9.3 Single Line-to-Ground (SLG) & Line-to-Line (LL) Fault Analysis

9.3 Single Line-to-Ground (SLG) & Line-to-Line (LL) Fault Analysis

Unsymmetrical short circuits represent over $95%$ of all faults occurring in electrical power systems. Because unsymmetrical faults destroy the natural phase balance of three-phase networks, they cannot be analyzed using simple single-phase per-phase equivalent circuits alone. Instead, we utilize Fortescue's Symmetrical Components Method, which transforms three coupled, unbalanced phase quantities into three decoupled, balanced symmetrical sequence networks: Positive-Sequence ($1$), Negative-Sequence ($2$), and Zero-Sequence ($0$).

On the NCEES PE Electrical and Computer: Power examination, mastering Single Line-to-Ground (SLG) and Line-to-Line (L-L) fault calculations requires deriving sequence current equations from boundary conditions, synthesizing sequence network interconnections, and calculating ground return currents and unfaulted phase voltage elevations.

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1. Symmetrical Components Transformation Review

Fortescue's transformation expresses physical phase quantities ($A, B, C$) in terms of symmetrical sequence components ($0, 1, 2$) using the complex phase rotation operator $a = 1\angle 120^\circ = -0.5 + j0.866025$ (where $a^2 = 1\angle 240^\circ = -0.5 - j0.866025$, and $1 + a + a^2 = 0$):

$\begin{bmatrix} V_a \\ V_b \\ V_c \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & a^2 & a \\ 1 & a & a^2 \end{bmatrix} \begin{bmatrix} V_{a0} \\ V_{a1} \\ V_{a2} \end{bmatrix} \quad \Longleftrightarrow \quad \begin{bmatrix} V_{a0} \\ V_{a1} \\ V_{a2} \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & a & a^2 \\ 1 & a^2 & a \end{bmatrix} \begin{bmatrix} V_a \\ V_b \\ V_c \end{bmatrix}$

$\begin{bmatrix} I_a \\ I_b \\ I_c \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & a^2 & a \\ 1 & a & a^2 \end{bmatrix} \begin{bmatrix} I_{a0} \\ I_{a1} \\ I_{a2} \end{bmatrix} \quad \Longleftrightarrow \quad \begin{bmatrix} I_{a0} \\ I_{a1} \\ I_{a2} \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & a & a^2 \\ 1 & a^2 & a \end{bmatrix} \begin{bmatrix} I_a \\ I_b \\ I_c \end{bmatrix}$

Sequence Thevenin Voltage Equations Seen at Fault Bus

• Positive-Sequence: $V_{a1} = V_f - I_{a1} Z_1$ (Contains internal prefault generated voltage $V_f$)
• Negative-Sequence: $V_{a2} = 0 - I_{a2} Z_2 = -I_{a2} Z_2$ (Passive network, no internal sources)
• Zero-Sequence: $V_{a0} = 0 - I_{a0} Z_0 = -I_{a0} Z_0$ (Passive network, no internal sources)

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2. Single Line-to-Ground (SLG / 1LG) Fault Analysis

The Single Line-to-Ground fault is the most common short-circuit event in power systems, accounting for approximately $70%$ to $80%$ of all faults (caused by insulator flashover, lightning strikes, tree branches, or animal contact).

               SINGLE LINE-TO-GROUND (SLG) FAULT TOPOLOGY

          Phase A  o----------------+--------o  Fault Current: I_a
                                    |
                                   +---+ 
                                   |   | Z_f (Fault Impedance)
                                   +---+ 
                                    |
                                   ===
          Phase B  o-------------------------o  I_b = 0 (Open)
          Phase C  o-------------------------o  I_c = 0 (Open)

Boundary Conditions (Fault on Phase A to Ground through $Z_f$)

1. $I_b = 0$
2. $I_c = 0$
3. $V_a = I_a Z_f$ (For a bolted fault, $Z_f = 0 \implies V_a = 0$)

Sequence Current Derivation

Substituting the boundary conditions $I_b = 0, I_c = 0$ into the inverse symmetrical component transformation:

