Two transformers are the same kVA and voltage. One is 3%Z, the other 6%Z. Which gives the higher available fault current?

B. The 3% one — about twice as much

C. They are the same, the kVA is the same

D. You cannot tell without the utility figures

Why is the real fault current at a panel down the hall lower than the infinite-bus figure at the transformer?

B. The utility source and every length of conductor add impedance in series with the transformer

A. The breaker limits it before it gets there

C. Fault current decreases with time

D. The neutral carries half of it

Why does available fault current matter when choosing a breaker or panelboard?

B. The equipment must be able to interrupt or withstand the current a fault could deliver at that point without blowing apart

C. It determines the conductor colour

D. It sets the ground-fault setting

You replace a transformer with one of the same kVA but lower %Z. What happens downstream?

B. Available fault current rises, so existing equipment ratings may no longer be adequate and need re-checking

A. Nothing changes — the kVA is the same

Quatrième année Lesson 9 of 12 · about 14 min

Available fault current

What you’ll learn

What limits the current in a short circuit, and why it can be tens of thousands of amps.
What a transformer’s %Z means, and the infinite-bus estimate that comes from it.
Why the real fault current is lower downstream, and what raises it.
Why the number matters: equipment must interrupt or withstand what is available where it sits.

Before this: Transformer basics · Overcurrent protection and coordination

In a bolted short circuit, the only things limiting the current are the impedances between the source and the fault — the utility, the transformer, and the conductors. Take those away and the current is limited by almost nothing. That current is the available fault current, and every breaker, fuse and panel in its path has to survive it.

Where the number comes from: %Z

On a normal load the load’s own impedance sets the current. In a short, the load is gone. For a service fed by a transformer, the biggest remaining impedance is usually the transformer’s own windings, and the nameplate states it as percent impedance, %Z: the percentage of rated voltage needed to push full-load current through the transformer with its secondary shorted.

If 5.75% of the voltage pushes full-load current, then 100% of the voltage pushes 100 ÷ 5.75 = 17.4 times full-load current. That is the infinite-bus estimate — the fault current at the transformer’s terminals if the supply behind it were infinitely strong:

FLA = kVA × 1000 ÷ (√3 × V)three-phase; ÷ V for single-phase
Multiplier = 100 ÷ %Ztimes full-load current
Isc = FLA × 100 ÷ %Zworst case at the terminals

The infinite-bus estimate counts only the transformer. The utility and every length of conductor add impedance, so the real figure is lower — and lower still the further downstream you go.

Worked: 500 kVA, 480 V, 5.75 %Z

Full-load current: 500,000 ÷ (1.732 × 480) = 500,000 ÷ 831.4 = 601 A.
Multiplier: 100 ÷ 5.75 = 17.39.
Infinite-bus fault current: 601 × 17.39 = about 10,460 A at the secondary terminals.
Same transformer at 208 V: FLA = 500,000 ÷ (1.732 × 208) = 1,388 A; Isc = 1,388 × 17.39 = about 24,100 A. Same kVA, lower voltage, far more fault current.

Allowing for the %Z tolerance

A nameplate %Z has a manufacturing tolerance, so the actual unit may be a little lower. Some engineers allow for that by using a reduced %Z — for example 90% of the plate value — which raises the estimate: 601 × 100 ÷ (0.9 × 5.75) ≈ 11,600 A. Use the allowance you are told to use; the calculator has a field for it.

Why the real figure is lower — and what raises it

The utility has impedance of its own, so a real transformer secondary sees less than the infinite-bus figure. Every run of conductor adds more: a panel at the end of a long feeder can see a fraction of what the main switchboard sees. Engineers work out the value at each point with a point-to-point calculation or study software.

Things that raise it: a bigger transformer, a lower-%Z transformer, a lower voltage for the same kVA, shorter or larger conductors, paralleled transformers, and running motors, which feed the fault for a few cycles. A service upgrade or a transformer swap is therefore a reason to re-check the equipment ratings downstream, not only the new gear.

Why it matters: interrupting and withstand ratings

A breaker or fuse is rated for the most current it can safely interrupt, and panelboards and other equipment for the short-circuit current they can withstand. Ask a device to clear more than it was built for and it can fail violently — a burst case, an arc that does not go out. So the equipment at each point must suit the fault current available there. Which ratings are required, how equipment is marked and how series ratings work are code-book and engineering questions; the idea you carry is simply: know the available fault current before you choose or accept the gear.

