Fault current calculator

Three-phase and line-to-ground fault current at a delta–wye-grounded transformer's secondary.

Result

Three-phase fault: 6.973kA
Line-to-ground fault: 7.844kA

Base current: 836.7A
Total Z₁: 0.12pu

Z₁ = Z_s + Z_t on a 20 MVA base, with Z_s = 0.04 pu and Z_t = 0.08 pu.

What it calculates

A bolted fault on the secondary bus of a delta–wye-grounded transformer, fed from a utility source. Everything is worked in per unit on the transformer's own rating, so the transformer impedance is simply its nameplate percent divided by 100, and the source impedance is the ratio of the two MVA values:

Z_s = \dfrac{S_\text{base}}{S_\text{sc}} \qquad Z_t = \dfrac{\%Z}{100}

The three-phase fault sees only the positive-sequence impedance:

I_{3\phi} = \dfrac{I_\text{base}}{Z_s + Z_t}

A line-to-ground fault puts the positive, negative and zero-sequence networks in series. The delta winding gives zero-sequence current no path back to the source, so the zero-sequence impedance is the transformer's alone:

I_\text{SLG} = \dfrac{3\,I_\text{base}}{2\,(Z_s + Z_t) + Z_t}

Assumptions

Prefault voltage is 1.0 pu and the fault is bolted, with no arc or ground resistance.
Positive and negative-sequence impedances are equal; the zero-sequence impedance is the transformer's.
Source and transformer impedances are added as magnitudes, as if they shared one angle.
No motor contribution and no DC offset: the result is the symmetrical RMS current.

Worked example

A 20 MVA transformer with 8% impedance and a 13.8 kV secondary is fed from a utility that can deliver 500 MVA of fault power at the primary. First the base current and the source impedance:

I_\text{base} = \dfrac{20\ \text{MVA}}{\sqrt{3} \times 13.8\ \text{kV}} = 836.7\ \text{A} \qquad Z_s = \dfrac{20}{500} = 0.04

Then the two fault currents:

I_{3\phi} = \dfrac{836.7}{0.04 + 0.08} = 6{,}973\ \text{A}

I_\text{SLG} = \dfrac{3 \times 836.7}{2\,(0.12) + 0.08} = 7{,}844\ \text{A}

Here the ground fault is larger than the three-phase fault. The source impedance sits only in the positive- and negative-sequence networks, so it counts at two-thirds weight in the ground fault but at full weight in the three-phase fault. With an infinite source the two currents are equal: tick “infinite source” and both rise to 10.46 kA, the most this transformer can deliver.

Questions

Why can the ground fault current be larger than the three-phase fault current?

At a delta–wye-grounded transformer the delta winding blocks the source's zero-sequence path, so the zero-sequence impedance is only the transformer's own. Any source impedance makes the positive-sequence impedance larger than the zero-sequence impedance, so the line-to-ground current exceeds the three-phase current. The gap narrows as the source gets stronger, and with an infinite source the two are equal.

What does an infinite source mean?

The utility is treated as having zero impedance, so only the transformer limits the current. It gives the largest fault current that transformer can deliver, and it is a common conservative assumption when the utility's available fault MVA is not known.

Does this include motor contribution or DC offset?

No. It gives the symmetrical RMS current of a bolted fault fed from the utility source alone. Motor contribution, arc resistance and asymmetrical current need a fuller short-circuit study.

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PhasorPrep is an independent study resource. It is not affiliated with, endorsed by, or sponsored by NCEES. Questions are original and written for practice; they are not actual exam questions. Calculators are for learning, not for engineering design.

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