---
title: "Fault at receiving end of 132 kV lines"
subject: "Power System Analysis II"
chapter: 3
key: "semester_5/power_system_analysis_ii:fault-132kv-receiving-end-2071"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/fault-132kv-receiving-end-2071
site: BCT Engineering
---

# Fault at receiving end of 132 kV lines

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

- Asked in: 2071 Chaitra, 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-132kv-receiving-end-2071
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3

## Question

Figure below shows power system fed by two generators. The rating and reactances of the equipments are shown. A 3-phase balanced short circuit fault occurs at the receiving end bus of 132 kV. Find (a) Fault level and fault current at receiving end (b) Fault current supplied by generators G1 and G2. Note that reactance of machines are based on their own rating. [Figure: G1 100 MVA, 11 kV, 15% - T1 100 MVA, 11/132 kV, X = 8%; G2 50 MVA, 11 kV, 10% - T2 50 MVA, 11/132 kV, X = 8%; both transformers feed the 132 kV bus; Line-1 and Line-2, each X = 0.2 Ω/km/phase, in parallel to the receiving end bus F (line length not given)]

## Short answer

Assuming 200 km lines (40 Ω each), on100 MVA base X_th =0.255 pu. Fault level =392 MVA, fault current =1.71 kA at132 kV. At11 kV, G1 supplies12.55 kA, G2 supplies8.02 kA.

## Answer

Fault level at a bus is $\text{MVA}_{base}/X_{th}$ (pu), where $X_{th}$ is the total reactance from the source EMFs (1 pu) to the fault.

**Assumption:** the line length is not given in the question, so a length of **200 km** is assumed. Each line then has $X = 0.2 \times 200 = 40\ \Omega$. G2 is 50 MVA, 11 kV, 10%; T2 is 50 MVA, 11/132 kV, 8%.

#### Step 1: Per unit values on 100 MVA base
Base: 100 MVA, 11 kV (generators), 132 kV (lines). $Z_{b,132} = 132^2/100 = 174.24\ \Omega$.

| Element | Calculation | X (pu) |
|---|---|---|
| G1 | 0.15 (own base = 100 MVA) | 0.15 |
| T1 | 0.08 | 0.08 |
| G2 | $0.10 \times 100/50$ | 0.20 |
| T2 | $0.08 \times 100/50$ | 0.16 |
| Each line | $40/174.24$ | 0.2296 |

#### Step 2: Reactance diagram and reduction
```text
 G1 0.15  T1 0.08
 (~)--------||---+               Line-1 0.2296
                 |-- 132 kV bus ====+=========+--- F
 (~)--------||---+               Line-2 0.2296
 G2 0.20  T2 0.16
```
$$
\begin{aligned}
X_1 &= 0.15 + 0.08 = 0.23, \quad X_2 = 0.20 + 0.16 = 0.36 \\
X_{gen} &= \frac{0.23 \times 0.36}{0.23 + 0.36} = 0.1403 \\
X_{line} &= \frac{0.2296}{2} = 0.1148 \\
X_{th} &= 0.1403 + 0.1148 = 0.2551\ \text{pu}
\end{aligned}
$$

#### (a) Fault level and fault current at the receiving end
$$
\begin{aligned}
I_f &= \frac{1}{0.2551} = 3.920\ \text{pu} \\
\text{Fault level} &= \frac{100}{0.2551} = 392.0\ \text{MVA} \\
I_{b,132} &= \frac{100\times10^6}{\sqrt3 \times 132\times10^3} = 437.4\ \text{A} \\
I_f &= 3.920 \times 437.4 = 1714\ \text{A} = 1.714\ \text{kA}
\end{aligned}
$$

#### (b) Currents supplied by G1 and G2
The fault current divides between the two generator branches inversely as their reactances:
$$
\begin{aligned}
I_{G1} &= I_f\,\frac{X_2}{X_1+X_2} = 3.920 \times \frac{0.36}{0.59} = 2.392\ \text{pu} \\
I_{G2} &= I_f\,\frac{X_1}{X_1+X_2} = 3.920 \times \frac{0.23}{0.59} = 1.528\ \text{pu}
\end{aligned}
$$
Base current on the 11 kV side: $I_{b,11} = 100\times10^6/(\sqrt3\times 11\times10^3) = 5248.6$ A.
$$
I_{G1} = 2.392 \times 5248.6 = 12.55\ \text{kA}, \qquad I_{G2} = 1.528 \times 5248.6 = 8.02\ \text{kA}
$$

**Answer (200 km lines): fault level ≈ 392 MVA, fault current ≈ 1.71 kA at 132 kV; G1 supplies ≈ 12.55 kA and G2 ≈ 8.02 kA at 11 kV.** For any other line length $l$ km, only $X_{line} = 0.2l/(2\times174.24)$ pu changes in Step 2.

## Check yourself

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

1. With the assumed 200 km length, each line's reactance and per-unit value are
   A. 0.2 ohm, 0.0011 pu
   B. 40 ohm, 40/174.24 = 0.2296 pu
   C. 40 ohm, 40/100 pu
   D. 200 ohm, 1.15 pu
   Answer: B. X = 0.2 x 200 = 40 ohm; Z_base = 132^2/100 = 174.24 ohm. (Key point: Line reactance assumption)

2. G2 (50 MVA, 10%) and T2 (50 MVA, 8%) on 100 MVA base are
   A. 0.10 and 0.08
   B. 0.05 and 0.04
   C. 0.20 and 0.16
   D. 0.10 and 0.16
   Answer: C. Multiply by 100/50 = 2: 0.20 and 0.16 pu. (Key point: Converting G2 and T2)

3. X_th = (0.23||0.36) + 0.2296/2 = 0.2551 pu. The fault level is
   A. 100/0.2551 = 392 MVA
   B. 100 x 0.2551 MVA
   C. 100/0.1403 MVA
   D. 100/0.1148 MVA
   Answer: A. Parallel lines halve to 0.1148 pu and add to the generator part 0.1403; MVA = 100/0.2551 = 392. (Key point: Fault level)

4. I_G1 = I_f x 0.36/0.59 = 2.39 pu, converted using
   A. The 132 kV base current 437.4 A
   B. The 11 kV base current 5248.6 A, giving 12.55 kA
   C. No conversion
   D. The 11 kV base current but divided by 3
   Answer: B. Generator currents flow at 11 kV, so multiply by 5248.6 A. (Key point: Generator contributions)

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