Calculating the Resistance of a Conductor = (K) (Approximate K & Exact K)

** Approximate K ** = 12.9
for copper.

** Approximate K ** = 21.2
for aluminum.

** Exact K ** =
Resistance multiplied by the Circular Mill and then divided by one thousand.
You will find resistance of a wire and the circular mill in NEC
Chapter 9 Table 8.

** Special Notes: ** In short for calculations during
testing always use exact K instead of approximate K to be safe. It does
not take that long to use exact K and this may save you from losing some
score points due to exact K being required without your knowledge.

**Example: Written Form **

“Exact K”of a # 8, uncoated, stranded, aluminum conductor, is found in the NEC chapter 9, table 8 . A # 8 stranded uncoated aluminum conductor is 1.28 ohms X 16,510 circular mills. Now, divide that by 1,000 = 21.1328 ohm. This is your “exact K”of the above mentioned # 8, uncoated, stranded, aluminum conductor.

Using
the NEC chapter 9 table
8 to find the exact K of a conductor, look for the
size column, then look down that column to the line stating the size for
Awg. # 8. You will find two lines with Awg. # 8. Now look for the column
for the stranding quantity. Then look down that column to # 8 again.
You will find a line for 1 strand, which would be for solid, and a line for
7 strands, which would be for stranded. You want stranded { *7strands *}
in this example. Now, go to the column for aluminum, and then look down that
column for a sub-column like the one under copper, coated or uncoated. You
will find none. Therefore, you want the only aluminum column present. Now,
look down that aluminum column to Awg. # 8 line, on the same line as the
Quantity of 7 strands, and then look back over to the aluminum column, and
you will find 1.28 ohm resistance, in the aluminum column across the same
line in the chart as the quantity of 7 strands. Now, look in the column for
the area in circular. mills, and look down that column to the Awg. # 8, with
the quantity of 7 strands, and you will find 16,510 mills. Now, multiply
the 1.28 ohms by the 16,510 circular. mills, the product of that calculation
equals 21,132.8 ohms. Now, divide 21,132.8 ohms by 1,000 to get the value
of exact K which equals 21.1328 K. The results of that calculation is your
exact K, on this Occasion.

**Special
Notes: ** The column for coated refers
to a coating found over the conductor that you probably noticed in shop
in junior high school when you built a motor by wrapping wires around an
iron core. This wire had a lacquer style coating “coated”

**Same Example: (Numeric Form)**

The “Approximate
K”of a # 8 uncoated aluminum stranded conductor = **21.2 “Approximate
K” **

The “Exact K”of a # 8 uncoated aluminum stranded conductor =

NEC Chapter 9 Table 8 =

1.28 X 16,510 divided by 1,000 = **21.1328 “Exact
K” **

** Example Results: **

Therefore, “approximate K”is different from “exact K”. You will not be sure when the authors of a test will want you to use “exact K”. Most of the time, you will find both answers for you to mark. Mark the wrong one, and you will lose points of your test score. If the properties of the conductor {cu. or al.} and the conductor size is known, “use exact K”.

Don’t take a chance. Use only “exact K”during testing !

The FPN Notes in the NEC are considered to be advisory, only. If you adhere to the NEC requirement considering voltage drop. The NEC requires the conductor to be able to carry the load, as per Article 220 PARTS B & C & D for feeders, and as per Article 210-19 for branch circuit conductors. Article 220 parts B, C, & D, calculates the load on a conductor at 100% for a non- continuous load, and at 125% for continuous load.

The NEC speaks nothing about giving a break for consideration of a loss of power, due to a voltage drop, that decreases the current carrying capacity of the conductor used. Therefore the NEC expects a full amount of power, [voltage], at “end of line”.

