Find the resistance between the top and bottom faces of the cylinder

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Homework Help Overview

The problem involves calculating the electrical resistance between the top and bottom faces of a right circular cylinder made of aluminum, where the diameter equals the height. The density and resistivity of aluminum are provided, and the task is to find the resistance at a specific temperature.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between mass, volume, and density to derive the height of the cylinder. There are attempts to substitute values into the resistance formula, and some participants express confusion about their calculations.

Discussion Status

There are multiple calculations presented, with participants questioning the correctness of their results. Some guidance is offered regarding the use of specific equations and the relationship between height and diameter. The discussion reflects a mix of attempts and uncertainty without a clear consensus on the correct approach.

Contextual Notes

Participants note the urgency of the homework deadline and express feelings of being lost in the calculations. There is mention of potential calculation mistakes and the need for clarification on the values used in the resistance formula.

mba444
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Homework Statement



A 361 g mass of aluminum (Al) is formed into
a right circular cylinder shaped so that its
diameter equals its height.
Find the resistance between the top and
bottom faces of the cylinder at 20◦C. Use
2700 kg/m3 as the density of Al and
2.82 × 10−8 OHM· m as its resistivity. Answer
in units of
OHM.

Homework Equations


R= raw *(L/A)
D=(m/v)
height = diameter ... therefore r will (h/2)
V=pi*r^2*h

The Attempt at a Solution




what i did is that i solved for the height by substituting it in the density problem getting D(given)= M(given)/(Pi*(h/2)^2*h) .. which after solving is h= (1.444/(2700*Pi))^(1/3)
after that i substitute it back into the resistance equation.


Im stuck ad i eed your help please
thanx i advance
 
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Hi mba444! :smile:

(have a pi: π and a rho: ρ and try using the X2 tag just above the Reply box :wink:)
mba444 said:
… what i did is that i solved for the height by substituting it in the density problem getting D(given)= M(given)/(Pi*(h/2)^2*h) .. which after solving is h= (1.444/(2700*Pi))^(1/3)
after that i substitute it back into the resistance equation.

Well, that looks ok so far …

what number did you get, and what did you get next? :smile:
 
i got for h .. 0.0554m
after substituting into the R equation i got 1.169e-7


but my answer is wrong !
 
I think you need to use the Callendar–Van Dusen equation [which is a relation between temperature and resistance]
 
mba444 said:
i got for h .. 0.0554m
after substituting into the R equation i got 1.169e-7

I get your h, but not your R …

how do you get 1.169e-7 ?
 
ignore the previous value i think its a calculation mistake
i calculated it again i got 3.65586e-8
is it right ??
 
mba444 said:
ignore the previous value i think its a calculation mistake
i calculated it again i got 3.65586e-8
is it right ??

how did you get it?
 
R= (2.82e-8)[(0.055)/(PI*(2*0.055)^2)]

radius = 2* h
 
mba444 said:
R= (2.82e-8)[(0.055)/(PI*(2*0.055)^2)]

radius = 2* h

uhh? :redface: radius = h/2

(and what happened to that π I gave you?)
 
  • #10
ya since in the question they told us that the height equal the diameter
 
  • #11
i really don't understand .. my home work is due today .. and i really feel lost
 
  • #12
mba444 said:

Homework Equations


R= raw *(L/A)

What values are you using for raw, L, and A?
 

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