Sur le chantier

Tips

Read the interrupting rating on every breaker you install; it is on the handle or the label.
Check the equipment label for the available fault current and the date it was calculated, where one is posted.
When the utility changes a transformer, ask for the new available fault figure at the service.
Use the infinite-bus figure as a quick worst case; ask for the study when the answer is close to a rating.

Common mistakes

Swapping in a breaker with the same trip rating but a lower interrupting rating.
Dividing by 100 instead of by %Z — or using the %Z as a decimal (0.0575) in the 100 ÷ %Z step.
Putting a √3 in a single-phase full-load current.
Assuming a smaller transformer always means less fault current — %Z and voltage matter as much as kVA.
Forgetting that fault current at the far end of a feeder is lower, and using the service figure everywhere (or the reverse).

Check yourself

0 right of 0 answered 7 questions · pick one answer each

1.A transformer is marked 5% impedance (%Z). On an infinite bus, a bolted fault at its secondary terminals draws about how many times its full-load current?
2.A 500 kVA, 480 V three-phase transformer has a 5.75% impedance. What is its infinite-bus fault current at the secondary?
3.Two transformers are the same kVA and voltage. One is 3%Z, the other 6%Z. Which gives the higher available fault current?
4.Why is the real fault current at a panel down the hall lower than the infinite-bus figure at the transformer?
5.Why does available fault current matter when choosing a breaker or panelboard?
6.You replace a transformer with one of the same kVA but lower %Z. What happens downstream?
7.A single-phase 50 kVA, 240 V transformer has 2%Z. What is its infinite-bus fault current?

Answers

B. 20 times — Multiplier = 100 ÷ %Z = 100 ÷ 5 = 20. %Z is the percentage of rated voltage that pushes full-load current through a shorted transformer, so full voltage pushes 100/5 = 20 times that current.
C. About 10,460 A — FLA = 500,000 ÷ (1.732 × 480) = 601 A. Isc = 601 × 100 ÷ 5.75 = 601 × 17.39 ≈ 10,460 A.
B. The 3% one — about twice as much — Fault current is inversely proportional to %Z. Halve the impedance and you double the fault current: 100/3 ≈ 33 times FLA against 100/6 ≈ 17 times.
B. The utility source and every length of conductor add impedance in series with the transformer — The infinite-bus estimate ignores everything except the transformer. The utility is not infinitely stiff, and each foot or metre of conductor adds impedance, so the available current drops as you move downstream.
B. The equipment must be able to interrupt or withstand the current a fault could deliver at that point without blowing apart — A device that is asked to interrupt more current than it is rated for can fail violently. Equipment interrupting and short-circuit ratings must suit the available fault current at their location — the required ratings and markings come from your code book and the engineer.
B. Available fault current rises, so existing equipment ratings may no longer be adequate and need re-checking — Lower %Z means more fault current. Equipment that was fine before may now be under-rated. This is why a transformer change is a reason to re-check the fault-current study, not just the nameplate kVA.
C. About 10,400 A — Single-phase FLA = 50,000 ÷ 240 = 208.3 A (no √3). Isc = 208.3 × 100 ÷ 2 = 208.3 × 50 ≈ 10,400 A.

Questions fréquentes

What does %Z on a transformer nameplate mean?

Short the secondary, then raise the primary voltage from zero until full-load current flows. The voltage it takes, as a percentage of rated voltage, is the %Z. A 5% transformer reaches full-load current at 5% of rated voltage — so at 100% voltage, into a dead short, it would pass 20 times full-load current.

Why call it an "infinite bus"?

It is the assumption that the supply behind the transformer can deliver any current without its voltage dropping, so only the transformer limits the fault. Real utilities have impedance too, so the real figure is lower. The assumption gives the worst case at the transformer terminals, which is a safe first number.

Do motors add to fault current?

Yes, briefly. When a fault drags the voltage down, running induction motors act as generators for a few cycles and feed current back into the fault. Detailed studies include it; the simple infinite-bus estimate does not.

Is fault current the same as arc-flash energy?

Related, not the same. Arc-flash incident energy depends on the arcing current (lower than the bolted fault current), how long the protection takes to clear it, and how far away you are. A fault-current study is one input to an arc-flash study, which is its own engineering job.

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