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resistivity

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can someone explain this to me? i get it's to do with the equation, and i can see that resistance is inversely proportional to area so area decreases by 1.6% but i don't understand how to find the % change for the diameter from there....
A = pi(d/2)2 but i dont understand/see how they got the final answer
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Original post by spofy
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Rnew=ρLAnewR_{new} = \dfrac{\rho L}{A_{new}}

Rold=ρLAoldR_{old} = \dfrac{\rho L}{A_{old}}

RnewRold=AoldAnew=1.016\dfrac{R_{new}}{R_{old}} = \dfrac{A_{old}}{A_{new}} = 1.016

  ⟹  RnewRold=π(Dold2)2π(Dnew2)2=(DoldDnew)2=1.016\implies \dfrac{R_{new}}{R_{old}} = \dfrac{\pi \left(\frac{D_{old}}{2} \right)^2}{\pi \left(\frac{D_{new}}{2} \right)^2} = \left( \dfrac{D_{old}}{D_{new}}\right)^2 = 1.016

  ⟹  DoldDnew=1.0079\implies \dfrac{D_{old}}{D_{new}} = 1.0079

  ⟹  DnewDold=0.992\implies \dfrac{D_{new}}{D_{old}} = 0.992

i.e. a 0.8% decrease in diameter.
Resistance R is given by:

R = ρL/A

since the resistivity and length are constant, R ~ 1/A, and the area is given by:

A = πr2 = π(d/2)2

So resistance is proportional to diameter as so:

R ~ π(d/2)-2
R ~ (d/2)-2
or R ~ 1/d2

hence, rearranging to make the diameter the subject:

d ~ R-1/2

If the resistance increases to 1.016 (+1.6%), then the diameter must increase by 1.016-1/2, this is 0.992 or 99.2%, i.e a decrease in 0.8%.

just note that for this method, when using the proportionality symbol ~ you can essentially take out any constants, this includes the resistivity, length, and scalar factors such as 1/2.
(edited 6 years ago)
Reply 3
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spofy
OP
12
Original post by testytesttest

R ~ (d/2)-2
or R ~ 1/d2


i dont know how you got from the first line to the second, i ended up with R ~ 4/d2 when i was trying to follow on paper :frown:
Original post by spofy
i dont know how you got from the first line to the second, i ended up with R ~ 4/d2 when i was trying to follow on paper :frown:


As I said at the bottom, I just took the constant factor of (1/2)-2, which is equal to 4, out.. so you're not incorrect, it's just for questions like this, the percentage increase is independent of constants like that, so you can just remove them, for the same reason you assumed that resistivity and length could be taken out because they're constant.

So:

R ~ (d/2)-2
R ~ const * d-2
or R ~ 1/d2
Reply 5
Avatar for spofy
spofy
OP
12
Original post by testytesttest
As I said at the bottom, I just took the constant factor of (1/2)-2, which is equal to 4, out.. so you're not incorrect, it's just for questions like this, the percentage increase is independent of constants like that, so you can just remove them, for the same reason you assumed that resistivity and length could be taken out because they're constant.

So:

R ~ (d/2)-2
R ~ const * d-2
or R ~ 1/d2


ah i see, thank you :smile:

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