$\begin{bmatrix} I_{a0} \\ I_{a1} \\ I_{a2} \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & a & a^2 \\ 1 & a^2 & a \end{bmatrix} \begin{bmatrix} I_a \\ 0 \\ 0 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} I_a \\ I_a \\ I_a \end{bmatrix}$

$I_{a0} = I_{a1} = I_{a2} = \frac{I_a}{3}$

Sequence Network Interconnection: SERIES Connection

Because all three sequence currents flowing out of the sequence networks are identical ($I_{a0} = I_{a1} = I_{a2}$), the Positive, Negative, and Zero sequence networks must be connected in SERIES at the fault point.

               SLG SEQUENCE NETWORK INTERCONNECTION (SERIES)

                   +---[ Z_1 ]---(~) V_f (Positive Sequence)
                   |               |
                   +---------------+ I_a1
                   |
                   +---[ Z_2 ]-----+ (Negative Sequence)
                   |               | I_a2
                   +---------------+ 
                   |
                   +---[ Z_0 ]-----+ (Zero Sequence)
                   |               | I_a0
                   +---------------+ 
                   |
                   +---[ 3*Z_f ]---+
                   |               |
                   +---------------+ Reference

Master SLG Fault Current Formula

Summing the voltages around the closed series loop:

$V_{a1} + V_{a2} + V_{a0} = V_a = I_a Z_f = 3 I_{a0} Z_f$ $(V_f - I_{a1} Z_1) + (-I_{a2} Z_2) + (-I_{a0} Z_0) = 3 I_{a0} Z_f$

Since $I_{a1} = I_{a2} = I_{a0}$:

$V_f - I_{a0} (Z_1 + Z_2 + Z_0 + 3 Z_f) = 0$

$I_{a0} = I_{a1} = I_{a2} = \frac{V_f}{Z_1 + Z_2 + Z_0 + 3 Z_f} \quad [\text{pu}]$

$\text{Total SLG Fault Current: } I_{fault} = I_a = 3 I_{a0} = \frac{3 V_f}{Z_1 + Z_2 + Z_0 + 3 Z_f} \quad [\text{pu}]$

$\text{Ground Return Neutral Current: } I_{ground} = I_n = I_a + I_b + I_c = I_a = 3 I_{a0}$

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3. High SLG Fault Current Phenomenon in Solidly Grounded Systems

One of the most critical and counter-intuitive phenomena tested on the PE Power examination is that Single Line-to-Ground fault currents can exceed Three-Phase symmetrical fault currents.

+---------------------------------------------------------------------------------------------------+
|                     SLG vs. THREE-PHASE FAULT CURRENT MAGNITUDE                                   |
+---------------------------------------------------------------------------------------------------+
| Three-Phase Symmetrical Fault:   I_f,3φ = V_f / Z_1 = (3 * V_f) / (3 * Z_1)                       |
| Single Line-to-Ground Fault:     I_f,SLG = (3 * V_f) / (Z_1 + Z_2 + Z_0)                          |
|                                                                                                   |
| In typical power networks, Z_2 ≈ Z_1. Substituting Z_2 = Z_1 into the SLG formula:                |
|                                                                                                   |
|   I_f,SLG = (3 * V_f) / (2 * Z_1 + Z_0)                                                           |
|                                                                                                   |
| CRITICAL COMPARISON:                                                                              |
|   - If Z_0 = Z_1:  I_f,SLG = (3 * V_f) / (3 * Z_1) = I_f,3φ  (SLG equals 3-Phase fault current)   |
|   - If Z_0 > Z_1:  I_f,SLG < I_f,3φ                          (SLG is LESS than 3-Phase)          |
|   - If Z_0 < Z_1:  I_f,SLG > I_f,3φ                          (SLG is GREATER than 3-Phase!)      |
+---------------------------------------------------------------------------------------------------+

[!IMPORTANT] Where Does $Z_0 < Z_1$ Occur? In solidly grounded systems near generating stations, large solidly grounded autotransformers, or grounded-wye/delta step-down transformers, the zero-sequence path has very low impedance ($Z_0 \approx 0.4$ to $0.8 \times Z_1$). In these locations, the SLG fault produces the highest short-circuit current in the entire facility, dictating breaker interrupting ratings and grounding grid sizing!

