In Article " Voltage drop calculations Part One ", I explained the following points:
 What is the Voltage Drop?
 Why to calculate the Voltage Drop?
 What are the factors affecting the voltage drop?
 NEC voltage drop Recommendations
Also, in this Article, I indicated that there are eight methods for Voltage Drop Calculations as follows:
1 Ohm’s Law Method,
2 European method,
3 U.S method which divided to:
 Circularmils method,
 Chapter (9) tables method.
5 Quick OnLine method,
6 AmpereFeet method,
7 Lookup tables,
8 Excel spreadsheets.
I explained the first two methods and today I will continue explaining other Voltage Drop Calculations Methods as follows.
3 U.S Method
This method can be divided to two submethods as follows:
 Circularmils method,
 Chapter (9) tables method.
3.1 CircularMils Method
Difference
between CircularMils Method and Ohms Law Method
This method is a
bit more involved than the Ohms Law method, but the big advantage is you can
use it for threephase or singlephase.

Rule#1: Using CircularMils Method for Voltage Drop Calculations
In this method,
the voltage drop is calculated by using the following formulas:
Where:
K =
DirectCurrent Constant.
K represents the
dc resistance for a 1,000circular mils conductor that is 1,000 ft long, at
an operating temperature of 75؛C. K is 12.9 ohms for copper and 21.2 ohms for aluminum.
I = Load Amperes
The load in
amperes at 100% (not at 125% for motors or continuous loads).
D = Distance
The distance the
load is from the power supply. When calculating conductor distance, use the
length of the conductor—not the distance between the equipment connected by
the conductor. To arrive at this length, add distance along the raceway route
to the amount of wire sticking out at each end. An approximation is good
enough. Where we specify distances here, we are referring to the conductor
length.
CM =
CircularMils
The circular
mils of the circuit conductor as listed in NEC Chapter 9, Table 8 (see below image).

Important!!!
For ac circuits
with conductors 2/0 AWG and larger, you must adjust the dc resistance
constant K for the effects of selfinduction (eddy currents). This is done by
using the Adjustment Factor ( Q ).
Q =
AlternatingCurrent Adjustment Factor
Calculate the “Q”
Adjustment Factor by dividing the ac ohmstoneutral impedance listed in
Chapter 9, Table 9 by the dc resistance listed in Chapter 9, Table 8.

Example#1:
A 3 phase 36kVA load rated 208 V is wired to the panelboard with 80 ft lengths of 1 AWG THHN aluminum. What is the approximate voltage drop of the feeder circuit conductors?
Solution:
Applying the threephase formula:
VD = 1.732 x K x I x D/CM
Where:
K = 21.2 ohms, aluminum
I = (36,000/(208 x 1.732)) = 100A
D = 80 ft
CM = 83,690 (Chapter 9, Table 8)
VD = 1.732 x 21.2 x 100A x 80/83,690 = 3.51 V
VD Percent = 3.51 / 208 = 1.69 %
Rule#1 Algebraic variations
Using basic
algebra, you can apply the same basic formula to find one of the other
variables (I, D or CM) if you already know the voltage drop.
Suppose you want to
know what size conductor you need to reduce the voltage drop to the desired
level. Simply rearrange the formula.
For threephase,
it would look like this:
CM (3
phase) = 1.732
x K x I x D/VD
For singlephase
calculations, you would use 2 instead of 1.732.

