Voltage Drop Formula

Look up or measure the resistance of the cable. Typically, conductor tables list the DC resistance at 20C in ohms/mile (or 1000 ft, ft) along with the AC resistance in the same units at 25, 50 and 75C. Select the resistance per applicable unit length and apply Ohm's Law.

Single phase voltage drop:

Vd = 2K x L x I / Cm or Cm = 2K x L x I / Vd

For both examples: Vd = voltage drop, K= 12.9 for resistance for a copper conductor, L = length of circuit in ft., I = current or amps of the load, Cm = area of the conductor in Circular Mills.

Three phase voltage drop:

Vd = 1.73K x L x I / Cm or Cm = 1.73K x L x I / Vd

The first examples give you the voltage drop if you have the length of the conductor, the current draw of the load in amps and the size of the wire in Circular Mils.

The "or" examples give you the size of of the wire in Circular Mils if you have the length of the conductor, the current draw of the load in amps and the voltage drop you are considering.

I think that the NEC allows a 3% drop for feeders and a 5% drop for branch circuits, but I would have to confirm this in the handbook.

Good Luck!

Jeff

The previous example is fraught with danger because it mixes measuring standards. You should NEVER MIX MEASUREMENT STANDARDS just ask NASA what happens when you do. Mixing measurement standards is a good way to loose a billion dollar spacecraft and a few thousand man years of work.

Here is a link that has all the information you needed to calculate resistance in cables.

Have a read of it and if you have any queries post them here.

I don't see any mix of measurement, unless you consider CM (circular mils ) to be centimeters

Cm = area of the conductor in Circular Mills.

This is a pretty atrocious statement all over but at any rate mili is the metric or SI prefix meaning one thousandth of multiply by 10^-3. The other thing is that you normally express area in square units not circular units.

My point is that the whole thing is open to being interpreted incorrectly because the units aren't stated in anything like a consistent way. Without getting into the arguments of which standard is best what ever you do it is critical that you stick to one standard throughout any calculation. NASA tried to mix units and it lost the a Mars explorer spacecraft worth something like a billion dollars.

Correct

But, depending on your location, the CM will be given in the measurements units you use. The poster with the formula is in the states, we still use the english system, the NEC is our code, based in English.

The NEC codebook gives tables, where all is calculated in the English system.

If the poster reside where SI is standard, they would naturally know that the distance etc are in SI

T Man is correct.

Here in the USA, we have to configure all electrical installs based on data that the NEC (National Electric Code) compiles throughout the years and publishes every 3 years. The next volume arrives in 2008, if you are interested.

All of their data is actually based on years of research, catastrophic and minor failures, good engineering practices and some input from our friends at the NFPA (National Fire Protection Association).

At times, I have had a few debates with state and county electrical inspectors concerning a few installations and the only thing that we agreed upon was "to dis agree". But at the end of the day, the inspectors are in control of passing or failing electrical installations based on data from the NEC. "At the end of the day", this is the formula that you would use.

With that said, the NEC is our "bible" here in the USA and this is what we have to comply with..........and that "bible" is where the voltage drop formula is posted..........if you are interested.

If further explanation of the formula is required, please do not hesitate to post here and I wll expand the details of the formula and provide examples.

Good luck and enjoy reading the NEC.

Jeff

Hi Jeff & T Man,

I understand all that and a similar system is used in Australia except the book is published by the Australian Standards Association and has an ASA-xxx number depending on which standard you are consulting.

The thing that seems strange is the line

Cm = area of the conductor in Circular Mills.

Which is the part I find confusing. It says circular mills and milli is a multiplier that needs to have a unit associated with it like millimeter, milliliter, millivolt, milliamp, milliwatt, etc. Since area is normally measured in square units i.e. mm², cm² m², in² etc. what dose the Circular Mills refer to? Is it the diameter of a circular conductor or the cross sectional area and what units dose it refer to mm, cm, in etc?

What I am trying to point out is that it is important to remember that anything you engineer may not always end up in the USA and you must always allow for others that don't use imperial units. Remember there are only three countries in the world that use imperial and two of them are tin pot regimes. Using the imperial units if fine and I don't have a problem with that but it's important to state the units to avoid confusion exactly this.

