Theoretical Amps/Volts question

As voltage goes up, amperage goes down. I got that.

So lets say I was running 100ft of wire to supply a 50 amp, 240 Volt electrical circuit.

I use #8 copper because it can handle that amount of amperage, and my voltage drop calculator tells me it will be about 2.6%, which is fine. So I'm all good.

But if I were to increase that distance with the same wire a stupid amount to say, 1000 feet, the voltage drop is a now a whopping 26%.

Beside severely effecting the device(s) that now aren't getting enough voltage to operate, will the amperage in the wire ALSO increase significantly (and potentially dangerously, possibly cooking the wire) since the voltage has been decreased so much?

 

It depends on the load. A simple load, like an incandescent lamp, will continue to reduce current demand as the voltage goes down. A more complex load, like a switchmode power supply, or an induction motor, will increase current demand until a point where the current will then reduce sharply.

 

What about, say, a toaster?

 

Kirchoff's voltage law states the voltage drops (current x impedance=IZ) around a closed circuit must equal the voltage source (E). Since the impedance of a toaster does not change with a decrease in source voltage, the current will also decrease so IZ=E.

 

Copy.

So then as long as the load isn't increasing current demand as the voltage drops, there really is no danger of burning up the wire due to excessive amperage. Correct?

telecom guy: what would be an example of a piece of equipment that uses an induction motor or a switchmode power supply that would do this?

 

This is not some rule that is obeyed by source and loads, generally.
Your toaster does exactly the opposite. Lower voltage takes more time to toast.
On your flatscreen tv, with something like an intelligent, switchmode power supply, it will raise the demand for current, to maintain a constant power input, within limits.
On your refrigerator, its compressor will strain and raise its current demand as the voltage goes down. But, that wasn't designed in, and is a byproduct of a rotary inductive device, where the rpm has a lot to say about current drain.

So, bottom line; loads can be non-linear, positive slope, negative slope, and even both, depending on where the operating point is. Driving a gas discharge lamp directly, with no ballast, results in a very interesting curve.

 

Thanks, that is very helpful. It clears up a lot for me

 

Note: For most heating equipment and incandescent lights, lowering the supply voltage does not reduce current (amperes) in proportion. At lower voltage the temperature attained is lower which for most equipment results in a lower resistance and therefore the amperes drawn would be greater than you might otherwise suspect. To compute the resistance you would want to know the characteristics of the metal or alloy used for the heating element or lamp filament. Alternatively you can measure the (manually varied) voltage (E) and the current (I) and the temperature and draw your resistance versus temperature graph that way.

For heating elements and other pure resistive loads, impedance equals resistance so here you would simply compute Z = E divided by I.