I need some explanation on this

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Homework Help Overview

The discussion revolves around a physics problem involving resistance in a circuit, specifically focusing on the configuration of a rectangle formed by nichrome wire and a rod with negligible resistance. Participants are examining the reasoning behind the addition of lengths in the context of calculating resistance.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the rationale behind adding the lengths of the rectangle's sides, specifically why L (height) and 2x (widths) are combined. There is also inquiry about the significance of the distance PQ in the context of the problem.

Discussion Status

Some participants have provided insights regarding the lengths that contribute to the resistance calculation, noting the negligible resistance of the rod. There is an ongoing exploration of the implications of the rod's length and its contribution to the overall resistance.

Contextual Notes

Participants are navigating assumptions regarding the dimensions of the rod and its role in the circuit, as well as the implications of the problem's constraints on resistance calculations.

flyingpig
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Homework Statement



[PLAIN]http://img23.imageshack.us/img23/949/unledto.png

Go to problem b)

[PLAIN]http://img850.imageshack.us/img850/5388/unledzb.png

The Attempt at a Solution



Now my question is why L + 2x? L is the height and x is the width, why are they adding them? I know [tex]\lambda[/tex] is the resistance per unit length, but I don't understand they are adding them. It would make sense if they added the entire arclength (perimeter) of the rectangle
 
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The problem says the rod has negligible resistance, which means that you only need to add up the lengths of the three sides of the rectangle that are formed by nichrome wire. This is one height (L) and two widths (2x).
 
cepheid said:
The problem says the rod has negligible resistance, which means that you only need to add up the lengths of the three sides of the rectangle that are formed by nichrome wire. This is one height (L) and two widths (2x).

What about the distance PQ?
 
flyingpig said:
What about the distance PQ?

The distance between P & Q is L.
 
OK!

Resistance is: R = λ(2x + L) + 0(L), because, as cepheid pointed out, the rod has negligible resistance.
 
Code:
  P--------------------
   |            x
   |  
   |
   | L
   |
   |
   |            x
  Q--------------------

Just to be clear, these are the three sides of the rectangle I was referring to. (The fourth side, made by the rod, is not shown). Adding their lengths up, we get:

L + x + x = L + 2x
 
SammyS said:
OK!

Resistance is: R = λ(2x + L) + 0(L), because, as cepheid pointed out, the rod has negligible resistance.

It's okay, OnlyMe explained it to me pretty well lol.

Also, the rod isn't length L, look closely...
 
flyingpig said:
Also, the rod isn't length L, look closely...

Maybe not, but the portion of it that conducts current and is part of the loop has length L...not that it matters since it has no resistance and hence you multiply it by 0 anyway. :rolleyes:
 

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