Short Circuit Current Calculator — Transformer Fault Types (3LG, LL, 1LG, 2LG)

Q: How do I get the zero-sequence impedance of a transformer?

Use the manufacturer zero-sequence test result or the transformer test report. The vector group gives a starting point: grounded star with a delta path has Z0 close to Z1; zig-zag windings have a lower Z0; star-star without a delta path has a much higher Z0. Presets are illustrative, not a substitute for test data.

A short-circuit calculation is only useful if it covers the fault types the protection actually sees. This calculator computes the four common faults — three-phase, line-to-line, single line-to-ground and double line-to-ground — at a transformer secondary, using symmetrical components and the transformer nameplate data.

Key takeaways

The four common short-circuit types are three-phase (3LG), line-to-line (LL), single line-to-ground (1LG) and double line-to-ground (2LG).
Three-phase current is the reference. With Z₂ = Z₁, the line-to-line current is about 0.866 × the three-phase current.
Ground-fault currents depend on the zero-sequence impedance Z₀, which is set by the transformer vector group and grounding.
Single line-to-ground is the most frequent fault on overhead and distribution systems, even though it is not always the largest current.
At the transformer secondary, the transformer impedance usually dominates; the upstream source impedance often changes the result by only a few percent.

1. What this calculator does — and what it does not

It calculates the initial symmetrical short-circuit current for four fault types at the secondary terminals of a two-winding transformer. The network is reduced to a positive-sequence impedance Z₁, a negative-sequence impedance Z₂ and a zero-sequence impedance Z₀, and the fault currents are solved with symmetrical components.

It is a single-source, bolted-fault reference. It does not model motor contribution, multiple sources, arc resistance, conductor impedance between the transformer and the fault, cable capacitance, transformer tap position, or the full IEC 60909 impedance-correction method. The peak current uses a single X/R ratio and the IEC 60909 factor κ. Where a formal study or the relay setting sheet gives different values, that document governs.

2. Short-circuit current calculator

Calculation basis

Transformer rating S (kVA)

Transformer impedance u_k (%)

Secondary voltage U₂ (kV)

Upstream short-circuit power S_sc (MVA) 0 or blank = infinite upstream bus.

X/R ratio Used for the impedance angle and the peak factor.

Voltage factor c IEC 60909: 1.05 for LV, 1.10 for MV/HV maximum current.

Z₂ / Z₁ 1.0 for transformers and lines.

Zero-sequence preset

Z₀ / Z₁ The most sensitive input for ground faults. Confirm with the zero-sequence test.

Initial symmetrical current (I_k″) for a bolted fault. No motor contribution, arc resistance, cable impedance or multiple sources. Peak current uses κ = 1.02 + 0.98 e^−3/(X/R) from IEC 60909. Confirm settings against the relay setting sheet and the formal fault study.

3. Common short-circuit types

Type	Phases involved	Sequence network	Typical share	What drives the current
Three-phase (3LG)	All three phases, with or without earth	Positive sequence only	5–10%	Z₁; the reference fault level. Symmetrical, no zero-sequence current.
Line-to-line (LL)	Two phases, no earth	Positive and negative in series	10–15%	Z₁ + Z₂; about 0.866 × 3LG when Z₂ = Z₁.
Single line-to-ground (1LG)	One phase to earth	Positive, negative and zero in series	65–80% on overhead systems	Z₁ + Z₂ + Z₀; strongly dependent on grounding.
Double line-to-ground (2LG)	Two phases to earth	Positive, negative and zero in parallel	5–15%	Z₁, Z₂ and Z₀; the two phase currents and the ground current differ.

Percentages are indicative field statistics, not a rule for a specific network. An open-phase condition is not a short circuit and is outside this calculator.

4. How to read the result

Three-phase current is the reference, not the worst case

For a bolted three-phase fault the three phase currents are equal and displaced by 120°. This is the value most fault-level tables quote. It is not automatically the highest current: with a very low zero-sequence impedance, a line-to-ground or double line-to-ground fault can exceed it.

Line-to-line current is lower when the sequence impedances match

For a b-c fault, the positive and negative networks are in series. With Z₂ = Z₁, the faulted-phase current is √3/2 ≈ 0.866 of the three-phase current. A machine or an unbalanced network with Z₂ ≠ Z₁ changes that ratio.

Ground-fault current is set by Z₀, not by the transformer kVA alone

A solidly grounded star winding with a delta path (Dyn11) gives Z₀ close to Z₁, so the single line-to-ground current is close to the three-phase current. A zig-zag winding gives a lower Z₀ and a higher earth-fault current. A star-star winding without a delta path has a high Z₀, so the earth-fault current is small. Use the zero-sequence test result or the manufacturer’s data; the presets are starting points, not settings.

