Resistance of wires in parallel

AI Thread Summary
The discussion revolves around the calculation of total resistance for wires in parallel, specifically in the context of a car's rear defroster system. The original poster questions the instructor's solution, which suggests a multiplicative approach to calculating total resistance, while they believe it should follow the standard formula for resistors in parallel. The correct formula indicates that the total resistance should be R/13 if each wire has the same resistance. Participants agree that the instructor's method contains an error, and the standard approach yields a more reasonable answer. This highlights the importance of correctly applying electrical principles in problem-solving.
Cataklyzm
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Homework Statement


My car has rear defrosters that are made of 13 wires embedded into the rear window. They can melt a thin layer of ice, roughly 2.25*10^-2 kg worth, in two minutes. These wires are all in parallel and are connected to the 12V battery of my car. Each is about 1.25m long and has a resistivity of 9.00*10^-8 Ω*m. What is the cross sectional area of each wire?


Homework Equations


R=ρL/A
Q=mLf
P=Q/t


The Attempt at a Solution


So, I submitted this homework a while back. My instructor finally posted solutions. In the solutions, he has written Rtotal=\frac{R^{12}}{13}, which he uses throughout his solution. My question is: why would the resistances be multiplicative? My thought is it would look like this: Rtotal=\frac{R}{13}. Thank you for your time.
 
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I agree with you about Rtotal = R/13 if R is the resistance of each wire.

Perhaps post his whole solution?
 
Here's his solution:
 

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I still think you are right. It should be just the standard equation for resistors in parallel ..

Rtotal = 1/(1/R1 + 1/R2... +1/Rn)

If R1 = R2 = RN = R

then

Rtotal = 1/(13/R) = R/13

Simples.
 
I haven't worked through the rest of the problem but I'm surprised using R13 gives a reasonable answer.

Time for bed here.
 
My solution to problem.
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Cataklyzm said:
Here's his solution:
This is an old thread, but it's interesting and instructive to find the error by the instructor...

1587574818672.png
 
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