---
title: "Fault at both ends of a 50 Ω line"
subject: "Power System Analysis II"
chapter: 3
key: "semester_5/power_system_analysis_ii:fault-points-p-q-line"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/fault-points-p-q-line
site: BCT Engineering
---

# Fault at both ends of a 50 Ω line

Power System Analysis II · Chapter 3: Power System Fault Calculation

- Asked in: 2069 Chaitra, 6+6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/fault-points-p-q-line
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3

## Question

The single line diagram of a power system is shown in figure below. Compute the fault current in p.u. and in absolute value if a 3 phase to ground fault occurs at (consider no fault impedance): i. the point P i.e. connection point between transmission line and HV side of transformer T1 ii. the point Q i.e. connection point between transmission line and HV side of transformer T2 (printed as T1). [Figure: G1 6 kV, 40 MVA, X = 10% - bus - T1 6.6/132 kV, 50 MVA, X = 10% - point P - line X = 50 Ω - point Q - T2 132/6.6 kV, 50 MVA, X = 10% - bus with load - G2 6 kV, 40 MVA, X = 10%]

## Short answer

Using 50 MVA, 132 kV base, per-unit reactances: G1,G2 0.1033, T1,T2 0.10, line 0.1435. At P, Thevenin reactance is parallel combination of (0.2033 and 0.3468) = 0.1282 pu. Fault current = 1/0.1282 = 7.80 pu = 1706 A (1.71 kA). Same at Q due to symmetry.

## Answer

For a bolted 3-phase fault, $I_f = 1/X_{th}$ pu, where $X_{th}$ is the parallel combination of the reactances from the fault point back to each generator (prefault voltage 1 pu, load current neglected).

#### Data and per unit values
Base: 50 MVA; 132 kV on the line, hence 6.6 kV on the generator buses (from the 6.6/132 kV transformers). $Z_{b,line} = 132^2/50 = 348.48\ \Omega$.

| Element | Calculation | X (pu) |
|---|---|---|
| G1, G2 (6 kV, 40 MVA, 10%) | $0.10 \times \frac{50}{40} \times \left(\frac{6}{6.6}\right)^2$ | 0.1033 each |
| T1, T2 (50 MVA, 10%) | on own base | 0.10 each |
| Line (50 Ω) | $50/348.48$ | 0.1435 |

Base current at 132 kV: $I_b = \dfrac{50\times10^6}{\sqrt3 \times 132\times10^3} = 218.7$ A.

```text
 G1      T1      P    Line     Q     T2     G2
(~)-0.1033-0.10--o--0.1435--o--0.10-0.1033-(~)
 |                                          |
 +============= reference (EMF 1 pu) =======+
```

#### i. Fault at point P
$$
\begin{aligned}
X_{left} &= X_{G1} + X_{T1} = 0.1033 + 0.10 = 0.2033 \\
X_{right} &= X_{line} + X_{T2} + X_{G2} = 0.1435 + 0.10 + 0.1033 = 0.3468 \\
X_{th} &= \frac{0.2033 \times 0.3468}{0.2033 + 0.3468} = 0.1282\ \text{pu} \\
I_f &= \frac{1}{0.1282} = 7.80\ \text{pu} \\
I_f &= 7.80 \times 218.7 = 1706\ \text{A}
\end{aligned}
$$
Contribution from G1 side $= 7.80 \times 0.3468/0.5501 = 4.92$ pu, from G2 side $= 2.88$ pu. Fault MVA $= 50/0.1282 = 390$ MVA.

#### ii. Fault at point Q
$$
\begin{aligned}
X_{left} &= X_{G1} + X_{T1} + X_{line} = 0.3468 \\
X_{right} &= X_{T2} + X_{G2} = 0.2033 \\
X_{th} &= \frac{0.3468 \times 0.2033}{0.5501} = 0.1282\ \text{pu} \\
I_f &= 7.80\ \text{pu} = 1706\ \text{A}
\end{aligned}
$$
Now G2 side supplies 4.92 pu and G1 side 2.88 pu.

The network is symmetrical about the middle of the line, so the fault current magnitude is the same at P and Q; only the sharing between the generators changes.

**Answer: At P and at Q, $I_f$ = 7.80 pu ≈ 1706 A (≈ 1.71 kA at 132 kV), fault level ≈ 390 MVA.**

## Check yourself

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

1. Why is the generator reactance multiplied by (6/6.6)^2?
   A. Because the generator rating is in MVA
   B. To correct 6 kV rating to the 6.6 kV base voltage of that zone
   C. Because the line voltage is 132 kV
   D. It is not needed
   Answer: B. The base on the generator bus is 6.6 kV (from the transformer ratio), but the machine is rated 6 kV, so X must be referred to the base voltage. (Key point: Voltage correction for generator reactance)

2. At P, the two paths to the sources are
   A. 0.2033 (G1 + T1) and 0.3468 (line + T2 + G2) in parallel
   B. 0.1033 and 0.10 in parallel
   C. All elements in series
   D. 0.1435 only
   Answer: A. X_th = 0.2033 x 0.3468/0.5501 = 0.1282 pu, giving I_f = 7.80 pu = 1706 A. (Key point: Thevenin reactance at P)

3. Why is the fault current at Q the same as at P?
   A. The line has no reactance
   B. The fault current is zero at Q
   C. The network is symmetrical about the middle of the line
   D. Because Q is closer to G1
   Answer: C. Both ends have identical total series reactances, so X_th = 0.1282 pu at both points; only the sharing between generators changes. (Key point: Fault at Q versus P)

4. The 7.80 pu is converted to 1706 A using
   A. The 6.6 kV base current
   B. I_b = 50 MVA/(sqrt(3) x 132 kV) = 218.7 A
   C. The 6 kV rated current
   D. The 50 ohm line current
   Answer: B. P and Q are on the 132 kV line, so the 132 kV base current applies: 7.80 x 218.7 = 1706 A. (Key point: Base current)

## Related in Power System Analysis II

- [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
- [Mid-line fault cleared at 60°, E = 1.04](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Amid-line-fault-50mva-cleared-60) — Power System Analysis II, chapter 6
- [No zero sequence in line-to-line fault](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-zero-sequence-absent) — Power System Analysis II, chapter 5, asked 2×
- [Mid-line fault, cleared at 60°](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Amid-line-fault-cleared-60) — Power System Analysis II, chapter 6
- [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-L fault at point B](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-point-b-transformers) — Power System Analysis II, chapter 5

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- [Fault current: two 50 MW generators, reactor inductance](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-two-50mw-inductance) — Power Plant Equipment, chapter 4
- [Maximum power with 132 kV at both ends](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Amax-power-132kv-both-ends) — Power System Analysis I, chapter 6
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