What Is a Fault Current and How Is It Different from a Load Current?

An overload is like cramming too many people through a doorway. The doorway still exists — people still pass through it — there are just more of them than the doorway was designed for. In electrical terms, the circuit is carrying more current than its rated capacity, but the current still flows through the intended path: conductors, connections, and the load. An overloaded cable heats up slowly, over minutes or hours, eventually damaging its insulation if the overload is not cleared.

A short circuit is fundamentally different. It is as if someone demolished the wall next to the doorway, creating a massive opening with no restriction. The current no longer flows through the load — it bypasses it entirely, finding a near-zero impedance path directly between live conductors (phase-to-phase fault) or between a live conductor and earth (phase-to-earth fault). The result is catastrophic: current surges to levels that can be 10 to 50 times the normal load current, and it happens in milliseconds.

An overloaded 100 A circuit might carry 130 A. The same circuit experiencing a short circuit might carry 10,000 A or more. The energy released at the fault point is proportional to I² — so a fault at 10,000 A releases 10,000 times more energy per unit time than the normal 100 A load. This is why short circuits cause explosions, fires, and arc flash injuries.

Ohm’s law explains everything: I = V / Z. The supply voltage (V) stays essentially constant during a fault — the utility grid is so large that a single fault barely affects its voltage. What changes dramatically is the impedance (Z) in the circuit.

Under normal conditions, the load impedance dominates the circuit. A motor, heater, or lighting circuit might present 2–10 Ω of impedance. The current is modest and controlled. When a short circuit occurs, the load impedance drops to nearly zero. The only impedance remaining is the source impedance: the transformer’s internal impedance plus the cable impedance between the transformer and the fault point.

Consider a 1000 kVA transformer with 5% impedance, supplying a 400 V system:

• Rated current: 1000 kVA / (√3 × 400 V) = 1,443 A
• Prospective fault current at transformer terminals: 1,443 / 0.05 = 28,868 A

That is 20 times the rated current, and this is at the transformer terminals where the cable impedance is zero. Further downstream, cable impedance reduces the fault current, but it can still be thousands of amperes at a final distribution board 50–100 metres from the transformer.

On larger supply systems — with multiple transformers in parallel or high-capacity utility supplies — fault currents can exceed 50,000 A or even 100,000 A at the main switchboard. Industrial sites fed by 2000 kVA or 3000 kVA transformers routinely see prospective fault currents in this range.

Engineers calculate two types of fault current, and understanding the difference is critical for safety:

The prospective short-circuit current (PSCC), also called the bolted fault current, assumes zero impedance at the fault point — as if the two conductors were solidly bolted together. This gives the maximum possible fault current and is used to select the interrupting rating of protective devices. It represents the worst case: the highest current the protective device might have to interrupt.

An arcing fault involves an electric arc between conductors, separated by a small air gap. The arc introduces additional impedance (typically 20–40 mΩ for low-voltage faults), which reduces the current below the bolted fault level. A bolted fault of 30,000 A might produce an arcing fault of only 15,000–20,000 A.

Here is the paradox: the arcing fault is more dangerous to people, even though the current is lower. Three reasons:

1. The arc generates extreme heat: An electric arc can reach temperatures of 15,000–20,000°C — hotter than the surface of the sun. This superheats the surrounding air, creating a pressure wave (arc blast) that can throw a person across a room and cause severe burns at distances of several metres.
2. Lower current may not trip protection quickly: If the arcing fault current is significantly lower than the bolted fault current, it may fall into the “time-delay” region of the protective device’s trip curve rather than the “instantaneous” region. This means the arc persists for longer — hundreds of milliseconds instead of tens — releasing far more total energy.
3. Arcing faults can be intermittent: The arc may extinguish and re-strike repeatedly, each time releasing a burst of energy. This intermittent behaviour can prevent the protective device from accumulating enough thermal energy to trip, allowing the fault to persist for seconds or even minutes. Intermittent arcing faults are a leading cause of electrical fires.

This is why arc flash studies (per IEEE 1584:2018) calculate incident energy using the arcing fault current, not the bolted fault current. The arcing current determines how long the fault persists, and duration is the primary driver of incident energy.

