Short-circuit current calculations

Every piece of equipment in an electrical system must be able to withstand and, where it is a protective device, interrupt the maximum fault current that could flow through it. If the available fault current exceeds a breaker's interrupting rating, the breaker may explode rather than clear the fault; if it exceeds a panelboard's short-circuit rating, the bus may distort and arc. Short-circuit calculations determine the available fault current at each point in the system, and they are also the starting point for coordination and arc flash studies. Engineers perform detailed studies with software, but electricians must understand the principles, perform quick estimates, and recognize when changes to a system invalidate existing ratings. This lesson covers fault current sources, symmetrical and asymmetrical current, transformer fault current, the ohmic and point-to-point methods for feeders, motor contribution, the types of fault, and how to apply the results.

Sources of fault current

When a short circuit occurs, current flows from every source connected to the system:

• The utility system, through the service transformer. Usually the largest contributor, limited by the transformer's impedance and the utility's own source impedance.
• On-site generators, which contribute a high current initially that decays as their internal fields respond.
• Motors: running induction and synchronous motors act briefly as generators when the voltage collapses, feeding current into the fault for a few cycles, typically about four times their full-load current for induction motors.
• Inverter-based sources such as UPS systems, solar inverters and batteries, which contribute limited fault current, typically 1.1 to 2 times rated current, determined by their controls.

The total available fault current at a point is the sum of these contributions, limited by the impedances between each source and the fault.

Symmetrical and asymmetrical current

At the instant a fault occurs, the current may contain a decaying DC component, depending on the point on the voltage wave where the fault starts and on the X/R ratio (reactance to resistance) of the circuit. The resulting asymmetrical current has a first peak much higher than the steady symmetrical current, up to about 2.6 times the symmetrical RMS value for high X/R ratios. Protective devices and equipment are rated in symmetrical RMS amperes, with test conditions that assume a particular X/R ratio; where the actual X/R ratio is higher, as close to large transformers and generators, the ratings may need to be derated, which the engineer's study accounts for. For most electrician calculations, the symmetrical RMS value is used.

Fault current at a transformer secondary

The simplest estimate assumes an infinite bus: the utility can supply unlimited current to the transformer primary, so only the transformer's impedance limits the fault. The transformer's percent impedance (%Z, on its nameplate, typically 2 % to 6 % for distribution transformers) is the percentage of rated primary voltage that drives full-load current through the secondary when it is short-circuited. The maximum three-phase fault current at the secondary terminals is therefore:

For example, a 1,500 kVA, 480 V transformer with 5.75 % impedance has a full-load current of 1,804 A and an infinite-bus fault current of 1,804 × 100 ÷ 5.75 = 31,380 A. The infinite-bus assumption gives a conservative (high) result; including the utility's source impedance lowers it. Note also that manufacturers' impedances have tolerances (often ±7.5 %), and conservative studies use the minimum impedance.

Fault current along a feeder

Conductor impedance reduces fault current along a feeder. The ohmic method treats the source as an impedance and adds the feeder impedance: the source impedance per phase is Z_S = V_L ÷ (√3 × I_SC at the source), the feeder impedance is the conductor's impedance per unit length times the length divided by the number of parallel sets, and the fault current at the end is:

Adding impedances arithmetically (rather than as complex numbers) gives a slightly conservative estimate suitable for field checks. Conductor impedances are taken from tables of resistance and reactance per 1,000 ft or per kilometer for each size, material and raceway type. A widely used alternative is the point-to-point method, which computes a factor f = (√3 × L × I_SC) ÷ (C × n × V), where C is a tabulated constant for the conductor and raceway, n the number of parallel conductors, and multiplies the starting fault current by M = 1 ÷ (1 + f). Both methods give similar results.

Motor contribution

Running motors add fault current for the first few cycles after a fault, which matters for the interrupting and momentary ratings of devices close to large motor loads. A common estimate for induction motors is four times the full-load current of all motors running at the time of the fault, added at the bus where they connect. For example, 500 A of running motor load adds roughly 2,000 A at its bus. Large motors and synchronous motors may contribute more; motors on variable frequency drives contribute little or nothing, because the drive's rectifier blocks reverse power flow.

Types of fault

The three-phase bolted fault (all three phases connected through zero impedance) usually gives the highest current and is used for equipment ratings. Other fault types include line-to-line faults (about 87 % of the three-phase value), line-to-ground faults (which in solidly grounded systems can approach or occasionally exceed the three-phase value near transformers, and in impedance-grounded systems are deliberately limited, Lesson 9), and arcing faults, whose current is lower than bolted fault current because of the arc's resistance (often 40 % to 90 % of bolted current at low voltage). The lower current of arcing faults is important: a protective device may trip slowly on an arcing fault that it would clear instantly as a bolted fault, which increases arc flash energy (Lesson 6). Minimum fault currents also matter, because a device must detect and clear the smallest fault at the end of its circuit.

Applying the results

Short-circuit results are used to:

• Verify that every device's interrupting rating and every assembly's short-circuit rating equal or exceed the available fault current at its location (Lesson 3).
• Verify that conductors, bus and equipment can withstand the fault until the protective device clears it.
• Provide input to coordination and arc flash studies (Lesson 6).
• Set protective device pickups so that minimum faults are detected.

When a system changes, results change. A utility transformer replaced with a larger or lower-impedance unit, added motor loads, a new generator, or shorter or larger feeders can all increase fault current and make previously adequate equipment underrated. Codes require equipment marking of available fault current at services in many cases, and updated studies when systems change. Electricians who notice changes, such as a new, larger utility transformer, should report them to the owner and engineer.

In the United States, NEC 110.9 and 110.10 require interrupting ratings and short-circuit current ratings adequate for the available fault current, and 110.24 requires service equipment in other than dwelling units to be field marked with the maximum available fault current and the date of the calculation, with the marking updated when modifications change the available fault current.

Using short-circuit results in the field

Short-circuit calculations are not just a designer's exercise. Electricians use the results every time they add a panelboard, replace a breaker or connect new equipment. The available fault current marked on the service equipment, and on downstream equipment where required, tells the installer the minimum interrupting rating of any device installed there. If a new breaker or piece of equipment has a lower rating than the available fault current, it cannot be installed at that location without further analysis, even if it fits physically.

Fault current changes over the life of a building. The utility may replace its transformer with a larger or lower-impedance unit, add a parallel supply, or reconfigure its network, increasing the fault current available at the service. On-site generators, large motors and battery systems also add fault current. For this reason, short-circuit studies should be reviewed whenever the system changes significantly, and codes in many places require the available fault current marking to be updated when it changes.

A quick estimate can still be useful in the field. Knowing that a transformer secondary's fault current is roughly its full-load current divided by its per-unit impedance allows an electrician to sense-check a label or drawing. If a 75 kVA transformer is labeled with 65 kA available at its secondary panel, something is wrong, because its own impedance limits the fault current to a small fraction of that. Spotting such errors early prevents costly equipment mismatches.

When in doubt, ask the designer or engineer of record for the short-circuit study results before selecting equipment. Proper documentation of available fault current protects both the installer and the owner, and it is often requested by inspectors.

Software tools now perform most short-circuit studies, modeling every cable, transformer, motor and source in the system. These programs can calculate symmetrical and asymmetrical currents, line-to-ground faults and motor contributions at hundreds of buses at once, and they feed directly into coordination and arc flash studies. Their accuracy depends entirely on the input data, so a hand calculation at a few key points remains a valuable check on the software results.