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I have a question - let’s say I have a very small 2 wire cable connected to a load that draws a large amount of current, I’ve been told ( I have yet to do my own testing) that the smaller cable will add more current draw to the circuit and there for trip a breaker or fuse quicker then a larger cable with less resistance.

From what I have learned and understand, a smaller more resistive conductor will exhibit a larger voltage drop to the load and cause current to be less then a larger lower resistance cable. How is it possible for a smaller conductor or adding more conductor to a circuit cause a breaker or fuse to trip quicker, seems as if I’m missing something .

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  • $\begingroup$ Perhaps you can tell us who "told" you this and in what context? $\endgroup$
    – Bob D
    Sep 6, 2019 at 19:48

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Based on your comment to @The Photon, it appears that the reference to a "smaller" cable was meant to be a lower cable AWG size. The lower the AWG number the larger the diameter of the conductors. Then there would be lower voltage drop across the conductors feeding the load increasing the current to the load.

However, it would still be somewhat far-fetched since the current is primarily dictated by the load. Increasing the supply conductor size should result in only a slight increase in current to the load, unless the conductors were severely undersized in the first place (which would result in the conductors overheating), or the length of the run were very long (which would result in an excessive voltage drop to the load). Both conditions would normally be precluded if the installation was in compliance with the NEC.

Finally, in order for the increase in current to trip a breaker or rupture a fuse, the normal load current would have to be pretty close to the breaker or fuse rating, or there are other loads on the circuit the combination of which are close to the rating. That's another possible NEC violation.

Other than that, I can't see any reason for the phenomenon, other than the possibility suggested by @The Photon.

Hope this helps.

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If your load is powered by a regulated switching power supply, then what you were told could be correct.

Internally, this device is trying to produce, for example, 5 V to power the actual circuits inside. And those circuits might need (as an example) 10 A. So the final load needs 50 W of power.

The switching power supply circuit is designed to adjust the current it draws from its input to whatever it needs to be to provide 5 V to its output.

So if you supply it 120 V, it will draw about 420 mA. If you use a too-small cable and only 100 V is available at the input to the power supply, then it will increase its current draw to 500 mA to be able to deliver 50 W to its load. (I'm ignoring the inefficiency that you'd find in any real power supply circuit that would make the required input power slightly higher than the output power)

Obviously I've used smaller numbers in my example than you are dealing with in your actual situation, since 500 mA is not going to throw a lot of breakers. But the general principle is the same. A higher voltage drop in the cabling to your device can lead to the device drawing more current.

If the load is a motor, again your advisor's comment could be correct, but for different reasons. If the supply wires drop too much voltage, then the motor will be more likely to stall, and when it stalls it will dramatically increase its current draw (which is likely to throw a breaker, as in your advisor's example). Lowering the voltage available to the motor could also increase the current draw if it's not stalled but operating with the same rpm and mechanical load (but I suspect whether it would maintain the same rpm depends on the kind of motor and the kind of controller used with it, about which I don't know enough to talk in detail).

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  • $\begingroup$ Re, "...whatever it needs to be to provide 5 V and 10 A to its output." I think your wording could confuse newbies. I'm sure that you intended the power supply in your example to maintain a constant 5V output, and the load happens to draw 10A at that voltage. But, I think the way you phrased it could make a newbie think that the power supply simultaneously controls both the voltage and the current. I've seen more than a few questions here, and on physics.stackexchange.com that hinge on that same misunderstanding. $\endgroup$ Sep 6, 2019 at 19:13
  • $\begingroup$ Thanks for the reply. My question is based on a simple resistive circuit let’s say a heating element. $\endgroup$
    – Scott
    Sep 6, 2019 at 19:29
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    $\begingroup$ @Scott, if the load is resistive (like a heating element), then more resistance in the supply line will reduce the current draw. $\endgroup$
    – The Photon
    Sep 6, 2019 at 19:39
  • $\begingroup$ @thephoton, that’s exactly what I thought, having a smaller AWG conductor supplying a load would net more of a voltage drop thus decreasing current at the load. Why I’ve been told by a technical advisor in my area the opposite is unknown. $\endgroup$
    – Scott
    Sep 6, 2019 at 19:51
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    $\begingroup$ It depends on the load. For a heater or incandescent lightbulb, no. For something with a motor (like a vacuum cleaner) there could be other things going on that lead to higher currents with lower delivered voltage. $\endgroup$
    – The Photon
    Sep 6, 2019 at 20:03

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