5. The positive sequence network, negative sequence network and zero sequence network are shown in Figure 3, consider a single-line-to-ground fault at point f(faulted phase is phase A, fault impedance is 0.05, (1)Draw the interconnected sequence network; (2)determine the sequence current and phase current.

5. The positive sequence network, negative sequence network and zero sequence network are shown in Figure 3, consider a single-line-to-ground fault at point f(faulted phase is phase A, fault impedance is 0.05, (1)Draw the interconnected sequence network; (2)determine the sequence current and phase current.

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

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…
Q3: - A three-phase fault occurs at bus-1 of the network shown in figure below. The pre-fault voltage is 1.05 PU and the pre-fault load current is neglected. (45%) i) determine the bus impedance matrix. ii) calculate the sub-transient fault current and the contribution to the fault current from the transmission line. T1 T2 T.L j0.1 PU 10.1 PU j0.105 PU G jXdg j0.15 PU M jXdm $0.2 PU
For the circuit shown, if a symmetrical three-phase fault occurs in node 2, determinea. The short-circuit current in all three phases. b. The voltage at node 1 during the fault.c. The current in generator 1 during the fault. Neglect the prefault currents and consider the reference voltage of the system at 10√3 ??.
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. V₁ = 120° p.u. V₂ = 120° p.u. jX(1) j0.1Y p.u. jX2)= j0.1Y p.u. jXko) j0.1X p.u. - 0 jX(1) = j0.2 p.u. 1JX(2) = 0.2 p.u. 2 jX1(0) = j0.25 p.u. jX2(1) j0.2 p.u. V₁=1/0° p.u. jX(2(2) = j0.2Y p.u. jX2(0) = j0.3X p.u. = V₂ = 120° p.u. jXT(1) j0.1X p.u. jXT(2) j0.1X p.u. JX3(1) j0.1Y p.u. JX3(2)=j0.1Y p.u. jXT(0) j0.1X p.u. JX3(0)=j0.15 p.u. 0- = 3 = Figure Q3. Circuit for problem 3b). For example, if your student number is c1700123, then: jXa(n) = j0.13 p. u., jXa(z) = j0.13 p. u., and jXa(o) = j0.12 p. u. 4 (i) Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from bus 4. (ii)…
The LLLG fault occurs at bus-8with pre-fault voltage as 1.02 p.u. If the positive sequence impedance at bus-8 is j0.15 and positive sequence current is -j1.125. which one of the following values will be the value of zero sequence voltage at bus 8? Select one: a. 1.3125 O b. 0 O c. 0.7587 O d. 0.3125
b) A fault occurs at bus 3 of the network shown in Figure Q4. 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, transformer and load are given in Figure Q4. V₁ = 120° p.u. V₂ = 120° p.u. V₂ = 1/0° p.u. V₂= 120° p.u. jXj0.1 p.u. JX2) 0.1 p.u. jX0j0.15 p.u. jXn-j0.2 p.u. 1 JX(2)-j0.2 p.u. 2 jX)=j0.25 p.u. JX20-10.15 p.u. jXa(z)-j0.2 p.u. 4 jX2(0)=j0.2 p.u. jXT(1) j0.1 p.u. jXT(2)=j0.15 p.u. jXT(0)=j0.1 p.u. Figure Q4. Circuit for problem 4b). = jXj0.1 p.u. j0.1 p.u. - JX(2) JXL(0) 10.1 p.u. = (i) Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from bus 3. (ii) Determine the positive sequence fault current for the case when a three- phase-to-ground fault occurs at bus 3 of the network. (iii) Determine the short-circuit fault current for the case when a one-phase- to-ground fault occurs at bus…
Q2. The single-line diagram of a simple three-bus power system is shown in Figure-2. Each generator is represented by an emf behind the sub-transient reactance. All impedances are expressed in per unit on a common MVA base. All resistances and shunt capacitances are neglected. The generators are operating on no load at their rated voltage with their emfs in phase. A three-phase fault occurs at bus 3 through a fault impedance of Zf = j0.19 per unit. (i) Using Th'evenin's theorem, obtain the impedance to the point of fault and the fault current in (ii) Determine the bus voltages per unit. ) j0.05 j0.075 j0.75 2 j0.30 j0.45 Figure-2: Single line diagram of the power system network for Q2 3
3.On a double line-to-ground faultImmersive reader A) The short-circuit currents in the faulted phases are equal. B) The voltages at the fault point on the faulted phases are zero. C) The phase not involved in the fault has a zero voltage at the fault point. D) There are no circulating currents through the neutrals to ground. E) N. A.
8.In a line-by-line failure we can expect Immersive reader A) Currents exist in the ground connections. B) The current values at the ground connections are zero. C) Short-circuit currents exist in all three phases. D) The voltages at the fault point with respect to ground are zero in the faulted phases. E) N. A.
1 Xine = j0.30 %3D G M X" = j0.15 X" = j0.20 Xiine = j0.30 Xiine = j0.30 %3D 1. Find the Thévenin equivalent looking into bus 3. 2. Calculate the subtransient fault current for a bolted three-phase fault at bus 3. Pre-fault voltage is VF 1Z°, and pre-fault current is neglected. 3. Distribute the fault current between the motor and generator.
The LLLG fault occurs at bus-6 with pre-fault voltage as 1.05 p.u. If the positive sequence impedance at bus-7 is j0.35 and positive sequence current is -j1.250. Find value of positive sequence voltage at bus-7.
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 r