This is exactly how the Mars explorer spacecraft was lost . The American programmers were suing imperial units and the European engineers SI. and a variable was passed from one program to another and the units didn't match up. The result was an incorrect deceleration burn and the loss of the craft. By suffixing any number with the units it avoids all this confusion.

Unit conversion can and has been an issue with the stuff I do. Fortunately for me, I was taught both english and metric in school, when the anticipation was that the states were going metric. That was 76. In college everything was SI, but since I work in Power, everything is still in english, that is why I keep that faded, torn, laminated conversion chart.

I still want my furlongs, slugs and inches.

Have a great day guys

So what dose the Circular Mills actually mean and how dose it relate to cable size? I have played with know figures and the SI units in an attempt to back track but I am stuffed it I can figure out how it relates to cable size.

used for the area of a round conductor, since conductors are not square, you cannot use square inches, meters, etc

Area = Pie R squared

the mil is a unit of measurement for length and is related to the inch by

1 mil = 1/1000 inches

1000 mils = 1 inch

Hi Guys,

I have been comparing the SI method formula from Wikipedia and SchoolScinece and there seem to be a discrepancy between the two methods that means the two methods show a difference of 1,000 fold

Ok lets have a look at the calculations for a 1 cm (0.394 inch) diameter copper cable that is 300 m (1,000 feet) long. The SI system gives

R = ρ Length ÷ Area

So for our copper conductor

R = 1.72 x 10-8 x 300 ÷ (π x 0.005m²) = 0.065699 Ω

R ≈ 66 mΩ

Now using your formula of and converting the measurements the length becomes 1,000 feet and the diameter is 0.3937 (394 Mills according to your posts) inches give us

R = 2K L ÷ Cm

R = 2 x 12.9 x 1000 / 394 = 65.44822 Ω

R ≈ 66 Ω

Now I have checked on two sites and the resistiveity of copper using SI units is definitely 1.72 x 10^-8 Ωm so what could cause the calculations to differ by a factor of 1,000? The only thing I can think of is that mills actually means millionths of an inch not thousandths of an inch which just confuses things even more because milli with SI means thousandth.

Something else that I find strange is that if you only use it as a diameter it falls over if the cable is stranded or not square like in a bus bar. I presume that the constant need to change with shape as well as material and to me this is very messy. It seem so much more logical to use the SI system of conductivity and calculate the cross sectional area.

Anyway getting back to the calculations do you agree with the calculations and if not why?

Using SI 300m length Diameter is .01 m

Area = (PI*.01^2)/4 = 8E-5

So R = p given in centimeters = 1.723 E-6 for meters = 1.723 E-8

R = (1.723E-8*300)/ 8E-5= 66m Ohms

For Imperial

P =10.37 (CM*Ohms)/ft 394 MILS

A CM = 394 Mils ^2 = 155236

R = (10.37* 1000)/155236 = .0668

66 milli OHMS

The numbers check out.

There are two differnt p for Copper

1 in SI, 1 in Imperial, you cannot take one and convert the other.

Hope this solves this issue.

Hi T Man,

Area = (PI*.01^2)/4 = 8E-5

Your equation uses the diameter of the cable 0.010 m instead of the radius 0.005 m but I guess this is a typo because the area is correct.

The second part of your answer is the imperial form of the one I used and yes I do realize that the conductivity constant varies with measuring systems but you are not using the equation that was given in post #2 which states

Vd = 2K x L x I / Cm or Cm = 2K x L x I / Vd

For both examples: Vd = voltage drop, K= 12.9 for resistance for a copper conductor, L = length of circuit in ft., I = current or amps of the load, Cm = area of the conductor in Circular Mills.

So what you are saying is that the phrase Cm = area of the conductor in Circular Mills. Means to square the diameter of the cable measured in thousandths of an inch. If that's the case then you need different values for conductivity if the cable isn't round. Another thing if you take the figures given in post #2 and divide by current to give resistance rather than voltage drop you get

Resistance_Round = 2 x 12.9 x 1,000 ÷ 394² = 0.0374 Ω

Resistance_Round = 37 mΩ

Which isn't the correct answer.

