Available fault current: what it means for the Texas exam

Use this diagram

Copy this HTML to show the diagram with a credit link<img src="https://bondingjumper.com/charts/bondingjumper/available-fault-current.svg" alt="A transformer at the left feeding a service panel and then a subpanel down the line, with a bar above each point: the tallest bar at the transformer terminals, a shorter one at the service and the shortest at the subpanel, because the impedance of each run of conductor adds up" width="100%"><p><a href="https://bondingjumper.com/guides/available-fault-current?utm_source=embed&amp;utm_medium=referral&amp;utm_campaign=figure_embed&amp;utm_content=available-fault-current.svg">The farther from the transformer, the more impedance, the less fault current. by Bonding Jumper</a></p>

What sets it

A bolted fault is as close to a perfect short circuit as you get: a solid connection with almost no resistance of its own. With the load gone, the only thing holding the current back is the impedance of the path.

• The transformer. Its nameplate impedance, given as a percent, is the share of rated primary voltage that pushes full-load current through a shorted secondary. The lower the percent, the higher the fault current at its terminals.
• The conductors. Each run between the transformer and the point you care about adds impedance. The fault current at a subpanel is lower than at the service, and lower again at the end of a long branch circuit.
• The utility source. A quick estimate treats the source feeding the transformer as unlimited. Real sources are not, so that estimate is an upper limit for the transformer terminals.

A worked example with invented numbers

A single-phase transformer is rated 48 kVA with a 240 volt secondary and a nameplate impedance of 2.5%.

1. Full-load current: 48,000 ÷ 240 = 200 amperes.
2. Percent to decimal: 2.5% is 0.025.
3. Fault current at the terminals with an unlimited source: 200 ÷ 0.025 = 8,000 amperes. Dividing by 0.025 is the same as multiplying by 40, so 200 x 40 = 8000.

Now give the same transformer a 5% impedance. 200 ÷ 0.05 is the same as 200 x 20 = 4000, so 4,000 amperes. Double the percent impedance and the fault current halves. For a three-phase transformer, find full-load current from kVA ÷ (volts × 1.732) first, then divide by the impedance in the same way.

Why it matters: interrupting rating

Every fuse and breaker has two different ratings. Its ampere rating sets the current it trips at and protects the conductor. Its interrupting rating is the largest fault current it can open safely. Under the overcurrent protection rules, the interrupting rating must be at least the fault current available where the device is installed, or the device can fail violently.

OSHA's electrical rules for general industry workplaces set the same requirement. 29 CFR 1910.303(b)(4) requires equipment meant to interrupt fault current to have an interrupting rating that covers the circuit voltage and the current available at its line terminals. 1910.303(b)(5) has the protective devices, the total impedance and the short-circuit current ratings of the components chosen together, so a fault clears without extensive damage.

With the example above, a device with an invented 5,000 ampere interrupting rating would be fine on the 5% transformer (4,000 amperes available) and not enough on the 2.5% one (8,000 amperes available).

Common mix-ups

• Thinking a higher percent impedance means more fault current. It means less.
• Using full-load current as the fault current. Full-load current is what the transformer delivers in normal use. Fault current is many times higher.
• Checking the breaker's ampere rating instead of its interrupting rating. A 20 ampere breaker can sit where thousands of amperes are available, if its interrupting rating covers them.
• Forgetting to convert the percent. 2.5% is 0.025, not 2.5.
• Assuming the number is the same everywhere. It falls with every run of conductor away from the transformer.

On the Texas exam

The PSI bulletin lists fault current in services, service equipment and separately derived systems, an NEC Knowledge area worth 11 scored items for master and 6 for journeyman. The same area in the master Calculations portion, worth 8, covers transformers.

Practice it

The transformer calculator gives full-load current from kVA on each side; divide its result by the percent impedance as a decimal to get the terminal fault current. Every term in this set is in the electrical terms glossary.

Sitting a master exam on the 2023 NEC outside Texas? Start with the master electrician exam outline or take the free NEC master check.

Know the idea? Take the free 10-question check

Take the free 10-question check for the Texas master exam. It needs no signup and explains every answer, so you see whether services, transformers or motors cost you the most points.

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