Unfaulted Phase Voltages During Bolted SLG Fault ($V_b, V_c$)

During an SLG fault on Phase A ($V_a = 0$), the healthy phase voltages ($V_b$ and $V_c$) shift depending on system grounding:

$V_{a0} = -I_{a0} Z_0 = -\frac{V_f Z_0}{Z_1 + Z_2 + Z_0}$ $V_{a1} = V_f - I_{a1} Z_1 = V_f - \frac{V_f Z_1}{Z_1 + Z_2 + Z_0} = \frac{V_f (Z_2 + Z_0)}{Z_1 + Z_2 + Z_0}$ $V_{a2} = -I_{a2} Z_2 = -\frac{V_f Z_2}{Z_1 + Z_2 + Z_0}$

• In Solidly Grounded Systems ($Z_0 \approx Z_1$): Healthy phase voltages remain near nominal ($|V_b| = |V_c| \approx 0.9$ to $1.0\text{ pu}$).
• In Ungrounded Systems ($Z_0 \to \infty$): Zero-sequence voltage rises to full phase voltage ($V_{a0} = -V_f$), causing the healthy phase voltages to elevate to full line-to-line voltage ($|V_b| = |V_c| = \sqrt{3} V_{LN} \approx 1.732\text{ pu}$). Insulation and surge arresters must be rated for line-to-line voltage.

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4. Line-to-Line (L-L / 2L) Fault Analysis

A Line-to-Line fault occurs when two phase conductors make physical contact or flashover across insulators without involving earth ground (e.g., Phase B shorted to Phase C).

                    LINE-TO-LINE (L-L) FAULT TOPOLOGY

          Phase A  o----------------------------------o  I_a = 0 (Unfaulted)

          Phase B  o----------------+-----------------o  Fault Current: I_b
                                    |
                                   +---+ 
                                   |   | Z_f (Fault Impedance)
                                   +---+ 
                                    |
          Phase C  o----------------+-----------------o  Fault Current: I_c = -I_b

Boundary Conditions (Fault between Phase B and Phase C through $Z_f$)

1. $I_a = 0$
2. $I_b = -I_c \implies I_b + I_c = 0$
3. $V_b - V_c = I_b Z_f$ (For a bolted fault, $Z_f = 0 \implies V_b = V_c$)

Sequence Current Derivation

Applying symmetrical component transformation:

$I_{a0} = \frac{1}{3} (I_a + I_b + I_c) = \frac{1}{3} (0 + I_b - I_b) = 0$

$I_{a1} = \frac{1}{3} (I_a + a I_b + a^2 I_c) = \frac{1}{3} (0 + a I_b - a^2 I_b) = \frac{a - a^2}{3} I_b = \frac{j\sqrt{3}}{3} I_b = \frac{j}{\sqrt{3}} I_b$

$I_{a2} = \frac{1}{3} (I_a + a^2 I_b + a I_c) = \frac{1}{3} (0 + a^2 I_b - a I_b) = \frac{a^2 - a}{3} I_b = -\frac{j\sqrt{3}}{3} I_b = -I_{a1}$

+---------------------------------------------------------------------------------------------------+
|                    L-L FAULT SEQUENCE CURRENT RELATIONSHIPS                                       |
+---------------------------------------------------------------------------------------------------+
| 1. Zero-Sequence Current is ZERO:   I_a0 = 0 (Zero-sequence network is completely ISOLATED/OPEN!)  |
| 2. Negative-Sequence opposes Pos.:  I_a2 = -I_a1                                                  |
| 3. Phase Fault Current:             I_b = -j*sqrt(3) * I_a1                                       |
+---------------------------------------------------------------------------------------------------+

Sequence Network Interconnection: PARALLEL Connection (Pos & Neg)

Because $I_{a1} = -I_{a2}$ and $I_{a0} = 0$, the Positive-Sequence and Negative-Sequence networks are connected in PARALLEL across the fault impedance $Z_f$, while the Zero-Sequence network is completely open.

