Mike Dolt Enterprises
Mike Dolt Enterprises
User: Nashua_Moe, Location: New Hampshire
Asked on: Saturday, June 7, 2025 at 10:55 am EDT
Answered on: Saturday, June 7, 2025 at 11:06 am EDT
Question: To question States a 240 volt 40 amp single phase load is located 150 ft from an existing junction box the junction box is located 50 free from the panel board and is wired with 4 AWG aluminum wire in a steel Raceway the total resistance of the two 4awg conductors from the panel board to junction box is approximately what?In the back of the book you give an answer of .05 ohmsBut no matter how you do the calculations with any number of voltage drop calculations that I was taught in school I cannot seem to come up with .05 ohms? I'm not even coming up with .05 ohms when I try to do the calculations you give in the back of the book.And on a side note there's been several other mistakes made in these books that I just don't have the time to bring up right now like for instance the answers to questions 46 to 49 are just completely missing
You are correct, there's likely an error in the answer provided in your book. Let's break down how to calculate the approximate resistance and why you're not arriving at 0.05 ohms.
Understanding the Problem
We need to find the total resistance of the 4 AWG aluminum wire running 50 feet from the panel board to the junction box.
Calculation Steps
1.
Find the Resistance per Foot
You'll need to consult a table of conductor properties. A common source is Chapter 9, Table 8 in the National Electrical Code (NEC). For 4 AWG Aluminum, the resistance per 1000 feet at 75°C is approximately 0.785 ohms. Resistance per foot = (Resistance per 1000 ft) / 1000 ft Resistance per foot = 0.785 ohms / 1000 ft = 0.000785 ohms/ft2.
Calculate the Resistance for One Conductor
We have 50 feet of wire. Resistance of one conductor = (Resistance per foot) (Length of conductor) Resistance of one conductor = 0.000785 ohms/ft 50 ft = 0.03925 ohms3.
Calculate the Total Resistance (Two Conductors)
Since we have two conductors (one for the hot leg and one for the neutral), we need to double the resistance of a single conductor. Total resistance = 2 (Resistance of one conductor) Total resistance = 2 0.03925 ohms = 0.0785 ohms
Therefore, the total resistance of the two 4 AWG aluminum conductors from the panel board to the junction box is approximately 0.0785 ohms.Why Your Book's Answer is Likely Incorrect
Rounding Errors
Even significant rounding wouldn't get you from 0.0785 ohms to 0.05 ohms.
Typo or Calculation Error
It's possible the book contains a typo or a calculation error. It happens!
Different Resistance Value
Maybe the book used a different value for the resistance of 4 AWG Aluminum from a different source. This is unlikely, but possible.
Recommendation
Trust your calculations. You've followed the correct procedure, and your answer is based on standard values for conductor resistance. Contact the publisher of the book. Explain the error and provide your calculations. They may issue an errata or correction. Double-check the resistance value in the NEC (Chapter 9, Table 8) to ensure you're using the correct value. Remember the temperature rating is also important as resistance increases as temperatures increase.
Regarding Other Errors:
I understand your frustration with finding other errors in the book. It's important to be vigilant and double-check everything. Making note of the errors, and discussing them with the publisher or instructor can help prevent the information from continuing to spread. Flag for review