---
title: "L-G fault at bus 1 through Zf"
subject: "Power System Analysis II"
chapter: 5
key: "semester_5/power_system_analysis_ii:lg-fault-bus1-parallel-lines"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/lg-fault-bus1-parallel-lines
site: BCT Engineering
---

# L-G fault at bus 1 through Zf

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

- Asked in: 2079 Bhadra, 8 marks
- Video: [Module 4 Lecture 7 - Problem 1- LG fault without fault impedance](https://www.youtube.com/watch?v=Cstkb5RieNk) — Sumitha T L
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/lg-fault-bus1-parallel-lines
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

Figure shows a power system network. Draw positive, negative and zero sequence network. If LG fault occurs at bus 1, find fault current. Assume fault impedance Zf = 0.05 p.u.

| Equipment | MVA rating | Voltage rating | X1 (p.u.) | X2 (p.u.) | X0 (p.u.) |
|---|---|---|---|---|---|
| G1 | 50 | 11 KV | 0.20 | 0.20 | 0.08 |
| G2 | 30 | 11 KV | 0.25 | 0.25 | 0.1 |
| Transformer T1 | 50 | 11/132 KV | 0.1 | 0.1 | 0.1 |
| Transformer T1 (T2) | 30 | 11/132 KV | 0.09 | 0.09 | 0.09 |
| Line L1 | 45 | 132 KV | 0.1 | 0.1 | 0.25 |
| Line L1 (L2) | 45 | 132 KV | 0.1 | 0.1 | 0.25 |

[Figure: G1 (Y grounded) - T1 (Δ on generator side, Y grounded on bus 1 side) - bus 1 - two parallel lines - bus 2 - T2 (Y grounded on bus 2 side, Y grounded through j0.03 on generator side) - G2 (Y grounded); fault F at bus 1]

## Short answer

Thevenin impedances at bus 1 are Z1=Z2=j0.2024 pu, Z0=j0.0845 pu. With fault impedance j0.05 pu, Ia1 = 1/(j0.6393) = -j1.564 pu. Fault current I_f = 3Ia1 = 4.692 pu = 1.03 kA at 132 kV.

## Answer

**Assumptions:** base 50 MVA, 11 kV (generators) and 132 kV (lines). The neutral reactance j0.03 of T2 is taken as already on the 50 MVA base. Pre-fault voltage 1 pu, no load.

#### Per-unit values on 50 MVA
| Element | $X_1 = X_2$ | $X_0$ |
|---|---|---|
| G1 | 0.20 | 0.08 |
| G2 ($\times 50/30$) | 0.4167 | 0.1667 |
| T1 | 0.10 | 0.10 |
| T2 ($\times 50/30$) | 0.15 | 0.15 |
| Each line ($\times 50/45$) | 0.1111 | 0.2778 |
| Two lines in parallel | 0.0556 | 0.1389 |

#### Sequence networks
```text
Positive (negative same, no emf):
 E1-j0.2-j0.1-(1)-j0.0556-(2)-j0.15-j0.4167-E2
               F
Zero:
 ref-j0.1-(1)-j0.1389-(2)-j0.15-[3x j0.03]-j0.1667-ref
 (G1 cut off by T1 delta; T2 is Yg-Yg, so G2
  zero seq path continues through T2)
```

#### Thevenin impedances at bus 1
$$
\begin{aligned}
Z_1 = Z_2 &= j0.30\ \|\ j(0.0556 + 0.15 + 0.4167) = j0.30\ \|\ j0.6222 = j0.2024\ \text{pu} \\
Z_0 &= j0.10\ \|\ j(0.1389 + 0.15 + 0.09 + 0.1667) = j0.10\ \|\ j0.5456 = j0.0845\ \text{pu}
\end{aligned}
$$

#### Fault current (L-G through $Z_f = 0.05$ pu)
$$
I_{a0} = I_{a1} = I_{a2} = \frac{V_f}{Z_1 + Z_2 + Z_0 + 3Z_f}
$$
Taking $Z_f$ as a reactance $j0.05$ (usual in these problems), $3Z_f = j0.15$:
$$
\begin{aligned}
I_{a0} &= \frac{1}{j(0.2024+0.2024+0.0845+0.15)} = \frac{1}{j0.6393} = -j1.564\ \text{pu} \\
I_f &= 3I_{a0} = -j4.692\ \text{pu}
\end{aligned}
$$
Base current at 132 kV:
$$
I_{base} = \frac{50\times10^6}{\sqrt3 \times 132\times10^3} = 218.7\ \text{A}
$$
$$
I_f = 4.692 \times 218.7 = 1026\ \text{A}
$$

**Answer: L-G fault current at bus 1 ≈ 4.69 pu ≈ 1.03 kA.**

(If $Z_f$ is taken as a pure resistance 0.05 pu, $|I_f| = 3/|0.15 + j0.4893| = 5.86$ pu.)

## Check yourself

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

1. Why is G1 cut off from the zero sequence network?
   A. G1 has Z0 = 0
   B. T1 has a delta on the generator side, which blocks zero sequence current
   C. G1 is ungrounded
   D. The lines have infinite Z0
   Answer: B. A delta winding blocks zero sequence flow to the generator side, so G1 does not appear in the zero network. (Key point: Zero sequence network from transformer connections)

2. How is the transformer neutral reactance j0.03 included in the zero sequence network?
   A. As 3 x j0.03 in series with the path
   B. As j0.03/3
   C. Ignored
   D. As j0.03 in parallel
   Answer: A. A neutral impedance carries 3I0, so it appears as 3Zn. (Key point: Neutral reactance in zero sequence)

3. In the series connection for an L-G fault with Zf, what is added?
   A. Zf
   B. Zf/3
   C. 3Zf
   D. Zf squared
   Answer: C. I = Vf/(Z1 + Z2 + Z0 + 3Zf), so 3Zf = j0.15 pu. (Key point: Fault impedance in the loop)

4. With Z1 = Z2 = j0.2024, Z0 = j0.0845 and 3Zf = j0.15, what is the fault current in pu?
   A. About 1.56 pu
   B. About 4.69 pu
   C. About 6.4 pu
   D. About 5.86 pu
   Answer: B. Ia0 = 1/j0.6393 = -j1.564, and If = 3 x 1.564 = 4.69 pu = about 1.03 kA. (Key point: Result)

## Related in Power System Analysis II

- [L-L fault current in kA at bus 3](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-bus3-ka) — Power System Analysis II, chapter 5
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- [Zero sequence Thevenin impedance at bus 2](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Azero-sequence-thevenin-bus2) — Power System Analysis II, chapter 4
- [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
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