Voltage drop is the voltage lost within an electrical circuit due to the resistance of the conductors. It represents the wasted electricity in a circuit. The maximum allowable voltage drop is typically 3% according to code. Voltage drop can be calculated using formulas involving current, resistance, conductor material, length and size. The NEC provides tables listing the resistance and ampacity of common wire gauges to determine the proper wire size for an electrical circuit.

1. Voltage Drop . . . What is it?
• Voltage drop is the voltage lost within the
circuit caused by the resistance of the
conductors making up the circuit
• in other words voltage drop is the wasted
electricity caused by line resistance of a
circuit
• So voltage drop is the wasted electricity
and watts are the used part of electricity

2. Voltage Drop Rules
• Unless specified otherwise all conductors
should be considered to be solid, copper and
uncoated.
• The maximum amount of voltage drop in a
branch circuit shall not exceed 3% or the
allowed amount as specified by the
equipment manufacturer.

3. Voltage Drop Notations
• I = amps R = resistance VD = voltage drop
VD permitted = 3% of voltage on any circuit
(120v x 3% = 3.6v)
• CM = size of conductor in circular mils (Table 8)
• D = distance of the circuit one way
• K = resistance of circular mil-foot wire (Table 8)
• K* = approximate K use 12.9 for copper and 21.2
for aluminum
• Power Loss = VD x I Exact K = R x CM / 1000

4. NEC Table 8
• In Chapter 9 of your NEC you will find Table 8.
• You will note that on the left most column there
are 2 entries for each size of wire. The forth
column shows the amount of strands in the
conductor (1 means solid, 7 means stranded).

5. NEC Table 8 cont.
• The third column from the left is where you get
the CM (circular mill) size of conductors.
• The fifth column from the right has the listings for
resistance of uncoated copper wire per 1000 feet

6. Voltage Drop Formulas
Voltage Drop
VD = I x R
VD = 2 x K x D x I
CM
Wire Size
CM = 2 x K* x D x I
VD permitted
Distance
D = CM x VD permitted
2 x K x I
Load
I = CM x VD permitted
2 x K x D

7. Calculator Trick for Long
Division
• Instead of doing a long division problem the old
way:
D= CM x VDper = 2550 x 3.6 = 9180 = 17.5
2 x K x I 2 x 13.1086 x 20
525.34
• Try the new easier way:
D=CM x VDper = 2550x3.6/2/13.1086/20 = 17.5
2 x K x I

8. Lets try a few problems
• As before, the problem will appear on the
screen first, try to work it out on your own.
• Then the solution, step by step will appear,
check your work to see if you followed the
right logic to answer it correctly.

9. Voltage Drop Problem #1
• What is the voltage drop in a branch circuit
to a siren that has a 50Ω load? The source
voltage is 6 volts, the distance is 40 feet, the
conductor is #14AWG.
• Ohm’s Law tells us I = E/R
So 6v/50Ω = .12amp
• Solve for K = 3.07 x 4110 / 1000 = 12.6177
• Now solve for VD = 2 x 12.6177 x 40x .12 /
4110 = 0.029472 volts

10. Voltage Drop Problem #2
• What is the total resistance of two #16AWG
conductors? Each is 85 feet in length and
they are connected in parallel.
• Table 8 for #16 = 4.89Ω per 1000 feet.
Divide by 1000 to get a foot.
• Multiply .00489 by 85 feet to get 0.41565Ω
per conductor.
• Now divide the resistance by 2 because they
are in parallel = 0.207825 Ω

11. Voltage Drop Problem #3
• What is the maximum load In amps the
code allows for a branch circuit using
#18AWG coated? The power source is 65
feet away and is 12 volts.
• First we find K.
• K = 8.08 x 1620 / 1000 = 13.0896
• Now I = 1620 x .36 / 2 / 13.0896 / 65 = .
3427amps

12. Voltage Drop Problem #4
• What size copper conductor is required for
a branch circuit to a horn that has a .25amp
load? The power source is 6 volts and is 50
feet away.
• CM = 2 x 12.9* x 50 x .25 / .18 = 1791
• the next larger conductor to 1791 in CM is
#16AWG

13. Voltage Drop Problem #5
• Find the approximate distance between the
source and the load if a #18AWG stranded
conductor is used and the total conductor
resistance is 1.05Ω ?
• #18 Stranded is 7.95Ω/1000 or .00795 per foot
• divide 1.05 by .00795 to get the total amount of
feet of wire at 132
• Now the question is how far apart are the source
and load so the distance is half of 132 or 66 feet.