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

# L-L fault at point q with two motors

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

- Asked in: 2071 Shrawan, 3+7 marks
- Video: [Example 3.4.  Line to line short circuit](https://www.youtube.com/watch?v=vx_TPt5qKWU) — Francisco Gonzalez-Longatt
- 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-q-motors
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

A single line diagram of a power system network is shown in figure below. The system data is given in the table below:

| Element | X1 (pu) | X2 (pu) | X0 (pu) |
|---|---|---|---|
| G | 0.1 | 0.12 | 0.05 |
| M1 | 0.05 | 0.06 | 0.025 |
| M2 | 0.05 | 0.06 | 0.025 |
| T1 | 0.07 | 0.07 | 0.07 |
| T2 | 0.08 | 0.08 | 0.08 |
| Line | 0.1 | 0.1 | 0.3 |

i) Draw sequence networks ii) Find fault current for a line-to-line fault on phase b and c at point q. Assume 1.0 pu pre-fault voltage throughout. [Figure: G - bus p - T1 (Δ/Y grounded) - bus q - line - bus r - T2 (Y grounded/Δ) - bus s, which feeds motors M1 (Y, ungrounded) and M2 (Y grounded)]

## Short answer

Positive/negative Thevenin impedances at bus q: Z1=j0.0929 pu, Z2=j0.0998 pu. For the L-L fault, Ia1 = -Ia2 = -j5.19 pu. Fault currents: Ib = -8.99 pu, Ic = 8.99 pu, Ia = 0.

## Answer

**Assumptions:** all reactances on a common base; pre-fault voltage 1 pu everywhere (no load current); bolted fault between phases b and c at q.

#### (i) Sequence networks
```text
Positive (emfs Eg, Em):
 Eg-j0.1-(p)-j0.07-(q)-j0.1-(r)-j0.08-(s)-+-j0.05-Em1
                    |                     +-j0.05-Em2
                    F
Negative: same layout, G j0.12, T1 j0.07,
 line j0.1, T2 j0.08, M1 j0.06, M2 j0.06, no emfs
Zero:
 ref-j0.07-(q)-j0.3-(r)-j0.08-ref
 G (j0.05) cut off by T1 delta; M1 ungrounded,
 M2 (j0.025) cut off by T2 delta
```
T1 is Δ/Yg with the grounded star at q, and T2 is Yg/Δ with the grounded star at r, so the zero network is the line between two transformer reactances to ground.

#### (ii) L-L fault at q
Motors in parallel: positive $0.05\|0.05 = 0.025$, negative $0.06\|0.06 = 0.03$.
$$
\begin{aligned}
Z_1 &= j(0.1+0.07)\ \|\ j(0.1+0.08+0.025) = j0.17 \| j0.205 = j0.0929\ \text{pu} \\
Z_2 &= j(0.12+0.07)\ \|\ j(0.1+0.08+0.03) = j0.19 \| j0.21 = j0.0998\ \text{pu}
\end{aligned}
$$
Sequence currents ($I_{a0} = 0$):
$$
I_{a1} = -I_{a2} = \frac{1}{j(0.0929+0.0998)} = \frac{1}{j0.1927} = -j5.190\ \text{pu}
$$
Fault current:
$$
\begin{aligned}
I_b &= -j\sqrt3 I_{a1} = -8.989\ \text{pu} \\
I_c &= -I_b = 8.989\ \text{pu}
\end{aligned}
$$

**Answer: fault current for L-L fault at q = 8.99 pu** (in phases b and c, opposite in sign; $I_a = 0$). The zero-sequence network does not take part in an L-L fault.

## Check yourself

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

1. Why is the zero network just two transformer reactances to ground with the line between?
   A. Motors have X0 = 0
   B. Delta windings cut off G, M1 and M2, leaving only the grounded stars of T1 and T2
   C. The line is open
   D. T1 and T2 are both delta
   Answer: B. T1's delta faces G, T2's delta faces the motors, and M1 is ungrounded, so only the grounded stars connect to reference. (Key point: Zero network cut-off)

2. Positive reactance of the two motors seen from bus s is:
   A. 0.10 pu
   B. 0.025 pu
   C. 0.05 pu
   D. 0.0125 pu
   Answer: B. Two 0.05 pu motors in parallel give 0.025 pu (negative: 0.06||0.06 = 0.03). (Key point: Motors in parallel)

3. The L-L fault current at q (Z1 = j0.0929, Z2 = j0.0998) is:
   A. 5.19 pu
   B. 10.4 pu
   C. 8.99 pu
   D. 3.0 pu
   Answer: C. Ia1 = 1/j0.1927 = 5.19 pu and I = sqrt3 x 5.19 = 8.99 pu; zero sequence is not used. (Key point: L-L current)

## Related in Power System Analysis II

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