Fault Analysis Part 3 - Symmetrical (Three-Phase) Fault Analysis

⚡ Welcome to Part 3 of our Fault Analysis series. Now that we've covered fault basics (Part 1) and the per-unit system (Part 2), we're ready for our first real fault calculation: the Symmetrical (Three-Phase) Fault — the simplest fault to analyze, and also the most severe.

What is a Symmetrical Fault?

A symmetrical (or three-phase) fault occurs when all three phases short together simultaneously — either directly, or through a common fault impedance, with or without a connection to ground. Because all three phases are affected equally, the system remains balanced even during the fault. This is exactly why it's the easiest fault type to analyze mathematically.

Why Only the Positive Sequence Network Matters

Since the system stays balanced during a three-phase fault, there's no unbalance to represent — which means:

• Negative sequence current = 0 (no unbalance to create it)
• Zero sequence current = 0 (no ground-return asymmetry)
• Only the positive sequence network carries current

This is a major simplification — you don't need symmetrical components at all for this fault type. A simple single-phase equivalent circuit is enough.

The Fault Current Formula

For a three-phase fault at a bus, with the system represented in per-unit:

I_f (p.u.) = E / Z1

Where E is the pre-fault voltage (usually taken as 1.0 p.u.) and Z1 is the Thevenin positive-sequence impedance seen from the fault point. That's it — no Z2, no Z0, no sequence network interconnection required.

Step-by-Step Approach

1. Convert all system impedances (generator, transformer, line) to a common base using the per-unit conversion formula from Part 2
2. Draw the positive sequence network only — reduce it to a single Thevenin impedance (Z1) at the fault point
3. Apply I_f = E / Z1 to get the fault current in per-unit
4. Convert back to actual amperes using I_base for that voltage level

Worked Example

A generator with Z1 = 0.15 p.u. (on system base) is connected directly to a fault point through a transformer with Z1 = 0.10 p.u. Find the fault current in per-unit if pre-fault voltage is 1.0 p.u.

Total Z1 = 0.15 + 0.10 = 0.25 p.u.
I_f = E / Z1 = 1.0 / 0.25 = 4.0 p.u.

If the system base current at that voltage level is, say, 500 A, the actual fault current would be 4.0 × 500 A = 2000 A.

Why This Fault Still Matters Despite Being Rare

Even though three-phase faults occur only 2–5% of the time, they produce the highest fault current of any fault type. This makes them the standard for sizing circuit breakers — a breaker must be rated to safely interrupt the worst-case (three-phase) fault current, even though it will spend most of its working life handling more common L-G faults.

Quick Revision Points

• Three-phase fault = balanced fault → only positive sequence network needed
• I_f (p.u.) = E / Z1
• Negative and zero sequence currents are zero for this fault type
• Produces the maximum possible fault current — used for circuit breaker rating
• Least frequent fault type in practice (~2-5%), but the most critical for equipment sizing

Practice MCQs

Q1. For a three-phase fault, which sequence network(s) are required?
a) Positive only
b) Positive and negative
c) Positive, negative, and zero
d) Zero only

Q2. The formula for three-phase fault current in per-unit is:
a) I_f = E × Z1
b) I_f = E / Z1
c) I_f = E / (Z1 + Z2 + Z0)
d) I_f = E / (Z1 + Z2)

Q3. Which fault type produces the highest fault current?
a) Line-to-ground
b) Line-to-line
c) Double line-to-ground
d) Three-phase

Q4. During a three-phase fault, the negative and zero sequence currents are:
a) Equal to positive sequence current
b) Zero
c) Maximum
d) Undefined

(Answers: 1-a, 2-b, 3-d, 4-b)

Coming up in Part 4: Symmetrical Components — the theory (Fortescue's theorem) that makes unsymmetrical fault analysis possible, and the foundation for every fault type we'll cover after this.

Practice this formula with a few different Z1 values before moving to Part 4 — it'll make the unsymmetrical fault math much easier to follow.