# voltage drop & resistivity calculation

I wanted to implement below formula into excel sheet. can someone suggest me with example.

cable equation example : system voltage =12v, sys current=8A, length in feet=50.

Resistance = resistivity* length/ area

length = 50 feet

if we consider 26 gauge wire

diameter of wire =0.0159 inch =0.001325 feet

resistivity =1.68*10^-8

therefore wire resistance= 0.0060

but data i got is not matching with value in table mention. can you some one suggest what mistake i am doing

voltage drop

wire gauge calculation

I am not saying my answer is correct . If you check the online calculator i posted it gives right choise of wire gauge.

i am trying to impliment same thing in excel sheet.

Help i need is example to explain the choose correct value & get answer similar to the online

length of cable= 50 feet

source voltage =12v

my intension to calculate re

1. resistance per length in ohm
2. voltage drop
3. % voltage drop
4. % loss
5. select area which is good for the application
• Units are a mess, and you've lost a decimal place in the 'feet radius'. Apart from that, it's ok!
– Chu
Aug 24 '15 at 9:42
• 1.68*10^-8 This seems like a metric resistivity. Oct 23 '15 at 20:30

26 AWG is 40 ohms / 1000 foot as per this table: -

A 50 foot length of this wire (2 ohms) with 8A will dissipate 128 watts i.e. it doesn't sound a sensible choice. The volt drop at 8A will be 16 volts.

As for your formula I can't tell whether it agrees or disagrees with whatever reference you have.

Ultimately you are combining metric resistivity(1.68*10^-8[$\Omega \cdot m$]), the wrong circular area (a = pi* radius^2) and english units.

From AWG Tables: #26 AWG has 40.81Ω / 1000ft @ 20°C and a diameter of 0.01594 inches.

The AWG is based on Circular Mils (CM) and a modified area calculation ($A = d^2$) to eliminate $\pi$.

$$1.72 \times 10^{-8} \Omega \cdot m = 10.37 Ω\cdot CM/ft\ @\ 20 ^\circ C$$ $$d = 0.01594 in = 0.01594 in × 1000 mils / in = 15.94 mils$$ $$A = d^2 = (15.94 mils)^2 = 254.08 CM$$ $$R = {ρ ℓ \over A} = {10.37Ω•CM/ft * 1000 ft \over 254.08 CM} = 40.81 @ 20°C$$

This agrees with the given values from the AWG table. This is essentially the math you need to go forward.

$$R = {ρ ℓ \over A} = {10.37Ω•CM/ft * 50 ft \over 254.08 CM} = 2.04 @ 20°C$$

$$V_{DROP} = I R = 6A * 2.04Ω = 12.2V$$

All of the potential of the 12V source is consumed by the wire. #26 is too small.

If you want to do a spreadsheet. Decide percentage lost to wire. For Example: 5%.

$$V_{Feeder} = 5\% \ of\ V_{Source} = 5\% × 12V = 0.6V$$

$$R_{Feeder} = {V_{Feeder} \over I_{Load}} = {0.6V \over 6A} = 0.1Ω$$

$$R = {ρ ℓ \over A}$$ Rearrange formula. $$A = {ρ ℓ \over R_{Feeder}} = {10.37Ω•CM/ft * 50 ft \color {red}{* 2}\over 0.1Ω } = 10,370 CM$$

The *2 comes from you have to get to load and back. Two conductors for DC.

Do a look up table to select a wire area > 10,370 CM. Same for actual area or Ω / 1000 ft. Calculate actual resistance. The rest should be easy to figure out.

• Wire size: #10 AWG
• Area = 10,381 CM
• $R_{Feeder} = 0.099 89\Omega$ for 100ft
• Maximum Current: 55 A

10,381CM is too close to design area of 10,370CM, so I'd go up to next size, which is #8 (#9 is not readily available to consumers).