Short Circuit Current Calculator
Compute the available short-circuit current (ASCC) at a fault point downstream of a transformer, accounting for transformer impedance and cable impedance to the fault location. Use this to verify that breakers and fuses have a sufficient interrupting rating.
Last updated: September 2026
Formula below · 2 sources (ieee.org, Wikipedia) · Updated Sep 2026
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About this calculator
Available short-circuit current determines the interrupting capacity required for protective devices. The formula combines transformer and cable impedances in per-unit on the transformer base, then divides the transformer’s full-load current by that impedance: ASCC (A) = (kVA × 1000) / (√3 × V × √(Zt² + Zc²)), where Zt = transformer impedance % / 100 and Zc = (R × L / 1000) × (kVA × 1000) / V². R is the cable resistance in ohms per 1,000 ft for the selected size (copper), L is the one-way length to the fault in feet, and V is the line-to-line voltage of the three-phase system. Combining Zt (almost all reactance) and Zc (mostly resistance) at right angles is the usual approximation. The calculation assumes an infinite utility source and ignores cable reactance and motor contribution, so it errs toward a higher fault current, which is the conservative direction for choosing interrupting ratings. The result must be compared to the interrupting rating (kAIC) on every breaker, fuse, and switchgear device in the circuit.
How to use
Given: 500 kVA transformer, 5% impedance, 480 V three-phase, 100 ft of 4 AWG copper (0.259 ohm per 1,000 ft) to the fault. Step 1 — transformer: Zt = 0.05. Step 2 — cable: 0.259 × 100 / 1,000 = 0.0259 ohm; base factor 500,000 / 480² = 2.17; Zc = 0.0259 × 2.17 = 0.0562. Step 3 — total: √(0.05² + 0.0562²) = 0.0752. Step 4 — ASCC = 500,000 / (1.732 × 480 × 0.0752) ≈ 8,000 A. At the transformer terminals (0 ft) it would be 500,000 / (1.732 × 480 × 0.05) ≈ 12,000 A, so equipment there needs at least a 14 kA or 22 kA rating.
Frequently asked questions
Why does transformer impedance percentage matter for short-circuit current calculations?
Transformer impedance (expressed as a percentage) is the primary factor limiting how much fault current can flow for a bolted three-phase short circuit at the transformer's secondary terminals. A lower impedance transformer delivers a higher short-circuit current because less voltage is dropped internally. For example, a 5% impedance transformer can theoretically supply 20 times its full-load current into a bolted fault. This is why specifying a low-impedance transformer for cost savings can require upgrading all downstream protective devices to higher interrupting ratings.
How does cable length to the fault location reduce available short-circuit current?
As current travels through a cable, the cable's resistance and reactance cause a voltage drop that effectively limits the fault current. Longer cables and smaller conductors have higher impedance, which adds to the transformer impedance in the denominator of the ASCC formula, reducing the computed fault current. This is beneficial for devices far from the transformer but means devices close to the source see much higher fault currents and require higher interrupting ratings. Engineers use this principle strategically in impedance-based protection coordination.
What is the difference between short-circuit current and interrupting rating on a breaker?
Short-circuit current (ASCC) is the actual fault current available at a specific point in the electrical system—it is a property of the power source and impedance path. Interrupting rating (kAIC, or ampere interrupting capacity) is the maximum fault current a protective device is designed to safely clear without being destroyed. The NEC (110.9) requires that every protective device have an interrupting rating at least equal to the available short-circuit current at its location. If a breaker's interrupting rating is exceeded, it can explode or weld shut, creating a serious arc-flash hazard.