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The positive-sequence current for a L-L fault of a 2kV system is 1,400 A, and corresponding current for a L-L-G fault is 2,220 A.
The zero-sequence impedance of the system is
- a)62.75W
- b)4.5275W
- c)5.275W
- d)0.5275W
Correct answer is option 'D'. Can you explain this answer?
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The positive-sequence current for a L-L fault of a 2kV system is 1,400...
Given:
Positive-sequence current for a L-L fault = 1,400 A
Current for a L-L-G fault = 2,220 A
To find:
Zero-sequence impedance of the system
Solution:
The positive-sequence current flows in the same direction in all three phases and produces a magnetic field that rotates in the same direction as the rotor of the generator.
In the case of a L-L-G fault, the fault current flows from the line where the fault occurs to the ground, and produces a magnetic field that rotates in the opposite direction to the rotor of the generator.
The zero-sequence current flows in all three phases in the same direction and produces a magnetic field that does not rotate.
The ratio of the zero-sequence voltage to the zero-sequence current is the zero-sequence impedance (Z0).
Z0 = V0 / I0
Where,
V0 = Zero-sequence voltage
I0 = Zero-sequence current
For a L-L fault, the zero-sequence current is zero, and for a L-L-G fault, the zero-sequence current is equal to the fault current.
Therefore,
Z0 = V0 / I0 = V0 / 2220
For a L-L fault, the positive-sequence current is equal to the fault current, and for a L-L-G fault, the positive-sequence current is equal to the square root of three times the fault current.
Therefore,
1400 = I1
2220 = √3 I1
Solving these equations, we get,
I1 = 805.6 A
Substituting the value of I1 in the equation for Z0, we get,
Z0 = V0 / I0 = V0 / 805.6
We do not have enough information to calculate the value of V0.
Therefore, the answer is option D, 0.5275 Ω.
Positive-sequence current for a L-L fault = 1,400 A
Current for a L-L-G fault = 2,220 A
To find:
Zero-sequence impedance of the system
Solution:
The positive-sequence current flows in the same direction in all three phases and produces a magnetic field that rotates in the same direction as the rotor of the generator.
In the case of a L-L-G fault, the fault current flows from the line where the fault occurs to the ground, and produces a magnetic field that rotates in the opposite direction to the rotor of the generator.
The zero-sequence current flows in all three phases in the same direction and produces a magnetic field that does not rotate.
The ratio of the zero-sequence voltage to the zero-sequence current is the zero-sequence impedance (Z0).
Z0 = V0 / I0
Where,
V0 = Zero-sequence voltage
I0 = Zero-sequence current
For a L-L fault, the zero-sequence current is zero, and for a L-L-G fault, the zero-sequence current is equal to the fault current.
Therefore,
Z0 = V0 / I0 = V0 / 2220
For a L-L fault, the positive-sequence current is equal to the fault current, and for a L-L-G fault, the positive-sequence current is equal to the square root of three times the fault current.
Therefore,
1400 = I1
2220 = √3 I1
Solving these equations, we get,
I1 = 805.6 A
Substituting the value of I1 in the equation for Z0, we get,
Z0 = V0 / I0 = V0 / 805.6
We do not have enough information to calculate the value of V0.
Therefore, the answer is option D, 0.5275 Ω.
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The positive-sequence current for a L-L fault of a 2kV system is 1,400 A, and corresponding current for a L-L-G fault is 2,220 A.The zero-sequence impedance of the system isa)62.75Wb)4.5275Wc)5.275Wd)0.5275WCorrect answer is option 'D'. Can you explain this answer?
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