VD Formula Question

[Jerramundi]

A question about Voltage Drop for you boys.
My inner perfectionist tends to nitpick things so I have ask to ensure I'm not crazy (n.b. the latter may be true).

Would I be correct in saying that the way the Voltage Drop formula(s) are structured, that we are effectively placing the Load Current at the END of the circuit? (i.e. "END" meaning the distance from the source to the load and back to the source).... as opposed to at the actual load itself (i.e. "a one-way trip").

Borrowing an example from a Mike Holt Newsletter on Voltage Drop Calculations...

What is the voltage drop of two No. 12 conductors that supply a 16 ampere, 120 volt load which is located 100 feet from the power supply (200 feet of wire)?

(a) 3.2 volts (b) 6.4 volts (c) 9.6 volts (d) 12.8 volts

Answer: (b) 6.4 volts

Voltage Drop = I x R

“I” is equal to 16 amperes

“R” is equal to 0.4 ohms (Chapter 9, Table 9: (2 ohm/1,000 feet) x 200 feet

Voltage Drop = 16 amperes x 0.4 ohms

Voltage Drop = 6.4 volts, (6.4 volts/120 volts = 5.3% volts drop)

Operating Voltage = 120 volts – 6.4 volts

Operating Voltage = 113.6 volts

Would it not be more accurate to say that the "Operating Voltage" of the load itself only sees HALF of the drop that we are calculating here?
I understand that this is "just how things are done" and I'd be better off just running with the industry standard, but I have to ask...

I feel like if I were to disconnect the neutral at the source (big no-no, I know) and take a potential difference measurement, I would get 113.6V between the very end of the neutral wire and neutral bus (that much checks out in my mind)...

...but at the load itself, the potential difference there or the "Operating Voltage" would be more like 116.8V?

 

[GoldDigger]

It causes no end of confusion in using online VD calculators, but for a single phase 120 V circuit the voltage at the hot terminal of the load will be reduced by the single wire voltage drop but the voltage seen by the load itself is what concerns us, and the rise in voltage at the neutral is just as large as the fall in voltage on the ungrounded lead.
On the other hand, if the load is a balanced 120V load with equal current on both polarities there will be no offset in the neutral voltage. So each 120V load will see a single wire voltage drop. A 240V load at the same location will see a two wire voltage drop but at double the voltage so the percentage drop will be the same.
A balanced three phase wye load will also see a percentage drop corresponding to the single wire drop.

 

[kwired]

common formula is 2 times length.....

that 2x ends up accounting for path from the source and path back to the source.

lets say you have 100 foot of conductor as net out and back, now you rearrange things somehow so that you have 25 feet out but 75 feet back, but still maintain same size the entire path, the end VD result would still be the same.

If you change conductor sizes anywhere along the way it gets more complicated than the basic formula is, usually need to do multiple point to point calculations and add each segment together to get the net drop of the whole circuit.

three phase we use 1.73 times the one way circuit length - presuming all three conductors are same size and length.

 

[Jerramundi]

...for a single phase 120 V circuit the voltage at the hot terminal of the load will be reduced by the single wire voltage drop

That much I got... and utilizing the load current in the VD formulas makes sense to me at this point in the circuit.
What doesn't make sense to me is why the value of the load current is utilized in combination with the FULL length of the circuit here.

I understand that we're accounting for the resistance across the entire length of wire from the source, to the load, and back to the source... but in my mind, the load current value is only going to see the resistance of a single length of wire before it gets to the load in a single A/C iteration. Obviously, the voltage will continue to drop as the single A/C iteration continues its' journey back to the panel along the neutral... but in this single iteration, I would think that what Mike calculates as the "Operating Voltage" is what you would see across the end of the neutral wire and the neutral bus... and not at the load itself.

 

[Jerramundi]

...the rise in voltage at the neutral is just as large as the fall in voltage on the ungrounded lead.

The "rise" in voltage at the neutral? I view a single A/C iteration across a 120V 2-wire branch circuit as a loop, a full circle from the source, through the ungrounded conductor, to the load, and back to the source along the grounded conductor... with the greatest amount of voltage drop occurring at the very end of that loop in a single iteration. I don't see how you could get a "rise in voltage at the neutral" unless you're talking about the fraction of a second between the first A/C iteration and the next one.

