---
title: "L-G, L-L and L-L-G faults at bus q"
subject: "Power System Analysis II"
chapter: 5
key: "semester_5/power_system_analysis_ii:lg-ll-llg-fault-bus-q"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/lg-ll-llg-fault-bus-q
site: BCT Engineering
---

# L-G, L-L and L-L-G faults at bus q

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

- Asked in: 2072 Chaitra, 16 marks
- Video: [Module 4  Lecture 10 - Problem 4 - LG,LL,LLG fault through fault impedance](https://www.youtube.com/watch?v=0_fZosB8ibc) — 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-ll-llg-fault-bus-q
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

Calculate the fault currents in each phase for the system shown in figure below if (i) L-G (ii) L-L (iii) L-L-G fault occurs at (q) bus. Neglect the fault impedances. Data for equipments are (in p.u.): G1: Xd'' = j0.16, X2 = j0.17, X0 = j0.06 (printed as X1); G2: Xd'' = j0.2, X2 = j0.22, X0 = j0.15 (printed as X1); T1: X1 = X2 = X0 = j0.1; T2: X1 = X2 = X0 = j0.1; Line: X1 = X2 = j0.11, X0 = j0.33. [Figure: G1 (Y grounded, E = 1∠0°) - bus p - T1 (Δ/Y grounded) - line - T2 (Y grounded/Δ) - bus q - G2 (Y grounded)]

## Short answer

At bus q, Z1=j0.1403, Z2=j0.1509, Z0=j0.15 pu. L-G: Ia=6.8∠-90° pu, Ib=Ic=0. L-L: Ib=5.95∠180°, Ic=5.95∠0° pu. L-L-G: Ib=6.96∠149.9°, Ic=6.96∠30.1° pu, ground current 6.98 pu.

## Answer

**Assumptions:** $E_1 = E_2 = 1\angle0^\circ$ pu, no pre-fault load, all values on a common base, $Z_f = 0$. The values printed as "X1" are the zero-sequence reactances ($X_0$).

#### Sequence networks seen from bus q
```text
Positive (negative similar, no emf):
 E1-j0.16-(p)-T1 j0.1-line j0.11-T2 j0.1-(q)-j0.2-E2
                                          |
                                          F
Zero:
 T1 delta on p side, T2 delta on q side
 -> line + transformers form a path that does not
    reach q.   q --j0.15-- ref  (G2 solidly grounded)
```

$$
\begin{aligned}
Z_1 &= j(0.16+0.1+0.11+0.1)\ \|\ j0.2 = j0.47 \| j0.2 = j0.1403\ \text{pu} \\
Z_2 &= j(0.17+0.1+0.11+0.1)\ \|\ j0.22 = j0.48 \| j0.22 = j0.1509\ \text{pu} \\
Z_0 &= j0.15\ \text{pu (only G2; the Δ of T2 blocks the rest)}
\end{aligned}
$$

#### (i) L-G fault (phase a)
$$
\begin{aligned}
I_{a1} = I_{a2} = I_{a0} &= \frac{1}{j(0.1403+0.1509+0.15)} = \frac{1}{j0.4412} = -j2.267\ \text{pu} \\
I_a &= 3I_{a1} = -j6.800\ \text{pu}, \quad I_b = I_c = 0
\end{aligned}
$$

#### (ii) L-L fault (phases b, c)
$$
\begin{aligned}
I_{a1} = -I_{a2} &= \frac{1}{j(0.1403+0.1509)} = -j3.435\ \text{pu}, \quad I_{a0} = 0 \\
I_b &= -j\sqrt3\,I_{a1} = -5.949\ \text{pu} = 5.949\angle180^\circ \\
I_c &= 5.949\angle0^\circ\ \text{pu}, \quad I_a = 0
\end{aligned}
$$

#### (iii) L-L-G fault (phases b, c to ground)
$$
\begin{aligned}
Z_2\|Z_0 &= \frac{0.1509\times0.15}{0.3009} = j0.0752 \\
I_{a1} &= \frac{1}{j(0.1403+0.0752)} = -j4.640\ \text{pu} \\
I_{a2} &= -I_{a1}\frac{Z_0}{Z_2+Z_0} = j2.313\ \text{pu} \\
I_{a0} &= -I_{a1}\frac{Z_2}{Z_2+Z_0} = j2.327\ \text{pu} \\
I_b &= I_{a0}+a^2I_{a1}+aI_{a2} = 6.960\angle{149.91^\circ}\ \text{pu} \\
I_c &= I_{a0}+aI_{a1}+a^2I_{a2} = 6.960\angle{30.09^\circ}\ \text{pu} \\
I_n &= 3I_{a0} = j6.980\ \text{pu}, \quad I_a = 0
\end{aligned}
$$

#### Summary
| Fault | $I_a$ (pu) | $I_b$ (pu) | $I_c$ (pu) | Ground current (pu) |
|---|---|---|---|---|
| L-G | 6.800∠−90° | 0 | 0 | 6.800 |
| L-L | 0 | 5.949∠180° | 5.949∠0° | 0 |
| L-L-G | 0 | 6.960∠149.9° | 6.960∠30.1° | 6.980 |

The L-G current is high because only G2's small $X_0$ (0.15) is in the zero-sequence path.

## Check yourself

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

1. Why does the zero-sequence impedance at bus q equal only G2's j0.15?
   A. The line has high X0
   B. T2's delta on the q side blocks the rest of the network
   C. G1 is ungrounded
   D. Fault impedance is zero
   Answer: B. The delta of T2 on bus q's side isolates the line and G1 from q for zero-sequence currents. (Key point: Zero network at q)

2. For the L-G fault, the three sequence networks are connected and give Ia as:
   A. In series; Ia = 3 x 1/(Z1+Z2+Z0) = 6.8 pu
   B. In parallel; Ia = 1/Z0
   C. In series; Ia = 1/(Z1+Z2+Z0)
   D. Positive and negative in parallel; Ia = 0
   Answer: A. Series connection gives Ia1 = Ia2 = Ia0 = 1/j0.4412, and Ia = 3Ia1 = 6.8 pu. (Key point: L-G fault connection)

3. In the L-L fault, which statement is correct?
   A. Zero sequence carries Ia0 = Ia1
   B. Ia1 = Ia2 and ground current is large
   C. Ib = -Ic = sqrt3 Ia1 = 5.95 pu, no ground current
   D. Ia = 5.95 pu
   Answer: C. Ia0 = 0, Ia2 = -Ia1 = -j3.435, giving |Ib| = |Ic| = 5.949 pu and Ia = 0. (Key point: L-L fault current)

4. In the L-L-G fault, the ground (neutral) current is:
   A. Ia1 = 4.64 pu
   B. 3Ia0 = 6.98 pu
   C. Ib = 6.96 pu
   D. Ia2 + Ia0
   Answer: B. Ground current In = Ib + Ic = 3Ia0 = j6.98 pu. (Key point: L-L-G ground current)

## Related in Power System Analysis II

- [SLG fault at bus 3, currents and voltages](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aslg-fault-bus3-three-generators) — 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
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- [L-G fault at bus 1 through Zf](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-bus1-parallel-lines) — Power System Analysis II, chapter 5
- [Fault through impedance at bus 2](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-impedance-bus2-actual-currents) — Power System Analysis II, chapter 3
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