---
title: "Fault through impedance at bus 2"
subject: "Power System Analysis II"
chapter: 3
key: "semester_5/power_system_analysis_ii:fault-impedance-bus2-actual-currents"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/fault-impedance-bus2-actual-currents
site: BCT Engineering
---

# Fault through impedance at bus 2

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

- Asked in: 2071 Shrawan, 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-impedance-bus2-actual-currents
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3

## Question

In the power system shown in figure below, the value marked are the per unit reactance taking 20 MVA and 11 kV as base values in the generator circuit. Both the transformers are rated for 11/110 kV. A three phase to ground fault with a fault impedance of 0.088 pu occurs at bus 2. Determine the actual values of fault current and the currents supplied by the generators. [Figure: G1 (XG1 = 0.15) - T1 (j0.1) - bus 1; G2 (XG2 = j0.15) - T2 (j0.225) - bus 3; line 1-3 j0.5; line 1-2 0.75; line 3-2 1.0]

## Short answer

Thevenin impedance at bus 2 is j0.5788 pu. With Z_f = j0.088 pu, fault current = 1.50 pu = 157.4 A at 110 kV. G1 supplies 0.883 pu (927 A) and G2 supplies 0.616 pu (647 A) at 11 kV, lagging by 90°.

## Answer

The fault current is found from the Thevenin (driving point) impedance at bus 2, $I_f = V_f/(Z_{22}+Z_f)$, with prefault voltage 1 pu. Bus voltages during the fault then give the generator currents.

**Assumptions:** all values are reactances (lines 1-2 = j0.75 and 3-2 = j1.0 pu), $Z_f = j0.088$ pu, no prefault load, prefault voltage 1 pu everywhere. Base: 20 MVA, 11 kV (generator side), so 110 kV on the line side.

#### Step 1: Reactance network
Generator + transformer branches: bus 1 to reference $= 0.15 + 0.10 = 0.25$; bus 3 to reference $= 0.15 + 0.225 = 0.375$.

```text
      (ref)               (ref)
        |                   |
      j0.25               j0.375
        |                   |
  bus 1 o------j0.5---------o bus 3
         \                 /
        j0.75           j1.0
           \             /
            o---- bus 2 (fault, Zf=j0.088)
```

#### Step 2: Bus admittance matrix and Z-bus (values of $-j$ / $j$ dropped, all reactive)
$$
Y_{bus} = -j\begin{bmatrix} 4+2+1.333 & -1.333 & -2 \\ -1.333 & 1.333+1 & -1 \\ -2 & -1 & 2.667+2+1 \end{bmatrix}
= -j\begin{bmatrix} 7.333 & -1.333 & -2 \\ -1.333 & 2.333 & -1 \\ -2 & -1 & 5.667 \end{bmatrix}
$$
Inverting:
$$
Z_{bus} = j\begin{bmatrix} 0.1884 & 0.1473 & 0.0925 \\ 0.1473 & 0.5788 & 0.1541 \\ 0.0925 & 0.1541 & 0.2363 \end{bmatrix}
$$
So the Thevenin impedance at bus 2 is $Z_{22} = j0.5788$ pu.

#### Step 3: Fault current
$$
\begin{aligned}
I_f &= \frac{1}{Z_{22}+Z_f} = \frac{1}{j(0.5788+0.088)} = -j1.4998\ \text{pu} \\
I_{b,110} &= \frac{20\times10^6}{\sqrt3 \times 110\times10^3} = 104.97\ \text{A} \\
|I_f| &= 1.4998 \times 104.97 = 157.4\ \text{A}
\end{aligned}
$$

#### Step 4: Bus voltages during fault ($V_i = 1 - Z_{i2} I_f$)
$$
\begin{aligned}
V_1 &= 1 - 0.1473 \times 1.4998 = 0.7791\ \text{pu} \\
V_2 &= 1 - 0.5788 \times 1.4998 = 0.1320\ \text{pu} \\
V_3 &= 1 - 0.1541 \times 1.4998 = 0.7689\ \text{pu}
\end{aligned}
$$
Check: $V_2 = Z_f I_f = 0.088 \times 1.4998 = 0.132$ pu.