 

[wwhitney]

Say the load is operating, and we measure 120V between the panel bus and the panel neutral bar, and let's define the neutral bar to be the zero point of voltage. And suppose the one way voltage drop to the load is 2V.

Then if the two circuit conductors are the same size and length, the voltages at the load will be 118V on the ungrounded conductor and 2V on the neutral conductor. So the load only sees 116V across it. And as voltage at the end of the neutral is higher than the voltage at neutral bar by 2V, we call that voltage rise.

Cheers, Wayne

 

[wwhitney]

PS Suppose there is a feeder from the service panel to the (sub)panel in the previous post, and suppose the one-way voltage drop on that feeder is 1V (and the two feeder conductors are the same size and length). And suppose we now define the zero point of voltage to be the service neutral bar (a more typical choice).

Then the voltage at the service panel would be 122V for the busbar, 0V for the neutral bar; at the subpanel it would be 121V for the busbar and 1V for the neutral bar; and at the load 119V for the ungrounded conductor and 3V for the neutral conductor. The voltage across the load is still 116V.

Cheers, Wayne

 

[GoldDigger]

That much I got... and utilizing the load current in the VD formulas makes sense to me at this point in the circuit.
What doesn't make sense to me is why the value of the load current is utilized in combination with the FULL length of the circuit here.

I understand that we're accounting for the resistance across the entire length of wire from the source, to the load, and back to the source... but in my mind, the load current value is only going to see the resistance of a single length of wire before it gets to the load in a single A/C iteration. Obviously, the voltage will continue to drop as the single A/C iteration continues its' journey back to the panel along the neutral... but in this single iteration, I would think that what Mike calculates as the "Operating Voltage" is what you would see across the end of the neutral wire and the neutral bus... and not at the load itself.

The load could generally not care less about the nominal operating voltage at its location as long as it is not excessive. But what does affect its operation is the voltage between its terminals, since that is where power is delivered to it.

 

[Jerramundi]

Say the load is operating, and we measure 120V between the panel bus and the panel neutral bar, and let's define the neutral bar to be the zero point of voltage. And suppose the one way voltage drop to the load is 2V.

Then if the two circuit conductors are the same size and length, the voltages at the load will be 118V on the ungrounded conductor and 2V on the neutral conductor. So the load only sees 116V across it. And as voltage at the end of the neutral is higher than the voltage at neutral bar by 2V, we call that voltage rise.

Cheers, Wayne

Correct me if I'm wrong, but one cannot measure both voltage rise and voltage drop during a singular A/C iteration. Whether it's a rise or a drop is dependent upon the direction of current flow, no?

 

[wwhitney]

No, voltage here refers to, say, the RMS voltage. So it is a measure of behavior over a full cycle.

Edit: the examples I gave are equally valid for AC or DC.

Cheers, Wayne

 

[winnie]

Doesn't matter if you call it a drop or a rise.

Because of wire resistance, the voltage of the hot terminal at the load is closer to the 'bus neutral' voltage by the _one way_ voltage drop. At the same time the voltage at the neutral terminal of the load is closer to the 'bus hot' voltage by the one way voltage drop.

In the OP example, if you had a meter with one lead on the panel neutral and measured the hot at the load, you would see 116.8V ... but if you measured the neutral at the load you would see 3.2V. And if you measured hot to neutral at the load you would get 113.6V.

The net result is that the load 'sees' the two way voltage drop.

Jon

 

[kwired]

Correct me if I'm wrong, but one cannot measure both voltage rise and voltage drop during a singular A/C iteration. Whether it's a rise or a drop is dependent upon the direction of current flow, no?

If the supply path is same size and length as the return path both will have same resistance.

Current is constant through the entire circuit, therefore each section of the circuit has resistance and an associated voltage across it.

