Voltage Drop Under Load

I've noticed that when 120V 20A AC circuits are heavily loaded, the voltage at their outlets drop somewhat. I thought that in a parallel
AC circuit, voltage remained constant but current changed. What is going on? Can someone point to a detailed discussion of this phenomenon (either in print or on the Web)? Thanks in advance.
-- Jack
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When current flows in a wire there is resistance due to the flow, when there is no current flowing there is no resistance. With no resistance you read the full line voltage... with resistance you read the lower voltage..
said another way, current flow creates the resistance. the more current flowing the more resistance. reducing the size of the conductor increases such resistance. If the conductor is large enough and the current flow is low enough there is no measurable resistance...but there will always be some resistance with current flow.
Phil Scott

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Phil, You do talk a lot of DRIVEL.
The resistance does NOT varey with the current. The voltage drop due to the current times the reistance increases with incrase in current (simple Math).
--
John G

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I don't care which way you say it, that explanation is just plain wrong.
Wire has resistance. Period. It is a function of the wire size, length, material, and temperature. It doesn't vary with current (before the purists jump on this, yes, it will change very slightly with the temperature which is a function of current, but that is insignificant for purposes of clarifying his statement).
When current flows through the wire resistance, it develops a voltage drop across it, which depends on the current by Ohm's law. Therefore, the voltage at the load is lower than at the supply. The more current, the lower the load end voltage.
Ben Miller
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Benjamin D. Miller, PE
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Thanks Ben You probably saved me from the Temp Coefficient flames which I left out because as you say it is insignificant to this discussion.
--
John G

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Wire no matter how fine, even 1 angstrom, microscopic in diameter has no resistance what so ever if there is no flow.
Sorry, thats just how it is. Resistance is a function of how many electrons you are trying to push through the wire (and the temperature of the wire... super cooled wire has very little resistance).
It is a function of the wire size, length,

You have right but you have your language confused... and you have missed on the issue the original poster was asking about..which is why does the voltage drop with a load applied. Indicating that he is getting a higher voltage across the feeders with no load, which will virtually always be the case.... than with a load applied, which voltage then is reduced by virtue of the resistance, which ohly shows up with current flow.
You are making an incorrect assumption that resistance stands separate from current flow. Ohms law which you cite is clear on that... drop the volts and amps (current) to zero....and resistance drops to zero.
Clear enough...no?
its in the algebra also.
Phil Scott

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If resistance is a function of "how many electrons you are trying to push through the wire", I must ask, how many electrons do I have to push through a 100 ohm resistor to make it 100 ohms?
Mike
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I think that is a superb question... you can also check the resistance of a motor winding with a hand held ohm meter using only 9 volts DC current... you can get an ohm reading of say 3 ohms on the run winding, using just the 9 volt meter...and that 3 ohms when plugged into Ohms law formula's along with the voltage supplied will give you the accurate measureable amp draw.... so the resistance appears to be completely independent of the current through the circuit in that case as you are saying.
So I am not in disagreement with you at all.... I think there are some semantics issues though..
and like you said you can purchase resistors with resistance coded on them.
Using ohms law however you can notice that the resistance in any *circuit is a function of voltage and amperage... if you input zero for both voltage and amperage, the formula will produce zero for resistance...because of course there is no flow to resist... I think it is this sort of application we are discussing when it can be seen that voltage drops a amperage on the line increases beyond the rated current carrying capacity of the feeders
same with house current at a 15 amp rated receptical for instance...it might read 110v with a test meter plugged into the receptical, and a lot less with a 15 amp electric heater plugged into it. you can see this effect in an inadequately wired house, as more lights are turned on, others dim... voltage drop induced by increasing resistance in the feeders.... that resistance a direct result of current flow.
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Phil Scott

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I have a problem with this-- A.C. motor, D.C. motor it doesn't matter, once it starts spinning you have inductance resisting A.C. or pulsing D.C. So I doubt your static D.C. measuement would be an accurate representation to give you a resistance to calculate the current draw of a motor.

Yes, you say there is no resistance without current flow, I say if a tree falls in the woods and no one is present, it still makes sound.

If you'll notice, ohms law is very linear, all the way to zero but not zero. There's just somthing funny about doing calculations with zero.

You don't need to go "beyond rated current carrying capacity of the feeders" to see the voltage drop caused by the resistance of the wire.

The voltage will drop as more current is drawn even in an adequately wired house, it's just a matter of degree.

The resistance doesn't increase because of increased current flow, it's the voltage drop that increases. Mike

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Those would be locked rotor numbers based on the static resistance of the motor windings alone... once it got going then there would be impedance I believe as you assert.... and the running amps would be much lower.

