---
title: "DLG fault sequence network connection"
subject: "Power System Analysis II"
chapter: 5
key: "semester_5/power_system_analysis_ii:dlg-sequence-network-connection"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/dlg-sequence-network-connection
site: BCT Engineering
---

# DLG fault sequence network connection

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

- Asked in: 2076 Chaitra, 6 marks; 2075 Chaitra, 6 marks; 2082 Bhadra (new course), 5 marks (asked 3 times)
- Video: [Double Line to Ground Fault on an Unloaded Generator/ LLG Fault](https://www.youtube.com/watch?v=_iN8y-FITcQ) — Dr.Nagaraja Kumari
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/dlg-sequence-network-connection
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

For a double line to ground fault (bolted, zero fault impedance) in an unloaded generator (at alternator terminal), obtain the sequence network and show that the positive sequence network is connected in series with the parallel connection of negative sequence and zero sequence networks.

## Short answer

Boundary conditions yield V_a1 = V_a2 = V_a0 and I_a0+I_a1+I_a2=0. Substituting network equations gives I_a1 = E_a / [Z_1 + (Z_2 Z_0)/(Z_2+Z_0)]. This corresponds to the positive sequence network in series with the parallel combination of negative and zero networks.

## Answer

In a double line to ground (L-L-G) fault on phases b and c, all three sequence voltages at the fault are equal and the sequence currents add to zero, so the positive sequence network is in series with the **parallel** combination of the negative and zero sequence networks.

#### Generator sequence equations (unloaded)
$$
V_{a1} = E_a - Z_1I_{a1}, \qquad V_{a2} = -Z_2I_{a2}, \qquad V_{a0} = -Z_0I_{a0}
$$
where $Z_0 = Z_{g0} + 3Z_n$.

#### Boundary conditions (bolted fault, b and c to ground)
```text
  a o-------------  Ia = 0
  b o-----+         Ib -->
  c o-----+         Ic -->
          |  If = Ib + Ic
        ground
```
$$
I_a = 0, \qquad V_b = 0, \qquad V_c = 0
$$

#### Sequence voltages
$$
\begin{bmatrix} V_{a0}\\V_{a1}\\V_{a2} \end{bmatrix} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}\begin{bmatrix} V_a\\0\\0 \end{bmatrix}
\quad\Rightarrow\quad V_{a0} = V_{a1} = V_{a2} = \frac{V_a}{3}
$$
Equal voltages across the three networks at the fault point means they are connected **in parallel** at that point.

#### Sequence currents
$$
I_a = I_{a0} + I_{a1} + I_{a2} = 0
$$
So the positive sequence current $I_{a1}$ divides into $-I_{a2}$ and $-I_{a0}$, i.e. it returns through the negative and zero sequence networks in parallel.

#### Solving
From $V_{a1} = V_{a2} = V_{a0}$:
$$
I_{a2} = -\frac{V_{a1}}{Z_2}, \qquad I_{a0} = -\frac{V_{a1}}{Z_0}
$$
Substituting in $I_{a0} + I_{a1} + I_{a2} = 0$ with $V_{a1} = E_a - Z_1I_{a1}$:
$$
I_{a1} = (E_a - Z_1I_{a1})\left(\frac{1}{Z_2} + \frac{1}{Z_0}\right)
\;\Rightarrow\;
I_{a1} = \frac{E_a}{Z_1 + \dfrac{Z_2Z_0}{Z_2 + Z_0}}
$$
This is exactly the current in a circuit with $E_a$ and $Z_1$ in series with ($Z_2 \parallel Z_0$), which proves the connection.
$$
I_{a2} = -I_{a1}\frac{Z_0}{Z_2 + Z_0}, \qquad I_{a0} = -I_{a1}\frac{Z_2}{Z_2 + Z_0}
$$
Fault (ground) current: $I_f = I_b + I_c = 3I_{a0}$.

#### Interconnection
```text
        Ia1 -->        F
  +--(Ea)--Z1----------o------+---------+
  |                           |         |
  |                          Z2        Z0
  |                           | Ia2     | Ia0
  N1 -------------------------+---------+
                              N2        N0
```
Positive sequence network in series with the parallel combination of negative and zero sequence networks. With fault impedance $Z_f$ to ground, $3Z_f$ is added in series with $Z_0$.

## Check yourself

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

1. For a bolted b-c-to-ground fault (Vb = Vc = 0), what is the relation between the sequence voltages?
   A. Va0 = Va1 = Va2 = Va/3
   B. Va0 + Va1 + Va2 = 0
   C. Va1 = 0
   D. Va2 = -Va1
   Answer: A. Transforming (Va, 0, 0) gives equal sequence voltages, which means the networks are in parallel at the fault point. (Key point: Sequence voltages)

2. Since Ia = 0, what follows for the sequence currents?
   A. Ia1 = Ia2 = Ia0
   B. Ia0 + Ia1 + Ia2 = 0, so Ia1 splits between the negative and zero networks
   C. Ia0 = 0
   D. Ia1 = 0
   Answer: B. Ia = Ia0 + Ia1 + Ia2 = 0 gives Ia1 = -(Ia2 + Ia0), so the positive current returns through the two other networks. (Key point: Current condition)

3. Which formula for Ia1 results?
   A. Ea/(Z1 + Z2 + Z0)
   B. Ea/(Z1 + Z2 + Z0)/3
   C. Ea/(Z1 + Z2 Z0/(Z2 + Z0))
   D. 3Ea/(Z1 + Z2 + Z0)
   Answer: C. Z1 in series with Z2 parallel Z0 gives Ia1 = Ea/(Z1 + Z2Z0/(Z2+Z0)). (Key point: Resulting connection)

## Related in Power System Analysis II

- [SLG fault: sequence networks in series](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aslg-sequence-networks-series) — Power System Analysis II, chapter 5, asked 5×
- [L-G fault of unloaded generator: equal currents](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-unloaded-generator-derivation) — Power System Analysis II, chapter 5
- [Sequence networks with parallel transformer banks](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Asequence-networks-parallel-transformers-2079) — Power System Analysis II, chapter 4
- [Fault at generator terminal feeding motor](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-generator-terminal-motor-new) — Power System Analysis II, chapter 3
- [No zero sequence in line-to-line fault](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-zero-sequence-absent) — Power System Analysis II, chapter 5, asked 2×
- [DLG fault on 30 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Adlg-fault-30mva-generator) — Power System Analysis II, chapter 5

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