The idea of having different conductivity constants for different shaped conductors is not one of the most intelligent ideas I have heard of. If you are worried about people not being able to calculate the cross sectional area of a conductor then include a table that shows cross sectional areas for different shapes but changing the conductivity constant sort of defeats the purpose of have a conductivity constant for a particular material in the first place.

You know I really hate working with imperial units, it seems to always end in confusion so I think I will just avoid it completely from now on in and stick with SI. No more giving answers in both, you will either need to learn the SI system or do your own conversions. (Just joking)

I think it is just a matter of division. The whole problem is divided.

Doesn't matter anyway, just a good exercise on watching your units, keeping them constant, and making sure your units are the same so your billion dollar Mars Explorer doesn't burn up.

I need to get down to Australia, never been there.

have a beer on me, mate.

I hope you have a great day

No worries, I will have a cold one waiting for you. If you do make it down under give me a shout and I will show you round Sydney.

When you plan your trip keep in mind Australia is about the same size as continental USA except there are only 20 million people here and most are in the 8 capital cities. I used to live in Adelaide but half of my work was in Darwin. That meant a 3 hour each way commute in a jet over totally unpopulated desert ever couple of weeks.

Seriously though it's a great place full of thing that will blow your mind so if you get the chance I would definitely recommend spending a few weeks here at least.

Resistance_Round = 2 x 12.9 x 1,000 ÷ 394² = 0.166

That gives two differences from your 66-Ohms: factor two for single-phase formula giving sum of live and neutral resistances.

Use of 12.9 instead of 10.37 - different temperature, or what?

BTW, this is a case ("Circular mils") where a unit was invented for use where/when measuring wire dimensions with callipers, calculators were not available, and wiremen needed to be able to check easily. Personally, I prefer SI any day - or at least a coherent set of units. Horses for courses?

Masu...

I think the concept was invented by non-mathematical bureaucrats to remind dummies that wires are round.

From a mathematical standpoint, the cross sectional surface area of a wire should be measured in square mills (.001 inches²), whether the wire is square, round, heptagonal, or in the shape of any irregular polygon. Dummies don't realize that the "square" has absolutely nothing to do with the shape of the wire. It has to do with

pi r squared.

Somehow, it became a "standard" over here, so we live with it. Why could they not have called it "beer"?? I think it would be alot more fun to work with.

Bill

Remember that the formulas rely on area.

Area for a circle is PI *R^2

For a Buss bar you must two step it

Say you have 5 inch wide bus, 1/2 in thick by 3 feet

The A CM is 5.0 in = 5000 mils

1/2 in = 500 Mils

A =(5000*500) =2.5 E6 sq mils

So 2.5E6 (4/PI CM)/(1 sq mil) = 3.185 E6 CM

Thats why they have tables, so you can forget what you learned in an entry level electrical class

Have a great day guys

Hello Mr. T MAN

i saw u r advertise so many times n i followed to them i wil b very happy if you please give me voltage drop formulas for underground & aboveground so this wil help me in future. I am new in this field just started.

please guide me.

Regards

E-MAIl - saifahmed.syed@hotmail.com

Dear guest,

A little word of advice.

Stop being lazy and give up using those annoying abbreviations when posting messages on CR4.

While it might save you a few seconds it wastes several minutes for the 10,000 plus CR4 participants that have to read it and many will skip over and not even bother to read posts like this.

CR4 is a fantastic web site for engineering type queries and can be very helpful so please put in that little bit extra and type your posts in full.

A circular mil is a unit of area, equal to the area of a circle with a diameter of one mil (one thousandth of an inch). A circular mil is a unit for referring to the area of the cross section of a wire or cable with a circular cross section, as the area in circular mils can be calculated without reference to pi.

Remember there are only three countries in the world that use imperial and two of them are tin pot regimes.

I am just curious -- What is a " Tin pot regime" and who are they?

A country that has yet to convert to Aluminum cans for Beer.

Thank You very much that surely satisfies my curiosity.

A circular mil is a unit of area, equal to the area of a circle with a diameter of one mil (one thousandth of an inch). A circular mil is a unit for referring to the area of the cross section of a wire or cable with a circular cross section, as the area in circular mils can be calculated without reference to pi.

