---
title: "L-L fault: I0 absent and I1 = −I2"
subject: "Power System Analysis II"
chapter: 5
key: "semester_5/power_system_analysis_ii:ll-fault-generator-i1-equals-i2"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ll-fault-generator-i1-equals-i2
site: BCT Engineering
---

# L-L fault: I0 absent and I1 = −I2

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

- Asked in: 2069 Chaitra, 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/ll-fault-generator-i1-equals-i2
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

With the help of suitable mathematical aid verify that, for a line to line fault in a synchronous generator the zero sequence component of current is absent and positive-sequence of current is equal to the negative sequence component of the current.

## Short answer

Boundary conditions I_a = 0, I_b = -I_c give I_a0 = 0, I_a1 = (1/3)(a - a^2)I_b, I_a2 = (1/3)(a^2 - a)I_b. Hence I_a2 = -I_a1, so the positive and negative sequence currents have equal magnitude and are 180° out of phase. The zero sequence component is absent because there is no ground path.

## Answer

For a line-to-line fault (phases b and c) on a synchronous generator, the line currents sum to zero, so $I_{a0} = 0$; and transforming the boundary conditions gives $I_{a2} = -I_{a1}$, i.e. the positive and negative sequence currents are **equal in magnitude** (opposite in phase).

#### Generator and fault
An unloaded generator produces balanced EMFs; its sequence equations are
$$
V_{a1} = E_a - Z_1I_{a1}, \qquad V_{a2} = -Z_2I_{a2}, \qquad V_{a0} = -Z_0I_{a0}
$$
#### Boundary conditions (fault between phases b and c through $Z_f$)
```text
  a o-----------  Ia = 0
  b o-----+        Ib = If -->
          Zf
  c o-----+        Ic = -Ib
```
$$
I_a = 0, \qquad I_b + I_c = 0, \qquad V_b - V_c = Z_fI_b
$$

#### Sequence currents
$$
\begin{aligned}
I_{a0} &= \tfrac13(I_a + I_b + I_c) = \tfrac13(0 + I_b - I_b) = 0 \\
I_{a1} &= \tfrac13(I_a + aI_b + a^2I_c) = \tfrac13(a - a^2)I_b \\
I_{a2} &= \tfrac13(I_a + a^2I_b + aI_c) = \tfrac13(a^2 - a)I_b
\end{aligned}
$$
Hence
$$
I_{a0} = 0, \qquad I_{a2} = -I_{a1}
$$

Since $a - a^2 = j\sqrt3$:
$$
I_{a1} = \frac{j\sqrt3}{3}I_b = \frac{j}{\sqrt3}I_b, \qquad I_{a2} = -\frac{j}{\sqrt3}I_b
$$
So $|I_{a1}| = |I_{a2}| = |I_b|/\sqrt3$: the positive and negative sequence currents have equal magnitude; they are 180° apart. The zero sequence current is absent because there is no ground connection and no neutral current flows ($I_n = 3I_{a0} = 0$).

#### Voltage condition
$$
\begin{aligned}
V_b - V_c &= (a^2 - a)V_{a1} + (a - a^2)V_{a2} = (a^2 - a)(V_{a1} - V_{a2}) \\
Z_fI_b &= Z_f\,(a^2I_{a1} + aI_{a2}) = Z_f(a^2 - a)I_{a1}
\end{aligned}
$$
so
$$
V_{a1} - V_{a2} = Z_fI_{a1}
$$
With $V_{a1} = E_a - Z_1I_{a1}$ and $V_{a2} = -Z_2I_{a2} = Z_2I_{a1}$:
$$
I_{a1} = -I_{a2} = \frac{E_a}{Z_1 + Z_2 + Z_f}
$$

#### Fault current
$$
I_f = I_b = -I_c = (a^2 - a)I_{a1} = -j\sqrt3\,I_{a1} = \frac{-j\sqrt3\,E_a}{Z_1 + Z_2 + Z_f}
$$
#### Interconnection of sequence networks
```text
        Ia1 -->           <-- Ia2
  +--(Ea)--Z1---o F1--Zf--o F2---Z2--+
  |                                  |
  N1 ------------------------------- N2

  Zero sequence network: not connected (Ia0 = 0)
```
The positive and negative sequence networks are connected **in parallel (in opposition)** through $Z_f$; the zero sequence network is left open.

Because the two networks form one series loop, the current entering the negative sequence network equals the current leaving the positive network, which is another way of seeing $I_{a1} = -I_{a2}$ in magnitude.

#### Fault-point voltages (bolted fault)
$V_{a1} = V_{a2} = E_a - Z_1I_{a1}$, $V_{a0} = 0$, so $V_a = 2V_{a1}$ and $V_b = V_c = -V_{a1}$.

## Check yourself

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

1. Why is Ia0 = 0 in an L-L fault on a generator?
   A. The neutral is isolated by design
   B. Ia + Ib + Ic = 0 since Ia = 0 and Ib = -Ic
   C. Z0 is zero
   D. Va0 = Ea
   Answer: B. Ia0 is one third of the sum of line currents, which is zero here. (Key point: Absence of Ia0)

2. From Ia1 = (a - a^2)Ib/3 and Ia2 = (a^2 - a)Ib/3, what is concluded?
   A. Ia1 = Ia2 in phase
   B. Ia2 = -Ia1, equal magnitudes, 180° apart
   C. Ia1 = 0
   D. Ia2 = 2Ia1
   Answer: B. The coefficients differ only by sign, so |Ia1| = |Ia2| = |Ib|/sqrt(3). (Key point: I1 and I2 relation)

3. From the voltage condition, Vb - Vc = Zf Ib reduces to:
   A. Va1 + Va2 = Zf Ia1
   B. Va1 = Va2
   C. Va1 - Va2 = Zf Ia1
   D. Va1 - Va2 = 0 always
   Answer: C. Both sides contain (a^2 - a), which cancels leaving Va1 - Va2 = Zf Ia1. (Key point: Voltage condition)

4. The L-L fault current in terms of Ia1 is:
   A. Ib = 3Ia1
   B. Ib = Ia1/sqrt(3)
   C. Ib = Ia1 + Ia2
   D. Ib = -j sqrt(3) Ia1
   Answer: D. Ib = (a^2 - a)Ia1 = -j sqrt(3) Ia1, with Ia1 = Ea/(Z1 + Z2 + Zf). (Key point: Fault current)

## Related in Power System Analysis II

- [L-L fault: I1 and I2 equal and opposite](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-positive-negative-opposite) — Power System Analysis II, chapter 5
- [Equal positive and negative sequence impedances](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Astatic-circuit-positive-negative-equal) — Power System Analysis II, chapter 4
- [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×
- [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
- [No zero sequence current in delta lines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Adelta-zero-sequence-absent) — Power System Analysis II, chapter 4, asked 2×
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