---
title: "L-L fault at point B"
subject: "Power System Analysis II"
chapter: 5
key: "semester_5/power_system_analysis_ii:ll-fault-point-b-transformers"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ll-fault-point-b-transformers
site: BCT Engineering
---

# L-L fault at point B

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

- Asked in: 2081 Bhadra, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ll-fault-point-b-transformers
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

For the power system network shown below, compute the fault current if a L-L fault occurs at point B. Assume pre-fault voltage at the point of fault is 1 p.u. [Figure: generator 20 MVA, 6 kV, Z1 = 10%, Z2 = 5%, Z0 = 2%, Zn = j2 Ω at bus A; transformer 50 MVA, 6/66 kV, Δ/Y, Y solidly grounded, X = 4%, between A and B; line X = 30 Ω between B and C; transformer 40 MVA, 66 kV/3.3 kV, Y/Δ, Y isolated, X = 4%, between C and D; machine 20 MVA, 3.3 kV, Z1 = 10%, Z2 = 5%, Z0 = 2%, Zn = j2 Ω at bus D]

## Short answer

Only positive and negative sequence networks are needed. On a 20 MVA base, Z1=j0.0800 pu, Z2=j0.0501 pu. L-L fault current magnitude = sqrt(3)/(Z1+Z2)=13.31 pu. Base current at 66 kV is 174.95 A, giving fault current 13.31*174.95 = 2.33 kA in phases b and c.

## Answer

**Assumptions:** base 20 MVA; base voltages 6 kV (generator side), 66 kV (line), 3.3 kV (motor side). The L-L fault needs only the positive and negative sequence networks, so zero-sequence data and the neutral impedances are not used. Pre-fault voltage $V_f = 1$ pu, no-load.

#### Per-unit reactances (20 MVA base)
$$
\begin{aligned}
X_{T1} &= 0.04 \times \frac{20}{50} = 0.016\ \text{pu} \\
Z_{base,66} &= \frac{66^2}{20} = 217.8\ \Omega,\quad X_{line} = \frac{30}{217.8} = 0.1377\ \text{pu} \\
X_{T2} &= 0.04 \times \frac{20}{40} = 0.02\ \text{pu} \\
\text{Generator: } X_1 &= 0.10,\ X_2 = 0.05;\quad \text{Machine at D: } X_1 = 0.10,\ X_2 = 0.05
\end{aligned}
$$

#### Sequence networks seen from B
```text
Positive:
 E-j0.1-A-j0.016-B-j0.1377-C-j0.02-D-j0.1-E
                |
           (fault point)
Negative:  same, with j0.05 for both machines, no emf
```
Thevenin impedances at B:
$$
\begin{aligned}
Z_1 &= j(0.10+0.016)\ \|\ j(0.1377+0.02+0.10) = j0.116\ \|\ j0.2577 = j0.0800\ \text{pu} \\
Z_2 &= j(0.05+0.016)\ \|\ j(0.1377+0.02+0.05) = j0.066\ \|\ j0.2077 = j0.0501\ \text{pu}
\end{aligned}
$$

#### L-L fault current (phases b and c)
$$
\begin{aligned}
I_{a1} &= -I_{a2} = \frac{V_f}{Z_1+Z_2} = \frac{1}{j0.1301} = -j7.687\ \text{pu} \\
|I_f| = |I_b| = |I_c| &= \sqrt3\,|I_{a1}| = \frac{\sqrt3}{0.1301} = 13.31\ \text{pu}
\end{aligned}
$$
Base current at B (66 kV):
$$
I_{base} = \frac{20 \times 10^6}{\sqrt3 \times 66 \times 10^3} = 174.95\ \text{A}
$$
$$
I_f = 13.31 \times 174.95 = 2329.5\ \text{A}
$$

**Answer: L-L fault current at B ≈ 13.31 pu ≈ 2.33 kA** (in phases b and c, opposite in direction).

## Check yourself

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

1. Which sequence networks are needed for the L-L fault at B?
   A. Positive, negative and zero
   B. Only positive
   C. Positive and negative only
   D. Only zero
   Answer: C. Ia0 = 0 for L-L faults, so zero sequence data and neutral impedances are not used. (Key point: Networks needed for L-L fault)

2. What is the per-unit reactance of T1 (50 MVA, X = 4%) on a 20 MVA base?
   A. 0.10
   B. 0.04
   C. 0.016
   D. 0.0016
   Answer: C. 0.04 x 20/50 = 0.016 pu. (Key point: Per-unit conversion)

3. How is Z1 at B found?
   A. Sum of all reactances in series
   B. Parallel of the generator side j(0.10 + 0.016) and the other side j(0.1377 + 0.02 + 0.10), about j0.080
   C. Only the generator reactance j0.10
   D. Parallel of the two machines only
   Answer: B. From B the two paths to the sources are in parallel: 0.116 || 0.2577 = 0.0800 pu. (Key point: Thevenin impedance at B)

4. What is the L-L fault current magnitude in pu?
   A. sqrt3 Vf/(Z1 + Z2) = 13.31 pu
   B. Vf/(Z1 + Z2) = 7.69 pu
   C. 3Vf/(Z1 + Z2) = 23.1 pu
   D. Vf/Z1 = 12.5 pu
   Answer: A. |If| = sqrt3/0.1301 = 13.31 pu, times 174.95 A = about 2.33 kA. (Key point: L-L fault current)

## Related in Power System Analysis II

- [Equivalent Z0, Z1, Z2 at bus 3](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Asequence-thevenin-z0-z1-z2-bus3) — Power System Analysis II, chapter 5
- [Generator-reactor system fault MVA](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Agenerator-reactor-fault-mva) — Power System Analysis II, chapter 3
- [SLG fault on 20 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aslg-fault-20mva-generator) — Power System Analysis II, chapter 5
- [L-L fault on 40 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-40mva-generator) — Power System Analysis II, chapter 5
- [DLG fault on 30 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Adlg-fault-30mva-generator) — Power System Analysis II, chapter 5
- [L-L fault currents and voltages, 30 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-30mva-currents-voltages) — Power System Analysis II, chapter 5

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