"Rise" on the neutral (grounded conductor) essentially means if the source end is what is grounded then there will be some voltage to ground on the load end of the grounded conductor.

for basic theory experimentation and explanation we ignore the conductors, and often they are short enough length their resistance is negligible.

reality is they do have resistance and if current is flowing there will be a voltage from one end to the other, when in series with a "conventinoal" load they are minor resistance in comparison to the main load but all the drops in voltage in the circuit must add up to the voltage applied to the circuit - this is Kirchoff's voltage law.

 

[Jerramundi]

that 2x ends up accounting for path from the source and path back to the source.

I get that, but what confuses me is the application of the load current across the entire length of the circuit when performing VD calculations.

I would imagine that as voltage decreases along the length of the circuit due to VD, that current would rise because Watts = Watts. I'm looking at the end of the circuit as pulling a slightly greater amperage than the beginning of the circuit (in a single A/C iteration) because of the inverse relationship of Volts and Amps. That's why applying the load current across the entire length of the circuit in these formulas seems odd to me.

Referring again to the above example. If the 120V load demands 16V and according to these formulas, sees a drop in 6.4V from the source voltage across the load itself, thus operating at 113.6V, the amperage being pulled at the load should be 16.9A... and even higher at the very end of the circuit, correct? I imagine this to be why GFCI's can cause nuisance tripping on very long circuits (i.e. because of an imbalance in current between the ungrounded and grounded conductors) and why fully loading a branch circuit isn't good practice, although not explicitly against the NEC (less instances of chord and plug connected devices).

 

[wwhitney]

Current is conserved, as our work does not involve making or destroying electrons. So if the current at one end of the wire is 1 amp, at the other end of the wire it will be 1 amp.

Energy (Power) is conserved, because the voltage drop that happens is a result of the resistance of the wire. The wire is a long, low power electric heater, the I^2 * R power loss in the wire is converted to heat. The result is that power output from the panel = power converted to heat in the wires + power used by the load.

Cheers, Wayne

 

[kwired]

I get that, but what confuses me is the application of the load current across the entire length of the circuit when performing VD calculations.

I would imagine that as voltage decreases along the length of the circuit due to VD, that current would rise because Watts = Watts. I'm looking at the end of the circuit as pulling a slightly greater amperage than the beginning of the circuit (in a single A/C iteration) because of the inverse relationship of Volts and Amps. That's why applying the load current across the entire length of the circuit in these formulas seems odd to me.

Referring again to the above example. If the 120V load demands 16V and according to these formulas, sees a drop in 6.4V from the source voltage across the load itself, thus operating at 113.6V, the amperage being pulled at the load should be 16.9A... and even higher at the very end of the circuit, correct? I imagine this to be why GFCI's can cause nuisance tripping on very long circuits (i.e. because of an imbalance in current between the ungrounded and grounded conductors) and why fully loading a branch circuit isn't good practice, although not explicitly against the NEC (less instances of chord and plug connected devices).

Current would only rise if there was an inductive load - a motor. So yes if the load in the circuit was a motor then current would rise some because of voltage drop, which would result in more voltage drop - but there is a point where it stabilizes or it would be a runaway with unlimited current possible. Motors try to achieve same output, so if voltage drops current must rise to maintain that output.

If the load were a resistance heating element, then voltage drop in conductors means less voltage at the load and less heat output from the load. Sum of voltage drops in the circuit will always equal voltage applied - also one of Kirchoff's laws.

Ohm's law says the current is proportional to voltage and inversely proportional to resistance. So total current is dependent on total resistance in the circuit, voltage measured across a portion of the circuit will be dependent on current (which will be same throughout a basic series circuit) and the resistance between those two points being measured.

Drop across the conductor(s) is nearly 100% resistance in nature and that alone will not cause any rise in current.

 

[wwhitney]

Referring again to the above example. If the 120V load demands 16A and according to these formulas, sees a drop in 6.4V from the source voltage across the load itself, thus operating at 113.6V, the amperage being pulled at the load should be 16.9A... and even higher at the very end of the circuit, correct?

No. Current in at one end of the wire is the same as current out at the other end of the wire. The "missing" 6.4V * 16A of power is being dissipated as heat in the wire. The supply end of the circuit is providing 120V * 16A of power to the circuit, but only 113.6V * 16A of power is arriving at the load.

