How do I find the total Resistivity if using AWG 18?

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Discussion Overview

The discussion revolves around calculating the total resistivity for a circuit using AWG 18 wire. Participants explore the relationships between resistivity, resistance, and wire dimensions, while addressing unit conversions and the use of different measurement systems.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a calculation for resistance using power and voltage, leading to a resistance value of 15.125 Ohms.
  • Another participant suggests consulting a table for wire sizes in the American Wire Gauge (AWG) system and provides links for reference.
  • A participant calculates resistivity, length, and area for AWG 18 wire, arriving at a resistance of 4.18 Ohms and a current of 4.68 Amps, questioning the necessity of unit conversion from meters to feet.
  • Concerns are raised about the area calculation for AWG 18, with a participant noting the difference between circular mils (CM) and centimeters (cm), and providing a reference for resistivity in Ohm-meters.
  • Clarifications are made regarding the definition of circular mils and the appropriate units for resistivity, with a participant stating that resistivity should be in Ohms * CM / ft.
  • Some participants express a preference for metric units over English/British units, discussing the challenges of mixed units in engineering education.

Areas of Agreement / Disagreement

Participants express differing views on unit conversions and the appropriate units to use for resistivity and area calculations. There is no consensus on a single method or approach to the problem, indicating ongoing debate and exploration of the topic.

Contextual Notes

Participants highlight limitations in their understanding of unit conversions and the definitions of terms like circular mils and centimeters, which may affect their calculations. The discussion reflects a mix of metric and English units, leading to potential confusion.

Who May Find This Useful

Students in engineering or physics courses, particularly those dealing with electrical circuits and wire specifications, may find this discussion beneficial.

JeeebeZ
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Pretty easy Circuit question...

So, I am in my first year ENGE Class second week in, and I am given this diagram

Homework Statement


Basically I am given this Diagram (attachment)

And told to find the total Resistivity if using AWG 18

So, i got

P=800W
E=100V
R=E^2/P = 15.125 Ohms

Homework Equations



So for the Resistivity R = p l/a.

The Attempt at a Solution



But I am completely unsure how to get p l & a.

This isn't even a assignment question it was a example done in class on the board, but my instructor just randomly says everything and doesn't explain anything.

So when alls said and done I have

R = 4.13
I = 110/23.5= 4.7 Amps

But absolutly no clue as to how to get it :(
 

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So,
p = resistivity of copper = 10.37 CM
l = length of wire = 200m / 0.3048 = 656.17 ft
a = area of 18 AWG = 1624 CM

R = 10.37 * 656.17 / 1624 = 4.18

So then to get I i got

I = E / R
I = E / R1 + R2 + R3
I = 110V / 4.18 + 4.18 + 15.125
I = 110/23.485
I = 4.68

Is there any way i could have done this without having to convert the meters to feet?
 
cm is a unit of length
CM is circular mil

you said they should all be in my text. Everything in my text is based off CM, i don't know why but it is >_<
 
A circular mil(or CM) is diameter is mills squared

OR

1 mil = 0.001 in.

That's what my book says...

Your resistivity(or rowe) should be in Ohms * CM / ft.
 
yah that sounds right, i just don't know how to make symbols on here so well... i didnt @_@
 
JeeebeZ said:
cm is a unit of length
CM is circular mil
Ah, my mistake. I should have recognized it, but its a unit I now use seldom. 1624 cmil is correct.

If resistivity is given in SI/metric units, one does not have to use feet. The problem in the US is that most engineering is still taught with English/British units, and there are sometimes mixed units.

I use both at work, but prefer metric.
 
I agree...prefer metric to work problems with...but still understand english better, when someone yells out a distance or figure or something...
 

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