Wire Gauge, Current Capacity vs Power Capacity?

[AG6QR]

Isn't this only about Ohms law --- If my motor will draw 10A at 12v I need a wire thickness that will not contribute much to the voltage loss due to wire heating.   So 10A at 12vdc dissipates much less at the same resistance (wire thickness) than 10A at 120vdc.

No. The voltage has no effect on the power dissipated by the wire. If your wire carries 12V 10A or 12 kV 10A, it'll still dissipate EXACTLY as much power*. The only thing that matters is the actual wire resistance (i.e. wire gauge).


(*: Assuming it doesn't arc over, at which point all bets are off. Different problem!)

I think you are confusing heat dissipation of the wire versus the total current capacity on the circuit.  Power = I * E ; so 12v * 10A   is not the same as 12KV * 10A.

ovnr is not confused.  You seem to be confusing power dissipated in the wire with power delivered to the load.  Power dissipated in the wire is I * E, where I is the current in the wire, and E is the voltage present from one end of the wire to the other end of that individual conducting wire.  The voltage across the load does not enter into it.  The voltage across the wire is just the current times the resistance of the wire (load resistance doesn't enter into it). 

Power delivered to the load is also written as I * E, but this time, the E represents the voltage presented across the load.  The equation is still P=I*E, but the letters represent different quantities.

Consider a circuit consisting of two wires carrying power from a source to a load.  The heating in a wire depends only on the current running through that wire, not on the voltage as measured from one wire to some different wire.   One individual wire "sees" only the current running through it, and the resulting voltage across the wire, as measured from one end of the wire to the other end of that same wire.  It doesn't "know" anything about the voltage of the other wire.  On the other hand, the power source and the load both see the voltage presented between the two wires.

[AG6QR]

All the way through this thread we have been trying to tell you that the voltage carried by the wire has no effect on the power dissipation. Somehow you are still not getting it.

Sorry for being dense but I guess that why this is the Beginners Section :-)  I will go back to my example question, under the conditions of a constant load (10A) at constant voltage (12V) would 18AWG be adequate or do I need 12 AWG.  The difference in wire resistance is an increase in total resistance of about 0.05 ohms (for 10 feet).    The voltage drop would be about 0.5vdc and the heat loss would be about 5W.  Is that too much for 18 AWG to handle safely?

It won't catch on fire and melt, if that's what you're asking.  But, assuming the run has a total of 20 feet (two conductors, each 10 feet long), you'll lose almost 10% of your power in that short transmission line.  What's the cost of oversizing your solar panel by 10%, versus the cost of increasing your wire gauge to lower the power loss?

[jstarr]

BIG thanks for all of your patience in helping me to understand the principles behind this.  I have one more question as I did not state the setup accurately.  Another beginner mistake.... my apologies 

The motor being supplied by the AGM battery is powered thru a switching supply.  The supply input is 12 VDC and the motor takes 5VDC at 10A peak.    I have read that switching supplies are near 85% efficiency, so am I correct that the draw on the battery will only be 4.2A  * 0.85 = 3.54 A peak??

[cowana]

So if the motor draws 10A at 5v:

Assume 100% efficiency converter:

10/(12/5) = 4.16A

However it's not quite that good - say 85% efficiency will increase the 12v current consumption slightly:

4.16/0.85 = 4.89A draw at 12v

[ovnr]

The motor being supplied by the AGM battery is powered thru a switching supply.  The supply input is 12 VDC and the motor takes 5VDC at 10A peak.    I have read that switching supplies are near 85% efficiency, so am I correct that the draw on the battery will only be 4.2A  * 0.85 = 3.54 A peak??

Switching supplies may range between <40% and 98% efficiency. Please tell us which PSU you're actually using.

And no, you've got it backwards. If your load is 5V @ 10A over a 85% efficient switcher, your battery will see a 12V @ 4.9A load.

Secondly, if possible put the PSU near the motor - you will have four times less power loss in your cables (4.9A vs 10A; assuming your cables are 0.25 ohms, that's 6W vs 25W).