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Fault Scenario Simulator

3Φ · L-L · L-G · L-L-G faults · Click any bus · Watch breaker trip sequence

Click a bus on the one-line to apply fault
1500 kVA13.8kV/480VCB-MAIN480V Main Bus0.48kVCB-FEEDER-ACB-FEEDER-BPanel MDP-A0.48kVPanel MDP-B0.48kVCB-BRANCH-1CB-BRANCH-2Load A0.48kVLoad B0.48kVClick any bus to apply fault

Fault Type

3Φ: Symmetrical fault — worst case fault current, equal in all three phases. Sets basis for short-circuit withstand ratings.

System Parameters

Transformer Rating1500 kVA
Utility Short-Circuit MVA100 MVA

Event Log

No fault applied — click a bus on the one-line diagram
IEEE 551 / NEC Reference: Short-circuit current ratings for equipment must exceed the available fault current (NEC 110.9). IEEE 551 (Violet Book) provides the standard calculation method. All OCPDs must have an interrupting rating ≥ available fault current at the point of installation.

About the Fault Scenario Simulator

This simulator models three-phase bolted, line-to-line, line-to-ground, and line-line-ground fault scenarios on a one-line diagram, calculating fault current and animating the protective device trip sequence. Engineers use it to understand short-circuit magnitudes and protection clearing times during system design reviews and equipment ratings verification.

How fault current calculation works

Short-circuit current is calculated using the impedance method: I_sc = V / (√3 × Z_total) for three-phase faults on a balanced system, where V is the line-to-line voltage in kV and Z_total is the total system impedance in ohms. In per-unit terms: I_pu = 1 / Z_pu, and the base current is I_base = kVA_base / (√3 × kV_base).

For unsymmetrical faults (line-to-line, line-to-ground), symmetrical component analysis is required. A three-phase fault produces the maximum symmetrical current (I_3Φ = V_LN / Z1). A line-to-line fault produces 86.6% of three-phase current (I_LL = √3 × V_LN / (Z1 + Z2) ≈ 0.866 × I_3Φ for Z1 = Z2). A line-to-ground fault can exceed three-phase current on solidly grounded systems when zero-sequence impedance Z0 is low: I_LG = 3V_LN / (Z1 + Z2 + Z0).

This simulator uses a simplified MVA method: the utility contribution is represented as a short-circuit MVA, and transformer impedance (typically 5.75% on transformer kVA base) is added in series. System buses deeper in the network see reduced fault current due to feeder impedances.

Applicable codes and standards

NEC 110.9 requires that all overcurrent protective devices (OCPDs) have an interrupting rating sufficient for the maximum available fault current at the point of installation. NEC 110.10 requires that conductors, switches, and buses have adequate short-circuit current ratings. IEEE 551 (Violet Book) provides the standard method for short-circuit calculation. IEEE 141 (Red Book) provides recommended practice for electric power distribution for industrial plants. ANSI C37.13 covers low-voltage AC power circuit breakers, including interrupting ratings. NEMA AB-1 covers molded-case circuit breaker ratings.

Design considerations

Equipment must be rated for the maximum available fault current at its point of installation, including asymmetrical (first-cycle) current. The asymmetrical factor is typically 1.6× symmetrical for X/R ratios typical of low-voltage systems. Circuit breakers and fuses are rated in symmetrical kA; switchgear buses and busbars must be rated for the total asymmetrical current (crest value = √2 × 1.6 × I_sym).

Fault current decreases as you move deeper into the system (more impedance in the fault path). The utility short-circuit MVA is the starting point — confirm this value with the serving utility, as it can vary significantly depending on system configuration and substation transformer size. High fault current demands fully rated interrupting capacity (not series-rated) in commercial and industrial installations.

How to use this simulator

Select the fault type (3-Phase is the worst-case symmetrical fault; L-G is the most common in field operations). Adjust the transformer kVA rating and utility short-circuit MVA to match your system. Click any bus on the one-line diagram to apply a fault at that bus — the simulator calculates the fault current in kA and animates the relay detect, breaker trip, and fault clear sequence with realistic timing. Reset the system and try different bus locations to understand how fault current magnitude varies throughout the distribution system.

Frequently asked questions

Which fault type produces the highest current?

For most systems, the three-phase bolted fault produces the highest symmetrical fault current. However, line-to-ground faults can exceed three-phase fault current on solidly grounded systems where zero-sequence impedance (Z0) is lower than positive-sequence impedance (Z1). This is why L-G fault current can be 110% or more of three-phase current in some configurations.

Why is the line-to-ground fault the most common type?

Approximately 70–80% of all power system faults are single line-to-ground (L-G) faults. They occur most frequently because a single insulation failure or conductor contact with a grounded structure creates the fault — it only requires one phase to be involved, unlike L-L or three-phase faults which require two or three phase conductors to contact each other.

What does the utility short-circuit MVA mean?

The utility short-circuit MVA (also called "available fault duty") is the equivalent source strength of the utility grid at the point of delivery. A higher MVA means lower source impedance, which means higher fault current at the service entrance. Typical values range from 50 MVA (weak rural feeders) to 2000+ MVA (large urban substations). Always verify with the utility.

What is the difference between symmetrical and asymmetrical fault current?

Symmetrical (RMS) fault current is the steady-state short-circuit value after the DC offset decays. Asymmetrical (first-cycle peak) current is higher due to the DC offset that occurs when a fault happens at a point other than the voltage zero crossing. The asymmetrical multiplying factor for low-voltage systems (X/R ≈ 6.6) is approximately 1.6×, so a 30 kA symmetrical fault has a first-cycle asymmetrical peak of about 68 kA.

What equipment ratings must be checked against fault current?

Circuit breaker interrupting ratings, fuse interrupting ratings, switchgear and MCC bus bracing, cable short-circuit withstand ratings (I²t), transformer secondary terminals, and busduct/wireway ratings must all be verified against available fault current. Equipment rated below the available fault current can catastrophically fail (explode) under fault conditions.

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