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

# L-L fault through Zf with two motors

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

- Asked in: 2083 Baishakh (new course), 7 marks
- Video: [LL Fault On Unloaded Synchronous Generator With Zf | Unsymmetrical Faults | Power System Analysis](https://www.youtube.com/watch?v=WvvaAo93uIQ) — Learn2Earn 4Life
- 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-motors-zf-new
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

For the power system network shown below, if Line to Line fault occurs at point 'F' then find the fault current, take fault impedance Zf = 0.1 p.u. For Generator: X1 = X2 = 20%, X0 = 5%; For Both Motor: X1 = X2 = 25%, X0 = 10%; For Both Transformer: X1 = X2 = X0 = 10%. The neutral reactance of generator G1 and motor M2 are in p.u. [Figure: G1 25 MVA, 11 kV, Y grounded through Xn = 0.1 - T1 30 MVA, 10.8/121 kV, Δ/Y grounded - line X1 = X2 = 100 Ω, X0 = 250 Ω - T2 30 MVA, 121/10.8 kV, Y grounded/Δ - motor bus with M1 (15 MVA, 10 kV, Y ungrounded) and M2 (7.5 MVA, 10 kV, Y grounded through Xn = 0.036); fault F at the motor bus]

## Short answer

On 25 MVA base, positive/negative Thevenin impedance at the motor bus is Z1 = Z2 = j0.1597 pu. With Zf = j0.1 pu, Ia1 = -j2.384 pu, and L-L fault current is √3 × 2.384 = 4.13 pu = 5418 A.

## Answer

**Assumptions:** base 25 MVA, 11 kV in the generator circuit; pre-fault voltage at F is 1 pu with no load current; $Z_f = j0.1$ pu. An L-L fault involves only positive and negative sequence networks, so the neutral reactances and zero-sequence data are not needed.

#### Base voltages
$$
V_{b,line} = 11 \times \frac{121}{10.8} = 123.24\ \text{kV}, \quad V_{b,motor} = 123.24 \times \frac{10.8}{121} = 11\ \text{kV}
$$

#### Per-unit reactances (25 MVA)
| Element | Calculation | $X_1 = X_2$ (pu) |
|---|---|---|
| G1 | given | 0.20 |
| T1, T2 | $0.1 \times \frac{25}{30} \times (\frac{10.8}{11})^2$ | 0.0803 each |
| Line | $100 / \frac{123.24^2}{25}$ | 0.1646 |
| M1 | $0.25 \times \frac{25}{15} \times (\frac{10}{11})^2$ | 0.3444 |
| M2 | $0.25 \times \frac{25}{7.5} \times (\frac{10}{11})^2$ | 0.6887 |

#### Positive (and negative) sequence network at F
```text
 Eg-j0.2-T1 j0.0803-line j0.1646-T2 j0.0803-(F)
                                             |
                               +-------------+
                               |             |
                            j0.3444       j0.6887
                              Em1           Em2
```
$$
\begin{aligned}
X_{gen\ side} &= 0.2 + 0.0803 + 0.1646 + 0.0803 = 0.5253 \\
X_{motors} &= 0.3444 \| 0.6887 = 0.2296 \\
Z_1 = Z_2 &= j(0.5253 \| 0.2296) = j0.1597\ \text{pu}
\end{aligned}
$$

#### L-L fault through $Z_f$
$$
\begin{aligned}
I_{a1} = -I_{a2} &= \frac{V_f}{Z_1 + Z_2 + Z_f} = \frac{1}{j(0.1597 + 0.1597 + 0.1)} = \frac{1}{j0.4195} = -j2.384\ \text{pu} \\
|I_f| = |I_b| = |I_c| &= \sqrt3 \times 2.384 = 4.129\ \text{pu}
\end{aligned}
$$
Base current at the motor bus (11 kV):
$$
I_{base} = \frac{25\times10^6}{\sqrt3 \times 11\times10^3} = 1312.2\ \text{A}
$$
$$
I_f = 4.129 \times 1312.2 = 5418\ \text{A}
$$

**Answer: L-L fault current at F ≈ 4.13 pu ≈ 5.42 kA.**

## Check yourself

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

1. Why are the neutral reactances and X0 data not needed?
   A. Motors are ungrounded
   B. An L-L fault involves only positive and negative networks
   C. Zf cancels them
   D. They are negligible
   Answer: B. Ia0 = 0 in an L-L fault, so the zero network is unused. (Key point: Sequence networks used)

2. How does Zf enter the L-L fault current?
   A. Ia1 = V/(Z1+Z2)
   B. Ia1 = V/(Z1+Z2+Zf)
   C. Ia1 = V/(Z1+Z2+3Zf)
   D. Ia1 = V/(Z1+Z2+Zf/2)
   Answer: B. The fault impedance is in series between phases, giving Ia1 = 1/j(0.1597+0.1597+0.1). (Key point: Fault impedance)

3. Which is used to convert a transformer or motor to the 25 MVA, 11 kV base?
   A. X x (new MVA/old MVA) x (old kV/new kV)^2
   B. X x (old MVA/new MVA)
   C. X x (new kV/old kV)
   D. X x (new MVA/old MVA) x (old kV/new kV)
   Answer: A. X_new = X_old x (MVA_new/MVA_old) x (kV_old/kV_new)^2, e.g. T1: 0.1 x 25/30 x (10.8/11)^2. (Key point: Base conversion)

## Related in Power System Analysis II

- [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
- [L-L fault at point q with two motors](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-point-q-motors) — Power System Analysis II, chapter 5
- [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
- [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
- [Fault at generator terminal feeding motor](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-generator-terminal-motor-new) — Power System Analysis II, chapter 3
- [Fault at both ends of a 50 Ω line](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-points-p-q-line) — Power System Analysis II, chapter 3

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- [Impedance diagram: 25 MVA generator, two motors](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aimpedance-diagram-25mva-two-motors) — Power System Analysis I, chapter 3
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