# How to figure the voltage drop at different points along a circuit?

• alan2023
alan2023
TL;DR Summary
How to figure the voltage drop at different points along a single circuit run connecting multiple loads along the way?
I'm trying to determine how to measure mathematically the voltage drop at various points along a circuit with loads at different points along the way. Let's assume all loads on 100% when the circuit is on. For a basic formula I use (for single phase): 2 * K-Value (12.9 Cu) * Amps * Length(ft)/Circular Mils of conductor. This works great for one load at one distance. How do you figure the wire size and voltage drop when it's one circuit with multiple loads along the circuit at various distances from the source? What I tried was just calculating each load's voltage drop at its distance. Then I added all the drops together to determine total drop and adjust wire size accordingly. My gut says I need to consider the other loads "upstream" from the load I'm calculating, as these are affecting the total amps at that point on the circuit. Am I overthinking this?

Welcome to PF.

Are you asking about how to factor in wire losses when connecting to multiple loads that are spaced at different distances from the source?

If so, that depends on the input characteristics of the loads. If they are purely resistive, then it's a pretty straightforward calculation with the wire segments replaced by their resistance.

If the loads are more like constant-power loads (like switching power supplies), things get way more complicated. A constant power load has two stable operating points in such situations: A high input voltage and low input current operating point, and a low input voltage and high input current operating point. If you have too much wire resistance so that the voltage from your source is drooping down near half of its original value for loads near the end of the wire, then those loads at the end will snap from the first operating point into the second operating point, which drags down the voltage on the wire even more. This can cause the whole power distribution network to essentially crowbar into current limit.

Let's say for this example all loads the same resistive load just different distances. All on at the same time.

The resistive loads will draw current, I = V/R, but the voltage will fall along the line.

Unless equal resistive loads are equally spaced along a ladder network, you will need to solve a set of simultaneous equations using matrices.

Alternatively, get a copy of SPICE to solve the network for you.

berkeman
Then just substitute resistors for the wire lengths and the loads and solve the KCL or KVL simultaneous equations.

(Oops, Baluncore beat me to it...)

 Cir. # Phase (1 or 3) (2 for 1 phase; 1.732 for 3 phase) COPPER/ALUM. (12.9/COPPER,21.2/ALUM.) Amps Distance (feet) Applied Voltage Size Wire 1 2 12.9 0.35 85 120 12 1 2 12.9 0.35 130 120 12 1 2 12.9 0.35 170 120 12 1 2 12.9 0.35 205 120 12 1 2 12.9 0.35 245 120 12 1 2 12.9 0.35 290 120 12 1 2 12.9 0.35 325 120 12 1 2 12.9 0.35 365 120 12 1 2 12.9 0.35 400 120 12 1 2 12.9 0.35 440 120 12 1 2 12.9 0.35 495 120 12 1 2 12.9 0.35 545 120 12 1 2 12.9 0.35 580 120 12

That table is from my example. All the same loads. Varying distances.

Was there supposed to be another column for the voltage seen at the load?

That's exactly what I want to figure out mathematically without measuring each one.

Well, you have two lengths of wire carrying the specified current, so just use the resistivity of the wire and the two lengths to figure out the voltage drop in the round trip of wire. Subtract that number from the applied 120Vrms source voltage to see what the voltage is reduced to at the load...

 Cir. # Phase (1 or 3) (2 for 1 phase; 1.732 for 3 phase) COPPER/ALUM. (12.9/COPPER,21.2/ALUM.) Amps Distance (feet) Applied Voltage Size Wire CM V. Drop 1 2 12.9 0.35 85 120 12 6530 0.12 1 2 12.9 0.35 130 120 12 6530 0.18 1 2 12.9 0.35 170 120 12 6530 0.24 1 2 12.9 0.35 205 120 12 6530 0.28 1 2 12.9 0.35 245 120 12 6530 0.34 1 2 12.9 0.35 290 120 12 6530 0.40 1 2 12.9 0.35 325 120 12 6530 0.45 1 2 12.9 0.35 365 120 12 6530 0.50 1 2 12.9 0.35 400 120 12 6530 0.55 1 2 12.9 0.35 440 120 12 6530 0.61 1 2 12.9 0.35 495 120 12 6530 0.68 1 2 12.9 0.35 545 120 12 6530 0.75 1 2 12.9 0.35 580 120 12 6530 0.80

This is the entire table. I just did the math for each load separately. Looks like total voltage drop would be 5.91 if this is the correct method.