Since fault current = voltage / total impedance, anything that adds impedance to the fault path reduces the fault current. The three main contributors are:

1. Transformer impedance (u_k%): This is usually the dominant factor. A transformer with 6% impedance limits the maximum fault current at its terminals to rated current / 0.06 = 16.7 times rated current. Higher impedance transformers (which are cheaper) produce lower fault currents but have worse voltage regulation. Lower impedance transformers give better voltage regulation but higher fault currents that require more expensive switchgear.
2. Cable impedance: Every metre of cable between the transformer and the fault point adds impedance. This is why fault current decreases with distance from the supply. A fault at a distribution board 100 metres from the transformer might see only 40–60% of the fault current at the transformer terminals. For very long cable runs, the cable impedance can exceed the transformer impedance, and the fault current drops significantly.
3. Fault resistance: The impedance at the fault point itself. For a bolted fault, this is essentially zero. For an arcing fault, it adds 20–40 mΩ. For a fault through water, soil, or a high-resistance path, the fault impedance can be substantial — producing a low-current fault that is difficult to detect but can still cause fires through localised heating.

There is also the utility source impedance upstream of the transformer. This represents the impedance of the entire upstream network (transmission lines, distribution transformers, generators). It is usually small compared to the local transformer but becomes significant in dense urban networks where the utility supply is very strong (low impedance, high fault level).

All of these impedances are added together, but the addition is not simple arithmetic — it is vector (complex number) addition because each impedance has both a resistive (R) and reactive (X) component. The total impedance magnitude determines the fault current level, while the R/X ratio determines the DC offset and peak asymmetric current.

Every circuit breaker, fuse, and protective device has an interrupting rating (also called breaking capacity or short-circuit rating), expressed in kA. This is the maximum fault current the device can safely interrupt without being destroyed.

When a circuit breaker opens under fault conditions, an arc forms between its separating contacts. The breaker must extinguish this arc within a few milliseconds. If the fault current exceeds the breaker’s interrupting rating, the arc may not be extinguished. The consequences are severe:

• Contact welding: The contacts fuse together and the breaker cannot open. The fault continues until upstream protection operates (if it can) or the cable is destroyed.
• Arc flash inside the breaker: The arc breaches the breaker’s enclosure, creating an external arc flash that can injure anyone nearby.
• Explosive failure: The energy released inside the breaker exceeds its containment capacity, and the device explodes, ejecting molten metal and hot gases.

This is why fault current calculations are not optional. Every protective device must have an interrupting rating equal to or greater than the prospective fault current at its point of installation. A domestic miniature circuit breaker (MCB) typically has a 6 kA or 10 kA rating — adequate for residential installations where fault currents rarely exceed 6 kA. Industrial moulded case circuit breakers (MCCBs) range from 16 kA to 150 kA. Air circuit breakers (ACBs) for main switchboards can handle 50–150 kA.

Common standards that define interrupting rating requirements include IEC 60898 and IEC 60947-2 for circuit breakers, IEC 60269 for fuses, and AS/NZS 3000:2018 and BS 7671:2018 for installation requirements.

The fundamental approach to fault current calculation is straightforward, even if the details get complex for large networks:

1. Determine the source impedance: Start with the utility supply fault level (usually provided by the utility in MVA or kA) and convert it to impedance. Add the transformer impedance. If the utility fault level is not known, assume an infinite bus (zero source impedance) for a conservative estimate that gives maximum fault current.
2. Add cable impedances: For each cable segment between the source and the fault point, calculate R and X per metre from cable data tables (provided in standards like AS/NZS 3008 or IEC 60364) and multiply by the cable length.
3. Sum impedances: Add all impedances vectorially: Z_total = √(R_total² + X_total²).
4. Calculate fault current: I_fault = V / (√3 × Z_total) for three-phase faults, or I_fault = V / (2 × Z_total) for single-phase faults (where Z_total includes the return path impedance).

The international standard for detailed short-circuit calculations is IEC 60909-0, which defines correction factors for voltage, motor contribution, and other effects that refine the basic calculation.

For most installations, the critical outputs are: the maximum fault current at each distribution board (for selecting protective device interrupting ratings), and the minimum fault current at the furthest point of each circuit (for verifying that the protective device will actually trip fast enough to protect the cable).

Understanding fault current is essential for safe electrical design. Every protective device must be rated for the prospective fault current at its installation point, and every cable must survive the fault current long enough for its protective device to clear the fault.

The ECalPro Short-Circuit Calculator computes prospective fault currents from transformer data and cable parameters, giving you the maximum and minimum fault levels at each point in the distribution system. Use the results to verify protective device ratings and cable fault withstand capacity.

For a detailed look at what happens inside a cable during a short circuit — the thermal race between energy input and insulation survival — read What Happens Inside a Cable During a Short Circuit in the Learn section.