---
title: "SLG fault at bus 3, currents and voltages"
subject: "Power System Analysis II"
chapter: 5
key: "semester_5/power_system_analysis_ii:slg-fault-bus3-three-generators"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/slg-fault-bus3-three-generators
site: BCT Engineering
---

# SLG fault at bus 3, currents and voltages

Power System Analysis II · Chapter 5: Unsymmetrical Faults on Power Systems

- Asked in: 2071 Chaitra, 10 marks
- Video: [Current Calculation For SLG Fault](https://www.youtube.com/watch?v=RtuomhJlF2Q) — Dina Mourad
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/slg-fault-bus3-three-generators
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

In a power system network shown in figure below, single line to ground (SLG) fault occurs at bus 3. i. Draw the positive, negative and zero sequence networks; ii. Determine phase currents in per units and amperes; iii. Phase voltages in per units and kilovolts. Specifications of the equipments are as under: G1: 100 MVA, 13.8 kV, X1 = X2 = 15% and X0 = 5%; G2: 100 MVA, 13.8 kV, X1 = X2 = 15% and X0 = 5%; G3: 100 MVA, 13.8 kV, X1 = X2 = 15% and X0 = 5%; T1: 100 MVA, 13.8/115 kV, X1 = X2 = X0 = 20%; T2: 100 MVA, 115/13.8 kV, X1 = X2 = X0 = 18%; Line: 100 MVA, 115 kV, X1 = X2 = 30% and X0 = 90%. [Figure: G1 (Y, ungrounded) and G2 (Y grounded) at bus 1 - T1 (Y grounded through j0.01 pu on bus 1 side / Δ) - bus 2 - line - bus 3 - T2 (Δ / Y grounded through 0.02 pu on bus 4 side) - bus 4 - G3 (Y grounded through j0.03 pu)]

## Short answer

Bus 3 lies between two delta windings, so Z0 is infinite. Consequently, SLG fault current is zero. Phase voltages: Va=0, Vb=1.732∠-150° pu (115 kV line-to-ground), Vc=1.732∠150° pu (115 kV).

## Answer

**Base:** 100 MVA; 13.8 kV (generator buses 1 and 4), 115 kV (buses 2 and 3). Pre-fault voltage 1 pu, no load. From the figure: T1 is Yg (through j0.01) on the bus 1 side and Δ on the bus 2 side; T2 is Δ on the bus 3 side and Yg (through 0.02) on the bus 4 side; G1 is ungrounded, G2 solidly grounded, G3 grounded through j0.03.

#### (i) Sequence networks
**Positive** (emfs at G1, G2, G3):
```text
 G1 j0.15 \
           (1)-j0.2-(2)-j0.3-(3)-j0.18-(4)-j0.15-G3
 G2 j0.15 /                   |
                              F
```
**Negative:** same reactances, no emfs.

**Zero:**
```text
 G2: ref-j0.05-(1)        G1: open (ungrounded)
 T1: (1)-[3(j0.01)+j0.2]-ref  ;  Δ side (2): open
 line: (2)-j0.9-(3)  -> floating, no ground path
 T2: Δ side (3): open ; (4)-[0.06+j0.18]-ref
 G3: (4)-[j0.05+3(j0.03)]-ref
 -> bus 3 has NO connection to the reference
```
Because bus 3 sits between the Δ windings of T1 and T2, the zero-sequence impedance seen from bus 3 is infinite: $Z_0 = \infty$.

Thevenin values: $Z_1 = Z_2 = j(0.075+0.2+0.3)\ \|\ j(0.18+0.15) = j0.575\|j0.33 = j0.2097$ pu.

#### (ii) Phase currents
$$
I_{a0} = I_{a1} = I_{a2} = \frac{1}{Z_1+Z_2+Z_0} = \frac{1}{\infty} = 0
$$
$$
I_a = I_b = I_c = 0\ \text{pu} = 0\ \text{A}
$$
No fault current flows: the 115 kV section between the two deltas is an ungrounded system (only a small capacitive current would flow in practice).

#### (iii) Phase voltages at bus 3
With no current, $V_{a1} = 1$, $V_{a2} = 0$, and the condition $V_a = 0$ forces $V_{a0} = -1$ pu (the whole neutral shifts):
$$
\begin{aligned}
V_a &= V_{a0}+V_{a1}+V_{a2} = 0 \\
V_b &= V_{a0} + a^2V_{a1} = -1 + 1\angle{-120^\circ} = 1.732\angle{-150^\circ}\ \text{pu} \\
V_c &= V_{a0} + aV_{a1} = -1 + 1\angle{120^\circ} = 1.732\angle{150^\circ}\ \text{pu}
\end{aligned}
$$
Base phase voltage at bus 3 $= 115/\sqrt3 = 66.40$ kV:
- $V_a = 0$ kV
- $|V_b| = |V_c| = 1.732 \times 66.40 = 115$ kV

#### Results
| Quantity | pu | Actual |
|---|---|---|
| $I_a, I_b, I_c$ | 0 | 0 A |
| $V_a$ | 0 | 0 kV |
| $V_b$ | 1.732∠−150° | 115 kV |
| $V_c$ | 1.732∠150° | 115 kV |

The healthy phases rise to full line voltage, showing why ungrounded sections need insulation for line voltage to ground.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is Z0 infinite seen from bus 3?
   A. Bus 3 is between the deltas of T1 and T2 with no path to ground
   B. The line X0 is very large
   C. G1 is ungrounded
   D. Neutral reactors are large
   Answer: A. Delta windings on both sides of the line leave bus 3 with no zero-sequence connection to reference. (Key point: Z0 infinite at bus 3)

2. The SLG fault current at bus 3 is:
   A. 4.77 pu
   B. 14.3 pu
   C. Very large
   D. Zero, since Ia = 3/(Z1+Z2+Z0) with Z0 = infinity
   Answer: D. Ia0 = Ia1 = Ia2 = 1/(Z1+Z2+Z0) = 0, so all phase currents are zero. (Key point: Fault current)

3. What happens to the voltages of phases b and c at bus 3?
   A. Stay at 1 pu
   B. Fall to zero
   C. Rise to 1.732 pu (full line voltage, 115 kV)
   D. Rise to 0.5 pu
   Answer: C. Va0 = -1 pu shifts the neutral, so |Vb| = |Vc| = 1.732 pu = 115 kV while Va = 0. (Key point: Healthy-phase voltages)

## Related in Power System Analysis II

- [SLG fault: sequence networks in series](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aslg-sequence-networks-series) — Power System Analysis II, chapter 5, asked 5×
- [SLG fault on bus with three alternators](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aslg-three-alternators-common-bus) — Power System Analysis II, chapter 5
- [L-G fault through Zf: currents and voltages](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-with-zf-derivation) — Power System Analysis II, chapter 5
- [LLG fault sequence currents at bus 3](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Allg-fault-bus3-sequence-currents) — Power System Analysis II, chapter 5
- [DLG fault sequence network connection](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Adlg-sequence-network-connection) — Power System Analysis II, chapter 5, asked 3×
- [L-G fault of unloaded generator: equal currents](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-unloaded-generator-derivation) — Power System Analysis II, chapter 5

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- [Delta load phase currents, power, line currents](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Adelta-load-100v-6-8-4-3-ohm) — Basic Electrical Engineering, chapter 8
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