A material of non-uniform resistivity ?=Az (z is the vertical direction) is in t

Question

A material of non-uniform resistivity ?=Az (z is the vertical direction) is in the

shape of a truncated right cone as shown. Assume that the radial distribution

of the current is uniform show that R =[SHOWN IN THE IMAGE BELOW] across the top and bottom surfaces.

A material of non-uniform resistivity ?=Az (z is the vertical direction) is in the shape of a truncated right cone as shown. Assume that the radial distribution of the current is uniform show that R =[SHOWN IN THE IMAGE BELOW] across the top and bottom surfaces.

Explanation / Answer

as from the given figure from the geometry of thelongitudinal section of the resistor

we get (b - r) / y = (b - a) / h

so the radius at a distance y from the base willbe

r = (a - b) (y / h) + b

for a disk shaped element ofvolume

dR = ? dy / ? r2

we get R = (? / ?)h?0 dy / [((a - b) (y / h) +b)2]

using the integral formula we get

? du / (au + b)2 = - 1 / a(au + b)

applying lower limit 0, upper limit h,

R =   Ah^2 {Ln(a/b) - (a-b)/a }/Pi (a-b)^2
as from the given figure from the geometry of thelongitudinal section of the resistor

we get (b - r) / y = (b - a) / h

so the radius at a distance y from the base willbe

r = (a - b) (y / h) + b

for a disk shaped element ofvolume

dR = ? dy / ? r2

we get R = (? / ?)h?0 dy / [((a - b) (y / h) +b)2]

using the integral formula we get

? du / (au + b)2 = - 1 / a(au + b)

applying lower limit 0, upper limit h,

R =   Ah^2 {Ln(a/b) - (a-b)/a }/Pi (a-b)^2

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