Vd = 2K x L x I / Cm so my interpretation of this formula is (voltage drop =2xK (12.9)xL(Distance of cable run)xI(load amperage)/(divided by?)cirmils?

The NEC allows a 2 percent drop to the feeders and 3 percent for branch circuits, 5 percent overall

210.19 ex:2 fpn 4:Maxtotal Vd on both feeders and branch ckts does not exceed 5 %

please send correct formula for correct voltage drop in single phase & three phase,Weather we are using Al or Cu cable

Interesting reading guys. Could anyone please help me with the mathematical calculation of voltage drop for a 3p 4 w busway with plug in units installed along the length.

I generallly use the mv/amp/metre voltage drop specifed my the manyfacturer and arrive at the voltage drop using the load current and length of bus.However with tap offs which take current off the main bus mess up my calculations. Help.

hello my frend, i have a litle idea how to calculate the voltage drop that i can share to you...

here is the formula, you should take first take the current by this formula

I = P(watts0) / 380 (1.732)(0.8 p.f) this for three phase

V.D, = Mv (as per table ) L (longest route of cable used) I (current) / 1000

V.D. % = (V.D. / 380 ) x 100

Here is a tough question that, so far, I have not found a credible reference.

In the voltage drop calculation

VD = (2 x K x L) / CM

Where does K come from? I have values that, so far, people have provided, but no reference where the values come from or an equation to find the value. Not even my EE books from college mention K or this version of Voltage Drop anywhere.

Any help is appreciated.

Gonzales FPE - I believe the "K" is a constant that varies depending on whether you're using aluminum or copper conductors. I haven't taken the time nor have I seen an explanation of how this constant is actually derived.

Tom Henry has an excellent video on this he explains how to find 12.9 for copper and 21.2 for aluminum for your "K" factors. Basically you would go to Table 8 in your code book. At the top be sure you are under the correct heading. On the right side of the page there is a Copper (Uncoated & Coated Column) and an Aluminum column. You will most all of the time use the uncoated column under the copper column for copper. Aluminum is self-explanatory. As you can see the tables are based on Ohms per 1000' and km. We will use feet. On the left side of the table are the wire sizes. Go down to 1000 wire size. Lay a ruler under that section all the way across. Look down the uncoated Copper column and look down the Aluminum column. For the copper you will see .0129 and the aluminum you will see .0212. Those are the answers. REMEMBER though that the table is based on 1000'. Move your decimal over. Voila. 12.9 for copper and 21.2 for aluminum. Tom Henry explains it way better.

That's the way an electrician would work it out.

On the other hand any engineer worth his money should be capable of working it out from the basic SI resistivity and temperature coefficients which are:

Cu 1.72 x 10^-8 at 20 °C temp coefficient 0.0039 °K^-1

Al 2.82 x 10^-8 at 20 °C temp coefficient 0.0039 °K^-1

You're right, I am an electrician. ( A Licensed Master Electrician) As far as theories are concerned, the "KISS" theory is my favorite. Do all of your math and equations if you'd like.

Even as an engineer, your designs still have to meet and/or exceed the NEC.

My last post was really a bit of a wind up and is a dig at both engineers and trades.

There is nothing wrong with trades and installers utilizing standardized tables to ascertain what specific item or component is required for a given installation.

However, engineers should be able to do the calculations necessary to do it from first principles rather than just referring to some arbitrary table. Unfortunately from the sort of posts that are turning up on CR4 from so called engineers this ability is being either not taught or forgotten.

I don't know how it works in other countries but in Australia we have the usual standards that govern electrical installations that like the material you referred to have similar tables, charts, graphs etcetera that can be used to calculate various parameters. However, it is possible for engineers to specify the use of materials and items that fall outside the norm provided they can show through calculations from first principles that the ultimate result will not fall outside the specified safety standards.

The other problem you have with standardized tables is their regional applicability. You can often get yourself in real hot water using tables for one area in another where they do not fully apply.

