# What is the *interrupting rating* of a GFCI receptacle?

Say, we have a UL943 compliant 15A GFCI receptacle on a garden-variety 15A grounded branch circuit. Some nitwit comes along and wirenuts the receptacle's load-side hot to the branch circuit ground, creating a bolted ground fault. When the power comes back on -- both the breaker and the GFCI are going to race to trip on what they both detect as faults.

If the GFCI "wins the race" and trips before the breaker does, how much fault current can it safely interrupt?

I checked a few GFCI data sheets and they all listed either a 10kA Short Circuit Current Rating or Max Interrupting Capacity.

I'm assuming that this is similar to the 10kA rating on common household circuit breakers where the device is required to break at least 10kA.

Cooper Industries has a PDF that describes short-circuit current ratings in more deail:

``````Article 100 Definitions
Short-Circuit Current Rating. The prospective
symmetrical fault current at a nominal voltage to which an
apparatus or system is able to be connected without
sustaining damage exceeding defined acceptance criteria.
``````
• Kind of makes me think it's a moot point - I can't imagine much past (or including) the transformer surviving too long under a 10kA dead short load. – Comintern Jan 23 '16 at 2:40
• @Comintern - Yeah, that's the reason for the rating, if a 10kA breaker were connected to something like a big 32kA distribution circuit, under a dead short the breaker could fuse closed and be unable to protect the circuit. – Johnny Jan 23 '16 at 3:24

`Short-circuit current` is the amount of short-circuit current the device can safely handle, so it should be greater than or equal to the short-circuit rating of the device protecting it. For example. If the GFCI has a `short-circuit current` rating of 10 kA. Then the short-circuit protection device protecting it (likely a circuit breaker), should have a short-circuit rating less than or equal to 10kA.

`Maximum interrupting capacity` is the maximum amount of current the device can interrupt. This value should be greater than or equal to the overload protection provided by the device protecting it. For example. If the GFCI has a `maximum interrupting capacity` rating of 20A. Then the overload protection device protecting it (likely a circuit breaker), should have an overload rating less than or equal to 20A.

As noted in the document cited by another answer.

“Short-circuit current rating” is not the same as “interrupting rating” and the two must not be confused.

# National Electrical Code 2014

## Chapter 1 General

### Article 100 Definitions

Interrupting Rating. The highest current at rated voltage that a device is identified to interrupt under standard test conditions.

Short-Circuit Current Rating. The prospective symmetrical fault current at a nominal voltage to which an apparatus or system is able to be connected without sustaining damage exceeding defined acceptance criteria.

As for which device would "win the race"... At current levels high enough to cause an instantaneous trip in a circuit breaker, the breaker should react in less than one cycle (16.67 milliseconds). I'm not sure what the reaction time of a GFCI device is, but I'd suspect it's slower than that.

I'd also like to note that depending on the length of the circuit, and size of conductors. A short-circuit might not be as much current as you'd expect.

• That's a good point about the wires limiting current, but in a worst case situation of a 12 gauge circuit 10 feet away from the panel + 50 feet of 2/0 to the pole, the wire would have around .042 ohms of resistance, so if you short out a 120VAC circuit at the peak of the waveform (170V), you could see up to around 4000 amps of instantaneous peak (not RMS) current in the circuit. – Johnny Jan 23 '16 at 5:45
• @Johnny Can you show your work on that 4000 amperes? I'm getting closer to half that, though I'm using 2/0 AL. for my calculation. – Tester101 Jan 23 '16 at 14:52
• Using AL instead of Cu for the 2/0 service feed makes a bit of difference in my numbers. Here's my calculation: 170V/(20 ft * .00159 ohms/ft + 100 ft * .000078 ohms/ft) = 4292A. Using 2/0 Al (.000128 ohms/ft) instead of Cu gives 3800A. I used these tables: Cu, Al – Johnny Jan 23 '16 at 16:40
• @Johnny I'm still not clear why you're using peak voltage, instead of RMS. Also the resistance values from those tables are slightly lower than the tables in NEC, but maybe that has to do with the difference between peak voltage and RMS. – Tester101 Jan 23 '16 at 17:30
• I chose peak voltage since that's the worst case the breaker contacts will see if they open near the peak of the waveform. I admittedly don't know if interrupting ratings are based on RMS or peak current. The NEC tables may be based on a different temperature. than the tables I used, the Copper table I used assumes 25 degrees C, higher temperatures would mean slightly higher resistance. – Johnny Jan 23 '16 at 19:45