               L-L SEQUENCE NETWORK INTERCONNECTION (PARALLEL)

                   +---[ Z_1 ]---(~) V_f (Positive Sequence)
                   |               |
             +-----+---------------+-----+  I_a1
             |                           |
             |     +---[ Z_2 ]-----+     |  I_a2 = -I_a1
             |     | (Negative Seq)|     |
             +-----+---------------+-----+ 
             |                           |
             +-----------[ Z_f ]---------+ (Fault Impedance)

             [ Z_0 Network ]  ===> OPEN CIRCUIT (I_a0 = 0)

Master Line-to-Line Fault Current Formula

From the parallel sequence loop:

$V_{a1} - V_{a2} = I_{a1} Z_f$ $(V_f - I_{a1} Z_1) - (-I_{a2} Z_2) = I_{a1} Z_f$ $V_f - I_{a1} Z_1 - I_{a1} Z_2 = I_{a1} Z_f$

$I_{a1} = -I_{a2} = \frac{V_f}{Z_1 + Z_2 + Z_f} \quad [\text{pu}], \quad I_{a0} = 0$

$\text{Phase B Fault Current: } I_b = -j\sqrt{3} I_{a1} = \frac{-j\sqrt{3} V_f}{Z_1 + Z_2 + Z_f} \quad [\text{pu}]$

$\text{Phase Fault Current Magnitude: } |I_{fault,LL}| = |I_b| = |I_c| = \frac{\sqrt{3} |V_f|}{|Z_1 + Z_2 + Z_f|} \quad [\text{pu}]$

The Standard $0.866$ Ratio for Bolted L-L Faults

When $Z_f = 0$ and the negative-sequence impedance equals the positive-sequence impedance ($Z_2 = Z_1$):

$|I_{fault,LL}| = \frac{\sqrt{3} V_f}{2 Z_1} = \frac{\sqrt{3}}{2} \cdot \left(\frac{V_f}{Z_1}\right) = \frac{\sqrt{3}}{2} I_{f,3\phi} \approx 0.866025 \times I_{f,3\phi}$

+---------------------------------------------------------------------------------------------------+
|              SUMMARY OF 3-PHASE, SLG, AND L-L FAULT CURRENT FORMULAS                              |
+---------------------------------------------------------------------------------------------------+
| Fault Type             | Sequence Interconnection | Sequence Currents      | Fault Current (I_fault)|
| :---                   | :---                     | :---                   | :---                   |
| **3-Phase (3Φ)**       | Positive Sequence Only   | $I_{a1} = I_a$         | $I_f = \frac{V_f}{Z_1}$ |
|                        |                          | $I_{a2}=0, I_{a0}=0$   |                        |
| **Single Line-to-      | Series Connection of     | $I_{a0} = I_{a1} = I_{a2}$ | $I_f = \frac{3 V_f}{Z_1 + Z_2 + Z_0 + 3Z_f}$ |
| **Ground (SLG)**       | Pos, Neg, & Zero        | $= I_a / 3$            |                        |
| **Line-to-Line (L-L)** | Parallel Connection of   | $I_{a1} = -I_{a2}$     | $|I_f| = \frac{\sqrt{3} V_f}{Z_1 + Z_2 + Z_f}$ |
|                        | Pos & Neg (Zero Open)    | $I_{a0} = 0$           | $= 0.866 \cdot I_{3\phi}$ (if $Z_1=Z_2$) |
+---------------------------------------------------------------------------------------------------+

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5. Step-by-Step Worked Mathematical Example

Problem Statement

A $13.8\text{ kV}$ (line-to-line), $100\text{ MVA}$ base substation bus is fed by a solidly grounded power system with the following equivalent Thevenin sequence impedances:

