A fault is any abnormal condition that makes current flow along an unintended path: a lightning flashover, a fallen tree branch, insulation failure in a cable or a broken conductor. Fault currents can be many times the normal load current, so engineers must calculate them to choose circuit-breaker ratings, set protective relays and design earthing. Unbalanced faults are analysed with symmetrical components. Both topics are core material for PGCB, BPDB, DPDC, DESCO, NESCO, BREB, BCS (EEE) and university power-system exams. This post explains fault types, the sequence operator “a”, sequence networks and the fault-current formulas, with an animation, a solved example and MCQs.
- Shunt faults: LG (most frequent, ≈ 70 %), LL (≈ 15 %), LLG (≈ 10 %), LLL (≈ 5 %, symmetrical, usually most severe).
- Fortescue: any unbalanced 3-phase set = positive + negative + zero sequence components.
- Operator a = 1∠120° = −0.5 + j0.866; 1 + a + a² = 0; a³ = 1.
- LG fault: If = 3Ea/(Z1 + Z2 + Z0 + 3Zf); LLL fault: If = Ea/Z1; LL fault: |If| = √3Ea/|Z1 + Z2|.
- Zero-sequence current needs a return path through the earth or neutral; In = 3Ia0.
1. Types of Faults
| Fault | Description | Approx. occurrence | Balanced? |
|---|---|---|---|
| Single line-to-ground (LG / SLG) | One phase touches earth | ≈ 70 % | Unsymmetrical |
| Line-to-line (LL) | Two phases short together | ≈ 15 % | Unsymmetrical |
| Double line-to-ground (LLG) | Two phases short together and to earth | ≈ 10 % | Unsymmetrical |
| Three-phase (LLL / LLLG) | All three phases short together | ≈ 5 % | Symmetrical |
These are shunt (short-circuit) faults. Series (open-conductor) faults, such as one or two broken conductors or a blown fuse, are also unsymmetrical. The percentages vary with voltage level and construction, but the order (LG most common, LLL least common) is standard exam knowledge. A three-phase fault keeps the system balanced, so it can be solved per phase using only the positive-sequence network. The others need symmetrical components.
2. Symmetrical Components
In 1918 C. L. Fortescue showed that any set of three unbalanced phasors can be replaced by the sum of three balanced sets:
- Positive sequence (1): three equal phasors 120° apart with the same phase order as the system (a-b-c).
- Negative sequence (2): three equal phasors 120° apart with the opposite order (a-c-b).
- Zero sequence (0): three equal phasors in phase with each other.
3. Operator “a” and Transformation Equations
a = 1∠120° = −0.5 + j0.866, a² = 1∠240° = −0.5 − j0.866, a³ = 1, 1 + a + a² = 0Synthesis (sequence → phase):
Va = Va0 + Va1 + Va2Vb = Va0 + a²Va1 + aVa2
Vc = Va0 + aVa1 + a²Va2
Analysis (phase → sequence):
Va0 = ⅓(Va + Vb + Vc)Va1 = ⅓(Va + aVb + a²Vc)
Va2 = ⅓(Va + a²Vb + aVc)
The same equations apply to currents. Two useful consequences:
- Neutral current In = Ia + Ib + Ic = 3Ia0. In a 3-wire system with no neutral or earth path, the zero-sequence current is zero.
- Line-to-line voltages never contain a zero-sequence component, because it cancels in the subtraction.
4. Sequence Impedances and Networks
| Element | Z1 (positive) | Z2 (negative) | Z0 (zero) |
|---|---|---|---|
| Transmission line | Z1 | = Z1 (static, symmetrical) | ≈ 2–3.5 × Z1 (earth return) |
| Transformer | Leakage Z | = Z1 | = Z1 or open circuit, depending on winding connection and earthing |
| Synchronous generator | Subtransient X″d (for fault calculation) | X2 ≈ X″d | Smallest, X0 < X2, plus 3Zn if neutral-earthed through Zn |
Only the positive-sequence network contains the source (generators produce balanced positive-sequence EMF). The negative and zero-sequence networks are passive. A delta winding lets zero-sequence current circulate inside the delta but blocks it from the line terminals, and an ungrounded star blocks it completely. This is why star–delta transformer connections strongly affect earth-fault currents.
5. Fault-Current Formulas
Ea = pre-fault phase voltage, Zf = fault impedance (0 for a bolted fault).
| Fault | Sequence networks | Fault current |
|---|---|---|
| LLL | Positive only | If = Ea/(Z1 + Zf) |
| LG (phase a) | All three in series | Ia1 = Ia2 = Ia0 = Ea/(Z1 + Z2 + Z0 + 3Zf); If = 3Ia0 |
| LL (b–c) | Positive and negative in parallel (opposed) | Ia1 = −Ia2 = Ea/(Z1 + Z2 + Zf); If = −j√3 Ia1 |
| LLG (b–c–g) | Negative ∥ zero, in series with positive | Ia1 = Ea/[Z1 + Z2(Z0 + 3Zf)/(Z2 + Z0 + 3Zf)]; In = 3Ia0 |
6. Solved Example
A generator has Ea = 1.0 pu, Z1 = Z2 = j0.2 pu, Z0 = j0.1 pu, with a solidly earthed neutral. Find bolted fault currents at its terminals. The base is 100 MVA, 132 kV.
