---
title: "Fault at far end of 33 kV line"
subject: "Power System Analysis II"
chapter: 3
key: "semester_5/power_system_analysis_ii:fault-far-end-33kv-line"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/fault-far-end-33kv-line
site: BCT Engineering
---

# Fault at far end of 33 kV line

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

- Asked in: 2078 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/fault-far-end-33kv-line
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3

## Question

For the system shown in figure below, determine the fault current, fault level if three phase balanced short circuit fault occurs at the far end F. [Figure: G1 11 kV, 20 MVA, 15% and G2 11 kV, 10 MVA, 10% on a common 11 kV bus; transformer 30 MVA, 5%, ratio 1:3 (to 33 kV); line of (3 + j5) Ω at 33 kV to the fault point F at the far end]

## Short answer

On 30 MVA, 33 kV base with pre-fault voltage 1 pu and neglecting load current, the three-phase fault current at the far end is 1605 A, and fault level is 91.8 MVA.

## Answer

Base: 30 MVA (transformer rating), 11 kV on the generator side and 33 kV on the line side (ratio 1:3). Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation).

#### Per-unit impedances
| Element | Calculation | pu |
|---|---|---|
| G1 | $0.15 \times 30/20$ | 0.225 |
| G2 | $0.10 \times 30/10$ | 0.300 |
| Transformer | 5% on 30 MVA | 0.05 |
| Line | $(3 + j5)/Z_{base}$ | 0.0826 + j0.1377 |

$Z_{base}$ at 33 kV $= 33^2/30 = 36.3\ \Omega$.

```text
 E=1--jX_G1--+
             +--j0.05--(3+j5 ohm)--F
 E=1--jX_G2--+
```

#### Total impedance up to F
$$
\begin{aligned}
X_G &= \frac{0.225 \times 0.300}{0.225 + 0.300} = 0.1286\ \text{pu} \\
Z_{th} &= jX_G + j0.05 + Z_L = j0.1286 + j0.05 + (0.0826 + j0.1377) \\
&= 0.0826 + j0.3163 = 0.3269\angle 75.36^\circ\ \text{pu}
\end{aligned}
$$

#### Fault current and fault level
$$
\begin{aligned}
I_f &= \frac{1}{|Z_{th}|} = \frac{1}{0.3269} = 3.0588\ \text{pu} \\
I_{base} &= \frac{30 \times 10^6}{\sqrt{3} \times 33 \times 10^3} = 524.86\ \text{A} \\
I_f &= 3.0588 \times 524.86 = 1605.4\ \text{A} \\
\text{Fault level} &= \frac{S_{base}}{|Z_{th}|} = \frac{30}{0.3269} = 91.76\ \text{MVA}
\end{aligned}
$$

**Answer:** fault current $\approx 1605$ A at 33 kV, fault level $\approx 91.8$ MVA.

## Check yourself

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

1. G1 is rated 20 MVA, 15%. On the 30 MVA base, what is its per-unit reactance?
   A. 0.10 pu
   B. 0.225 pu
   C. 0.15 pu
   D. 0.300 pu
   Answer: B. X_new = X_old x S_new/S_old = 0.15 x 30/20 = 0.225 pu. (Key point: Per-unit base conversion)

2. How is the line's per-unit impedance found?
   A. Divide the ohmic value by 33^2/30 = 36.3 ohm
   B. Divide the ohmic value by 11^2/30 ohm
   C. Multiply the ohmic value by 36.3
   D. Divide by 33/30 ohm
   Answer: A. The line is in the 33 kV zone, so Z_base = kV^2/MVA = 33^2/30 = 36.3 ohm, and Z_pu = Z_ohm / Z_base. (Key point: Base impedance of the line)

3. How are the elements combined to get Z_th seen from the fault F?
   A. All elements in parallel
   B. G1 and G2 in series, then in parallel with the line
   C. G1 in parallel with G2, then in series with the transformer and line
   D. Only the line and transformer are used; generators are ignored
   Answer: C. Both generators feed the same bus, so they are in parallel; the transformer and line then lie in series between that bus and F. (Key point: Network reduction)

4. With |Z_th| = 0.3269 pu on a 30 MVA base, what is the fault level?
   A. 30 x 0.3269 = 9.8 MVA
   B. 30/0.3269 = 91.8 MVA
   C. 1/0.3269 = 3.06 MVA
   D. 30/(0.3269)^2 = 281 MVA
   Answer: B. Fault level = S_base/|Z_th| = 30/0.3269 = 91.76 MVA. (Key point: Fault level)

## Related in Power System Analysis II

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- [Fault at receiving end of 132 kV lines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-132kv-receiving-end-2071) — Power System Analysis II, chapter 3
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## Related in other subjects

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- [Reactance diagram: 20 MVA generator, 15 kV/25 MVA base](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-20mva-15kv-25mva) — Power System Analysis I, chapter 3
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