Fault Analysis

Short circuits drive currents many times normal load. Learn the per-unit system, symmetrical three-phase faults, fault MVA, DC offset, and unsymmetrical faults with symmetrical components — the basis for rating breakers and setting protection.

When insulation fails — a tree touches a line, a cable is dug up, lightning flashes over — current takes a low-impedance path and rises to tens of kiloamps. Equipment must withstand it and breakers must interrupt it within a few cycles.

Fault studies calculate those currents. They size switchgear, set protection relays, design earthing and check that equipment survives the mechanical and thermal stress.

An analogy

A short circuit is like a dam breach: normal flow is set by the turbines and channels (the load); a breach lets water rush through limited only by the dam's own opening (the source impedance).

Types of fault

Faults are classified by which conductors are involved.

• Single line-to-ground (SLG): ~70–80 % of faults.
• Line-to-line (LL): ~15 %.
• Double line-to-ground (LLG): ~10 %.
• Three-phase (LLL): ~5 % but usually the most severe; balanced, so easiest to calculate.

The per-unit system

Expressing quantities as fractions of chosen base values makes transformer ratios disappear (a transformer's pu impedance is the same on both sides) and keeps numbers in a familiar range. Choose S_base once for the whole system and V_base in each voltage zone set by transformer ratios.

Fault level

S_sc = S_base / Z_pu (total pu impedance to the fault). A 20 MVA transformer with 10 % impedance fed from an infinite grid can supply at most 200 MVA into a fault on its secondary. Transformer impedance is the main thing limiting distribution fault levels.

DC offset and breaker duties

Because current in an inductive circuit can't change instantly, a fault starting at a voltage zero includes a decaying DC offset with time constant τ = X/(ωR). The first peak can reach ~2.5× the symmetrical RMS value. Breakers are rated for symmetrical breaking current and for peak making current.

Symmetrical components

An unbalanced three-phase set can be split into positive-, negative- and zero-sequence sets. Each sequence has its own network; the fault type decides how they connect: SLG puts all three in series. Zero-sequence current flows only if a path to earth exists, so transformer winding connections (Δ blocks zero sequence) and earthing strongly affect earth-fault currents.

Formulas

• Base current: I_base = S_base / (√3 · V_base)
• Change of base: Z_new = Z_old × (S_new/S_old) × (V_old/V_new)²
• Three-phase fault: I_f = E / Z₁, S_sc = S_base / Z_pu
• Single line-to-ground fault: I_f = 3E / (Z₁ + Z₂ + Z₀)
• Line-to-line fault: I_f = √3·E / (Z₁ + Z₂)

Worked example: Fault at a distribution transformer's secondary

Given: Infinite grid; 20 MVA, 33/11 kV transformer, Z = 10 %

Find: Fault MVA and current at the 11 kV bus.

• S_sc = 20 / 0.10 = 200 MVA
• I_f = 200e6 / (√3 × 11e3) = 10.5 kA

Answer: 200 MVA, 10.5 kA

Worked example: Single line-to-ground vs three-phase

Given: E = 1 pu; Z₁ = Z₂ = j0.1 pu; Z₀ = j0.05 pu (solidly earthed); I_base = 1050 A

Find: SLG and three-phase fault currents.

• LLL: I_f = 1/0.1 = 10 pu = 10.5 kA
• SLG: I_f = 3/(0.1 + 0.1 + 0.05) = 12 pu = 12.6 kA

Answer: SLG (12.6 kA) exceeds LLL (10.5 kA) here because Z₀ < Z₁

Common mistakes

• Wrong: Forgetting the upstream grid impedance. — Right: Include the source fault level; an 'infinite bus' overestimates fault current, which is conservative for ratings but wrong for protection sensitivity.
• Wrong: Assuming the three-phase fault is always the worst. — Right: With solidly earthed neutrals near transformers, SLG faults can exceed three-phase faults.

Applications

• Switchgear rating: Every breaker and busbar is chosen so its short-circuit rating exceeds the maximum prospective fault current at its location.
• Protection settings: Minimum fault currents (e.g. high-resistance earth faults at the end of a feeder) decide whether relays and fuses will detect a fault at all.
• Arc-flash studies: Fault current and clearing time determine incident energy (cal/cm²), which sets the PPE workers must wear.

Key takeaways

• Per-unit simplifies multi-voltage calculations; S_sc = S_base/Z_pu.
• Transformer impedance mostly limits distribution fault levels.
• DC offset raises the first peak; breakers are rated for making and breaking duty.
• Symmetrical components solve unbalanced faults; SLG: I = 3E/(Z₁+Z₂+Z₀).

Going further

• IEC 60909: The standard method for short-circuit calculation uses an equivalent voltage source c·V_n/√3 (c = 1.1 for maximum currents) and impedance correction factors, giving results that are reproducible across utilities and software.
• Motor and inverter contribution: Running induction motors feed fault current for a few cycles from their stored flux. Inverter-based resources contribute only ~1.1–1.5× rated current, which weakens renewable-dominated grids and challenges overcurrent protection.
• Fault current limiters: Where fault levels exceed switchgear ratings, utilities use series reactors, network splitting, or superconducting fault-current limiters that become resistive only during a fault.