---
title: "L-L fault: I1 and I2 equal and opposite"
subject: "Power System Analysis II"
chapter: 5
key: "semester_5/power_system_analysis_ii:ll-fault-positive-negative-opposite"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ll-fault-positive-negative-opposite
site: BCT Engineering
---

# L-L fault: I1 and I2 equal and opposite

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

- Asked in: 2071 Chaitra, 6 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-positive-negative-opposite
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

Show that positive and negative sequence currents are equal in magnitude but out of phase by 180° in a line to line fault on a power system network. Draw a diagram showing inter-connection of sequence networks for this type of fault.

## Short answer

From I_a=0 and I_b=-I_c, sequence currents become I_a0=0, I_a1 = (j/√3)I_b, I_a2 = -(j/√3)I_b. Thus I_a2 = -I_a1: equal magnitude, 180° phase difference. The positive and negative sequence networks connect in parallel (opposition) through the fault impedance; the zero sequence network is left open.

## Answer

In a line-to-line fault, $I_{a2} = -I_{a1}$: the positive and negative sequence currents have equal magnitude and are 180° out of phase, while $I_{a0} = 0$.

#### 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
$$
I_{a0} = 0, \qquad I_{a2} = -I_{a1}
$$

Now $a - a^2 = j\sqrt3$ and $a^2 - a = -j\sqrt3$, so
$$
I_{a1} = \frac{j\sqrt3}{3}I_b = \frac{I_b}{\sqrt3}\angle90^\circ, \qquad
I_{a2} = \frac{-j\sqrt3}{3}I_b = \frac{I_b}{\sqrt3}\angle{-90^\circ}
$$
Both have magnitude $|I_b|/\sqrt3$ and differ in angle by 180°: they are **equal and opposite**.

#### 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 $V_{a1} = V_f - Z_1I_{a1}$ and $V_{a2} = -Z_2I_{a2} = Z_2I_{a1}$:
$$
I_{a1} = -I_{a2} = \frac{V_f}{Z_1 + Z_2 + Z_f}
$$

Fault current: $I_f = I_b = (a^2 - a)I_{a1} = -j\sqrt3\,I_{a1} = \dfrac{-j\sqrt3\,V_f}{Z_1 + Z_2 + Z_f}$.

#### 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.

The positive network drives $I_{a1}$ out of F1 and the same current enters F2 of the negative network, so $I_{a2} = -I_{a1}$.

## Check yourself

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

1. For a line-to-line fault between phases b and c, which set of boundary conditions applies?
   A. Ia = 0, Ib + Ic = 0, Vb - Vc = Zf Ib
   B. Ib = Ic = 0, Va = Zf Ia
   C. Ia = 0, Ib = Ic, Vb = Vc = 0
   D. Va = Vb = Vc, Ia + Ib + Ic = 0
   Answer: A. Phase a is healthy so Ia = 0, the fault current leaves b and returns in c (Ic = -Ib), and the voltage difference across the fault is Zf Ib. (Key point: Boundary conditions of L-L fault)

2. Why is Ia0 = 0 in an L-L fault?
   A. The neutral is always isolated
   B. Zero sequence impedance is infinite
   C. Ia0 = (Ia + Ib + Ic)/3 and Ia = 0, Ic = -Ib, so the sum is zero
   D. Positive and negative currents cancel the zero sequence
   Answer: C. The three phase currents sum to zero, so there is no zero sequence component and no ground path is involved. (Key point: Zero sequence current is absent)

3. Which step shows that Ia2 = -Ia1?
   A. Ia1 = (1/3)(a - a^2)Ib and Ia2 = (1/3)(a^2 - a)Ib, which are negatives of each other
   B. Both equal Ib/3 with the same sign
   C. Ia1 = Ib and Ia2 = -Ib directly from Ic = -Ib
   D. Va1 = -Va2 from the voltage condition
   Answer: A. Substituting Ia = 0 and Ic = -Ib gives Ia1 = (a - a^2)Ib/3 = j Ib/sqrt3 and Ia2 = -j Ib/sqrt3, equal in magnitude and 180° apart. (Key point: Ia2 = -Ia1)

4. How are the sequence networks connected for an L-L fault?
   A. All three in series
   B. Positive and negative in parallel (in opposition) through Zf, zero network open
   C. Positive, negative and zero all in parallel
   D. Positive and zero in series, negative open
   Answer: B. Ia0 = 0 leaves the zero network open, and Va1 - Va2 = Zf Ia1 with Ia2 = -Ia1 joins the positive and negative networks across Zf. (Key point: Network interconnection)

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