Therefore the NEC requires that there will be no voltage drop allowance at all. You will find in ARTICLE 90-4 in the NEC that “judgment calls”are up to the Authority Having Jurisdiction. Therefore, the Authority Having Jurisdiction has the authority to allow the use of the FPN note speaking of 3% on a branch, or feeder, and with a total voltage drop of 5% on the feeder, and branch combined. This NEC FPN Note can be used as an allowance for you to use that is within the authority of his judgment. Otherwise the NEC flat says that the conductor must be able to carry the load as computed in Article 220.

**Special
Notes: ** In your voltage drop calculations you will find the number
2 substituted by 1.732. When changing from a single phase to a three phase
calculation. The 2 [single phase] represents the wire going to the load
plus the wire coming back from the load to the grounded source. [Not grounding]
The same principle is used in the 1.732 [three phase] calculation. The
1.732 figure representing the three wires used to run a three phase motor
is the square root of the 3 conductors.

** Power Loss = ** Voltage
drop x load in amps **. **

**Example: (Written Form) **

120 volts with a feeder voltage drop of 3 % as per the NEC allowance in the FPN note NEC 215-2-b FPN note 2 =116.4 volts

** Same Example: (Numeric Form) **

120 volts x 3% VD allowed=3.6 volts of power loss; 120 volts - 3.6 volts=116.4 volts

120 volts at source - 3.6 volts at end of line due to the

power loss = 116.4 volts applied at end of line

**Example Results**

Due to resistance of the conductor we threw away 3.6 volts of power in this occasion, yet we have to pay for that voltage we did not get due to the voltage loss.

If we had a 1/2 horsepower motor designed to run on 120 volts, nominal, with a F.L.C. of 9.8 amp as per NEC Table 430-148 , and we lost the 3.6 volts due to a long distance run causing the above calculated voltage drop this motor would now be pulling at the full load amps as follows;

**Special
Notes:** You must change the Amps into Volt-Amps, in order to
accomplish the conversion, in the change of voltage, due to the voltage
drop calculated above.

9.8 AMPS x 120 VOLTS = 1,176 VOLT AMPS

1,176 VA DIVIDED BY 116.4 VOLTS AT END OF LINE = 10.103 AMPS

Formula to change amps into approximate volt - amps = amps x volts = volt amps

The results of the increase in motor load Amps and the loss of horse power due to the loss of voltage that is required to do the same work as the motor is designed to do without the voltage loss would cause an increase in cost, heat build up of the motor, loss of horse power, loss of life expectancy of the motor, etc.

As you can see voltage drop plays a big part in the

“Art of Electricity ”

“Calculating Conductor Properties as Relating to Voltage Drop ”

** [ VD ] { single
phase } = 2 [ **

Then,
divide the total of all of the above answer, by the Circular Mill {wire size
}. This { *C M * } using (1 for solid or 7 for stranded) is found in the NEC Chapter
9

Table 8 which is wire size of the feeder, or branch circuit being calculated.

FORMULA TO FIND VOLTAGE DROP { *V
D *} ;

{exact}

2x
K x L x I

CM

** Voltage
Drop [ VD ] { three
phase }= 1.732 [ **

Then,
divide the total answer of all of the above, by the Circular Mill {wire size
}. This { *C M * } using (1 for solid or 7 for stranded) is found in the NEC Chapter
9

Table 8 which is the wire size of the feeder, or branch circuit being calculated.

** FORMULA TO FIND VOLTAGE DROP { V
D } ; **

{exact} **
1.732 x K x L x I
CM **

** Voltage
Drop [ wire size ] { single
phase } { Circular Mill / CM }
= 2 [ **

Then,
divide the total all of the above answer, by the voltage drop permitted.
This { *VD Permitted *} is found in the NEC 215-2-B
FPN Note 2 which is 3% allowed on a feeder circuit
only or 3% allowed on a branch circuit only, or 5% total for the feeder,
and branch circuits combined.

