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In this question, many commenters and answerers pointed out that a device that would trip the circuit breaker corresponding to an outlet by shorting the outlet would be extremely dangerous. It would indeed trip the relevant circuit breaker if that circuit breaker was working, but if not, it could start a fire.

My question is the following: suppose your wiring is connected to a 10 A circuit breaker. Would it be dangerous to have a device that tripped that circuit breaker by shorting the circuit using an 11 A circuit breaker? That way, the current could go above 10 A, tripping the original circuit breaker, but guaranteeing that no more than 11 A of current could go through the circuit, thus precluding a fire.

I am not proposing to actually implement this -- I am just curious about any possible dangers that could arise from such a scheme, since in my understanding the dangers suggested in the original answer would be mitigated by the use of an additional circuit breaker.

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    Circuit breakers aren't that precise; you'd have to have one that was rated substantially higher than the built-in one. (Beyond that, it doesn't seem like a good idea.)
    – Daniel Griscom
    Commented Jan 2, 2017 at 12:01
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    The point of tripping a standard breaker for test purposes would only be to see if it is tripping at proper current, this would be done with an actual commercially available circuit breaker tester (either at the panel or on the bench top). To trace a line to the circuit breaker they make a variety of tracers for that purpose.
    – spicetraders
    Commented Jan 2, 2017 at 15:43
  • Aww, I thought this was going to be about that question's thing (which is just a switch, not a CB). The main concerns with a direct short are: damage to the wire and devices, electrocution, fire, arc flash and the resulting molten metal that will shoot out at you, or all of the above.
    – Mazura
    Commented Jan 4, 2017 at 5:19

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In theory, that sounds great. In reality, maybe not so much. You could be talking about putting a crap-ton of current through the house wiring, and a device in your hand!

Looking at a 20 ampere circuit, where the receptacle is 20' (along the wire) away from the breaker. That's 40' of wire in the wall. According to NEC, 12 AWG wire has a resistance of 0.00193 ohms per foot. Assuming the wiring for the device is 12 AWG, and about a foot in length. That's a total of 41' of wire, at 0.00193 ohms per foot.

Using Ohm's Law, we can calculate the fault current.

I = V/R
I = 120 v / 0.07913 ohms
I = 1,516.492 amperes

That's more than 75 times the rated current of the 20 ampere breaker, so it should be within the instantaneous trip range of the breaker. If the house breaker does not open, you've got 1500 amperes flowing through the house wiring, and a device in your hand. Hopefully the breaker in your device works, and opens the circuit in less than a cycle. Otherwise, that wire is going to get really hot, really fast.

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    If you absolutely insist on being stupid, you would want a device which produced a current-limited flow for a very limited time. Putting a second breaker in doesn't do that; you would need load resistors/inductors adjusted to just over the breaker's liad rating, and something which kept the circuit from being held closed any longer than required to establish that the breaker had failed. And it would still be an incompetent's solution to the wrong problem. Better answers are available cheaper and with less effort. Stop trying to make this work.
    – keshlam
    Commented Jan 2, 2017 at 17:30
  • Liad => load,, of course.
    – keshlam
    Commented Jan 2, 2017 at 18:41
  • I'm evidently not trying to make it work -- that was written right in the question.
    – ajd
    Commented Jan 2, 2017 at 21:00

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