2.5 Transformer Ratios, Connections, Taps & Available Fault Current

Why transformer theory is tested

DLI's knowledge-area table lists motor and transformer connections (single-phase, three-phase, taps, wye and delta, 115/230 and 230/460 V) and transformers (autotransformers and short-circuit current available at the secondary) for the journeyworker exam. Its exam guide includes worked examples for both full-load current and fault current. These questions are pure arithmetic once you know the formulas.

Ratios and the ideal transformer

A transformer changes voltage by the ratio of its turns. For an ideal transformer, power in equals power out, so current changes in the opposite direction from voltage:

$a = \frac{N_P}{N_S} = \frac{E_P}{E_S} = \frac{I_S}{I_P}$

DLI's ratio example: a 480/120 V single-phase transformer has a ratio of $480 \div 120 = 4$, or 4:1. If the secondary carries 40 A, the primary carries $40 \div 4 = 10$ A. DLI's guide states that, for general calculations, the kVA of the primary is considered equal to the kVA of the secondary. Transformers are rated in kVA, not kW, because their heating depends on current and voltage regardless of power factor.

Full-load current

Always use line-to-line voltage in the three-phase formula. A common error is dividing by 120 or 277 V instead of 208 or 480 V.

Common connections

In a high-leg (wild-leg) delta, one winding is center-tapped to give 120 V from two phases to neutral, and the third phase measures about $120 \times 1.732 = 208$ V to neutral. NEC 110.15 requires that leg to be marked orange, and 408.3(E)(1) puts it in the B position.

Dual-voltage primaries and taps

Distribution transformers usually offer primary taps in 2.5% steps above and below nominal (often labeled FCAN and FCBN, full-capacity above and below normal). Moving the primary connection to a tap with fewer turns raises the secondary voltage, because the ratio $N_P \div N_S$ gets smaller. If a 480 V primary is fed with only 456 V (5% low), connecting to the −5% tap restores nominal secondary voltage.

Dual-voltage windings follow the same rule as motors: windings in series for the higher voltage and in parallel for the lower voltage. A 120/240 V secondary with two 120 V windings is connected in series, center-tapped, for 120/240 V, or in parallel for 120 V only at twice the current.

Autotransformers and buck-boost units

An autotransformer uses a single, tapped winding shared by primary and secondary, so the two sides are not electrically isolated. Buck-boost transformers are small isolation transformers wired as autotransformers to raise (boost) or lower (buck) voltage by about 5 to 20%, such as boosting 208 V to about 230 V for equipment rated 230 V.

Key NEC points:

• 210.9: Branch circuits may not be supplied by autotransformers unless the circuit supplied has a grounded conductor electrically connected to a grounded conductor of the supply system. An exception allows an autotransformer to extend or add an individual branch circuit without that connection when it changes 208 V to 240 V or 240 V to 208 V.
• 215.11: The same rule applies to feeders supplied by autotransformers.
• 450.4: An autotransformer of 1000 V or less needs overcurrent protection in each ungrounded input conductor rated not more than 125% of rated input current (with next-higher-size rounding allowed where 125% is not standard and the current is 9 A or more), or 167% where input current is less than 9 A.

Available short-circuit (fault) current

DLI's guide explains that the available fault current must be known to select equipment ratings, and that most calculations start with the maximum bolted fault at the transformer's load terminals, assuming an infinite primary source:

$I_{SC} = I_{FLA} \times \text{multiplier}, \qquad \text{multiplier} = \frac{100}{\%Z}$

The transformer impedance (%Z) is printed on the nameplate.

DLI adds that these are line-to-line faults; line-to-neutral or line-to-ground faults can be 10 to 15% higher. The calculation ignores conductor impedance (which lowers the fault current downstream) and motor contribution (which raises it).

Why the number matters

• NEC 110.9: Equipment intended to interrupt current at fault levels must have an interrupting rating at least equal to the available fault current at its line terminals.
• NEC 110.10: Overcurrent devices, total circuit impedance, equipment short-circuit current ratings (SCCR), and other characteristics must be coordinated so faults are cleared without extensive damage.
• NEC 110.24: In other than dwelling units, service equipment must be legibly field-marked with the maximum available fault current and the date the calculation was performed. The marking must be updated when modifications change the available fault current.
• 2026 NEC 408.6: In other than one- and two-family dwellings, switchboards and panelboards must be field-marked with the available fault current, the calculation date, and their short-circuit current rating. The calculation must be documented and redone after changes that affect it.

Worked example

A 112.5 kVA, 480 V delta to 208Y/120 V transformer has 3.5% impedance. Find the secondary full-load current and the maximum available fault current.

1. $I_{FLA} = 112{,}500 \div (208 \times 1.732) = 112{,}500 \div 360.26 = 312.3$ A
2. Multiplier $= 100 \div 3.5 = 28.57$
3. $I_{SC} = 312.3 \times 28.57 = 8{,}922$ A

A panelboard fed directly from this transformer needs breakers with an interrupting rating of at least 10,000 A (the common 10 kA level exceeds 8,922 A). Remember DLI's note that a line-to-ground fault may run 10 to 15% higher; designers often allow for that margin.