• Positive-Sequence Impedance: $Z_1 = j0.12\text{ pu}$
• Negative-Sequence Impedance: $Z_2 = j0.12\text{ pu}$
• Zero-Sequence Impedance: $Z_0 = j0.06\text{ pu}$ (Solidly grounded source with low zero-sequence impedance)
• Prefault Voltage: $V_f = 1.0\angle 0^\circ\text{ pu}$

Calculate:

1. The base current ($I_{base}$) at $13.8\text{ kV}$.
2. The bolted Three-Phase fault current ($I_{3\phi}$) in per-unit and Amperes.
3. The bolted Single Line-to-Ground (SLG) fault current ($I_{SLG}$) in per-unit and Amperes, and determine the percentage by which $I_{SLG}$ exceeds $I_{3\phi}$.
4. The SLG fault current if a fault impedance of $Z_f = 0.50,\Omega$ is present in the ground return path.
5. The bolted Line-to-Line (L-L) fault current ($I_{LL}$) in per-unit and Amperes, and verify the $86.6%$ ratio.
6. The voltages of the healthy phases ($V_b$ and $V_c$) during the bolted SLG fault on Phase A.

=========================================================================================
CALCULATION WORKFLOW & DETAILED STEP-BY-STEP SOLUTION:
=========================================================================================

Step 1: Calculate System Base Quantities
  Base Power: S_base = 100 MVA
  Base Voltage: V_base_LL = 13.8 kV
  Base Current:
    I_base = (100 * 10^6 VA) / (sqrt(3) * 13,800 V) = 4,183.70 A
  Base Impedance:
    Z_base = (V_base_LL)^2 / S_base = (13.8 kV)^2 / 100 MVA = 1.9044 ohms

Step 2: Three-Phase Symmetrical Fault Current
  I_f,3φ,pu = V_f / Z_1 = (1.0 /_ 0°) / (j0.12) = -j8.3333 pu = 8.3333 /_ -90° pu
  I_f,3φ,actual = 8.3333 * 4,183.70 A = 34,864.2 A ≈ 34.86 kA rms

Step 3: Bolted Single Line-to-Ground (SLG) Fault Current (Z_f = 0)
  Total Sequence Impedance:
    Z_total,SLG = Z_1 + Z_2 + Z_0 = j0.12 + j0.12 + j0.06 = j0.30 pu

  Sequence Currents:
    I_a0 = I_a1 = I_a2 = V_f / Z_total,SLG = (1.0 /_ 0°) / (j0.30) = -j3.3333 pu

  Total Phase A Fault Current:
    I_a = 3 * I_a0 = 3 * (-j3.3333 pu) = -j10.0000 pu = 10.0000 /_ -90° pu

  Actual SLG Fault Current:
    I_f,SLG,actual = 10.0000 * 4,183.70 A = 41,837.0 A ≈ 41.84 kA rms

  Comparison Ratio:
    Ratio = I_f,SLG / I_f,3φ = 10.0000 pu / 8.3333 pu = 1.200 (120% of 3-Phase fault!)
    --> The SLG fault current is 20.0% GREATER than the 3-phase fault current!

Step 4: SLG Fault with Fault Impedance Z_f = 0.50 ohms
  Convert Z_f to per-unit:
    Z_f,pu = 0.50 ohms / 1.9044 ohms = 0.26255 pu
    3 * Z_f,pu = 3 * 0.26255 = 0.78765 pu

  Total Series Loop Impedance:
    Z_loop = (3 * Z_f,pu) + j(Z_1 + Z_2 + Z_0) = 0.78765 + j0.30000 pu
    |Z_loop| = sqrt( (0.78765)^2 + (0.30000)^2 ) = sqrt( 0.62039 + 0.09000 ) = 0.84285 pu

  Sequence Current Magnitude:
    |I_a0| = 1.0 / 0.84285 = 1.18645 pu

  Total Fault Current Magnitude:
    |I_a| = 3 * 1.18645 pu = 3.55935 pu
    I_f,actual = 3.55935 * 4,183.70 A = 14,891.3 A ≈ 14.89 kA rms