[And since this is the one way voltage drop, only 107.2V * 16A of power is being used at the load; the remaining 6.4V * 16A of pwoer is being dissipated as heat in the second circuit coductor.]

Cheers, Wayne

 

[synchro]

I get that, but what confuses me is the application of the load current across the entire length of the circuit when performing VD calculations.

I would imagine that as voltage decreases along the length of the circuit due to VD, that current would rise because Watts = Watts. I'm looking at the end of the circuit as pulling a slightly greater amperage than the beginning of the circuit (in a single A/C iteration) because of the inverse relationship of Volts and Amps. That's why applying the load current across the entire length of the circuit in these formulas seems odd to me.

When making voltage drop and load calculations, for simplicity we assume that the current drawn by a load is a fixed value independent of the amount of voltage drop. This is fine and a standard procedure as long as the voltage drop is kept relatively small like 5% or less.

In reality there are very few loads that maintain a relatively constant current draw vs. applied voltage. With resistive loads if you have a 5% voltage drop then the current draw will be 95% of what it would be without any voltage drop. And so in reality, the actual voltage drop across the conductors will be 95% of what's calculated with a fixed current that's independent of the voltage at the load (and so it'll be a 4.75% drop instead of 5%). This is not of any real significance given the tolerances of the conductors and loads.

Loads like drivers for LED fixtures may draw an approximately constant wattage vs. applied input voltage, along the lines of what you mentioned. This is because they have an internal switching supply that can convert a wide range of input voltages while maintaining the drive to the LEDs at a fixed power level. And so the input power to the driver is also fixed because power is conserved as was mentioned previously. With such a driver the current would increase about 5% if there was a 5% voltage drop. So there'd be a 5.25% voltage drop instead of 5%, again not really significant.

 

[NewtonLaw]

A question about Voltage Drop for you boys.
My inner perfectionist tends to nitpick things so I have ask to ensure I'm not crazy (n.b. the latter may be true).

Would I be correct in saying that the way the Voltage Drop formula(s) are structured, that we are effectively placing the Load Current at the END of the circuit? (i.e. "END" meaning the distance from the source to the load and back to the source).... as opposed to at the actual load itself (i.e. "a one-way trip").

Borrowing an example from a Mike Holt Newsletter on Voltage Drop Calculations...

What is the voltage drop of two No. 12 conductors that supply a 16 ampere, 120 volt load which is located 100 feet from the power supply (200 feet of wire)?

(a) 3.2 volts (b) 6.4 volts (c) 9.6 volts (d) 12.8 volts

Answer: (b) 6.4 volts

Voltage Drop = I x R

“I” is equal to 16 amperes

“R” is equal to 0.4 ohms (Chapter 9, Table 9: (2 ohm/1,000 feet) x 200 feet

Voltage Drop = 16 amperes x 0.4 ohms

Voltage Drop = 6.4 volts, (6.4 volts/120 volts = 5.3% volts drop)

Operating Voltage = 120 volts – 6.4 volts

Operating Voltage = 113.6 volts

Would it not be more accurate to say that the "Operating Voltage" of the load itself only sees HALF of the drop that we are calculating here?
I understand that this is "just how things are done" and I'd be better off just running with the industry standard, but I have to ask...

I feel like if I were to disconnect the neutral at the source (big no-no, I know) and take a potential difference measurement, I would get 113.6V between the very end of the neutral wire and neutral bus (that much checks out in my mind)...

...but at the load itself, the potential difference there or the "Operating Voltage" would be more like 116.8V?

Consider an ideal series circuit that is supplied by 120 volts. The current flow goes through R1 (phase wire) through R(load) (your load at the assumed end of the circuit) then through R2 (the neutral wire) to the real end of the circuit at the source voltage of 120 volts.

This is what you are considering when you do the calculation. If you open the path to R2, then the voltage between the R2 side of R(load) will be equal to 120 volts if measured to R2. This same measurement taken from the R1 side of R(load) will also give 120 volts, thus the voltage across R(load) will be zero.