Example#2:
Suppose you have a 3 phase, 15 kVA load rated 480V and 390 ft of conductor. What size conductor will prevent the voltage drop from exceeding 3% percent?
Solution:
Applying the threephase formula:
CM (3 phase) = 1.732 x K x I x D/VD
Where:
K = 12.9 ohms, copper
I = (15,000/(480 x 1.732)) = 18 A
D = 390 ft
VD = 480V x 0.03 = 14.4V
CM = 1.732 x 12.9 x 18 x 390/14.4V = 10,892
From Chapter 9, Table 8, you will find that the conductor that has CM equal to or next larger than 10,892 is 8 AWG.
Example#3:
What is the maximum length of 6 AWG THHN you can use to wire a 480V, 3 phase, 37.5 kVA transformer to a panelboard so voltage drop does not exceed 3% percent?
Solution:
Applying the threephase formula:
D (3 phase) = CM x VD/(1.732 x K x I)
Where:
CM = 26,240, for 6 AWG Chapter 9, Table 8
VD = 480V x 0.03 = 14.4 V
K = 12.9 ohms for copper
I = (37,500/480 x 1.732)) = 45A
D = 26,240 x 14.4/1.732 x 12.9 x 45 = 376 ft
Sometimes, the only method of limiting voltage drop is to limit the load. Again, we can rearrange the basic formula algebraically: I = CM x VD/1.732 x K x D.
Example#4:
An installation contains 1 AWG THHN conductors, 300 ft long in an aluminum raceway, to a 3 phase 460/230V power source. What is the maximum load the conductors can carry, without exceeding the NEC recommendation for voltage drop?
Solution:
Applying the threephase formula:
I = CM x VD/(1.732 x K x D)
Where:
CM = 83,690 for (1 AWG), Chapter 9, Table 8
VD = 460V x 0.03 = 13.8V
K = 12.9 ohms, for copper
D = 300 ft
I = 83,690 x 13.8/1.732 x 12.9 x 300 = 172A
But as per NEC 110.14(C) and Table 310.15(B)(16), the maximum load permitted on 1 AWG THHN at 75C is 130A.
So, as shown in the above example, You maintain the voltage drop within 3% to meet NEC Recommendation But you violate NEC Requirements for conductor Ampacity.
Important!!!
In any
conductors/cables selection or design process, verify the NEC requirements
first for Conductor amapcity and others then perform the recommended check
for voltage drop to ensure best performance and low cost.

3.2 Chapter (9) Tables Method
Rule#2: Using Chapter (9) Tables Method for Voltage Drop Calculations
In this method,
the voltage drop is calculated by using the following formulas:
Where:
VD: The
voltage drop (V),
L : The
length of conductor (m),
R: The
resistance value from Chapter 9, Table 8 (ohm/km) or (ohm/kFT)
The value
R is determined from the National Electrical Code (NEC), Chapter 9, Table 8
column Direct Current Resistance at 75 degrees C/ Copper/ Uncoated.
I: The
load current (A).

Example#5:
Determine the voltage drop of a 380V, 3 phase circuit with a current of 100A and a length of 150 m and a conductor size of 3 AWG.
Solution:
This is a secondary service feed so, recommended voltage drop = 3%
Applying the threephase formula:
VD = 1.732 x L x R x I / 1000
Where:
L = 150 m
R = 0.802 for (3 AWG), Chapter 9, Table 8
I = 100 A
VD = 1.732 x 150 x 0.802 x 100 /1000 = 20.834 V
But, the maximum voltage drop allowed in this case = 0.03 x 380 V = 11.4 V.
The voltage drop value must be minimized.
Example#6:
For example#5, determine the size of cable that will be required to meet the voltage drop requirement (3%).
Solution:
Since the maximum voltage drop allowed in this case = 0.03 x 380 V = 11.4 V.
So, you need to determine the value of R that will meet the following formula:
11.4 = 1.732 x 150 x R x 100 / 1000
R = 11.4 x 1000/(1.732 x 150 x 100) = 0.438 ohm/km
Referencing the NEC Chapter 9, Table 8 indicates that the cable size with a voltage drop of 0.438 ohm/km or less is 1/0 AWG (70 mm) cable with a resistance of 0.399 ohm/km.
Calculating the voltage drop for the 1/0 AWG (70mm) cable results in:
VD = 1.732 x L x R x I / 1000 = 1.732 x 150 x 0.399 x 100 / 1000 = 10.36 V
So, Percentage Voltage Drop = 10.36 x 100 / 380 = 2.73 %
Therefore, in order to transmit a 3 phase current of 100A per phase over a length of 150 m, with a total voltage drop equal to or less than the maximum 11.4 volts, a 1/0 AWG (70 mm2) cable is needed.
In the next Article, I will continue explaining Chapter (9) Tables Method for Voltage Drop Calculations. Please keep following.
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