---
title: "L-G fault at bus 3, parallel lines"
subject: "Power System Analysis II"
chapter: 5
key: "semester_5/power_system_analysis_ii:lg-fault-bus3-two-generators"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/lg-fault-bus3-two-generators
site: BCT Engineering
---

# L-G fault at bus 3, parallel lines

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

- Asked in: 2078 Bhadra, 16 marks
- Video: [Numericals on L-G Fault | Part 3](https://www.youtube.com/watch?v=aqjNE4ghZCk) — Dr. Afroz Alam
- 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-bus3-two-generators
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

Determine the Fault current when a line to ground fault occurs at Bus 3 as shown in figure below. G1, G2: 100 MVA, 11 kV, X1 = X2 = 15%, X0 = 5%, Xn = 6%; T1, T2: 100 MVA, 11 kV/220 kV, Xleak = 10%; L1, L2: X1 = X2 = 10%, X0 = 10% on a base of 100 MVA. [Figure: G1 (Y grounded through Xn) - bus 1 - T1 (Y grounded/Y grounded) - bus 2 - parallel lines L1 and L2 - bus 3 - T2 (Y grounded/Y grounded) - bus 4 - G2 (Y grounded through Xn)]

## Short answer

Thevenin Z1=Z2=j0.1364 pu, Z0=j0.1766 pu at bus 3. For bolted L-G fault, sequence currents in series: Ia1 = -j2.2255 pu; fault current I_f = 3Ia0 = 6.676 pu = 1.75 kA at 220 kV.

## Answer

**Base:** 100 MVA; 11 kV on generator side, 220 kV on line side. All data are already on this base. $X_n = 0.06$ pu for each generator. Pre-fault voltage $E = 1\angle0^\circ$ pu, system unloaded.

#### Single line diagram
```text
 G1        T1                 L1            T2        G2
 (~)--(1)--)(--(2)====================(3)--)(--(4)--(~)
  |       Yg Yg     ====== L2 =======      Yg Yg     |
 Xn                                                  Xn
 |                                                   |
gnd                      F (L-G at bus 3)           gnd
```

#### Per-unit reactances
| Element | $X_1$ | $X_2$ | $X_0$ |
|---|---|---|---|
| G1, G2 | 0.15 | 0.15 | 0.05 + 3(0.06) = 0.23 |
| T1, T2 | 0.10 | 0.10 | 0.10 |
| L1 ∥ L2 | 0.05 | 0.05 | 0.05 |

#### Positive sequence network
```text
 E1-j0.15-(1)-j0.1-(2)-j0.05-(3)-j0.1-(4)-j0.15-E2
                            |
                            F
```
$$
Z_1 = j(0.15+0.1+0.05)\ \|\ j(0.1+0.15) = j0.30\ \|\ j0.25 = j0.1364\ \text{pu}
$$

#### Negative sequence network
Same as positive, without the sources:
$$
Z_2 = j0.1364\ \text{pu}
$$

#### Zero sequence network
Both transformers are Yg–Yg, so zero-sequence current passes through them; generator neutrals are grounded through $X_n$, which appears as $3X_n$.
```text
 ref-j0.23-(1)-j0.1-(2)-j0.05-(3)-j0.1-(4)-j0.23-ref
                             |
                             F
```
$$
Z_0 = j(0.23+0.1+0.05)\ \|\ j(0.1+0.23) = j0.38\ \|\ j0.33 = j0.1766\ \text{pu}
$$

#### Fault current (L-G at bus 3, $Z_f = 0$)
Sequence networks in series:
$$
\begin{aligned}
I_{a1} = I_{a2} = I_{a0} &= \frac{E}{Z_1+Z_2+Z_0} = \frac{1}{j(0.1364+0.1364+0.1766)} = \frac{1}{j0.4494} = -j2.2255\ \text{pu} \\
I_f &= 3I_{a0} = -j6.676\ \text{pu}
\end{aligned}
$$
Base current at 220 kV:
$$
I_{base} = \frac{100\times10^6}{\sqrt3 \times 220\times10^3} = 262.43\ \text{A}
$$
$$
I_f = 6.676 \times 262.43 = 1752\ \text{A}
$$

#### Phase currents at the fault
- $I_a = I_f = 6.676\angle{-90^\circ}$ pu $= 1.752$ kA
- $I_b = I_c = 0$ (healthy phases, unloaded system)

#### Fault MVA (for reference)
Fault MVA $= |I_f| \times$ base MVA (with $V_f = 1$) $= 6.676 \times 100 = 667.6$ MVA.

**Answer: L-G fault current at bus 3 ≈ 6.68 pu ≈ 1.75 kA.**

Note: $Z_0$ (0.1766) is larger than $Z_1$ (0.1364) mainly because of the neutral reactances, so here the L-G current (6.68 pu) is less than the three-phase fault current ($1/0.1364 = 7.33$ pu).

## Check yourself

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

1. What zero sequence reactance does each generator branch have with X0 = 0.05 and Xn = 0.06?
   A. 0.05
   B. 0.11
   C. 0.23
   D. 0.06
   Answer: C. X0 = 0.05 + 3(0.06) = 0.23 pu. (Key point: Generator zero sequence reactance)

2. How is Z1 at bus 3 obtained?
   A. Sum of all reactances
   B. Parallel of the left path j0.30 and right path j0.25, giving j0.1364
   C. Parallel of the two generators only
   D. j0.15 + j0.15
   Answer: B. Left: 0.15 + 0.1 + 0.05 = 0.30; right: 0.1 + 0.15 = 0.25; 0.30 || 0.25 = 0.1364 pu. (Key point: Positive sequence Thevenin impedance)

3. Why does zero sequence current flow through T1 and T2?
   A. They are delta-delta
   B. They are Yg-Yg
   C. Their neutrals are open
   D. The lines are ungrounded
   Answer: B. Both transformers are grounded star on both sides, passing zero sequence; Z0 = 0.38 || 0.33 = 0.1766 pu. (Key point: Zero sequence Thevenin impedance)

4. What is the L-G fault current at bus 3?
   A. Ia1 = 2.2255 pu, with If = 6.68 pu = 1.75 kA
   B. If = 2.2255 pu
   C. If = 7.33 pu
   D. If = 3.3 pu
   Answer: A. Ia1 = 1/j0.4494 = 2.2255 pu, If = 3Ia1 = 6.676 pu, times 262.43 A = 1752 A. (Key point: Fault current)

5. Why is the L-G current (6.68 pu) less than the three-phase current (7.33 pu) here?
   A. Fault impedance is nonzero
   B. Z0 (0.1766) is larger than Z1 (0.1364) due to the neutral reactances
   C. Z2 is zero
   D. Transformers block zero sequence
   Answer: B. L-G exceeds 3-phase only when Z0 < Z1; here Z0 > Z1 because the neutral reactances count 3Xn. (Key point: L-G versus 3-phase)

## Related in Power System Analysis II

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- [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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