Testing for voltage drop

I have an idea of how I would do this, but I'd like to see if I can get suggestions for a better way. How would you go about testing the amount of voltage drop from the source at the breaker panel to a distant outlet? Also, if I'm trying to run a 110 volt welder from a 110 volt outlet, why does it matter that I use a circuit that is fed through a 20 amp breaker rather than a 15 amp one? If I don't draw enough current to trip the breaker, what reduction could a lower rated breaker cause? (I think I know the answer to this one, too, but I'd like to see if I really understand it.)

Stan

Q1: The easiest way is to remove all loads from the circuit and measure the voltage from hot to neutral at the outlet. (With no load, your outlet voltage will be the same as the panel voltage.) Then apply the load and re-measure the voltage. Any difference between the two measurements is the voltage drop. If you have significant voltage drop, you should check to see if the voltage is dropping at the panel, which would mean there's a drop upstream of the panel.

Q2: If you are using a 15 amp circuit, then the circuit most likely has 14 gauge conductors. A 20 amp circuit has 12 gauge conductors. The heavier conductors will have lower voltage drop, and a 120 volt welder doesn't have a lot of voltage to spare under the best of conditions.

Im pretty well with Mark on this one. Its NOT the size of the overload protection device that makes the difference in voltage drop, ITS THE SIZE OF THE CONDUCTORS AND CURRENT DRAW AND DISTANCE. Typically a 15 amp branch circuit uses 14 gauge conductors while a 20 amp uses 12 gauge and the bigger the wire the less the voltage drop in the conductors all else the same.

Use the same meter and measure voltage at the source and then at the load with it on and drawing current tells you how much voltage you're dropping in the circuit from source to load. Easier and more practical then using longgggggggg leads to actually measure the voltage drop across the conductors end to end.

John T

Mark;

I figured out the answer to Q2 just as I finished writing it. I was about to delete it, but then decided to leave it to see if there was more to it than that.

Your answer to my Q1 raises an issue I don't understand. What I read is that there would be no voltage drop at a distant location until a load is applied. My understanding (or possibly my misunderstanding) was that the calculated voltage drop at the end of a conductor of a certain length and a certain gauge would be shown on a voltage tester. For instance, an online voltage drop calculator I tried said that 120 volts of single phase AC current run through 1000' of 18 gauge copper wire would measure 107.23 volts at the distant end. Your answer seems to say that it will read 120 volts until a load is applied. In my experience on this site, you are usually right. So, what am I not understanding?

Stan

Stan, that load table is assuming some current, probably whatever current an 18 gauge wire is good for, a few amps. No current, no voltage drop.

Stan, THERE IS NO VOLTAGE DROP UNTIL THERES AN ACTUAL LOAD AND CURRENT FLOWS

Think about it, Ohms Law, V = IR so if I = 0 V = 0 x R or Voltage drop = ZERO

John T Long retired Electrical Engineer and rusty but think Ohms law is still valid????

Was until a bunch of lawyers got to monkeying with it.

Simple, the bigger the cable the better. Think of your wires as pipes and the supply of electricity as a water tank. Now what would a pressure gauge read at each end? Start as the same and when you start running the water the tank is still the same but the other end the pressure drops. Bigger pipe=less pressure drop. pretty much the same. One thing that is very funny and TRUE is the copper. Copper wire is just about the best conductor of electrcity. It is also about the most restrictive type of water pipe. Lots of friction. No kidding.

Geeeeeeeeeeeeeee now Ohms law is illegal lol

John T

Voltage drop is caused by R (resistance) in the circuit. In this case it is copper loss and is significant due to length of conductor. Calculation for that is: V ( source to load line drop for this summary) = I (current) time R (resistance). In Mark's first instance I = 0 amps. Therefore there is no drop, just like if you are running water through a pipe vs static pressure at the hydrant before you open it and have flow, it's the same physics, just a different critter.

The wire has a certain resistance usually measured per foot. Using Ohms law you find the amount. If the line is running near rated load, you have to factor in the fact that the wire, usually copper, will heat and the resistance will go up which will raise your drop over time more than the initial amount. If the wire can radiate the heat, like power utility overhead lines, you can run more amperage because it runs cooler and the drop remains manageable.

My reference books use 2% drop as the full load rating for a conductor. Tables are available numerous places that give you the working current for conductors, both copper and aluminum, for a given length and in some instances where and how they are packaged...conduit, buried, airborne. Obvious reason for that is the wire resistance is usually measured per foot or 1000' for commercial applications so the more feet the more the resistance and the more the drop.

Course insulation type and temp rating is also part of that. I remember one installation where weight was at a premium. Kapton (Dupont Polyamide Film) insulated wire was the answer and the wire ran at about 300F due to the small diameter for the required current. Lossy ? Yes, but the application didn't care about that. It had other priorities.

Your breaker size has nothing to do with line drop. Not enough length to make a difference.

