# Main breaker not tripping

We just connected the power company's meter to our new construction and got an immediate humming sound which the meter technician said should not happen. We later found that one of our electrical boxes was wired incorrectly—a hot wire was connected to the neutral bar along with the neutral wire (we've rewired it and the humming has stopped).

But the thing is, this didn't trip the main 400A breaker at the meter, it only produced a humming sound.

• The breaker model we have is Jxd22b400 Product Details - Industry Mall - Siemens USA.
• At 240V, the UL interrupt rating is 65KA.
• The instantaneous trip setting is 2000A-4000A
• The magnetic adjustment is at low (which should be 2000A?)

The length of the wire to the breaker and the sub panel is about 400 ft (there and back) and is 750 kcmil aluminum The voltage between the hot and neutral is 120V. The 400 ft of wire has a resistance of about .0125 ohms. The amps produced based on our calculation is 120/.0125 = 9,600A

Questions:

1. Should that current not be enough to trip the breaker?
2. Is it because the short circuit current is much lower than the interrupt rating?
3. Does a short circuit trip a breaker immediately or does it take time to trip it?
4. Did it not trip because the problem already existed upon connecting the power?
• Is the feed from the power company capable of delivering 400A? May 3, 2022 at 21:07
• Circuit breakers are generally slow. "400A" and "instantaneous trip 2000A" - Anywhere between 401 - 2000A will cause it to trip eventually; more = faster. This data should be outlined as a chart in the datasheet/tech specs. May 3, 2022 at 21:25
• The hot and neutral of one circuit (subpanel feed? regular circuit? please clarify) were connected together on one end on a neutral bar. What was connected on the other end? May 3, 2022 at 21:34
• The power company is supplying power via a transformer, right? If so, if it very unlikely that the supplied transformer can deliver the current you calculated. Sure 400 A. Sure, more than that. But how much more? It has its own internal resistances, both primary and secondary, so there are likely unaccounted for limitations on the actual current you are predicting.
– jonk
May 3, 2022 at 23:29
• Disagree that this is "DIY". The question asks about the electrical engineering aspects of the problem, and both (so far) answers respond with engineering analysis. The question is not "how do I do this" or "does this comply with electrical code"
– Daniel Chisholm
May 4, 2022 at 10:00

Keep in mind the current needed to make the round trip. So .0125 ohms (sounds about right) for one way, then ??? on the return trip, as the neutral may be smaller and the ground is definitely smaller.

Then on top of that you have the hot wire going to the subpanel where the short happened, and the round-trip back on a fourth wire presumably neutral.

Oh, and then you need to account for the service drop from the utility, it will have its own wire size and length, and you'll need to calculate its resistance as well, coming and going, that is 5 and 6.

Also, at the poletop it joins a distribution wire from the transformer that has some length, so that will be resistances 7 and 8. And then we have the transformer's own effective resistance, that'll be resistance 9.

You'll need to sum all those resistances plus any resistance in the connections... before you plug the cumulative resistance into Ohm's Law.

You could test these cumulative resistances by attaching a known load to the panel where the fault occurred and measuring the resultant voltage sag. E.G. if a 36A/120V load (3 space heaters) sags voltage from 122V to 121.4V, then voltage drop is 0.6V, and plug into Ohm's Law V=IR..... 0.6=36*R solve for R.

Breaker trip curves generally call for instant trip at 600-1050% of rating (manufacturing tolerance), so we're talking 2400-4200 amps here. You could get below that pretty easily, which would lock out instant-trip and be down to thermal/inverse-time trip. That could be a second or two or longer. You can google "breaker trip curve" for examples.

Should that current not be enough to trip the breaker?

One would certainly hope so, yes... but given your long distances it may not be a fast trip.

Remember the purpose of the breaker is to trip before wires burn (i.e. get hot enough to become degraded and unfit for service going forward). So it is a race condition between the breaker's inverse-time thermal trip device and the ability of the wire to absorb heat.

Heat is stored in atoms not mass (i.e. 1 copper atom stores the same heat as 1 aluminum atom) and aluminum wire of a given ampacity has more atoms. So it can absorb more thermal energy for the same temperature rise.

Is it because the short circuit current is much lower than the interrupt rating?

No, that has nothing to do with it. This solves a different problem and a lot of engineering goes into that.

The "interrupt rating" (10kA, 22kA, 65kA, those are kiloamps so 10,000, 22,000 etc) means the limit beyond which the breaker cannot interrupt, and sits there arcing. If you used a 10kA interrupt rating, and the power company had low resistance wires that could actually deliver 18,000 amps to your meter, and you had a dead short right there, **the disconnect would see 18k amps. And the 10,000A breaker would throw up its hand and say "I can't interrupt this, best get yourself some marshmallows and graham crackers". Whereas a 22kA breaker would work.

This is decided by the service drop and distribution and transformer (resistances 5-9 above), and is the subject of a pair of rules. In residential connections, the power company must use thin enough enough wires that even a dead short at the meter cannot exceed 10kA (inclusive of transformer impedance). The other rule requires the consumer to use a 10KA rated disconnect for most residential services. (so 240 = 10000 * R, solve for R=.024 ohms, so service wiring must be at least 0.024 ohms on their side). If you ever see an installation where you look at the service wires and go "hey wait! I had to use 4/0 wire, how did they get away with #1 on the service drop?" They're not getting away with it, it's mandatory to guarantee <10kA!

However on a large 400A service, guaranteeing <10kA may not be achievable, so the power company would guarantee a higher interrupt rating such as 22kA (again 22,000 amps), and in turn you'd be required to use a 22kA rated breaker.

Or your breaker could have a 65kA rating simply because of market conditions. 400A residential services are almost always done a different way (involving dual 200A breakers), so there isn't enough demand for residential 400A breaker for economies of scale to kick in. As such, the extant 400A breakers are all industrial gear and sized for industrial applications. That's why it was so darned expensive.

Does a short circuit trip a breaker immediately or does it take time to trip it?

Like I say about trip curves... yes.

Did it not trip because the problem already existed upon connecting the power?

No, they're not that smart.

Online voltage drop calculators show the voltage would drop to about 55 volts at the 2000 amp presumed magnetic trip setting. The calculator shows that there is too much impedance for the source to drive 9600 amps through a short at the breaker. Likely impedance upstream of the meter causes additional limitation of the prospective short-circuit current at the breaker. At the 400 amp breaker rating, the voltage drop is a bit more that it should be, about 7.6%.

1. Should that current not be enough to trip the breaker?

Probably not immediately considering the probable prospective short-circuit current at the breaker. The thermal trip would probably have tripped eventually.

Is it because the short-circuit current is much lower than the interrupt rating?

No.

Does a short circuit trip a breaker immediately or does it take time to trip it?

It should trip immediately, but the prospective short-circuit current is not high enough for that to happen.

Did it not trip because the problem already existed upon connecting the power?

No.