Just adding all the voltage drops together in the column "V. Drop"

alan2023 said:
Just adding all the voltage drops together in the column "V. Drop"
Sorry, that makes no sense to me. 12AWG copper wire has a resistivity of about 1.6 Ohms/1000 feet, so for your 495 foot distance for the two sections of wire you will get about 1.6 Ohms * 0.35A = 0.56V of drop. So the 120Vrms source voltage is dropped to 119.44Vrms.

I think we're getting off track. I know there are different methods of figuring voltage drop. You are using the resistivity of the copper wire. I'm using the 2KIL/CM method. That's a topic for another conversation. I guess I really just wanted to understand the science of how the voltage drop could change in different points on the circuit based on its distance from source and the loads after and before.

alan2023 said:
I'm using the 2KIL/CM method.
What's that?

alan2023 said:
I guess I really just wanted to understand the science of how the voltage drop could change in different points on the circuit based on its distance from source and the loads after and before.
As Baluncore and I said, draw your circuit with the resistances and solve the KCL simultaneous equations. If all you have is a source connected by a wire pair to a single load, it is a trivial calculation like the one I did above.

Sounds good, thank you for your time.
Mike Holt Voltage Drop This is a copy/paste from one source on that method. The chaper and tables are referencing the National Electric Code book.
Voltage Drop Using the Formula Method

When the circuit conductors have already been installed, the voltage drop of the conductors can be determined by using one of the following formulas:

VD = 2 x K x Q x I x D/CM - Single Phase

VD = 1.732 x K x Q x I x D/CM - Three Phase

“VD” = Volts Dropped: The voltage drop of the circuit conductors as expressed in volts.

“K” = Direct Current Constant: This is a constant that represents the direct current resistance for a one thousand circular mils conductor that is one thousand feet long, at an operating temperature of 75º C. The direct current constant value to be used for copper is 12.9 ohms and 21.2 ohms is used for aluminum conductors. The “K” constant is suitable for alternating current circuits, where the conductors do not exceed No. 1/0.

“Q” = Alternating Current Adjustment Factor: Alternating current circuits No. 2/0 and larger must be adjusted for the effects of self-induction (skin effect). The "Q" adjustment factor is determined by dividing alternating current resistance as listed in NEC Chapter 9, Table 9, by the direct current resistance as listed in Chapter 9, Table 8.

“I” = Amperes: The load in amperes at 100 percent, not 125 percent for motors or continuous loads.

“D” = Distance: The distance the load is located from the power supply, not the total length of the circuit conductors.

“CM” = Circular-Mils: The circular mils of the circuit conductor as listed in Chapter 9, Table 8.

I'd say this is what separates the electrical engineers from the electricians. If you have a long pair of wires with taps along the length of it, each tap drawing currents, you have to solve this by drawing out a circuit including the resistance of the wire. You will end up with a network of resistors that is solvable using basic circuit analysis.

DV1=Z1*SUM(I1,I2...In)
I1=S1/V2 and so on.
DV2=Z2*(I2,I3...In)
Usually we know Vn and then we may build the voltage line as Vn-1=Vn+DVn
DVn=Zn*In
In=Sn/Vn

If we only know V1, we must first consider all V2 to Vn equal to V1 and iterate the voltage drops as many times as we need to achieve the proposed error limit.

The result may be like this:

All currents are assumed to be equal to 0.35 A and the voltage variation at each point does not produce current change.
If the circuit consists of two conductors of the same type - forward and reverse - we will have to multiply by 2 -in the table I did not!.

Let's consider the power injected in each point is constant P(i)=I(i)*V(i)=K P=120*0.35=42 W
In this case first we take V1=V2=...Vn=120 V
I(i)=42/V(i) and finally -after 2 iterations-V13=115.0779V [the error is 0.000313%]

All currents are assumed to be equal to 0.35 A and the voltage variation at each point does not produce current change.
If the circuit consists of two conductors of the same type - forward and reverse - we will have to multiply by 2 ...
If the supply voltage was 120 VDC; and assuming 12 AWG copper, for both the supply line, and the return line; and the loads sink a constant current of 350 mA DC;
Then 115.212 volts DC would appear across the terminals of the final current sink.

Do the loads sink 350 mA RMS, or a constant current of 350 mA, alternating with supply polarity ?
Or maybe have a resistance of; 120 V / 0.342 A = 342.857 ohm.
There can be many answers to a poorly defined problem.

This is actually a real world install. The .35A loads are LED bollard lights installed along a walkway. The #12 is THHN insulated wire installed 24" underground in PVC conduit. (I believe we actually installed #10 THHN) The 120v is AC nominal voltage,(could be anywhere from 115-125v, let's figure 120v). It's one circuit connecting all the loads. Being 120v, it would be two conductors. One grounded and one ungrounded conductor. (Hot and Neutral) In my table the "distance" is the approximate distance each light is from the source (the breaker panel).

