UOM Converter
Transformer fault current
Full-load current and the maximum fault current at a transformer's secondary terminals, from kVA and nameplate impedance.
Try an example
Convert from
More optionsThree-phase
1,000 kVA, 480 V three-phase, 5.75%Z
How it's calculated
- Full-load amps = kVA × 1,000 ÷ (√3 × volts).
- Fault current = full-load amps ÷ (%Z ÷ 100) = 1,202.8 ÷ 0.0575.
Assumes an infinite source (zero utility impedance), so this is the most the transformer can deliver; real fault current is lower. Nameplate %Z has a manufacturing tolerance, and a lower real impedance means more fault current. Not a substitute for a short-circuit study.
Worked examples
1,000 kVA, 480 V three-phase, 5.75%Z
- Maximum fault current: 20.9 kA
- Full-load current: 1,202.8 A
75 kVA, 208 V three-phase, 4%Z
- Maximum fault current: 5.2 kA
- Full-load current: 208.2 A
More in Electrical
kVA · kW · ampsTransformer, generator and UPS sizing. Three-phase uses line-to-line volts.- Voltage dropVoltage drop on a run, and the smallest wire that keeps it within your target.
- Power factor correctionCapacitor kvar needed to raise a load's power factor, and what it does to kVA and current.
- Motor speed & slipSynchronous speed from frequency and poles, slip from the nameplate speed, and the same motor on the other frequency.
- Line ↔ neutral voltageThree-phase wye systems: line-to-neutral = line-to-line ÷ √3.
- Ohm's lawEnter any two of volts, amps, ohms and watts to get the other two.
- Energy costWhat a load costs to run, and what a retrofit saves.