MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Calculating Resistance Truncated Cone Solution
Consider a material of resistivity ρ in a shape of a truncated cone of altitude h, and radii a
and b, for the right and the left ends, respectively, as shown in the Figure 6.4.2. Assuming
that the current is distributed uniformly throughout the cross-section of the cone, what is
the resistance between the two ends? You may find the following integral useful (where
! and ! are constants).
du
# (! u + " )
2
=$
1
.
! (! u + " )
1
Solution: Consider a thin disk of radius r at a distance x from the left end.
From the geometry illustrated in the figure above, we have
b! r b! a
=
.
x
h
We can solve for the radius of the disk
x
r = (a ! b) + b .
h
The resistance R is related to resistivity ! by R = !l / A , where l is the length of the
conductor and A is the cross section. The contribution to the resistance from the disk
having a thickness dy is
! dx
! dx
.
dR =
=
2
"r
" [b + (a # b)x / h]2
Straightforward integration then yields
R=
$
h
0
! dx
!h
,
=
2
" [b + (a # b)x / h] " ab
where we have used
du
# (! u + " )
2
=$
1
.
! (! u + " )
Note that if b = a , then the cross sectional area is A = ! a 2 , and set h = l . The resistance
is then R = ! l / A , the result for a cylindrical or rectangular shaped resistor.