---
title: "DLG fault at bus 2, fault MVA"
subject: "Power System Analysis II"
chapter: 5
key: "semester_5/power_system_analysis_ii:dlg-fault-bus2-sequence-networks"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/dlg-fault-bus2-sequence-networks
site: BCT Engineering
---

# DLG fault at bus 2, fault MVA

Power System Analysis II · Chapter 5: Unsymmetrical Faults on Power Systems

- Asked in: 2072 Kartik, 10 marks
- Video: [Power Systems | Tutorial - 16 | Solved Problems on Unsymmetrical Faults](https://www.youtube.com/watch?v=sOQBAKTLM64) — Engineering Devotion
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/dlg-fault-bus2-sequence-networks
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

For the given power system network, draw sequence networks for DLG fault at bus no 2 and determine the fault current, short circuit MVA. G1: 100 MVA, 15.75 kV, X1 = X2 = 0.15 pu, X0 = 0.05 pu; G2: 100 MVA, 15.75 kV, X1 = X2 = 0.2 pu, X0 = 0.1 pu; T1 = T2: 100 MVA, 15.75/138 kV, X1 = X2 = X0 = 0.1 pu; Line: X1 = X2 = 25 Ω, X0 = 70 Ω. [Figure: G1 (Y grounded) - bus 1 - T1 (Δ/Y grounded) - bus 2 - line - bus 3 - T2 (Y grounded/Δ) - bus 4 - G2 (Y grounded)]

## Short answer

For a DLG fault at bus 2, sequence networks connect in parallel. Thevenin impedances: Z1 = Z2 = j0.1583 pu, Z0 = j0.0824 pu. Fault current If = 3I0 = j9.287 pu = 3886 A (ground); short-circuit MVA = 928.7 MVA.

## Answer

**Base:** 100 MVA; 15.75 kV (generators), 138 kV (line). Pre-fault voltage 1 pu, no load, bolted fault ($Z_f = 0$).
$$
Z_{base,line} = \frac{138^2}{100} = 190.44\ \Omega, \quad X_{1,line} = \frac{25}{190.44} = 0.1313, \quad X_{0,line} = \frac{70}{190.44} = 0.3676\ \text{pu}
$$

#### Sequence networks
**Positive** (emfs $E_1$, $E_2$):
```text
 E1-j0.15-(1)-j0.1-(2)-j0.1313-(3)-j0.1-(4)-j0.2-E2
                   |
                   F
```
**Negative:** same reactances ($X_2 = X_1$), no emfs.

**Zero:**
- T1 is Δ (bus 1 side)/Yg (bus 2 side): bus 2 to reference through $j0.1$; G1 isolated.
- T2 is Yg (bus 3 side)/Δ (bus 4 side): bus 3 to reference through $j0.1$; G2 isolated.
```text
 ref-j0.1-(2)-j0.3676-(3)-j0.1-ref
           |
           F      (G1 j0.05, G2 j0.1 isolated by deltas)
```

#### Thevenin impedances at bus 2
$$
\begin{aligned}
Z_1 = Z_2 &= j0.25\ \|\ j(0.1313 + 0.1 + 0.2) = j0.25\ \|\ j0.4313 = j0.1583\ \text{pu} \\
Z_0 &= j0.1\ \|\ j(0.3676 + 0.1) = j0.1\ \|\ j0.4676 = j0.0824\ \text{pu}
\end{aligned}
$$

#### DLG fault (b, c to ground): networks in parallel
$$
\begin{aligned}
Z_2\|Z_0 &= \frac{0.1583 \times 0.0824}{0.2406} = j0.0542 \\
I_{a1} &= \frac{1}{j(0.1583 + 0.0542)} = -j4.707\ \text{pu} \\
I_{a2} &= -I_{a1}\frac{Z_0}{Z_2+Z_0} = j1.612\ \text{pu} \\
I_{a0} &= -I_{a1}\frac{Z_2}{Z_2+Z_0} = j3.096\ \text{pu}
\end{aligned}
$$
Phase currents:
$$
I_b = 7.177\angle{139.68^\circ}\ \text{pu}, \quad I_c = 7.177\angle{40.32^\circ}\ \text{pu}
$$
Fault (ground) current:
$$
I_f = 3I_{a0} = j9.287\ \text{pu}
$$
$$
I_{base} = \frac{100\times10^6}{\sqrt3\times138\times10^3} = 418.4\ \text{A} \Rightarrow I_f = 9.287 \times 418.4 = 3886\ \text{A}
$$
(Each faulted line carries $7.177 \times 418.4 = 3003$ A.)

#### Short-circuit MVA
$$
\text{SC MVA} = |I_f|_{pu} \times V_f \times \text{MVA}_{base} = 9.287 \times 1 \times 100 = 928.7\ \text{MVA}
$$

**Answer:** $I_f \approx 9.29$ pu $= 3.89$ kA; short-circuit level ≈ 929 MVA.

## Check yourself

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

1. The 25 ohm line reactance on 100 MVA, 138 kV base is:
   A. 0.25 pu
   B. 0.1313 pu
   C. 0.0824 pu
   D. 0.3676 pu
   Answer: B. Zbase = 138^2/100 = 190.44 ohm, so 25/190.44 = 0.1313 pu. (Key point: Line reactance in pu)

2. In the zero-sequence network, why are G1 and G2 isolated?
   A. They are ungrounded
   B. Their X0 is too small
   C. The line is open
   D. The transformer delta windings next to them block zero-sequence current
   Answer: D. Delta windings on the generator sides of T1 and T2 give no path for zero-sequence current to the generators. (Key point: Delta blocks zero sequence)

3. Z0 at bus 2 is:
   A. 0.1 in parallel with (0.3676+0.1) = 0.0824 pu
   B. 0.1 + 0.3676 + 0.1
   C. 0.1 only
   D. 0.05 + 0.1
   Answer: A. T1's ground path (j0.1) is in parallel with line plus T2 path (0.3676 + 0.1). (Key point: Thevenin zero impedance)

4. How is short-circuit MVA found from the fault current?
   A. Ibase x 100
   B. Ia1 x Vf x MVAbase
   C. |If| pu x Vf x MVAbase = 9.287 x 100
   D. sqrt3 x If x 138
   Answer: C. SC MVA = If(pu) x Vf x MVA base = 9.287 x 1 x 100 = 929 MVA, using If = 3Ia0. (Key point: Short-circuit MVA)

## Related in Power System Analysis II

- [L-L fault current in kA at bus 3](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-bus3-ka) — Power System Analysis II, chapter 5
- [LLG fault sequence currents at bus 3](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Allg-fault-bus3-sequence-currents) — Power System Analysis II, chapter 5
- [SLG fault at bus 3, currents and voltages](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aslg-fault-bus3-three-generators) — Power System Analysis II, chapter 5
- [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
- [DLG fault sequence network connection](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Adlg-sequence-network-connection) — Power System Analysis II, chapter 5, asked 3×
- [Equivalent Z0, Z1, Z2 at bus 3](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Asequence-thevenin-z0-z1-z2-bus3) — Power System Analysis II, chapter 5

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