Ibase = 100 × 10⁶ / (√3 × 132 × 10³) = 437.4 A- LLL: If = 1/0.2 = 5.0 pu = 2.19 kA
- LG: Ia0 = 1/(0.2 + 0.2 + 0.1) = 2 pu; If = 3 × 2 = 6.0 pu = 2.62 kA (greater than LLL because Z0 < Z1)
- LL: |If| = √3 × 1/(0.2 + 0.2) = 4.33 pu = 1.89 kA (= 0.866 × ILLL when Z1 = Z2)
- LLG: Z2 ∥ Z0 = 0.2 × 0.1/0.3 = 0.0667; Ia1 = 1/(0.2 + 0.0667) = 3.75 pu; Ia0 = −Ia1 × Z2/(Z2 + Z0) = −2.5 pu; earth current In = 3|Ia0| = 7.5 pu = 3.28 kA
Fault MVA (three-phase) = MVAbase/Z1(pu) = 100/0.2 = 500 MVA, which is the minimum breaking capacity for a breaker at that point.
7. Why Fault Analysis Matters
- Circuit-breaker selection: breaking and making capacity (kA, MVA) must exceed the maximum fault level.
- Relay settings and coordination: overcurrent, earth-fault and distance relays need the minimum and maximum fault currents. The negative-sequence current is used to detect unbalance and to protect generators against rotor overheating.
- Earthing design: the earth-fault current sets the size of the earth grid and the step and touch voltages.
- Equipment withstand: busbars, CTs and cables must survive the thermal and mechanical stress of short-circuit current (I²t).
- Fault limiting: series reactors, neutral earthing resistors and bus splitting reduce excessive fault levels in growing grids.
8. Practice MCQs
1. The most frequently occurring fault in power systems is: (a) LLL (b) LL (c) LG (d) LLG
2. The only symmetrical fault is: (a) LG (b) LL (c) LLG (d) LLL
3. The value of 1 + a + a² is: (a) 1 (b) 0 (c) 3 (d) −1
4. For an LG fault, the sequence networks are connected in: (a) Parallel (b) Series (c) Only positive (d) Positive and negative only
5. Zero-sequence current is absent in: (a) Earthed star systems (b) Three-wire systems without neutral/earth path (c) LG fault (d) LLG fault
6. Neutral current equals: (a) Ia1 (b) 3Ia0 (c) Ia2 (d) √3 Ia1
7. For an LL fault, which sequence component is zero? (a) Positive (b) Negative (c) Zero (d) None
8. For a transmission line, Z1 and Z2 are: (a) Equal (b) Z2 > Z1 always (c) Z2 = 0 (d) Z2 = 3Z1
9. E = 1 pu, Z1 = Z2 = Z0 = j0.1 pu. LG fault current is: (a) 3.33 pu (b) 10 pu (c) 30 pu (d) 1 pu
10. Symmetrical components were introduced by: (a) Kirchhoff (b) Fortescue (c) Thevenin (d) Steinmetz
9. বাংলায় দ্রুত রিভিশন
- ফল্টের ধরন: LG (সবচেয়ে বেশি, ≈ ৭০%), LL, LLG, LLL (একমাত্র সিমেট্রিক্যাল, ≈ ৫%)।
- সিমেট্রিক্যাল কম্পোনেন্ট (ফোর্টেস্কিউ, ১৯১৮): পজিটিভ (a-b-c), নেগেটিভ (a-c-b), জিরো (তিনটি একই ফেজে)।
- অপারেটর a = 1∠120°; 1 + a + a² = 0; নিউট্রাল কারেন্ট = 3Ia0।
- LG ফল্ট: তিন সিকোয়েন্স নেটওয়ার্ক সিরিজে; If = 3E/(Z₁ + Z₂ + Z₀ + 3Zf)।
- LLL: If = E/Z₁; LL: |If| = √3E/|Z₁ + Z₂|; LL ফল্টে জিরো সিকোয়েন্স কারেন্ট নেই।
- ফল্ট বিশ্লেষণে সার্কিট ব্রেকারের রেটিং, রিলে সেটিং ও আর্থিং ডিজাইন নির্ধারিত হয়।
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References: J. Duncan Glover et al., Power System Analysis and Design; Hadi Saadat, Power System Analysis; V. K. Mehta, Principles of Power System; C. L. Fortescue, AIEE Transactions (1918). Images and animations: original work by Totalecer.
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