---
title: "Fault at receiving end of 220 kV lines"
subject: "Power System Analysis II"
chapter: 3
key: "semester_5/power_system_analysis_ii:fault-220kv-two-generators"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/fault-220kv-two-generators
site: BCT Engineering
---

# Fault at receiving end of 220 kV lines

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

- Asked in: 2074 Asoj, 10 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-220kv-two-generators
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3

## Question

Figure below shows a generating station feeding a 220 kV system. Determine the total fault current, fault level and fault current supplied by each generator for a three phase fault at the receiving end of the line. G1: 11 kV, 100 MVA, X''g1 = j0.15; G2: 11 kV, 75 MVA, X''g2 = j0.125; T1: 100 MVA, XT1 = j0.10, 11/220 kV; T2: 75 MVA, XT2 = j0.08, 11/220 kV. [Figure: G1 through T1 and G2 through T2 to a 220 kV busbar; two parallel lines of j42 Ω each from the busbar to the fault point F]

## Short answer

On 100 MVA base, total fault current is 1509 A at 220 kV (574.8 MVA); G1 supplies 15.76 kA and G2 supplies 14.41 kA at 11 kV.

## Answer

Base: 100 MVA, 11 kV (generators) and 220 kV (lines). Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation).

#### Per-unit reactances (100 MVA)
| Element | Calculation | pu |
|---|---|---|
| G1 | 0.15 | 0.15 |
| T1 | 0.10 | 0.10 |
| G2 | $0.125 \times 100/75$ | 0.1667 |
| T2 | $0.08 \times 100/75$ | 0.1067 |
| Each line | $42/(220^2/100) = 42/484$ | 0.0868 |

```text
 G1-j0.15-T1-j0.10-+          +-j42 ohm-+
                   +-220 kV --+         +-- F
 G2-jX_G2-T2-jX_T2-+          +-j42 ohm-+
```

#### Thevenin reactance at F
$$
\begin{aligned}
X_a &= 0.15 + 0.10 = 0.25, \quad X_b = 0.1667 + 0.1067 = 0.2733 \\
X_{gen} &= \frac{0.25 \times 0.2733}{0.25 + 0.2733} = 0.1306 \\
X_{lines} &= 0.0868/2 = 0.0434 \\
X_{th} &= 0.1306 + 0.0434 = 0.1740\ \text{pu}
\end{aligned}
$$

#### Total fault current and fault level
$$
\begin{aligned}
I_f &= \frac{1}{0.1740} = 5.7484\ \text{pu} \\
I_{base,220} &= \frac{100 \times 10^6}{\sqrt{3} \times 220 \times 10^3} = 262.43\ \text{A} \\
I_f &= 5.7484 \times 262.43 = 1508.6\ \text{A} \\
\text{Fault level} &= \frac{100}{0.1740} = 574.84\ \text{MVA}
\end{aligned}
$$

#### Current supplied by each generator
The fault current divides between the two generator branches in inverse ratio of their reactances:

$$
\begin{aligned}
I_{G1} &= I_f\frac{X_b}{X_a + X_b} = 5.7484 \times \frac{0.2733}{0.5233} = 3.0023\ \text{pu} \\
I_{G2} &= I_f\frac{X_a}{X_a + X_b} = 5.7484 \times \frac{0.25}{0.5233} = 2.7460\ \text{pu}
\end{aligned}
$$

On the 11 kV side, $I_{base} = \dfrac{100 \times 10^6}{\sqrt{3} \times 11 \times 10^3} = 5248.6$ A:

| Quantity | pu | Amperes (11 kV side) | Amperes (220 kV side) |
|---|---|---|---|
| $I_{G1}$ | 3.0023 | 15758 | 787.9 |
| $I_{G2}$ | 2.7460 | 14413 | 720.7 |
| Total $I_f$ | 5.7484 | – | 1508.6 |

**Answer:** total fault current $\approx 1509$ A at 220 kV, fault level $\approx 574.8$ MVA; G1 supplies about 15.76 kA and G2 about 14.41 kA at 11 kV.

## Check yourself

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

1. Each 220 kV line is j42 ohm. On 100 MVA base its per-unit value is
   A. 42 x 484 pu
   B. 42/484 = 0.0868 pu
   C. 42/220 pu
   D. 42/(11^2/100) pu
   Answer: B. Z_base = 220^2/100 = 484 ohm, so 42/484 = 0.0868 pu. (Key point: Line per-unit)

2. How is X_th found?
   A. (0.25 || 0.2733) in series with the two lines in parallel (0.0434)
   B. 0.25 + 0.2733 + 0.0868
   C. (0.25 || 0.2733) in series with one line (0.0868)
   D. 0.25 || 0.2733 || 0.0868
   Answer: A. Generator branches are in parallel (0.1306); the two identical lines are in parallel (0.0434); both groups are in series, giving 0.1740 pu. (Key point: Network reduction)

3. Why does G1 (branch reactance 0.25) supply more current than G2 (0.2733)?
   A. It has a higher MVA base
   B. Current divides inversely with branch reactance, so the smaller reactance carries more
   C. Its transformer has a higher voltage
   D. Current divides in direct proportion to reactance
   Answer: B. I_G1 = I_f x X_b/(X_a + X_b) = 3.00 pu versus 2.75 pu for G2. (Key point: Current sharing)

4. The pu generator currents are converted to amperes using
   A. The 220 kV base current (262 A) for the 11 kV side currents
   B. The 11 kV base current (5248.6 A) for the currents at the generators
   C. Any base current, as the answer is the same
   D. The line current only
   Answer: B. At the generator terminals the voltage is 11 kV, so the base current is 100 MVA/(sqrt(3) x 11 kV) = 5248.6 A, giving about 15.76 kA and 14.41 kA. (Key point: Converting generator currents)

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