary phase current (in Amp) will bea)2.22b)22.21Correct answer is between '2.22,22.21'. Can you explain this answer?.
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A 11/6.6/0.5kV, star/star/delta, three winding transformer has a magnetizing current of 3A. At rated voltages, the secondary supplies a balanced load of 300 kVA at 0.8 pf lag and tertiary supplies a balanced load of 150 kW at 0.9 pf lead. The magnitude of primary phase current (in Amp) will bea)2.22b)22.21Correct answer is between '2.22,22.21'. Can you explain this answer?, a detailed solution for A 11/6.6/0.5kV, star/star/delta, three winding transformer has a magnetizing current of 3A. At rated voltages, the secondary supplies a balanced load of 300 kVA at 0.8 pf lag and tertiary supplies a balanced load of 150 kW at 0.9 pf lead. The magnitude of primary phase current (in Amp) will bea)2.22b)22.21Correct answer is between '2.22,22.21'. Can you explain this answer? has been provided alongside types of A 11/6.6/0.5kV, star/star/delta, three winding transformer has a magnetizing current of 3A. At rated voltages, the secondary supplies a balanced load of 300 kVA at 0.8 pf lag and tertiary supplies a balanced load of 150 kW at 0.9 pf lead. The magnitude of primary phase current (in Amp) will bea)2.22b)22.21Correct answer is between '2.22,22.21'. Can you explain this answer? theory, EduRev gives you an
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