So if given this information, what would be a formula(s) for calculating what we should expect to see for voltage at any point along the circuit knowing the distances and loads before and after?

(I'm tempted to just go there and turn them on and test at different points, but I want to understand the science of this out of pure interest in understanding.)

alan2023 said:
The .35A loads are LED bollard lights installed along a walkway.
Is that 0.35 amp RMS sinewave, or a fixed continuous current?
Do they each have PF control?
Can each lamp be modelled as a 120 / 0.35 = 342.86 ohm load?

Input 120 VDC. End of the line; 115.367 VDC

Baluncore said:
Is that 0.35 amp RMS sinewave, or a fixed continuous current?
Do they each have PF control?
Can each lamp be modelled as a 120 / 0.35 = 342.86 ohm load?
Yes, I believe 0.35 amp RMS sinewave. Just typical 120v AC 60 hz power you would find in the USA.
No PF control.
For simplicity, yes, just figure 120/.35=342.86 ohms.

Using Baluncore's 1m6 ohms per foot for 12 ga, and a constant current draw for each load, the resulting voltage at each load point can be calculated, with an end-of-the line value of 115.212 volts. This is hardly different from the result with a constant resistance for each load.

I suspect the result with a constant power (for constant brightness) load would also be hardly different.

For a constant, polarity dependent, current draw from a supply of, 120 VDC or 120 VACrms, without reservoir capacitor:
Baluncore said:
Then 115.212 volts DC would appear across the terminals of the final current sink.

As above but with resistive loads:
Baluncore said:
Input 120 VDC. End of the line; 115.367 VDC

The Electrician said:
Using Baluncore's 1m6 ohms per foot for 12 ga, and a constant current draw for each load, the resulting voltage at each load point can be calculated, with an end-of-the line value of 115.212 volts. This is hardly different from the result with a constant resistance for each load.
I blame berkeman for the 1m6 ohms per foot for 12 ga.
berkeman said:
12AWG copper wire has a resistivity of about 1.6 Ohms/1000 feet,...
Someone, sometime, had better check that number, and that it is resistance and not resistivity per foot. If we all use that number, we will all agree to six digits, accepting that reality may be different, due to alloy composition, work hardening, and absolute temperature.

The Electrician said:
I suspect the result with a constant power (for constant brightness) load would also be hardly different.
I agree.

A clever PF corrected, LED controller, would supply a constant current to the LED from a reservoir capacitor, that was charged through a rectifier from the supply, with a charge current proportional to instantaneous supply voltage. That is the 120 VDC, or 120 VACrms, resistive (342.86 ohm) load model, giving an end of the line; 115.367.

It seems that actually it is a little more complicate.

The supply voltage is AC -as usual- and a driver AC/DC is provided in order to supply DC. Then 350 mA it could be the DC output [18-42 V].Let’s say Vdc=20 V Pdc=0.35*20=7 W.

In the interval of 115 to 125 Vac the LED current is 350 mA, but the Iac current is variable from 7/115=0.06087 to 7/125=0.056 A neglecting the losses and power factor.

If we consider 0.8 as power factor and 95% as efficiency we get

S=7/0.8/0.95=9.21VA and the currents Iacmin=9.21/125=0.07368 A and 36.87 degrees

Iacmax=9.21/115=0.08 A and 36.87 degrees.

Last edited by a moderator:
The Electrician said:
@alan2023, are you still around?
From their profile page:
Last seen:
Yesterday, 10:03 AM

Yes, I'm still here.

Without getting into all the nitty-gritty with the inner workings of an LED driver/fixture, I believe I'm starting to understand a bit more how the loads after a point on a circuit affect the voltage drop.

Looking at this original question from an electrical install perspective, my challenge to someone would be how to build a spreadsheet to plug in the distances and any other necessary values (voltage, wire size/resistance, amps, etc.). As an electrician, I use custom calculators usually in a spreadsheet to determine voltage drop, conduit fill, cable tray fill. I first understand the code/math requirements, then build the sheet/formulas.

Wanting to add the feature of multiple loads on the same circuit, (which is something very common, and voltage drop becomes an issue in long distance applications like exterior lighting), I believe I could just build the circuits including the wire as a resistor. So basically a combo series/parallel circuit. I know there are probably some great software tools you guys use, but I believe this could built in a spreadsheet and in doing so, I have a calculator to speed up future calculations, but also understand how it works.

Thank you all for the input, and I'm sorry if sent us down a rabbit hole.

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