One example of regional variations are the US 60 Hz 220 Vac and Australian 50 Hz 240 Vac domestic power systems. On the outset they don't look to be greatly different apart from the 10% increase in voltage and slightly lower frequency. However the US system utilizes a split phase system where each line give 110 Vac to ground and is 180° out of phase with the other line. On the other hand the Australian system uses a three phase 240 Vac to ground system with each phase 120° out of phase with the others giving you a 415 Vac phase to phase voltage. In the US system you will get a 110 Vac belt from either line but in Australia you get a 240 Vac belt but only from the active as the neutral line is tied to ground.

Anyway, I waffling again and there is nothing wrong with utilizing tables but the ability to work it out from basic principles is always nice to have and if you put the two together you end up with a much more reliable result.

Here (USA), we size wire based on the ampacity. We also have our derating lists for ambient temperature, and numbers of current carrying conductors installed in the same raceway.

You are correct. After having lived in Germany for a few years, I perfectly understand what you're talking about with the voltage and hz difference. The voltage does affect the ampacity.

I was just trying to answer Youngin as to one known location the K factor is derived. You also have another way for him to get the answer. Like you said, if he uses them both, the results are more reliable.

By the way, Youngin, the K factor is not a constant. If you are attempting to size the wire for the circuit and don't know which wire size to use, then 12.9 Cu and 21.2 Al.

If you already have the wire size, use the K factor given for the appropriate wire size. (A #12 THHN has a lot more resistance than a 4/0 THHN.)

so with all the comment you have and all the debates about CM, what is the formula of voltage drop? in complete details

G'day ampirahe,

I don't know how others do it, but here's how I would go about the problem.

1. Resistivity: The first step is to find out the resistivity (ρ) for the conductor.
2. Area: Next up you need to find out the diameter of the cable and use it to calculate the cross sectional area (A_C) of the conductor
3. Length: The next step is to calculate or ascertain the length (L_C)
4. Current Draw: The next factor you need to find is the current (I_C)the cable will be carrying.

Now that you have all the information you need to calculate the resistance (R_C) of the cable by using the following formula;

Now you know the resistance of the cable you can calculate the voltage drop (V_C)using good old Ohm's law as shown below;

However, you have to keep in mind this will only give you the voltage drop on one conductor. If you need to calculate the total voltage drop on both conductors you will have to multiply the result by two.

Examples are always better so lets have a go. We need to know the resistance and voltage drop over a cable that supplies 10 A to an outbuilding that is 30 m away with a cable that has conductors with a diameter of 1.00 mm.

1. Resistivity: For copper the resistivity ρCu is 1.68 x 10^-8 Ωm
2. Area: Given the diameter of 1.00 mm the cross sectional area would be;

3. Length: Since we are talking about an out and back conductor we need to multiply the length of 30 m by 2 to give the total length of 60 m.
4. Current Draw: This is simple in this case as it's 10 A, however, in the real world where loading varies with time you will need to use the maximum current draw.

Calculating Resistance Now w calculate the resistance as shown below;

Voltage Drop is now a simple matter of applying Ohm's law as follows;

Ok, so the example wouldn't be acceptable in the real world as the voltage drop would be too high, but it was only meant as an example.

There are a couple of points that I thing need to be emphasized. First of is using normalized units. You will note that when I calculated the area I expressed it as 785 x 10-9 metres² rather than 0.785 millimetres². The reason for this is that unless otherwise specified constants like resistivity are expressed in Whole base units like metres, Amps, Volts, etcetera not scaled units like millimetres, milliamps, kilovolts, etcetera.

Secondly, since we know the resistance of the overall cable you can use Ohm's law to calculate the voltage drop for different current draws rather than having to redo the calculation from scratch or looking it up in a table.

Does that help?

Regards, masu

want to know why this formula is used.... with sin and cos both......?

G'day guest,

You asked:

"want to know why this formula is used.... with sin and cos both......?"

I don't know where the references to trigonometric ratios came from so I guess you are referring to the equations used to calculate the resistance and then the voltage drop.

The first one:

is used to calculate the resistance R_C of a conductor given its conductivity ρ_C and cross sectional area A_C and length L_C. If you want to know more you may find this link to the Wikipedia article on resistivity worth reading.

The second equation:

is just good old Ohm's law and is used to calculate the voltage drop V_C given the current I_C and resistance R_C that was calculated from the first equation.

Is that what you wanted to know or is it something else?

Regards, masu