Ground Fault Current Calculator
Estimate prospective and factored ground-fault current from entered voltage, impedance, and screening factor for preliminary review.
- Prospective fault current
- A
- Factored fault current
- A
- Fault voltage used
- V
- Loop impedance used
- ohm
Calculation details
- Calculation basis
- Boundary
Recent results
Formulas
- prospective fault current = fault voltage / loop impedance
- factored current = prospective current x factor / 100
A ground fault current calculator estimates the current that could flow through an entered fault loop by dividing Fault voltage by total Fault-loop impedance. The result is the Prospective fault current. Applying the entered Screening factor produces a second planning value, the Factored fault current.
This calculation supports preliminary electrical design and installation review where a fault-loop estimate helps identify whether conductor routing, raceway length, connection quality, or source-to-load distance could materially affect fault current. It can be useful alongside feeder layout, branch-circuit routing, voltage-drop review, equipment planning, and troubleshooting of unusually long or high-impedance fault paths.
It does not establish available fault current at equipment terminals or determine interrupting rating, overcurrent protective-device performance, coordination, arc-flash results, or NEC compliance.
Fault-Loop Impedance
A ground fault requires a complete current path back to the source. Depending on the system and fault condition, that path can include the ungrounded conductor, equipment grounding conductor or bonding path, terminations, raceway sections, transformer winding impedance, and other connected impedance.
The calculator uses one combined value: Fault-loop impedance in ohms.
A lower entered loop impedance produces a higher calculated fault current. A higher loop impedance produces a lower calculated fault current.
For example, if the fault voltage remains 480 V:
| Fault-loop impedance | Prospective fault current |
|---|---|
| 0.06 ohm | 8,000 A |
| 0.12 ohm | 4,000 A |
| 0.24 ohm | 2,000 A |
The calculation is sensitive to the quality of the impedance value. A value taken from a simplified conductor-length estimate is not equivalent to a complete engineering fault-current study. Conductor material, AWG or kcmil size, conductor length, parallel paths, conductor temperature, raceway continuity, transformer characteristics, source capacity, and connection resistance can all affect an actual fault path.
Calculator Inputs
| Field | Unit | Electrical meaning |
|---|---|---|
| Fault voltage | V | Voltage used by the calculator in the fault-current equation |
| Fault-loop impedance | ohm | Total impedance entered for the assumed fault-current loop |
| Screening factor | % | Explicit percentage applied to prospective fault current |
Fault voltage is entered exactly as the voltage to be used in the screen. It is not automatically derived from nominal system voltage, phase configuration, transformer secondary voltage, or a line-to-line versus line-to-ground relationship.
Fault-loop impedance must represent the total loop impedance intended for the screening calculation. Entering only a branch-circuit conductor resistance while omitting the return path, source impedance, or bonding path changes the meaning of the result.
Screening factor is an explicit arithmetic factor. It reduces or scales the prospective value by the percentage entered. It is not a conductor adjustment factor, ampacity correction factor, NEC derating value, voltage-drop allowance, utility contribution factor, or protective-device setting.
Fault Current Formula
The calculator applies the following formulas:
\(\displaystyle \text{prospective fault current} = \frac{\text{fault voltage}}{\text{loop impedance}}\)
\(\displaystyle \text{factored current} = \text{prospective fault current} \times \frac{\text{factor}}{100}\)
Where:
- Prospective fault current is expressed in amperes.
- Fault voltage is expressed in volts.
- Loop impedance is expressed in ohms.
- Factor is the entered Screening factor percentage.
The first formula follows Ohm’s law, I = V/Z, using the entered total fault-loop impedance rather than a single conductor resistance.
Calculation Example
Enter the following values:
| Input | Entered value |
|---|---|
| Fault voltage | 480 V |
| Fault-loop impedance | 0.12 ohm |
| Screening factor | 80% |
First, calculate prospective fault current:
\(\displaystyle \frac{480\text{ V}}{0.12\text{ ohm}} = 4{,}000\text{ A}\)
Then apply the screening factor:
\(\displaystyle 4{,}000\text{ A} \times \frac{80}{100} = 3{,}200\text{ A}\)
The screen reports:
| Result | Value |
|---|---|
| Prospective fault current | 4,000 A |
| Factored fault current | 3,200 A |
| Fault voltage used | 480 V |
| Loop impedance used | 0.12 ohm |
The 4,000 A value is the direct result of the entered voltage and loop impedance. The 3,200 A value is the same result after application of the entered 80% screening factor.
Use in Electrical Layout Review
A preliminary fault-current estimate can help identify installations needing closer review before finalizing a feeder or branch-circuit layout. Long conductor runs, undersized equipment grounding paths, poor bonding continuity, and high-resistance connections can increase loop impedance and reduce estimated fault current.
That review may overlap with other electrical calculations, but each calculation has a different purpose:
| Related review | Primary question |
|---|---|
| Conductor ampacity | Can the selected AWG or kcmil conductor carry the calculated load after applicable adjustment and correction factors? |
| Voltage drop | Does conductor size and run length maintain acceptable operating voltage at the load? |
| Raceway fill | Does the selected conduit accommodate the installed conductors under applicable fill rules? |
| Motor calculation | Are conductors, overload protection, and feeder components selected for motor load characteristics? |
| Ground fault current screening | What current results from the entered fault voltage and total loop impedance? |
A conductor can have adequate ampacity yet still create a high-impedance fault loop if the route is long or the return path is poorly bonded. Conversely, a low impedance result does not establish that a conductor, raceway, breaker, fuse, disconnect, panelboard, or other component has the required rating for the installation.
Field Verification
Use values that reflect the planned or observed electrical path. Confirm the circuit configuration, source, conductor route, terminations, metallic raceway continuity where applicable, equipment grounding conductor path, and bonding connections before relying on an impedance assumption.
For a final installation decision, separately verify the applicable equipment ratings, conductor ampacity, terminal rating, insulation temperature rating, current-carrying conductor adjustment, ambient-temperature correction, voltage drop, protective-device characteristics, and AHJ requirements.
Boundary: This worksheet performs fault-current screening only. It does not provide an available fault current study, interrupting-rating determination, protective-device coordination analysis, arc-flash calculation, or code result.
FAQs
Is this an available-fault-current study?
No. It uses only the voltage and impedance entered on the page and omits source, transformer, X/R, and system modeling.
Does the factored result set an interrupting rating?
No. Interrupting ratings and protective coordination require a separate system study and equipment review.