Mark

Well, John, Ohm's Law was never really a law in the first place, in the sense of the laws of motion or law of gravity. It's really just the definition of resistance, and there's no way to directly measure resistance; it's always measured indirectly through voltage and resistance. Compare that to Newton's Second Law of Motion, F = ma. Force, mass and acceleration are all things which can be directly measured.

Let's not forget there are plenty of cases where Ohm's Law doesn't hold true at all, particularly with semiconductors. Try measuring the "resistance" of a diode, for example, and you'll find it varies greatly depending on the applied voltage.

Maybe Bubba can white to his congressman and get it repealed. Oh! I remember Bubba does not write reading!

For sure Mike, as BOTH an Electrical Engineer and an Attorney, I well knew what's referred to and known over the world as "Ohms Law" is NOT either Common or Statutory Law lol. Its something you typically learn by studying engineering and electricity and electronics, NOT in law school, although believe it or not I do recall its discussion in some rare case out there involving an electrical accident, but as a part of electricity and physics NOT law. The attorney didn't rest his case on "Ohms Law" but on the laws of Tort and Negligence.

HOWEVER I was trying to explain to Steve how there is no voltage drop in a circuit until current flows through its conductors. In so doing, I cited "OHMS LAW" which states V = I x R. Now, if I = 0 in such a circuit then V or voltage drop also must equal ZERO. In this case "OHMS LAW" does apply, yields the correct answer to his question.

Anyway, I still stand by my answer and "Ohms Law" and other physical explanations in answering Steve's good question trying to explain why he wouldn't experience any voltage drop in his circuit unless a load was attached and there was current flow NOT ZERO CURRENT. As far as I know whatever laws or electricity or physics you want to use to explain it to him, I STILL SAY IF THERES NO CURRENT FLOW THERES NO VOLTAGE DROP AND I STILL SAY V = I x R and if I = 0 then V = 0

That's my story and Ima stickin to it lol until, someone can prove different and there is voltage drop even if theres no current flow ??????????

Best wishes, hope this helps, take care and I always enjoy fun sparky chat

John T Retired Electrical Engineer and Attorney at Law "Long live Ohms Law" lol

Hey based on some of Bubbas wiring out there, I think maybe he should do that lol

Fun chattin with yall

John T

Great explanation......... When I was practicing (long retired now) we referred to a conductors Ampacity and what many non electricians and non engineers didn't comprehend was it had to do with NOT ONLY current, but temperature and the insulation and whether or not the conductors were in free air or jacketed or in conduit and if so how many or buried etc. etc. Lay persons may use figures such as 12 gauge is good for 20 amps while 14 gauge is good for 15 etc etc not realizing they need to also consider all those and more factors I outlined above.

I also cited Ohms Law (electricians and engineers are familiar with it) as one method to understand why if there's no current there's no voltage drop since if V = I x R and I is zero then V = 0.

Then to expand it further if you get into the NEC and ampacity considerations I also used the 80% rule in which case I never designed for more then 80% continuous load when I sized the conductors. In other words if the ampacity was say 20 amps, my max sustained continuous full load current was not to exceed 80% of 20 or 16 amps NOT 20.

Love sparky chattin with ya

John T Longggggggg retired Electrical Engineer and a bit rusty but still believe Ohms Law to be valid lol

An example of what voltage can do to lower work out put.

I have a spot welder that I bought used. It has maybe an 8' rather light cord feeding it. I had it on a 20 amp circuit, about 50' of #12 wire.
I use it building sheet metal projects, 18 to 30 gauge, mostly 26 gauge.
It has always worked well. Sometimes it would trip that 20 amp breaker. Sometimes when using it a lot the tongs would get rather hot, so I would dip them in a 5 gallon pail of water.
A few years ago I was repairing the fenders on my 20' flat bed trailer. The fenders are maybe 18 gauge, and I wanted to fasten some 1/8th inch angle to the flat side of the fenders. I figured it would take some doing to spot them but I would give it a try. I plugged a 20' #10 extension cord in right at a sub panel.
I was surprised and very pleased that the spot welder worked very well.
We moved to a new place, and I built a new shop. In this shop I used #10 romex on 20 amp breakers to feed the receptacles.
Now when using that spot welder I have to be carful or it burns holes in the sheet metal.

Dusty

Ampacity is a new term around here. Course my profession was electronics and interfacing with power sourcing rather than managing the sourcing. Maybe that is why the term is not that common till lately. However, all that has to be taken into consideration and surely Ampacity is a good way of indicating that there is more to amps flowing down a conductor than coulombs.
I too enjoy these exchanges as I have been retired since Jan. '05 and forgot a lot. The mind stimulation is comforting.

Mark

Way back when during my practice years involving AC Distribution (I was more into AC power then electronics), we had to attend NEC classes every so often where I had instructors the likes of Joe McPartland and Mike Holt. For sure the term "ampacity" was what was used then. I still enjoy sparky chat.

John T