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Control Panels

Transformer Sizing and Full-Load Current

Primary and secondary full-load current for a transformer, plus the approximate available fault current at the secondary from the impedance.

Inputs

From the transformer nameplate. A lower impedance produces a higher available fault current.

Calculated reference results

Primary full-load current
54.1A
Secondary full-load current
124.9A
Approximate available fault current at the secondary
2,498A

Assumes an infinite primary source, so the real figure will be lower. Use it as a conservative upper bound for SCCR, not as a study result.

Approximate fault current
2.5kA

The arithmetic

  1. Constant = 1.732 (three phase)
  2. Primary FLA = 45.00 kVA x 1000 / (1.732 x 480 V) = 54.1 A
  3. Secondary FLA = 45.00 kVA x 1000 / (1.732 x 208 V) = 124.9 A
  4. Fault current = 124.9 A / 0.0500 = 2,498 A

What this does not account for

  • The fault current figure assumes an infinite primary source and ignores primary and conductor impedance. It is deliberately conservative. A real short-circuit study accounts for the utility contribution, conductor impedance, and motor contribution, and produces a lower number.
  • Do not use this figure to justify a panel short-circuit current rating without a proper study. NEC 409.110 requires the panel to be marked with an SCCR, and 110.24 requires the available fault current at the service to be documented.
  • Transformer secondary conductor and overcurrent protection rules are in Article 450 and 240.21(C), and are not covered here.

How this is calculated

Transformer full-load current is the kVA rating multiplied by 1000, divided by the voltage, and divided again by 1.732 for a three-phase transformer. Approximate available fault current at the secondary is the secondary full-load current divided by the per-unit impedance, which assumes an infinite primary source and therefore gives a conservatively high figure.

Formulas

FLA = (kVA x 1000) / (1.732 x V) for three phase

FLA = (kVA x 1000) / V for single phase

I_fault = FLA_secondary / (%Z / 100)

Assumptions and limitations built into this calculator

  • Fault current assumes an infinite primary source, which overstates the real value.
  • Ignores conductor impedance between the transformer and the point of interest.
  • Ignores motor contribution to fault current, which adds to the total during the first cycles.

Frequently asked questions

Is this a short-circuit study?
No. It is a conservative first-pass estimate. A study accounts for the utility source, conductor impedance, and motor contribution, and is what an engineer seals when one is required.
Why does lower impedance mean higher fault current?
Impedance is what limits current during a fault. A 4% transformer lets through more fault current than a 6% transformer of the same rating, which is why the nameplate impedance matters for equipment ratings.

Direct contact

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