Voltage drop

[EnriqueB]

What’s the voltage drop of 125 ft. of 12 AWG wire (0.24 ohms) on a balance 3-wire, 120/240V multiwire circuit supplying a 16A load?

Why the voltage drop answer form the Mike Holt's book is 6.70V dropped on phase wire.

 

[Carultch]

What’s the voltage drop of 125 ft. of 12 AWG wire (0.24 ohms) on a balance 3-wire, 120/240V multiwire circuit supplying a 16A load?

Why the voltage drop answer form the Mike Holt's book is 6.70V dropped on phase wire.

We'd be glad to help, but we'd like to see your participation. Please share your work so far, and we'll offer feedback to help you the rest of the way.

 

[EnriqueB]

We'd be glad to help, but we'd like to see your participation. Please share your work so far, and we'll offer feedback to help you the rest of the way.

Step 1: Evd = I x R.

Step 2 : I =16A.

Step 3 : R = (1.93 Ω per 1,000ft/1,000) x 125 ft x 1 wire. [1.93 Ω – 12 AWG from Table 8, Chapter 9]

Step 4 : R = 0.00193 Ω x 125 ft x 1 wire

Step 5 : R = 0.24125 Ω ≈ 0.24 Ω

Step 6 : Evd = 16A x 0.24 Ω

Step 7 : Evd = 6.70V dropped on phase wire ?

 

[Tulsa Electrician]

What if you used a different formula.

 

[Tulsa Electrician]

Look at the question closely
I come up with 6.68552

 

[wwhitney]

Step 6 : Evd = 16A x 0.24 Ω

Step 7 : Evd = 6.70V dropped on phase wire ?

You're correct up to Step 6, as long as you notice that you used the 1-way length of the circuit. So 16A x 0.24 Ω is the voltage drop on just one of the #12 conductors of the circuit. But current will be returning on the other line of the circuit (no current on the neutral as it is described as a balanced circuit), so that #12 conductor will also have voltage drop.

That makes the total voltage drop 2 x 16A x 0.24 Ω.

Cheers, Wayne

 

[EnriqueB]

You're correct up to Step 6, as long as you notice that you used the 1-way length of the circuit. So 16A x 0.24 Ω is the voltage drop on just one of the #12 conductors of the circuit. But current will be returning on the other line of the circuit (no current on the neutral as it is described as a balanced circuit), so that #12 conductor will also have voltage drop.

That makes the total voltage drop 2 x 16A x 0.24 Ω.

Cheers, Wayne

Step 7 is from Mike Holt's book:

What’s the voltage drop of 125 ft. of 12 AWG wire (0.24 ohms) on a balance 3-wire, 120/240V multiwire circuit supplying a 16A load?

• 2.50V.
• 3.40V.
• 4.80V.
• 6.70V.

 

[EnriqueB]

Look at the question closely
I come up with 6.68552

Step 7 is from Mike Holt's book:

What’s the voltage drop of 125 ft. of 12 AWG wire (0.24 ohms) on a balance 3-wire, 120/240V multiwire circuit supplying a 16A load?

• 2.50V.
• 3.40V.
• 4.80V.
• 6.70V.

 

[wwhitney]

Step 7 is from Mike Holt's book:
. . .
6.70V.

Hmm, 2 * 16A * 0.24 = 7.68V, which is not 6.7V.

Perhaps it is more appropriate to use Table 9's "Effective Z at 0.85 PF for Uncoated Copper Wires". Which is 1.7 ohms/kft, not 1.93 ohms/kft.

Using 1.7 gives 2 * 16A * 1.7 * 125/1000 = 6.8V. Which isn't 6.7V, but is close enough for a multiple choice question.

Cheers, Wayne

 

[EnriqueB]

Look at the question closely
I come up with 6.68552

How you come out with this number?

 

[Tulsa Electrician]

It's kinda a trick question.
For me the ansawer came from the verbage.

Hint 120/240
Balanced 16 amp load
VD on phase wire


You can use an variation of the VD= IxR if you correctly calculate the R

Calculate R is the

Who knows I may be wrong. I do things kinda weird.

See if that clicks
I also used exact K for #12 solid
12.6029 for one formula
And used another number for VD=
Is has an effect on footage.
R= .417857257

 

[Tulsa Electrician]

Step 1: Evd = I x R.

Step 2 : I =16A.

Step 3 : R = (1.93 Ω per 1,000ft/1,000) x 125 ft x 1 wire. [1.93 Ω – 12 AWG from Table 8, Chapter 9]

Step 4 : R = 0.00193 Ω x 125 ft x 1 wire

Step 5 : R = 0.24125 Ω ≈ 0.24 Ω

Step 6 : Evd = 16A x 0.24 Ω

Step 7 : Evd = 6.70V dropped on phase wire ?

Are these your steps or what is in the book?

 

[EnriqueB]

Yes

Are these your steps or what is in the book?

Are these your steps or what is in the book

 

[EnriqueB]

Yes

 

[EnriqueB]

100 from

Yes

100% from the book

 

[Tulsa Electrician]

If these were the steps in The book.
The ansawer has to be wrong and my equation and idea how to solve is incorrect.
16*.24 = 3.84.

This may be one of those that has to sent to Mike holt for the correct answer.

Sorry for any confusion.

 

[Mr. Serious]

Chapter 9, table 8 resistance (1.93Ω/kft) is for wire at 75 degrees C, rather hot. The answer given seems to agree with the total voltage drop of both #12 wires at a lower temperature. Under normal conditions it'll run at a much lower temperature than 75° with 16 amps of current. Perhaps the answer was calculated using a lower resistance value of #12 at a lower temperature, and the rest of the calculation is a misprint.

The wording of the question seems to be asking for the voltage drop along only one wire, though, not both wires of the circuit.

 

[EnriqueB]

If these were the steps in The book.
The ansawer has to be wrong and my equation and idea how to solve is incorrect.
16*.24 = 3.84.

This may be one of those that has to sent to Mike holt for the correct answer.

Sorry for any confusion.

There is some way to send this question back to him?

 

[Tulsa Electrician]

I have read that if you e mail them they will get back with you.
I did find this info with email address

 

[Tulsa Electrician]

I look forward to any ansawer you get. I may learn something from it.
Thank you