---
title: "L-G fault in phase B"
subject: "Power System Analysis II"
chapter: 5
key: "semester_5/power_system_analysis_ii:lg-fault-phase-b"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/lg-fault-phase-b
site: BCT Engineering
---

# L-G fault in phase B

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

- Asked in: 2073 Shrawan, 8 marks
- Video: [Single Line to Ground Fault on an Unloaded Generator/LG Fault](https://www.youtube.com/watch?v=WtJvn4aLcTE) — 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/lg-fault-phase-b
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

For an unloaded 3-phase, ABC phase sequence, synchronous generator, starting from the boundary condition draw and justify the interconnection of sequence networks if a L-G fault occurs in phase B.

## Short answer

For an L-G fault on phase B, the phase-B sequence currents are equal: I_b0=I_b1=I_b2=I_b/3. The three sequence networks connect in series, with the positive network containing E_b. In phase-a reference, this requires I_a1 = aI_a0, I_a2 = a^2I_a0. The fault current is I_f = 3E_b/(Z1+Z2+Z0).

## Answer

For an L-G fault on phase B, the sequence currents **of phase B** are equal, $I_{b0} = I_{b1} = I_{b2} = I_b/3$, so the three sequence networks are connected in series. In terms of phase-a reference components they are joined through phase shifts: $I_{a1} = aI_{a0}$, $I_{a2} = a^2I_{a0}$.

#### Generator sequence equations (phase a reference)
$$
V_{a1} = E_a - Z_1I_{a1}, \qquad V_{a2} = -Z_2I_{a2}, \qquad V_{a0} = -Z_0I_{a0}
$$

#### Boundary conditions (bolted fault on B)
$$
I_a = 0, \qquad I_c = 0, \qquad V_b = 0
$$

#### Sequence currents
$$
\begin{aligned}
I_{a0} &= \tfrac13(0 + I_b + 0) = \tfrac13I_b \\
I_{a1} &= \tfrac13(0 + aI_b + 0) = \tfrac13aI_b = aI_{a0} \\
I_{a2} &= \tfrac13(0 + a^2I_b + 0) = \tfrac13a^2I_b = a^2I_{a0}
\end{aligned}
$$
Phase-b components are $I_{b0} = I_{a0}$, $I_{b1} = a^2I_{a1} = I_{a0}$, $I_{b2} = aI_{a2} = I_{a0}$. So
$$
I_{b0} = I_{b1} = I_{b2} = \frac{I_b}{3}
$$
The same current flows in all three networks when they are written for phase b, which means a **series** connection.

#### Voltage condition
$$
V_b = V_{a0} + a^2V_{a1} + aV_{a2} = V_{b0} + V_{b1} + V_{b2} = 0
$$
Substituting:
$$
-Z_0I_{a0} + a^2(E_a - Z_1aI_{a0}) - aZ_2a^2I_{a0} = 0
\;\Rightarrow\;
I_{a0} = \frac{a^2E_a}{Z_0 + Z_1 + Z_2} = \frac{E_b}{Z_0 + Z_1 + Z_2}
$$
since $E_b = a^2E_a$. Hence
$$
I_f = I_b = 3I_{a0} = \frac{3E_b}{Z_1 + Z_2 + Z_0}
$$

#### Interconnection of sequence networks
Drawn with phase b as reference (EMF $E_b$ in the positive network):
```text
      Ib1 -->
  +--(Eb)--Z1--F1  N2--Z2--F2  N0--Z0--F0--+
  |             |__|        |__|           |
  N1 ---------------------------------------+
     Ib0 = Ib1 = Ib2 = Ib/3
```
With phase a as reference the same series loop is used, but the currents entering the networks are linked by phase shifts: $I_{a1} = aI_{a0}$ and $I_{a2} = a^2I_{a0}$ (ideal phase-shifting transformers of $+120^\circ$ and $+240^\circ$ in the positive and negative networks).

**Justification:** the series connection satisfies both conditions: equal phase-b sequence currents and zero total phase-b voltage. The result is the same as an L-G fault on phase a, only shifted by $-120^\circ$, as expected for a symmetrical machine.

## Check yourself

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

1. Which are the boundary conditions for a bolted L-G fault on phase B?
   A. Ib = 0, Va = Vc = 0
   B. Ia = 0, Ic = 0, Vb = 0
   C. Ia = Ib = Ic, Vb = 0
   D. Ia = 0, Ic = 0, Vb = Vc
   Answer: B. The healthy phases carry no current and the faulted phase B is at ground potential. (Key point: Boundary conditions for fault on B)

2. Which statement is correct for this fault?
   A. Ia0 = Ia1 = Ia2 = Ib/3
   B. Ib0 = Ib1 = Ib2 = Ib/3, with Ia1 = a Ia0 and Ia2 = a^2 Ia0
   C. Ia1 = a^2 Ia0 and Ia2 = a Ia0
   D. Ib1 = Ib2 = -Ib0
   Answer: B. Ia1 = aIb/3 and Ia2 = a^2 Ib/3, so relative to phase a the components are shifted, but referred to phase b they are all equal to Ib/3. (Key point: Equal phase-b sequence currents)

3. Why are the networks connected in series for a fault on B?
   A. Because phase a is the reference
   B. Because the phase-b sequence currents are equal and Vb0 + Vb1 + Vb2 = 0
   C. Because the fault current is zero
   D. Because Z0 = Z1 = Z2
   Answer: B. Equal sequence currents (phase-b basis) plus zero total phase-b voltage is exactly a series loop. (Key point: Series connection justification)

4. What is the fault current for the L-G fault on phase B?
   A. 3Ea/(Z1 + Z2 + Z0)
   B. Eb/(Z1 + Z2 + Z0)
   C. 3Eb/(Z1 + Z2 + Z0) with Eb = a^2 Ea
   D. 3Eb/(Z1 + Z2 + Z0 + 3Zf) with Eb = a Ea
   Answer: C. Ia0 = a^2Ea/(Z0+Z1+Z2) = Eb/(Z0+Z1+Z2), and If = Ib = 3Ia0. (Key point: Fault current)

## 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
- [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
- [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
- [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×
- [Sequence networks decoupled for static circuit](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Astatic-circuit-sequence-decoupled) — Power System Analysis II, chapter 4

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