**Formula to Find Circular Mill (Wire Size):**

{approx.}

2x {K} x L x
I

** Voltage Drop Permitted **

** Voltage
Drop [ wire size ] { three
phase } { Circular Mill / CM }
= ** Multiply

Then,
divide the total answer of all of the above, by the voltage drop permitted.
This { *VD Permitted *} is found in the NEC 215-2-B
FPN Note 2 which is 3 % allowed on a feeder circuit
only, or 3 % allowed on a branch circuit only, or 5% total for the feeder,
and branch circuits combined.

**Formula To Find Circular Mill (Wire Size):**

{approx.}

** 1.732x {K} x L x
I
Voltage Drop Permitted **

In order to find the maximum load allowed considering voltage drop

Use the following ;

** Voltage
Drop [ load ] { single
phase } { load / I } ** =
[

Then, divide the total answer of all of the above by the total answer of the following;

Multiply **2 [ ** *representing single
phase *] { *which is the wire going out to the load <hot wire>,
and the wire going back to the source <grounded wire>120 volt, or
two hot wires going to the load 240 volt, either style 120/240v making
up the referance“2 *”}, That 2 is multiplied by the resistance
{ *K *} [ *using exact {K} for copper, or for aluminum * *calculated,
by you, using the NEC * * chapter
9 table 8 {normally uncoated if copper} * ].
That is then, multiplied by the length { *L *} [ *The total distance
of the circuit from the source to the load *].

** CM x Voltage Drop Permitted
2 x K x L
**

** ** ** Voltage
Drop [ load ] { three
phase } { load / I } = ** CM
[

Then, divide the total answer of all of the above by the total answer of the following;

Multiply **1.732 [ ** *representing
three phase *] { *which is the square root of the three wires going
out to the load <hot wires>, making up the 1.732 *}, multiplied
by the resistance { *EXACT *}{ *K *} [ *calculated, by you, using
the NEC * * chapter
9 table 8 {normally uncoated} * ].
That is then, multiplied by the length { *L *} [ *The total distance
of the circuit from the source to the load *] *. *

CM x Voltage Drop Permitted

1.732
x K x L

{exact}

In order to find the maximum distance allowed considering

voltage drop Use the following ;

** Voltage
Drop [ distance ] { single
phase } { L } = ** CM
[

Then, divide the total answer of all of the above by the total answer of the following;

Multiply **2 [ ** *representing single
phase *] { *which is the wire going out to the load <hot wire>,
and the wire going back to the source <grounded wire> 120 volt, or
two hot wires going to the load 240 volt, either style 120/240v, making
up the reference “2 *”}, multiplied by the resistance { *EXACT *}{ *K *}
[ *using exact {K} for copper, or for aluminum * *calculated,
by you, using the NEC * * chapter
9 table 8 {normally uncoated if copper} * ].
That is then, multiplied by the load { *I * } [ *The total load on
the circuit in amps *].

CM x Voltage Drop Permitted

2
x K x I

** Voltage
Drop [ distance ] { three
phase } { I } = ** CM
[

Then, divide the total answer of all of the above by the total answer of the following;

Multiply **1.732 [ ** *representing
three phase *] { *which is the square root of the three wires going
out to the load <hot wires>, making up the 1.732 *}, multiplied
by the resistance { *K *} [ *using exact {K} for copper, or for
aluminum * *calculated, by you, using the NEC * * chapter
9 table 8 {normally uncoated if copper} * ].
That is then, multiplied by the load { *I * } [ *The total load on
the circuit in amps *].

CM
x Voltage Drop Permitted

1.732
x K x I

Find a Pre-Screened Electrician

THIS ARTICLE IS PROVIDED 'AS IS' WITH NO WARRANTY OF ANY KIND. THE AUTHOR, THE SITE OWNER AND ITS AFFILIATES ASSUME NO LIABILITY FOR ERRORS OR OMISSIONS CONTAINED THEREIN OR FOR ANY USE OF THE INFORMATION CONTAINED IN THIS DOCUMENT. The article is for informational purposes only and is not a substitute for professional advice.