Maximum allowable current in a 12-gauge aluminum wire

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Discussion Overview

The discussion revolves around determining the maximum allowable current for a 12-gauge aluminum wire, comparing it to the established maximum for copper wire, and addressing the calculations involved in reaching that conclusion. The context includes theoretical and practical aspects of electrical wiring, specifically focusing on resistivity and power delivery in different materials.

Discussion Character

  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant states that the maximum allowable current for 12-gauge copper wire is 20A and questions what it should be for aluminum wire if power per unit length is to remain the same.
  • Another participant proposes a method to calculate the resistances of copper and aluminum using their respective resistivities and the cross-sectional area of the wire, arriving at a maximum current of approximately 15.53A for aluminum.
  • The same participant notes that the greater the resistance in a wire, the lesser the maximum current when comparing wires of the same gauge made from different metals.
  • A later reply humorously suggests that the original question may be outdated, indicating a lapse of time since it was posted.
  • Another participant clarifies that the question was part of their physics homework, implying that others may also benefit from the discussion.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the maximum allowable current for aluminum wire, though one participant suggests it is approximately 15.53A based on calculations. There is acknowledgment of the different resistivities affecting current capacity, but no definitive agreement on the final value.

Contextual Notes

The discussion relies on specific resistivity values and assumptions about wire gauge and insulation type. There may be limitations in the calculations due to variations in material properties and environmental factors that are not fully addressed.

Who May Find This Useful

This discussion may be useful for students studying electrical engineering or physics, particularly those interested in wiring materials and their properties, as well as individuals preparing for electrical codes and standards.

yanyin
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Aluminum wiring has sometimes been used instead of copper for economy. According to the National Electrical Code, the maximum allowable current for 12-gauge copper wire with rubber insulation is 20A. What should be the maximum allowable current in a 12-gauge aluminum wire if the power per unit length delivered to the resistance in the aluminum write is the same as that delivered in the copper wire?
Please help me to figure out the steps. the answer is 15.5A.
 
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Although I am late with an answer for the original author someone might benefit.
You need to calculate the respective Resistances for Copper and Aluminium using their resistivities. The problem also stated that the wire was 12 gauge- I found the xs Area to be 3.310 x 10^-6 m2 (looked this up on internet) and the Power used by each wire to be equal while the Imax for the copper wire was 20A.
Starting with Copper the resistivity is 1.7 x 10^-8 ohm/m.
Using the formula R/l= resistivity/ xs Area I found that R/l= 5.14 x 10^-3 ohm/m
The using V=IR I found V= 1.03 x 10^-1 V.
Calculate the Power using P=I2R or P=IV you get 2.056W.

Next work out the copper story by firstly finding the R/l= 2.82 x 10^-8 ohm.m/ 3.310 m2
and that works out to be 8.52 x 10^-3 ohm/m.
Using the Power value above substitute the value for R(aluminium) and Power (copper) in P=I2R and you get Imax for Aluminium = 15.53A.

It also agrees with the theory that the greater the resistance in a wire the lesser the Imax for the wire when comparing two different metal wires of the same gauge.

This is my FIRST POST ANSWER!
 
Last edited:
drmermaid said:
...This is my FIRST POST ANSWER!

Unfortunately, I think it's 5 years too late for this fellow's NEC exam (or whatever it was for)!
 
It was in my physics homework so who knows who else wants to figure it out.
 

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