Resistance of Wires: 2L & A vs 2a & A

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Homework Help Overview

The discussion revolves around comparing the resistance of two wires, both of length 2L, where Wire 1 has a constant cross-sectional area A, and Wire 2 has a cross-sectional area that starts at 2a and gradually decreases to A. The problem is situated within the context of electrical resistance and material properties, specifically focusing on how varying cross-sectional areas affect resistance.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between resistance, length, and cross-sectional area using the formula R = kL/A. There is discussion about whether calculus is necessary to solve for the resistance of Wire 2, with some participants questioning the assumptions about the wire's shape and the implications of its varying cross-section.

Discussion Status

The discussion is ongoing, with participants sharing resources and expressing varying levels of understanding. Some have suggested that a reasonable argument could suffice to compare the resistances without detailed calculations, while others emphasize the need for clarity regarding the shape of Wire 2 to arrive at a solution.

Contextual Notes

There is mention of the class composition, indicating a mix of pre-calculus and calculus students, which may influence the approaches discussed. Additionally, the problem's wording is noted as potentially vague, complicating the ability to derive a definitive answer without making assumptions about the wire's shape.

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Homework Statement


Two wires both 2L. Wire 1 has a cross sectional area of A. Wire 2 cross sectional area starts at 2a but gradually gets thinner till it is A. Both have the same resitivity constant. Which one has more resistance?


Homework Equations


R=kL/A


The Attempt at a Solution


since constants the same I started with just l/a. Wire one has 2l/a resitivity, but for wire twos have no clue on how to solve for it.
 
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Thanks. However, is their any way to do this without calculus? I understand what it is saying but my class is mainly comprised of pre-calculus students witha few calculus students like me. Is their even a way to do this without calculus?
 
The question simply asks which has more resistance, not the exact answer. So you can use a reasonable argument, if you want to avoid doing the actual calculation.
 
Yeah but I kinda want to know how to solve it
 
hmm, I can only think of using calculus. There is no other way I know to get the exact answer.

Also, the question says "Wire 2 cross sectional area starts at 2a but gradually gets thinner till it is A." And wukunlin has taken this to mean that the wire is a truncated cone. This is the most simple, and most likely possibility. But we don't actually know what the exact shape is. So the question is too vaguely worded to be solved anyway. It is only when we make the assumption about the shape, then it can be solved.
 
that pdf i linked actually did all the calculus for you, the result is

R = \frac{ \rho l }{ \pi r_1 r_2 }

where as the typical cylindrical wire gives you

R = \frac{ \rho l }{ \pi r^2 }



you can think of it this way, cut the resistor into tiny little sections, each section will have a portion of the original resistor. when all the section's resistances are added together you get the original resistance. If you cut the resistors into so many sections that the one with varying cross section effective turn into bits of cylinders with different widths, what will be their individual resistance be like comparing to the cylindrical resistor if you cut it into sections that are equally fine?
 

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