b) A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodalvoltages throughout the network are of 1 p.u. and the impedance of the electricarc is neglected. Sequence impedance parameters of the generator,transmission lines, and transformer are given in Figure Q3, where X and Y arethe last two digits of your student number.jX(1) j0.1Y p.u.jX2)= j0.1Y p.u.jXko) = j0.1X p.u.V₁ = 120° p.u. V₂ = 120° p.u.(i)(ii)0jX(1) = j0.2 p.u.1 jx(2) j0.2 p.u. 2jX1(0) = j0.25 p.u.jXT(1)jXT(2)종 3j0.1X p.u. JX3(1)j0.1Y p.u.j0.1X p.u. JX3(2)j0.1Y p.u.jXT(0) j0.1X p.u. JX3(0)=j0.15 p.u.0=x = 1,jX2(1) j0.2Y p.u. V₁=1/0° p.u.jX(2(2) = j0.2Y p.u.jX2(0) = j0.3X p.u.=V3 = 120° p.u.Figure Q3. Circuit for problem 3b).For example, if your student number is c1700123, then:y = 7==jXa(r) = j0.13 p.u., jXa(z) = j0.13 p. u., and jXa(o) = j0.12 p. u.Assuming a balanced excitation, draw the positive, negative and zerosequence Thévenin equivalent circuits as seen from bus 4.4Determine the positive sequence fault current for the case when a three-phase-to-ground fault occurs at bus 4 of the network.

x=1y=7Note: reactance of element right to bus 4 is not given. Hence we will not consider that in…

A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, and transformer are given in Figure Q3, where X and Y are the last two digits of your student number. 10.av

A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, and transformer are given in Figure Q3, where X and Y are the last two digits of your student number. 10.av

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Concept explainers

Three-Phase Transformers

Three-segment transformers are a type of transformer used to transform voltages of electrical systems into three ranges. Two type transformers are shell-type transformer and core type transformer. In brief, it could be described because of the exquisite kinds of configurations.

Transformer

Ever since electricity has been created, people have started using it in its entirety. We see many types of Transformers in the neighborhoods. Some are smaller in size and some are very large. They are used according to their requirements. Many of us have seen the electrical transformer but they do not know what work they are engaged in.

Question

b) A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodal
voltages throughout the network are of 1 p.u. and the impedance of the electric
arc is neglected. Sequence impedance parameters of the generator,
transmission lines, and transformer are given in Figure Q3, where X and Y are
the last two digits of your student number.
jX(1) j0.1Y p.u.
jX2)= j0.1Y p.u.
jXko) = j0.1X p.u.
V₁ = 120° p.u. V₂ = 120° p.u.
(i)
(ii)
0
jX(1) = j0.2 p.u.
1 jx(2) j0.2 p.u. 2
jX1(0) = j0.25 p.u.
jXT(1)
jXT(2)
종 3
j0.1X p.u. JX3(1)
j0.1Y p.u.
j0.1X p.u. JX3(2)
j0.1Y p.u.
jXT(0) j0.1X p.u. JX3(0)=j0.15 p.u.
0
=
x = 1,
jX2(1) j0.2Y p.u. V₁=1/0° p.u.
jX(2(2) = j0.2Y p.u.
jX2(0) = j0.3X p.u.
=
V3 = 120° p.u.
Figure Q3. Circuit for problem 3b).
For example, if your student number is c1700123, then:
y = 7
=
=
jXa(r) = j0.13 p.u., jXa(z) = j0.13 p. u., and jXa(o) = j0.12 p. u.
Assuming a balanced excitation, draw the positive, negative and zero
sequence Thévenin equivalent circuits as seen from bus 4.
4
Determine the positive sequence fault current for the case when a three-
phase-to-ground fault occurs at bus 4 of the network.

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Step 5: ii) Positive sequence fault current for three phase tp ground fault

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