#### Step 5: Generator currents
$$
\begin{aligned}
I_{G1} &= \frac{1 - V_1}{0.25} = \frac{0.2209}{0.25} = 0.8834\ \text{pu} \\
I_{G2} &= \frac{1 - V_3}{0.375} = \frac{0.2311}{0.375} = 0.6163\ \text{pu}
\end{aligned}
$$
Check: $0.8834 + 0.6163 = 1.4997 \approx I_f$.

Generator currents flow on the 11 kV side, base current $I_{b,11} = 20\times10^6/(\sqrt3 \times 11\times10^3) = 1049.7$ A:
$$
I_{G1} = 0.8834 \times 1049.7 = 927.4\ \text{A}, \qquad I_{G2} = 0.6163 \times 1049.7 = 647.0\ \text{A}
$$

**Answer: Fault current at bus 2 = 1.50 pu = 157.4 A (110 kV side); G1 supplies 0.883 pu = 927 A and G2 supplies 0.616 pu = 647 A (11 kV side), all lagging the prefault voltage by 90°.**

## Check yourself

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

1. With fault impedance Zf at bus 2, the fault current is
   A. 1/Z22
   B. 1/(Z22 + Zf)
   C. 1/(Z22 - Zf)
   D. 1/Zf
   Answer: B. The Thevenin impedance Z22 and the fault impedance are in series: I_f = V_f/(Z22 + Zf) = 1/j0.6668 = 1.50 pu. (Key point: Fault current with fault impedance)

2. Bus 2 voltage during the fault equals
   A. Zero, as in a bolted fault
   B. 1 pu
   C. Zf x I_f = 0.088 x 1.4998 = 0.132 pu
   D. Z22 x I_f
   Answer: C. The fault impedance still has a voltage drop across it, so V2 = Zf I_f. Also V2 = 1 - Z22 I_f gives the same value. (Key point: Bus voltage during fault)

3. How is G1's current found?
   A. I_f divided equally between the generators
   B. I_G1 = (1 - V1)/0.25, the drop across its generator and transformer reactance
   C. I_G1 = V1/0.25
   D. I_G1 = (1 - V2)/0.15
   Answer: B. The EMF is 1 pu and bus 1 voltage is V1 = 0.7791, so I_G1 = 0.2209/0.25 = 0.883 pu. (Key point: Generator currents)

4. Which base currents are used for the fault current and the generator currents?
   A. 1049.7 A for both
   B. 104.97 A (110 kV) for the fault; 1049.7 A (11 kV) for the generators
   C. 104.97 A for both
   D. 1049.7 A for the fault; 104.97 A for the generators
   Answer: B. Bus 2 is on the 110 kV side (base 104.97 A); the generators are on the 11 kV side (base 1049.7 A). (Key point: Base current for amperes)

## Related in Power System Analysis II

- [Generator-reactor system fault MVA](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Agenerator-reactor-fault-mva) — Power System Analysis II, chapter 3
- [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
- [Short circuit MVA of feeder circuit breaker](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afeeder-cb-short-circuit-mva) — Power System Analysis II, chapter 3, asked 2×
- [Neutral reactance for 15 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aneutral-reactance-15mva) — Power System Analysis II, chapter 5
- [L-L fault through Zf with two motors](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-motors-zf-new) — Power System Analysis II, chapter 5
- [Fault MVA at bus 5 with inductive lines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-mva-bus5-mh-lines) — Power System Analysis II, chapter 3, asked 2×

## Related in other subjects

- [Per unit circuit of three-generator system, 30 MVA base](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Apu-circuit-30mva-g1-g2-g3) — Power System Analysis I, chapter 3
- [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
- [Per unit values: 100 MVA generator, two motors](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aper-unit-100mva-6kv-two-motors) — Power System Analysis I, chapter 3
- [Reactance diagram: 300 MVA, 20 kV generator, two motors](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-300mva-20kv) — Power System Analysis I, chapter 3