Voltage Drop, One-Way Distance Vs. Out and Back Distance

I have read other threads concerning this question but It still wasn't made clear to me.

Say you have a load connected in parallel 50 feet from the source. The code recommends not more than a 3% voltage drop in a branch CIRCUIT, that is to say out and back, so in calculating the R portion of VD=IR you would use 100 feet. 50 feet out to the load and 50 feet back to the breaker.

My question is, Why does the voltage drop on the line RETURNING from the load matter? wouldn't you only be concerned with the voltage SUPPLYING the load?

I have theories as to why the Out and Back length is used as opposed to the one way length.

It might have something to do with the rules for "E" in a parallel circuit; the voltage across one load is equal to the total voltage of the circuit. But even knowing that I'm not sure how that works.

I'm coming at it from the perspective of the load, if it gets enough voltage coming to it, it will work properly. How would the voltage lost after it went across the load matter?

Also,? would you still have to double the one way distance for a series circuit?
 

hurk27

Senior Member
I have read other threads concerning this question but It still wasn't made clear to me.

Say you have a load connected in parallel 50 feet from the source. The code recommends not more than a 3% voltage drop in a branch CIRCUIT, that is to say out and back, so in calculating the R portion of VD=IR you would use 100 feet. 50 feet out to the load and 50 feet back to the breaker.

My question is, Why does the voltage drop on the line RETURNING from the load matter? wouldn't you only be concerned with the voltage SUPPLYING the load?

I have theories as to why the Out and Back length is used as opposed to the one way length.

It might have something to do with the rules for "E" in a parallel circuit; the voltage across one load is equal to the total voltage of the circuit. But even knowing that I'm not sure how that works.

I'm coming at it from the perspective of the load, if it gets enough voltage coming to it, it will work properly. How would the voltage lost after it went across the load matter?

Also,? would you still have to double the one way distance for a series circuit?
Google Kirchoff's laws of a series circuit.

For any circuit to function you must have a complete circuit path to and from the source, if the load pulls 5 amps then you will have 5 amps to the load as well as from the load, this means any resistance in the conductor to the load is added to the resistance of the conductor from the load and the total is what creates the voltage drop, if you have a higher resistance in the return conductor than the resistance of the supply conductor it still gets added together with the resistance of the supply conductor to get the total resistance of the circuit, but since current in a series circuit is equal in all parts of the circuit then the voltage must change proportionally to the resistance difference in each part of the circuit, this is called a voltage divider, in a parallel circuit the voltage is the same in all parts of the circuit but the current changes proportionally to the resistance in each part of the circuit, each part of a circuit is called a Node,
 

GoldDigger

Moderator
Staff member
Location
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Another quick way to look at it is that the load sees only the voltage across its terminals. It does not care how that voltage is derived.
The voltage across the load is the source voltage minus the sum of the two voltage drops.
The load does not know what is happening on the neutral after the current leaves the load. But it certainly knows what the resulting voltage at its neutral terminal is!

Tapatalk!
 

fmtjfw

Senior Member
I have read other threads concerning this question but It still wasn't made clear to me.

Say you have a load connected in parallel 50 feet from the source. The code recommends not more than a 3% voltage drop in a branch CIRCUIT, that is to say out and back, so in calculating the R portion of VD=IR you would use 100 feet. 50 feet out to the load and 50 feet back to the breaker.

My question is, Why does the voltage drop on the line RETURNING from the load matter? wouldn't you only be concerned with the voltage SUPPLYING the load?

I have theories as to why the Out and Back length is used as opposed to the one way length.

It might have something to do with the rules for "E" in a parallel circuit; the voltage across one load is equal to the total voltage of the circuit. But even knowing that I'm not sure how that works.

I'm coming at it from the perspective of the load, if it gets enough voltage coming to it, it will work properly. How would the voltage lost after it went across the load matter?

Also,? would you still have to double the one way distance for a series circuit?
I'm unclear as to what you mean by parallel versus series in your questions.

For the following circuit: hot to black wire, black wire 50 feet, load, white wire 50 feet, white to neutral, you really have 3 resistances in series: the black wire, the load, and the white wire. It is reasonable to assume the same resistance for the black wire and the white wire (assuming they are the same size and same length), so you double the voltage drop of one to make the calculation simpler.

The voltage drop in the black wire (E = I*R) and the voltage drop in the white wire are added together to determine the voltage drop to the load VDblack + VDwhite = VDtotal (which is the same as doubling one value). The voltage at the load is Voriginal - VDtotal.

If you were to measure the voltage difference between the beginning of the black wire and its end, then measure the same for the white wire beginning to end, you would see that they are the same. In the same way if you measure the voltage across the line at the beginning of the black and white wires and compare that reading with the voltage between the black and white wires at the load, you would see the difference was 2*VD in one wire.

There is nothing magic about the white wire.

If you don't already have an "Uglies" electrical handbook, I recommend you get one. It should be available at one or more of the electrical wholesalers in you area, or you can order one from Amazon. It is chocked full of useful information and the first section will help you with calculations like this.

If you will explain your parallel versus series circuits, I'll address them.
 
Much Clearer Now

Much Clearer Now

I think these replies got it straightened out for me. With voltage being defined as a difference in potential between two points, the portion of the circuit leaving the load is just as relevant as the portion supplying the load. Now that I think about what y'all said, this may have been a stupid question.

@FMTJFW, Thank you for your reply, What you said about the voltage being the same from the beginning to end of the white and black wires was eye opening; Voriginal-VDtotal cleared several things up. This helped me to realize that doubling the distance is a mathematical operation that takes into consideration the length of conductor supplying the load, and the length of conductor returning from the load, both of which are relevant in a voltage drop situation.

@golddigger, Thank you for your reply, What you said about the load seeing only the voltage ACROSS the terminals, as opposed to what I was thinking; the voltage on the supply side only and the voltage on the return side only, helped a lot.

@Hurk27, Thank you for your reply, your referral to kirchoffs law of series circuits was relevant to my question.

Thanks again to you all! I take my exam june 10th.
 
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