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A 2 MVA, 3.3 kV, three-phase generator as the following reactance’s on its own base:
+ve sequence reactance, x1 = j0.10 p.u.
-ve sequence reactance, x2 = j0.10 p.u.
Zero sequence reactance, x0 = j0.05 p.u.
Reactance from neutral to ground, xn = j0.05 p.u.
The generator supplies a 3-phase, 1.5 MVA balanced passive load through a 3-phase, 2.0 MVA, 3.3 kV/400 V, delta-star transform with leakage reactance of 10%, as shown in the figure. A single-line-to-ground fault takes place at the low voltage side of the transformer. The current flowing from the transformer to the fault is (Assume base MVA is 1.5 MVA)
  • a)
    5773.5 A
  • b)
    12371.8 A
  • c)
    14433.8 A
  • d)
    17320.5 A
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A 2 MVA, 3.3 kV, three-phase generator as the following reactance&rsqu...
The generator supplies a 3-phase, 1.5 MVA balanced passive load.
so, the system works under a base load of 1.5 MVA
The generator reactance on a base load of 1.5 MVA are as follows.
+ve sequence reactance x
-ve sequence reactance x=  
zero sequence reactance x0 = 
Reactance form neutral to ground xn
leakage reactance of transformer x
= j0.075 pu
Positive sequence network for the above system is,

X1eq = x+ xT
= j0.075 + j0.075
= j0.15 pu
Negative sequence network for above system is
X2eq = x2 + xT = j0.075 + j0.075
= j0.15 pu
Zero sequence network for above system is
X0eq = xT = j 0.075 pu
For a line to ground fault

 
= 2165.06 A
Fault current (If) = (Ib) × (If(pu))
= 2165.06 × 8
= 17320.5 A
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A 2 MVA, 3.3 kV, three-phase generator as the following reactance’s on its own base:+ve sequence reactance, x1= j0.10 p.u.-ve sequence reactance, x2= j0.10 p.u.Zero sequence reactance, x0= j0.05 p.u.Reactance from neutral to ground, xn= j0.05 p.u.The generator supplies a 3-phase, 1.5 MVA balanced passive load through a 3-phase, 2.0 MVA, 3.3 kV/400 V, delta-star transform with leakage reactance of 10%, as shown in the figure. A single-line-to-ground fault takes place at the low voltage side of the transformer. The current flowing from the transformer to the fault is (Assume base MVA is 1.5 MVA)a)5773.5 Ab)12371.8 Ac)14433.8 Ad)17320.5 ACorrect answer is option 'D'. Can you explain this answer?
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A 2 MVA, 3.3 kV, three-phase generator as the following reactance’s on its own base:+ve sequence reactance, x1= j0.10 p.u.-ve sequence reactance, x2= j0.10 p.u.Zero sequence reactance, x0= j0.05 p.u.Reactance from neutral to ground, xn= j0.05 p.u.The generator supplies a 3-phase, 1.5 MVA balanced passive load through a 3-phase, 2.0 MVA, 3.3 kV/400 V, delta-star transform with leakage reactance of 10%, as shown in the figure. A single-line-to-ground fault takes place at the low voltage side of the transformer. The current flowing from the transformer to the fault is (Assume base MVA is 1.5 MVA)a)5773.5 Ab)12371.8 Ac)14433.8 Ad)17320.5 ACorrect answer is option 'D'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 2 MVA, 3.3 kV, three-phase generator as the following reactance’s on its own base:+ve sequence reactance, x1= j0.10 p.u.-ve sequence reactance, x2= j0.10 p.u.Zero sequence reactance, x0= j0.05 p.u.Reactance from neutral to ground, xn= j0.05 p.u.The generator supplies a 3-phase, 1.5 MVA balanced passive load through a 3-phase, 2.0 MVA, 3.3 kV/400 V, delta-star transform with leakage reactance of 10%, as shown in the figure. A single-line-to-ground fault takes place at the low voltage side of the transformer. The current flowing from the transformer to the fault is (Assume base MVA is 1.5 MVA)a)5773.5 Ab)12371.8 Ac)14433.8 Ad)17320.5 ACorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2 MVA, 3.3 kV, three-phase generator as the following reactance’s on its own base:+ve sequence reactance, x1= j0.10 p.u.-ve sequence reactance, x2= j0.10 p.u.Zero sequence reactance, x0= j0.05 p.u.Reactance from neutral to ground, xn= j0.05 p.u.The generator supplies a 3-phase, 1.5 MVA balanced passive load through a 3-phase, 2.0 MVA, 3.3 kV/400 V, delta-star transform with leakage reactance of 10%, as shown in the figure. A single-line-to-ground fault takes place at the low voltage side of the transformer. The current flowing from the transformer to the fault is (Assume base MVA is 1.5 MVA)a)5773.5 Ab)12371.8 Ac)14433.8 Ad)17320.5 ACorrect answer is option 'D'. Can you explain this answer?.
Solutions for A 2 MVA, 3.3 kV, three-phase generator as the following reactance’s on its own base:+ve sequence reactance, x1= j0.10 p.u.-ve sequence reactance, x2= j0.10 p.u.Zero sequence reactance, x0= j0.05 p.u.Reactance from neutral to ground, xn= j0.05 p.u.The generator supplies a 3-phase, 1.5 MVA balanced passive load through a 3-phase, 2.0 MVA, 3.3 kV/400 V, delta-star transform with leakage reactance of 10%, as shown in the figure. A single-line-to-ground fault takes place at the low voltage side of the transformer. The current flowing from the transformer to the fault is (Assume base MVA is 1.5 MVA)a)5773.5 Ab)12371.8 Ac)14433.8 Ad)17320.5 ACorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE). Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
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