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Current
Submitted by: I.D. 037706835
The problem:
The space between two coaxial cylinders, r1 > r2 , is filled with a conductor with the resistivity ρ.
What is the resistance between the inner and outer surfaces. The length of the cylinders is l.
The solution:
The resistance is given by the product of resistivity and length of the conductor, divided by the
area that is tangential to the current. Looking at an infinitesimal layer of the cylinder:
dR = ρ ·
dl0
2πl0 · l
(1)
Integrate both sides to get the total resistance:
Z r1
ρ
1 0
ρ
r1
R =
·
dl =
ln
0
2πl r2 l
2πl r2
(2)
1
Electric current
Submitted by: I.D. 200475846
The problem:
Find the resistance between the bases of a circle conic with radii a and 2a, with a height L and
filled with material with a conductivity σ.
The solution:
dx
1. The equation we shall use to find the resistance is: dR = S(x)σ
. σ is given to us, so to find the
total resistance we need to find the area S(x).
In order to find the radius of the cone as a function of L we shall use basic geometry and we will
come up with the expression:
r(x) =
a(L + x)
.
L
(1)
Now all we need to do is to substitute S(x) = πr2 (x) into dR =
dx
S(x)σ .
L2 dx
σπa2 (L + x)2
dR =
(2)
Now we shall perform the integral. The integral boundaries are from 0 to L.
Z
R=
0
L
L
L2 dx
=
σπa2 (L + x)2
2σπa2
(3)
1
Conductivity
Submitted by: I.D. 123456
The problem:
A resistor is built with two concentric conducting cylindrical shells of a height h and radii R1 > R2 .
The space between the shells is filled with material with constant resistivity ρ.
1. If constant current I flows in the circular direction, what is the resistance of the resistor?
2. The resistor is connected to the voltage source(as in the picture). What is the current through
the source?
The solution:
1. We think of the cylindrical resistor as many resistors of an infinitecimal width connected in
parallel.
Such an infinitecimal resistor is a cylindrical shell at a radius r. Since the current is circular,
2πr
dR = ρ
hdr
Z
Z R2
1
1
hdr
h
R2
=
=
=
ln
R
dR
2πρr
2πρ
R1
R1
(1)
(2)
Then,
R=
2πρ
2
h ln R
R1
(3)
2. By connecting the source in the pictured way, we have actually two resistors connected in parallel
R1 =
R2 =
R
4
3R
4
(4)
(5)
(here R1 , R2 are resistances) since they are built from the respective part of the cylindrical resistor.
Then the total resistance is
Rt =
R1 R2
3R
=
R1 + R2
16
(6)
Then
2
8V h ln R
V
16V
R1
I=
=
=
Rt
3R
3πρ
(7)
1