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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
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.
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
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).
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
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
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
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.
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
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
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
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
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.

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LE Calculations 2

  • 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.