A copper resistor of resistivity ρ is in the shape of a cylinder of radius b and length L1
appended to a truncated right circular cone of length L2 and end radii b and a as shown in
Figure 6.7.1.
Figure 6.7.1
(a) What is the resistance of the cylindrical portion of the resistor?
(b) What is the resistance of the entire resistor? (Hint: For the tapered portion, it is
necessary to write down the incremental resistance dR of a small slice, dx, of the resistor
at an arbitrary position, x, and then to sum the slices by integration. If the taper is small,
one may assume that the current density is uniform across any cross section.)
(c) Show that your answer reduces to the expected expression if a = b.
(d) If L1
= 100 mm, L2
= 50 mm, a = 0.5 mm, b = 1.0 mm, what is the resistance?
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PHYSICS
In Figure, current is set up through a truncated right
circular cone of resistivity 731Ω.m, left radius a 2.00 mm, right radius b=2.30mm, and length L=1.94cm. Assume that the current density is uniform across any cross-section taken perpendicular to the length. What is the resistance of the cone?
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ANSWER
(a) The current i is shown in Figure entering the truncated cone at the left end and
leaving at the right. This is our choice of positive x-direction. We make the assumption
that the current density J at each value of x may be found by taking the ratio i/A where A=πr
2
is the cone’s cross-section area at that particular value of x. The direction of
J
is identical to that shown in the figure for i (our +x direction). Using Eq., we then
find an expression for the electric field at each value of x, and next find the potential
difference V by integrating the field along the x-axis, in accordance with the ideas of
Chapter 25. Finally, the resistance of the cone is given by R=V/i. Thus,
J=
πr
2
i
=
ρ
E
where we must deduce how r depends on x in order to proceed. We note that the radius
increases linearly with x, so (with c
1
and c
2
to be determined later) we may write
r=c
1
+c
2
x.
Choosing the origin at the left end of the truncated cone, the coefficient c
1
is chosen so
that r=a (when x=0); therefore, c
1
=a. Also, the coefficient c
2
must be chosen so that
(at the right end of the truncated cone) we have r=b (when x=L); therefore,c
2
=(b−a)/L . Our expression, then, becomes
r=a+(
L
b−a
)x
Substituting this into our previous statement and solving for the field, we find
E=
π
iρ
(a+(
L
b−a
x))
−2
Consequently, the potential difference between the faces of the cone is
V=−∫
0
L
Edx=
π
iρ
∫
0
L
(a+
L
b−a
x)
−2
dx=
π
iρ
b−a
L
(a+
L
b−a
x)
−1
∣
∣
∣
∣
∣
∣
0
L
=
π
iρ
b−a
L
(
a
1
−
b
1
)=
π
iρ
b−a
L
ab
b−a
=
πab
iρL
The resistance is therefore
R=
i
V
=
πab
ρL
=
π(2.00×10
−3
m)(2.30×10
−3
m)
(731m)(1.94×10
−2
m)
=9.81×10
5
Ω
Note that if b=a, then R=ρL/πa
2
=ρL/A, where A=πa
2
is the cross-sectional area of the cylinder.
Answered By=shreyash. .....
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