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Basic current question

My thinking, when saying what I said, came from something along these lines:

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I apologize if I misrepresented that idea.

I measured roughly a 15V drop, running a power saw in a room at the far end of a home. The run of wire to that end of the home, including various loops around the walls and via various connectors, appears to have been around 100 feet each way. The wiring in the walls was 14 gauge aluminum. Not sure what the actual draw was from the saw, but it was enough to drop about 15V. The reason I bothered with any measurements at all was because the lights dimmed quite substantially. Bugged the hell out of me.

I can't imagine trying to understand a schematic by first replacing every wire with some complex representation of its impedance and then trying to fathom it. We will just have to disagree with each other on this point.

By the way, I took courses on the Tektronix campus back in the 1970's and this is how they also taught me to examine schematics. It's a thinking method not of my own concoction.

Jon

Reply to
Jonathan Kirwan
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From the wire table in the NEC, #14 Al is about .51 ohms/100 feet. You know the rest, given that spec, but I'll spell it out in case it is helpful to somebody.

200 feet wire totals ~ 1 ohm, so a 15V drop equates to 15 amps drawn. Seems about right (mathematically) - I suppose it depends on your saw. But electrically - you want to keep the drop a lot smaller, and using a 15 amp saw on a 15 amp circuit will give you dimming and is not too kind to the saw. #12 copper runs about .2 ohms per 100 feet, so you would have about 6 volt drop on a 100' circuit wired with it. (200 feet total wire). That's acceptable (barely) whereas 15 v drop is not. 20 amp circuits call for #12. So it is better to use a 20 amp circuit for the saw. Actually, one could pick a nit about it being acceptable - but set that aside. The idea is to plug the power saw in to a 20 amp circuit instead of a 15 amp circuit, when possible.

Ed

Reply to
ehsjr

Agreed. However... there was only _one_ circuit to that room. Which is the usual case.

Jon

Reply to
Jonathan Kirwan

On Fri, 16 Jun 2006 01:30:32 GMT, in message , Jonathan Kirwan scribed:

Yes, you misrepresented the idea. Maximum power transfer matches load to source, so that the source impedance and the load impedance are equal. This load impedance is of the entire transmission system to which the source is attached. Theoretically the connections used in a transmission line will be perfect, and thus lossless. In reality, of course, there will be some loss, but it will likely be measured in the tenths or hundredths of dB.

Thus power is not halved at every connection; however power will be halved if 1:2 branching is done, e.g. using a coax line splitter. In that case, The signal will be reduced by about 3dB for each leg. The more branches, the more loss when splitting. This is because each load cannot take 100% of the power the source is delivering; rather each of two loads, if matched properly, will get 50% of the source.

Reply to
Alan B

On Mon, 12 Jun 2006 22:15:43 +0100, in message , Pooh Bear scribed:

I agree absolutely. It's a nice shortcut to simplify the math in engineering classes, but really, every engineer should understand that wires and connectors are components also, and have electrical characteristics that may have critical effect on a circuit design. I could tell a hundred stories....

It's not just the "layout guys." A *lot* of engineers, left without adult supervision, would run wild with poor wire design.

Reply to
Alan B

Thanks. That clears up my muddled thinking and verbiage. :)

Jon

Reply to
Jonathan Kirwan

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