Fault Current Calculator

About the Fault Current Calculator

When a short circuit happens, current is limited only by the impedance of the transformer and wiring between the source and the fault. That current can reach thousands or tens of thousands of amps in a fraction of a second. Every breaker, fuse, panel and switch in the path must be able to withstand it and, for protective devices, safely interrupt it. Electrical codes require the available fault current to be known so equipment with an adequate rating can be selected, and at many services it must be marked on the equipment.

This fault current calculator estimates available short-circuit current using the widely taught point-to-point method. It first works out the fault current at the transformer secondary from its kVA, voltage and percent impedance, assuming a strong utility source. It then reduces that figure for a run of copper conductors to the point of interest, such as a panel or disconnect.

How to Use the Fault Current Calculator

Choose three-phase or single-phase.

Enter the transformer rating in kVA, the secondary voltage and the percent impedance from the nameplate.

Enter the conductor run length, the conductor size and the number of parallel sets.

The Formulas

  full-load amps  = kVA × 1,000 ÷ (√3 × V)       three-phase
                  = kVA × 1,000 ÷ V              single-phase
  secondary Isc   = full-load amps × 100 ÷ %Z
  f               = √3 × L × Isc ÷ (C × n × V)   three-phase
                  = 2 × L × Isc ÷ (C × n × V)    single-phase
  M               = 1 ÷ (1 + f)
  fault current   = Isc × M

C is the conductor constant for copper in steel conduit, n the number of parallel sets.

Step-by-Step Example

A 75 kVA, 208 V three-phase transformer with 2% impedance, 50 ft of 1/0 copper.

  full load   75,000 ÷ (1.732 × 208) = 208 A
  secondary   208 × 100 ÷ 2 = 10,409 A
  f           1.732 × 50 × 10,409 ÷ (8,925 × 1 × 208) = 0.49
  M           1 ÷ 1.49 = 0.67
  at panel    10,409 × 0.67 ≈ 7,007 A

Transformer Impedance

Percent impedance is the voltage, as a percentage of rated voltage, needed to drive full-load current through the transformer with its secondary short-circuited. A 2% transformer lets fifty times full-load current flow in a bolted fault; a 5% transformer twenty times. Lower-impedance transformers are more efficient and regulate voltage better, but deliver higher fault currents, which need higher-rated equipment downstream. Actual impedance can be up to about 10% lower than nameplate, so engineers often reduce it slightly for a conservative result.

The Point-to-Point Method

The method reduces fault current step by step from the source to each point in the system. The factor f compares the conductor's impedance with the impedance already in the system. Long runs, small conductors and high available current give a large f and a big reduction. Short runs of large conductors barely reduce the current at all, which is why panels close to large transformers face high fault currents. The C values used here are for copper in steel conduit at 600 volts; aluminum, non-magnetic conduit and busway have different values.

Why It Errs High

The calculation assumes the utility system feeding the transformer can supply unlimited current, an "infinite bus". In reality the utility's impedance reduces fault current somewhat. Motors running at the moment of the fault also feed current back into it, which adds to the total. The infinite-source assumption usually outweighs motor contribution in small systems, giving a conservative answer, but large motor loads should be added according to the engineering method used.

Equipment Ratings

Breakers and fuses have an interrupting rating, often 10,000, 22,000, 42,000 or 65,000 amps for low-voltage equipment. Panels and other equipment have a short-circuit current rating. Each must be at least the available fault current where it is installed, or protected by a tested series-rated combination. Equipment that is underrated can explode during a fault.

Arc Flash

Available fault current is one input to an arc-flash study, which estimates the energy released by an arc fault and sets the protective equipment workers need. Arc-flash studies require detailed data and methods beyond this calculator, and must be performed by qualified people.

Single-Phase Services

Most homes are fed from a single-phase transformer on a pole or pad. Residential transformers of 25 to 100 kVA commonly have impedances of about 1.5 to 3%, which can put 10,000 to 40,000 amps at the service entrance. Utilities can provide the available fault current on request, and many now print it on the meter base or provide it with new service approvals. Their figure includes the actual source impedance and is the best number to use where available.

Changes Over Time

Available fault current is not fixed. When a utility replaces a transformer with a larger or lower impedance unit, fault current at existing buildings can rise above the ratings of installed equipment. The NEC requires the marked value to be updated when changes occur. Rechecking after any change to the supply is good practice.

Understanding Your Result

The headline gives the available fault current at the end of the run.

The transformer line gives the fault current at the transformer terminals.

The conductor run line gives the f and M factors.

The equipment rating line states the minimum rating needed.

When Should You Use This Calculator?

Use it to check whether panel and breaker ratings are adequate.

Use it to estimate the fault current to mark on new services.

Use it to see how cable length affects fault levels.

Common Mistakes

Ignoring the conductor run. Long runs reduce fault current.

Using line-to-neutral voltage. Use the transformer's line voltage.

Assuming a low result is safe. The utility can change transformers.

Using this for arc-flash labels. That needs a full study.