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# Electrical Theory Question

#1
10-14-06, 02:28 AM
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Electrical Theory Question

I've wondered for some time about the following question, and although I've reviewed several electrical books, I haven't found a satisfactory answer. It may sound like a simple question, but I'd still like to understand the answer.

Question: If you take a wire with a 120 volt plug on it, and connect the hot and neutral wires (no load), then plug it into a 120 volt outlet, I'd expect (I know better than to try it) the circuit breaker to trip because this is a short circuit. But, if you plug in any electrical device, no short circuit. Yet, it seems to me the flow of electricity is essentially the same, i.e., from hot wire to neutral wire to breaker boxes ground wire to the ground. Then, why does "straight wiring" the circuit trip the breaker? Is it because without any load (i.e. resistance) the amps gets too big for the breaker? If so, then will any size resistance do, like could you just hook the ends of the wires to say a 100 ohm resistor and not trip the breaker?

Maybe it's an electrical rule that a circuit must have a load, but I've never seen this in writing.

Hope this isn't too elementary for you electricians.

Thanks,

#2
10-14-06, 04:33 AM
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If you complete a circuit as you suggested you've made a short circuit so yes you will pop the fuse or breaker. This is due to the fact that the only resistance in the circuit comes from the conductors and the recepticale which depending on the length of the run may only be very few ohms. So lets assume 2 ohms for real easy numbers 120/2 = 60 amps now if you inserted a properly rated 100 ohm resisitor you would have 120/100 = 1.2 amps . Just remeber ohms law is E=IR and you can work out whatever parameter from that assuming you know the other two. Things get more complicated when you enter into inductive and capacitive loads since you start dealing with in rush current, power factors and phase shift but thats a little beyond what you are asking.

Last edited by DIYaddict; 10-14-06 at 07:20 AM. Reason: removed quote as it's unnecessary to quote the entire post that's directly above yours
#3
10-14-06, 05:13 AM
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Fred,
You are right to not experiment with this at home. Some community colleges have testing benches if you want to set up tests. If you are doing this at home stick to paper and pencil

If you have a 120 v circuit with a 50 ohm reisitor you will be using 2 amps. Those two amps will be converted to some other kind of energy. In this case Heat. So, the resistor is very likely to burn up fast, even though it is not necessarily overloaded.

#4
10-14-06, 06:08 AM
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Resistance

Remember Ohm's law. The resistance must be sufficient to keep the current at a level below the breaker rating (15 or 20 amps.)

#5
10-14-06, 07:28 AM
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Originally Posted by jwhite
Fred,
You are right to not experiment with this at home. Some community colleges have testing benches if you want to set up tests. If you are doing this at home stick to paper and pencil
I know of several electricians who carry one of these tools in their bag in order to find the breaker feeding a receptacle- plug it in, and look for the tripped breaker! I accidentally did something this myself in a high school lab- made a big shower of sparks and tripped the breaker, heh. I still have no idea why the hell our teacher thought it was a good idea to let high school kids play with live 120V AC power...

Kick-It-Up: I know E=IR is the way we engineers write it, but no need to confuse the laymen with our obtuse nomenclature!

Volts = Amps x Ohms is a lot easier for the DIYs out there to understand A copper wire has extremely little resistance- it's not 0 Ohms, but it might as well be. Now, when you take that copper wire and wind it into a motor or stuff it into a heater, you add resistance. The method of resistance gets complicated to explain when you get into inductance and magnetic flux, but sufficed to say that a 60 Watt light bulb is roughly equivilent to a 240-Ohm resistor. That is, Volts = Amps x Ohms, or Amps = Volts / Ohms; 120V / 240 Ohms = 0.5 Amps flowing through that light bulb.

Normally, we ignore the line resistance of the copper wires as negligible, but in truth, they DO offer SOME resistance. So when you short circuit, you don't get 120V / 0 Ohms = Infininite Amps, but rather 120V / 0.1 Ohms = 1200 Amps or something like that. And the breaker will VERY quickly trip!

Ohms Law: Volts = Amps x Ohms
Watts Law: Watts = Volts x Amps

#6
10-14-06, 08:35 AM
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Originally Posted by grover
I know of several electricians who carry one of these tools in their bag in order to find the breaker feeding a receptacle- plug it in, and look for the tripped breaker!
As far as I know there are no testers that are listed (by an approved testing and listing labratory like UL) to test breakers by causing them to open on overload. By this I mean for field use, not as equipment in a testing labortory enviornment.

I did run into a guy on internet forums who is from MA, USA who was in the process of getting such a tool listed. This was last fall and I have not heard back to see if he had any success.

#7
10-14-06, 08:57 AM
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Just want to thank each of you who responded because now I get it: when you straight wire a circuit there is virtually no resistance. Under Ohm's Law, since there is no resistance the amps shoots way up, far exceeding the breaker's amp limit, and pop goes the breaker. The example below of the 60 watt light bulb is a good one for me because it tells me that even when the only thing using power on the circuit is a 60 watt light bulb, the amps is down around .5, which is why the breaker doesn't trip.

Thanks again.

#8
10-15-06, 06:28 AM
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If you want to dig a little deeper, here is and excellent reference.

For the specific question you asked, look here.

#9
10-15-06, 07:05 AM
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Think of one of the other terms for voltage: Electomotive Force. Voltage represents potential energy, but no current flows and no WORK is done, until a PATH exists for current flow. When a path is completed, then the FORCE and the RESISTANCE determine how much work is done. (Ohms Law).

The WORK will be the potential energy converted to HEAT energy and in the case of a bulb some of it becomes LIGHT energy).

#10
10-15-06, 07:34 AM
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594tough: Thanks for the info. I went there and found it informative; so, I added it to my "favorites" so I can go back and read through the other stuff there in the future.

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