---
title: "DLG fault on 30 MVA generator"
subject: "Power System Analysis II"
chapter: 5
key: "semester_5/power_system_analysis_ii:dlg-fault-30mva-generator"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/dlg-fault-30mva-generator
site: BCT Engineering
---

# DLG fault on 30 MVA generator

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

- Asked in: 2074 Chaitra, 9 marks
- 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-30mva-generator
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

A double line to ground fault occur at generator terminal of 30 MVA, 11 kV and having Z1 = Z2 = j0.2 pu and Z0 = j0.05 pu. Find the line currents, fault current and line to neutral voltages under fault condition.

## Short answer

Sequence networks parallel: I_a1=-j4.167 pu, I_a0=j3.333 pu. Line currents: Ia=0, Ib=Ic=6.614 pu=10.42 kA. Fault current If=3I_a0=10 pu=15.75 kA. Line-neutral voltages: Va=0.5 pu (3.18 kV), Vb=Vc=0.

## Answer

**Data:** 30 MVA, 11 kV, $Z_1 = Z_2 = j0.2$, $Z_0 = j0.05$ pu, solidly grounded, unloaded ($E_a = 1\angle0^\circ$). Fault: phases b and c to ground.

$I_{base} = \dfrac{30\times10^6}{\sqrt3\times11\times10^3} = 1574.6$ A; $V_{base,ph} = 11/\sqrt3 = 6.351$ kV.

#### Fault conditions
$I_a = 0$, $V_b = V_c = 0$, giving $V_{a0} = V_{a1} = V_{a2}$: the three sequence networks are in **parallel**.

#### Sequence currents
$$
\begin{aligned}
Z_2 \| Z_0 &= \frac{(j0.2)(j0.05)}{j0.25} = j0.04 \\
I_{a1} &= \frac{E_a}{Z_1 + Z_2\|Z_0} = \frac{1}{j0.24} = -j4.1667\ \text{pu} \\
I_{a2} &= -I_{a1}\frac{Z_0}{Z_2+Z_0} = j4.1667 \times 0.2 = j0.8333\ \text{pu} \\
I_{a0} &= -I_{a1}\frac{Z_2}{Z_2+Z_0} = j4.1667 \times 0.8 = j3.3333\ \text{pu}
\end{aligned}
$$

#### Line currents
$$
\begin{aligned}
I_a &= I_{a0} + I_{a1} + I_{a2} = 0 \\
I_b &= I_{a0} + a^2I_{a1} + aI_{a2} = 6.614\angle{130.89^\circ}\ \text{pu} \Rightarrow 6.614 \times 1574.6 = 10\,415\ \text{A} \\
I_c &= I_{a0} + aI_{a1} + a^2I_{a2} = 6.614\angle{49.11^\circ}\ \text{pu} \Rightarrow 10\,415\ \text{A}
\end{aligned}
$$

#### Fault current (to ground)
$$
I_f = I_b + I_c = 3I_{a0} = j10.0\ \text{pu} \Rightarrow 10 \times 1574.6 = 15\,746\ \text{A}
$$

#### Line-to-neutral voltages
$$
\begin{aligned}
V_{a1} = V_{a2} = V_{a0} &= E_a - Z_1I_{a1} = 1 - (j0.2)(-j4.1667) = 0.1667\ \text{pu} \\
V_a &= 3V_{a1} = 0.5\ \text{pu} = 0.5 \times 6.351 = 3.175\ \text{kV} \\
V_b &= V_c = 0
\end{aligned}
$$

#### Results
| Quantity | pu | Actual |
|---|---|---|
| $I_a$ | 0 | 0 |
| $I_b$, $I_c$ | 6.614 | 10.41 kA |
| Fault current $I_f$ | 10.0 | 15.75 kA |
| $V_a$ | 0.5 | 3.18 kV |
| $V_b$, $V_c$ | 0 | 0 |

## Check yourself

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

1. For a double line-to-ground fault, the sequence networks are connected:
   A. In series
   B. Positive alone, others open
   C. Positive and negative in parallel only
   D. In parallel, with Va0 = Va1 = Va2
   Answer: D. Fault conditions Ia = 0 and Vb = Vc = 0 give Va0 = Va1 = Va2, so the networks are in parallel. (Key point: DLG network connection)

2. What is Ia1 for Z1 = Z2 = j0.2 and Z0 = j0.05?
   A. -j4.1667 pu
   B. -j2.5 pu
   C. -j5 pu
   D. -j20 pu
   Answer: A. Ia1 = 1/(j0.2 + j0.04) = 1/j0.24 = -j4.1667 pu, where j0.04 is Z2 in parallel with Z0. (Key point: Positive sequence current)

3. The current flowing to ground in this fault is:
   A. Ib alone
   B. Ia1 + Ia2
   C. 3Ia0 = 10 pu
   D. Ia0 = 3.33 pu
   Answer: C. If = Ib + Ic = 3Ia0 = 3 x j3.333 = j10 pu, about 15.75 kA. (Key point: Ground fault current)

4. What is the voltage of the healthy phase a?
   A. 0.1667 pu
   B. 1 pu
   C. 0 pu
   D. 0.5 pu
   Answer: D. Va = 3Va1 = 3 x 0.1667 = 0.5 pu, while Vb = Vc = 0. (Key point: Faulted-phase voltage)

## Related in Power System Analysis II

- [L-L fault currents and voltages, 30 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-30mva-currents-voltages) — Power System Analysis II, chapter 5
- [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
- [L-G fault and 3-phase fault, 30 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-30mva-voltages-3phase) — Power System Analysis II, chapter 5
- [Generator and motor currents for terminal fault](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-current-generator-motor-2080) — Power System Analysis II, chapter 3
- [SLG fault on 20 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aslg-fault-20mva-generator) — Power System Analysis II, chapter 5
- [L-L fault on 40 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-40mva-generator) — Power System Analysis II, chapter 5

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