For the power system shown at right the impedance matrix 2 is j0.5 j0.2 3 j0.4 0.145 0.120 0.0692 Zbus=j 0.120 0.246 0.101 | 0.0692 0.101 0.164 (3) j0.2 j0.25 Pre-fault all bus voltages are 1 pu. Then a three-phase fault to ground occurs at bus 3. What are the post-fault per-unit bus voltages? (Express in polar phasor form |V|Z0° ) ~) 1pu 1pu (~ V, = 0.578 Lo° V;= 0.384 L0° V;= %3D feuetot bus -j6.098 Z33 AVI 0.o692 0.4220 AV2 jG.098 (j) 0.101 0.l6159 D.164 %3D

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Can someone please explain how to convert V1=-0.4220 to poloar phasor form? (V1=0.578<0)

For the power system shown at right the impedance matrix 2
is
j0.5
j0.2 3
j0.4
0.145
0.120 0.0692
(1)
0.246 0.101
bus
0.0692 0.101 0.164
j0.2
j0.25
Pre-fault all bus voltages are 1 pu. Then a three-phase fault
to ground occurs at bus 3. What are the post-fault per-unit
bus voltages? (Express in polar phasor form |v|Z0°)
1pu
1pu
V, = 0.578 Lo°
V;= 0.384 L0°
V;=
Fr fault ot bus 3 If
-j6.098
Z33
jo.
AV
0.o692
0.4220
j6098 (G) 0.101
0.l6159
D.164
Transcribed Image Text:For the power system shown at right the impedance matrix 2 is j0.5 j0.2 3 j0.4 0.145 0.120 0.0692 (1) 0.246 0.101 bus 0.0692 0.101 0.164 j0.2 j0.25 Pre-fault all bus voltages are 1 pu. Then a three-phase fault to ground occurs at bus 3. What are the post-fault per-unit bus voltages? (Express in polar phasor form |v|Z0°) 1pu 1pu V, = 0.578 Lo° V;= 0.384 L0° V;= Fr fault ot bus 3 If -j6.098 Z33 jo. AV 0.o692 0.4220 j6098 (G) 0.101 0.l6159 D.164
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New voltage at a bus = old bus voltage + change in voltage

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