The reactance data for the power system shown in below figure in per unit on a common base is as the following:xx 10.1G₁j0.25TiSelect one:0 à. - j 1,660 puO b. - j 5,369 pu0 c j 4,669 puO d. No right answer2 j0.3#A1 °j0.05j0.25| Assume that the pre fault voltage at bus 2 = 1.15 pu.For a bolted single line to ground fault (bus 2), the fault current in pu =j0.53j0.25T,34Aj0.25j0.1Gj0.05

Answered: Image /qna-images/answer/1df8177f-5abb-458e-8964-38a29507c308.jpg

The reactance data for the power system shown in below figure in per unit on a common base is as the following: ਮਾ X-X² j0.1 G₁ to j0.25 Select one: O a.-j 1.660 pu O b.-j 5.369 pu O c.-j 4.669 pu O d. No right answer Ti A+ j0.05 Assume that the pre fault voltage at bus 2 = 1.15 pu. For a bolted single line to ground fault (bus 2), the fault current in pu = 2 j0.3 + j0.25 j0.5 3 j0.25 T₂ 38 A j0.25 j0.1 G₂ 1 j0.05

The reactance data for the power system shown in below figure in per unit on a common base is as the following: ਮਾ X-X² j0.1 G₁ to j0.25 Select one: O a.-j 1.660 pu O b.-j 5.369 pu O c.-j 4.669 pu O d. No right answer Ti A+ j0.05 Assume that the pre fault voltage at bus 2 = 1.15 pu. For a bolted single line to ground fault (bus 2), the fault current in pu = 2 j0.3 + j0.25 j0.5 3 j0.25 T₂ 38 A j0.25 j0.1 G₂ 1 j0.05

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

See similar textbooks

Similar questions

19
The speed of circuit breaker is defined as the time between the occurrences of the fault to the extinction of the arc. True False
machine The per unit impedance of a synchronous is 0.242. If the base voltage is increased by 1.1 times, the per unit value will be a) 0.266 c) 0.220 b) 0.242 d) 0.200
Single line to ground fault of a solidly grounded generator positive sequence current is j2.5 p.u, then the current flow in the neutral during the fault is O j1.05 p.u O jo.25 p.u O j2.5 p.u O j7.5 p.u
(Q.2) For the following network, calculate the fault currents (IF1, IF2, IF3 and IF.) in pu and in kA, and fault MVA (ie. short-circuit level) at the indicated locations. Internal reactance of generators are j0.9 pu. Base quantities: 11 kV, 400 MVA on the generator side. 113.3 kV 4 MVA, 7% j0.5 2 j0.4 2 300 A 200 A 11 KV 250 MVA
Q.3 When a line-to-ground fault occurs, the current in faulted phase 'a' is 100A. The zero-sequence current in phase 'c' is
33
The negative sequence current of a solidly grounded generator for single line to ground fault is j 2.5 p.u. Then the positive sequence current during the fault is equal to o j7.5 pu o jo.75 pu o j2.5 pu o jo.25p.u
please solve it before 10 min
In the shown below power system, both generators G1 and G2 emf’s are equal.All the system data are shown on the system diagram.a) Draw the system diagram in per unit using (MVA)base = 75 and (KV)base =20 KV in the generators side.b) Calculate the three-phase balanced short circuit current (If) at busbar 3 inAmpere and the fault level (MVA)sc in MVA.c) If the fault at busbar 3 is a line-to-line fault, calculate the fault current inAmpere and the fault level (MVA)sc in MVA.