What is available fault current?

Available fault current is the maximum current that can flow at a point in a system during a bolted fault. NEC 110.9 requires all equipment to have an interrupting rating (AIC) at least equal to the available fault current at its terminals.

What does NEC 110.9 require for fault current?

NEC 110.9 requires equipment that interrupts current at fault levels to have an interrupting rating sufficient for the available fault current at its terminals. Circuit breakers and fuses must be rated at or above the available short-circuit current — AIC rating must equal or exceed fault current.

How do I use this fault current calculator?

Select single-phase or three-phase, enter the system voltage, then enter transformer KVA and percent impedance (%Z) or a known upstream fault current. Optionally enter conductor size and one-way length to get the reduced fault current at the panel using the point-to-point method. The calculator outputs available fault current and the minimum AIC rating required.

What is the point-to-point method for fault current calculation?

The point-to-point method calculates how fault current decreases along conductor runs due to impedance. Available fault current at any downstream point equals the source voltage divided by the combined source impedance plus conductor impedance up to that point — the same method used by Eaton, Square D, and Cooper Bussmann.

What is AIC rating on a circuit breaker?

AIC (Ampere Interrupting Capacity) is the maximum fault current a breaker or fuse can safely interrupt. A 10 kAIC breaker can interrupt up to 10,000 A of fault current. Per NEC 110.9, the AIC rating must equal or exceed the available fault current at the device's terminals.

How does transformer percent impedance affect fault current?

Transformer %Z directly limits maximum fault current at the secondary. Lower %Z means higher fault current. A 100 kVA 240 V single-phase transformer at 2% %Z yields approximately 20,833 A; at 5% %Z it yields approximately 8,333 A at the secondary terminals.

Frequently asked questions about fault current calculations

The point-to-point method accounts for conductor impedance between source and equipment. Always use maximum available fault current for equipment selection per NEC 110.9. Fault current at the panel is always less than at the transformer secondary due to conductor impedance. NEC 110.10 also requires coordination so fault clearing does not cause extensive circuit damage.

⚡ Electrical Calculator

Fault Current Calculator

NEC 110.9 · Available Short Circuit Current · Point-to-Point Method · AIC Rating

⚙️ System Configuration

System Type

System Voltage (V)

🔌 Upstream Source

Transformer KVA

Transformer %Z (impedance)

📐 Conductor Run to Equipment

Wire Material

Conductor Size

One-Way Length (ft)

Set length to 0 to calculate fault current at the transformer secondary only (no conductor run reduction).

🔍 Equipment AIC Check — optional

Installed Equipment AIC Rating (leave blank to skip check)

Please fill in all required fields.

RESULT

—

At Transformer
Secondary (kA)

—

At Panel /
Equipment (kA)

—

Min AIC Rating
Required (kA)

💡

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Transformer secondary (3Φ): I = (kVA × 1,000) / (√3 × V × Z%) | Single-phase: I = (kVA × 1,000) / (V × Z%)
Point-to-point (3Φ): Z_cable = R(Ω/kft) × L / 1,000 per phase; I_fault = V_LN / (Z_source + Z_cable)
Point-to-point (1Φ): Z_cable = 2 × R(Ω/kft) × L / 1,000; I_fault = V / (Z_source + Z_cable)
Conductor resistance from NEC Chapter 9, Table 9 (75°C, PVC conduit). Bolted fault (worst-case) assumption. Always verify with a licensed electrical engineer before installation.

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Fault Current Calculator

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