DC Voltage Drop Calculation

Hi
I am working on some voltage drop calculations - but with a DC voltage. The calculation is very simple, a 2 core cable with 5% drop
allowance. I know from many sources that I need to double the length of the cable when calculating the resistance because of the two wires.
My question is why do I double the length? I would have thought that according to Kirchoffs Voltage Law the voltage drop at the load would have only been the voltage drop along one wire?
Thanks in advance.
Ethean
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snipped-for-privacy@yahoo.co.uk wrote:

Ah, the "ground is ground" assumption. Common in new companies who's products have odd troubles with noise.
According to Kirchoff's Voltage Law the voltage around the loop is the drop from wire 1 plus the drop from the load plus the drop from wire 2. Since wire 2 is an actual wire and not a line on a piece of paper it does have a real resistance, and hence a real voltage drop.
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Tim Wescott
Wescott Design Services
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snipped-for-privacy@yahoo.co.uk wrote:

Let's say that the supply voltage is a 100-volt battery. One side of it is tied to ground, and a wire goes from each side to the load. (Both wires are the same length and diameter.) The current and wire resistance is enough to cause a 5% drop along one wire, so that the high side of the load is 95 volts above ground. What is the voltage with respect to ground of the low side of the load? What is the voltage across the load?
Jerry
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Thanks Jerry
5 volts with respect to ground at the low side of the load and 90 volts across the load.
So this would be a 10% drop with respect to the consumers terminals?
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Ethean wrote:

That's exactly right. I'm glad you understand it now. I hope I helped by not giving the answer up front.
Jerry
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Indeed you did. Very much appreciated.
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The current has to fight it's way there, through the load, and then fight it's way back. If the path there and back are the same then ...
snipped-for-privacy@yahoo.co.uk wrote:

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