Step 5: Bolted Line-to-Line (L-L) Fault Current (Z_f = 0)
  Sequence Currents:
    I_a0 = 0
    I_a1 = -I_a2 = V_f / (Z_1 + Z_2) = (1.0 /_ 0°) / (j0.12 + j0.12) = 1.0 / (j0.24) = -j4.1667 pu

  Phase B Fault Current Magnitude:
    |I_b| = sqrt(3) * |I_a1| = sqrt(3) * 4.1667 pu = 7.2169 pu

  Actual L-L Fault Current:
    I_f,LL,actual = 7.2169 * 4,183.70 A = 30,193.3 A ≈ 30.19 kA rms

  Verification of 0.866 Ratio:
    |I_b| / I_f,3φ = 7.2169 pu / 8.3333 pu = 0.86603 = sqrt(3)/2 (CONFIRMED)

Step 6: Healthy Phase Voltages (V_b, V_c) During Bolted SLG Fault on Phase A
  Sequence Voltages at Fault Point:
    V_a0 = -I_a0 * Z_0 = -(-j3.3333) * (j0.06) = -0.2000 pu
    V_a1 = V_f - I_a1 * Z_1 = 1.0 - (-j3.3333)*(j0.12) = 1.0 - 0.4000 = 0.6000 pu
    V_a2 = -I_a2 * Z_2 = -(-j3.3333) * (j0.12) = -0.4000 pu

  Check Phase A Voltage: V_a = V_a0 + V_a1 + V_a2 = -0.2000 + 0.6000 - 0.4000 = 0.00 pu (CONFIRMED)

  Phase B Voltage:
    V_b = V_a0 + a^2 * V_a1 + a * V_a2
        = -0.2000 + (0.6000 /_ 240°) + (-0.4000 /_ 120°)
        = -0.2000 + 0.6000(-0.5 - j0.86603) - 0.4000(-0.5 + j0.86603)
        = -0.2000 - 0.3000 - j0.51962 + 0.2000 - j0.34641
        = -0.3000 - j0.86603 pu
    |V_b| = sqrt( (-0.3000)^2 + (-0.86603)^2 ) = sqrt( 0.0900 + 0.7500 ) = sqrt(0.8400) = 0.9165 pu

  Phase C Voltage:
    V_c = V_a0 + a * V_a1 + a^2 * V_a2
        = -0.2000 + (0.6000 /_ 120°) + (-0.4000 /_ 240°)
        = -0.3000 + j0.86603 pu
    |V_c| = sqrt( (-0.3000)^2 + (0.86603)^2 ) = sqrt(0.8400) = 0.9165 pu
=========================================================================================

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6. Common PE Exam Traps & Tactical Pitfalls

• Assuming Three-Phase Fault Current is Always the Maximum: Assuming $I_{3\phi}$ is always greater than $I_{SLG}$. In solidly grounded systems with $Z_0 < Z_1$, the SLG fault current is strictly higher than the 3-phase fault current. Always check $Z_0$ relative to $Z_1$ before declaring the maximum fault level.
• Omitting the Factor of 3 on Fault Impedance ($3 Z_f$): Placing $Z_f$ directly in series with the sequence networks instead of $3 Z_f$. Because $I_a = 3 I_{a0}$, the actual voltage drop across $Z_f$ is $V_f = I_a Z_f = 3 I_{a0} Z_f$. In the sequence loop where current $I_{a0}$ circulates, the effective impedance is $3 Z_f$.
• Forgetting that Ground Return Current is $3 I_{a0}$: Calculating ground/neutral current as $I_{a0}$ instead of $3 I_{a0}$. Neutral overcurrent relays (50N/51N) connected in the residual CT circuit measure $I_n = I_a + I_b + I_c = 3 I_{a0}$.
• Attempting to Circulate Zero-Sequence Current in Line-to-Line Faults: Including $Z_0$ in L-L fault calculations. Because an L-L fault does not involve ground, the sum of phase currents is zero ($I_b + I_c = 0$), forcing $I_{a0} = 0$. The zero-sequence network is completely isolated and plays no role.