Resistance of a Copper Wire with initial and final radius

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Homework Help Overview

The problem involves determining the resistance of a copper wire with a radius that varies along its length, described by the function R(x) = a*e^x + b. The initial and final radii are given as 0.45 mm and 9.67 mm, respectively, with a total length of 38 cm.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using the resistance formula R = pL/A and consider integrating over the length of the wire to account for the varying radius. There are questions about the correct area formula to use and how to set up the integral for total resistance.

Discussion Status

Some participants have offered insights on setting up the problem, including the use of a small wire element and the integration approach. There is ongoing clarification regarding the values of the radius at specific points along the wire and how to express these in the context of the problem.

Contextual Notes

Participants are working with specific values for the radius at the endpoints of the wire and are encouraged to convert measurements to SI units for consistency. There is some confusion regarding the interpretation of the radius function and its application in the resistance calculation.

jawhnay
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Homework Statement



A Copper wire has a shape given by a radius that increases as R(x)= aex+ b. Its initial radius is .45 mm and final radius is 9.67 mm and its horizontal length is 38 cm. Find its resistance.

Homework Equations


R = pL/A where p = resistivity of copper
A = ∏r2
L = length of copper wire

resistivity of copper = 1.7 x 10-8

1/Rtotal = 1/Rinitial + 1/Rfinal

The Attempt at a Solution



My attempt at the solution was finding the resistance of the wire when the radius is initially at .45 mm and then finding the resistance of the wire at 9.67 mm using the formula R = pL/A. Then I tried adding those resistances together by using the Resistance formula for parallel circuits which is 1/Rtotal = 1/Rinitial + 1/Rfinal
 
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Think of a small wire element of length dl and find its resistance dR, then integrate over the length to find the total resistance.
 
so you're saying R = pL/A ... dR = pdr/A what would my area be? Would it be A = ∏r2 or A = 2∏rL How exactly do I go about doing this?
 
Can anyone else offer any insight on how to do this problem?
 
So you're given:
r(x) = a*e^x + b
r(0 m) = 0.45e-3 m
r(38e-2 m) = 9.67e-3 m

A(x) = π*r(x)^2

You could imagine summing up the resistances of a bunch of thin slices of wire with width dx as a heuristic to help you set up the integral:
<br /> R(x_e) = \int_0^{x_e} \frac{\rho}{A(x)} \mathrm{d}x<br />
 
where are you getting r(0 m) = 0.45e-3 m and r(38e-2 m) = 9.67e-3 m ?
 
Let x = 0 cm and x = 38 cm represent the endpoints with the small and large radius, respectively. You're then given the values that r(x) should take at these endpoints.
 
How do you know that r(0) = .45 x 10-3 and r(.38) = 9.67 x 10-3 ? I'm kind of confused... Can anyone make this clearer for me?
 
Imagine putting a coordinate axis down the length of the wire. You could call this the x-coordinate axis and x would represent a specific position along the length of the wire.

You know the length of the wire is 38 cm and the radii of the thin and thick ends are 0.45 mm and 9.67 mm, respectively. You also know how the radius of the wire changes from one end to the other.

If r(x) is the radius at position x, you could let x = 0 represent the thin end of the wire and x = 0.38 could then be the thick end (it's usually good practice to convert everything to SI base units, meter [m] for length).

Does that help?
 

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