The upstream source often matters less than expected

At the transformer secondary, the transformer impedance is referred to the low-voltage side as Z_T = u_k × U₂² / (100 × S). For a 1000 kVA, 0.4 kV, 6% transformer this is 0.0096 Ω. An upstream source of 200 MVA contributes only 0.0008 Ω — less than 10% of the total. For a weak source, the upstream term becomes significant and must be included.

Peak current is a separate check

The peak value i_p = κ √2 I_k″ is what the switchgear and cable withstand ratings see. It depends on the X/R ratio, not only on the symmetrical current. A high X/R ratio gives a higher peak for the same rms current.

5. Formulas and sequence networks

Quantity	Expression	Note
Transformer impedance at the secondary	Z_T = u_k × U₂² / (100 × S)	U₂ in kV, S in MVA, Z in Ω.
Upstream source at the secondary	Z_sc = U₂² / S_sc	S_sc in MVA; 0 means infinite bus.
Phase voltage	E = c U₂ / √3	c from IEC 60909.
Three-phase	I₁ = E / Z₁; I_k = \|I₁\|	Positive sequence only.
Line-to-line	I₁ = E / (Z₁ + Z₂); I_k = √3 \|I₁\|	Faulted phases b and c.
Single line-to-ground	I₁ = I₂ = I₀ = E / (Z₁ + Z₂ + Z₀); I_k = 3 \|I₁\|	Faulted phase a.
Double line-to-ground	I₁ = E / (Z₁ + Z₂Z₀/(Z₂+Z₀)); I₂ = −I₁Z₀/(Z₂+Z₀); I₀ = −I₁Z₂/(Z₂+Z₀)	I_b = a²I₁ + aI₂ + I₀; I_c = aI₁ + a²I₂ + I₀.
Peak current	i_p = κ √2 I_k″; κ = 1.02 + 0.98 e^−3/(X/R)	IEC 60909 peak factor.

a = e^j120°. All impedances are complex and use the same X/R ratio in this calculator. For a formal study, use the actual R and X of each element and the IEC 60909 impedance-correction factors.

6. Worked example — 1000 kVA, 11/0.4 kV, u_k = 6%

Transformer 1000 kVA, secondary 0.4 kV, u_k = 6%, upstream short-circuit power 200 MVA, X/R = 10, Z₂ = Z₁, Z₀ = Z₁ (Dyn11), c = 1.05.

Z_T = 0.06 × 0.4² / 1 = 0.0096 Ω. Z_sc = 0.4² / 200 = 0.0008 Ω. Z₁ = 0.0104 Ω. E = 1.05 × 0.4 / √3 = 0.2425 kV.

Fault type	Current	Peak (κ = 1.75)
Three-phase (3LG)	23.3 kA	57.6 kA
Line-to-line (LL)	20.2 kA	49.9 kA
Single line-to-ground (1LG)	23.3 kA	57.6 kA
Double line-to-ground (2LG)	23.3 kA phase, 23.3 kA ground	57.6 kA

With Z₀ = Z₁, the single line-to-ground and double line-to-ground currents reach the same value as the three-phase current. Change Z₀ to 0.2 (zig-zag) and the earth-fault currents rise; change it to 10 (no delta path) and they fall sharply. That is why the zero-sequence input deserves the most attention.

7. Frequently asked questions

Which short-circuit type gives the highest current?

Three-phase is the usual reference and is often the highest, but not always. With a very low zero-sequence impedance, a single line-to-ground or double line-to-ground fault can exceed the three-phase current. Always calculate all four types for the network in question.

What is the difference between 2LG and 2L-G?

They are the same fault: two phases connected to earth, also written double line-to-ground. The notation varies between standards and relay manufacturers. This calculator calls it 2LG.

How do I get the zero-sequence impedance of a transformer?

Use the manufacturer’s zero-sequence test result or the transformer test report. If it is not available, the vector group gives a starting point: Dyn11 and other grounded-star-with-delta arrangements have Z₀ close to Z₁; zig-zag windings have a lower Z₀; star-star without a delta path has a much higher Z₀. Presets are illustrative, not a substitute for test data.

Should I use the transformer’s own impedance or the upstream source impedance?

Use both. The fault current at the secondary is driven by the series combination. At the transformer terminals the transformer impedance usually dominates, but a weak upstream source adds a significant term. The calculator adds the magnitudes and applies the X/R ratio to the total.

Does this calculator include motor contribution?

No. Motors add current for the first few cycles and can raise the peak and the initial symmetrical current. Include motor contribution from the formal fault study when motors are a significant part of the load.

8. Where to go next

For relay test methods, see secondary injection testing and overcurrent relay testing. The VAE-430 and VAE-660 relay test systems provide the current and voltage channels needed to verify the settings derived from these fault levels; for a configuration matched to your protection schemes, request a quote.