voltage drop

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sparkease

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using a 1 phase circuit at 208 volts is the length of wire used to figuire v. drop twice the linear distance from source to load. load in this case being 20 amps at 600 linear feet or total conductor lenght 1200 feet using length out and back.
 
You lose voltage on the way out, and on the way back. Whatever is not lost in the wires, is available at the load. So whatever formula you use to calculate the voltage drop, be sure it includes the total distance of a "round trip" (i.e., twice the linear distance from source to load).
 
sparkease said:
using a 1 phase circuit at 208 volts is the length of wire used to figuire v. drop twice the linear distance from source to load. load in this case being 20 amps at 600 linear feet or total conductor lenght 1200 feet using length out and back.
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Using 208 or 240 volts, you only use the distance one way. If the voltage is 120 volts, you have a return on the neutral so you use 2 x the distance.
 
bob said:
Using 208 or 240 volts, you only use the distance one way. If the voltage is 120 volts, you have a return on the neutral so you use 2 x the distance.
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The online formula I linked to shows one way either scenario. I know there is more than one way to figure VD but....
 
bob said:
Using 208 or 240 volts, you only use the distance one way. If the voltage is 120 volts, you have a return on the neutral so you use 2 x the distance.
Click to expand...

Careful here. As Charlie notes, you have current on both 'sides' of the circuit, and thus you have voltage drop on both sides of the circuit. One way or the other, you have to account for the voltage drop on all of the conductor length of the circuit.

However many voltage drop formulas take this into account. You give the formula 600 feet, and built into the formula is a calculation for 1200 feet worth of conductor. Or the formula takes into account only 600 feet worth of conductor, but takes the percentage voltage drop relative to the line-neutral voltage, or in some other way hides what is going on. But no matter what, you are still dealing with voltage drop in all of the conductors of the circuit.

-Jon
 
Huh?

Huh?

bob said:
Using 208 or 240 volts, you only use the distance one way. If the voltage is 120 volts, you have a return on the neutral so you use 2 x the distance.
Click to expand...

I don't think so. E = IxR whether or not we have a neutral. Consider a 240V circuit, L1, N, and L2. With a 240V load, current flows in L1 and L2. You must consider the drop in each leg. Now for the same currents, the percentage drop at 240V will be half the percentage at 120V.
 
Just the links alone that I have obtained from this forum more than pay for my membership. OH YEAH IT'S FREE.....
 
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