I think the issue is not that the actual resistance NUMBER goes up, but that when the amperage rises, multiplied by the static resistance the total wattage of resistance in the wire increases, it gets hotter.
As the wattage lost to heat in the wire goes up, with resistance stable, the voltage in the wire must go down.
This algorithm may or may not be entirely relevant to that. http://www.angelfire.com/pa/baconbacon/page2.html

that has to be predominantly correct... so what happens then when you reduce the load or increase the load, will not the wattage lost in the wire change as the current in the wire changes...and that wattage is derived from volts x amps...so as amps rise in the wire of constant resistance, voltage must drop..... is that not the basis of ohms law as applied to both loads and losses in any circuit?

agreed... the resistance number remains exactly the same..so resistance does not incease as I suggested.... the heat loss in the wire increases, and for that to happen voltage must drop as amperage increases... I was attributing that correctly to resistance in the wire, but resistance had not increased just the wattage lost to the resistance had increased...
Phil Scott
Phil Scott

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----------------------------

------------ Ben Miller has it right. Resistance is a property of the material.
Please note that, for a pure ohmic resistance, ignoring thermal effects we have R=V/I. You are saying that for I= 0 then R= 0. However for a non-zero voltage V the relationship implies that R must be infinite if I is 0. This is contradictory to your statement. If the voltage is also 0 then the statement becomes R=0/0 which is NOT necessarily 0 and could be any value (and of no interest) as 0/0 is undefined.
Now you give an example of a motor which measures 3 ohms with a hand held ohmmeter - which does NOT have a 9VDC current-but a 9VDC voltage source behind an internal resistance- the current is a few ma. The actual current drawn in the case of a motor is not determined by the resistance of 3 ohms but by the mechanical load on the motor as reflected to the electrical side- the motor presents a generated or back emf in series with the winding resistance- it is an active load. The ohmmeter result does not allow determination of the current by Ohm's Law because, with an active load, such as a motor, Ohm's Law is useless garbage. Ohm's Law, strictly speaking, is true for a motor only when the shaft is locked so it cannot turn. In such a situation, the motor is useless. I suggest that you review Ohm's Law definition and applicability.
Your last paragraph has poor math as 0/0 is not defined. It could be 0 or could be infinite or anywhere in between. Secondly, the last statement in this paragraph appears somewhat garbled. Any current, whether above or below the rating of the conductor will cause a voltage drop between source and load. The larger the current, the higher the drop - not because the resistance increases but because there is a V=IR drop in the conductor even for R constant- leading us back to the original statements given correctly by Ben and others.
--

Don Kelly @shawcross.ca
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Resistance is a physical constant of the wire. You can look up "resistance per foot" in reference tables for a given size wire at a given temperature. You can have wire-wound resistors sitting on your bench, with the resistance clearly marked on them. You can measure the resistance of a wire with 1 mA flowing or 10A flowing, and it is the same!

I didn't miss it. I explained it. The resistance is there, in the wire. Voltage drop occurs when current flows.

I must have slept through that class. As I have explained several times, resistance DOES stand alone.

Not clear at all. Could you cite some reference material that explains your theory?
Ben Miller

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Benjamin D. Miller, PE
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That is surely the case. there are many examples.. on the other hand resistance according to Ohms law is zero when voltage and amperage are zero.... and we can see voltage dropping on an overloaded parallel lighting circuit as fixtures are added and the wattage starts to exceed the current carrying capability of the wire... that was the OP's original question... then I spun it to resistance...perhaps incorrectly... as the extra light bulbs were added...
... then the resistance would have dropped in the *load circuits (the combined light bulbs), causing more amps to flow in the supply circuit... more amps than the size of wire would permit without a voltage drop...that is also in the NEC tables, voltage drop with increasing amperage on a certain size wire.
the only thing that will cause the voltage to drop as I see it (and I could be missing something here) is resistance in the wire... that is also seen in too small house feeders...low voltage at the house under full load etc.
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Phil Scott

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Phil said, "more amps than the size of wire would permit without a voltage drop..." Wire has X amount of resistance per length and the current times that resistance equals the voltage drop. EX. A fifty foot run of #12 wire with 10 amps flowing. 50 feet must be multiplied by two ( the current must return to the source) So 100 feet of #12 wire has about .16 ohms of resistance. .16 ohms x 10amps = 1.6 volts That 50 foot run would have a voltage drop of 1.6 volts with a 10 amp load.

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amdx wrote:

Okay. Thanks to everybody who answered my question. I have a much clearer picture of what is going on. I'm going to start taking a space heater around with me so I can measure the length of the runs to the outlet I'm at just for the fun of it :)
-- Jack
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source)
load.
Do you know the wire size? Also you need to know the voltage at the circuit breaker panel with the heater plugged into the outlet and then at the outlet to calculate the wire length.

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All he needs to do is measure the voltage at the receptacle without the heater, and that is the voltage at the panel. Then plug in the heater and note the value under load. The difference is the voltage drop.
Ben Miller
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Benjamin D. Miller, PE
B. MILLER ENGINEERING
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Yes, that will work, good call. ( although, if you know the voltage is equal at both ends, then you know nothing else is drawing current from that run.) Mike
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Jack Nerad wrote:

It's not quite that easy! You need to know the resistance of the space heater, since it will not draw the same current at which it was rated, with any voltage drop. It's resistance is its rated voltage squared divided by its rated wattage. This will be a constant, (in spite of what Phil Scott says), neglecting the slight difference in temperature at the lower voltage. Then, the resistance of the wiring to the outlet is simply the difference in the loaded voltage and the unloaded voltage, divided by the loaded voltage, all times the heater resistance.
In the USA, almost all outlets are wired with #12, (20A), or #14, (15A) wire. Copper wire of these sizes has a resistance per thousand feet of 1.62 or 2.57 ohms. You can calculate the run lengths from that.
--
Virg Wall, P.E.

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Thats rather clear...thanks... the voltage drop is a function of the static reistance and the amps.
Phil Scott

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