Current, Resistance, and Electromotive Force
2515
But
2
1
r
h
r
β
+
=
1
2
2
1
1
2
1 2
1 2
1
1
h
r
r
h
R
r
r
r
r
rr
rr
ρ
ρ
ρ
πβ
π
π
⎡
⎤
⎛
⎞⎛
⎞
−
=
−
=
=
⎜
⎟⎜
⎟
⎢
⎥
−
⎣
⎦
⎝
⎠⎝
⎠
(b) E
VALUATE
:
Let
1
2
.
r
r
r
=
=
Then
2
2
/
/
where
and
.
R
h
r
L A
A
r
L
h
ρ
π
ρ
π
=
=
=
=
This agrees with
Eq. (25.10).
25.64.
I
DENTIFY
:
Divide the region into thin spherical shells of radius
r
and thickness
dr
.
The total resistance is the
sum of the resistances of the thin shells and can be obtained by integration.
S
ET
U
P
:
/
I
V
R
=
and
2
/4
J
I
r
π
=
, where
2
4
r
π
is the surface area of a shell of radius
r
.
E
XECUTE
:
(a)
2
2
1
1
1
.
4
4
4
4
4
b
b
a
a
dr
dr
b
a
dR
R
r
r
r
a
b
ab
ρ
ρ
ρ
ρ
ρ
π
π
π
π
π
−
⎛
⎞
⎛
⎞
=
⇒
=
= −
=
−
=
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
∫
(b)
4
(
)
ab
ab
V
V
ab
I
R
b
a
π
ρ
=
=
−
and
2
2
4
.
(
)4
(
)
ab
ab
I
V
ab
V ab
J
A
b
a
r
b
a r
π
ρ
π
ρ
=
=
=
−
−
(c)
If the thickness of the shells is small, then
2
4
4
ab
a
π
π
≈
is the surface area of the conducting material.
2
1
1
(
)
4
4
4
b
a
L
L
R
a
b
ab
a
A
ρ
ρ
ρ
ρ
π
π
π
−
⎛
⎞
=
−
=
≈
=
⎜
⎟
⎝
⎠
, where
.
L
b
a
=
−
E
VALUATE
:
The current density in the material is proportional to
2
1/
.
r
25.65.
I
DENTIFY
and
S
ET
U
P
:
Use
E
J
ρ
=
to calculate the current density between the plates. Let
A
be the area of each
plate; then
I = JA.
E
XECUTE
:
0
0
and
E
Q
J
E
K
KA
σ
ρ
=
=
=
P
P
Thus
0
0
and
,
Q
Q
J
I
JA
KA
K
ρ
ρ
=
=
=
P
P
as was to be shown.
E
VALUATE
:
0
0
/
and
/
/
C
K
A d
V
Q C
Qd
K
A
=
=
=
P
P
so the result can also be written as
/
.
I
VA d
ρ
=
The
resistance of the dielectric is
/
/
,
R
V
I
d
A
ρ
=
=
which agrees with Eq. (25.10).
25.66.
I
DENTIFY
:
As the resistance
R
varies, the current in the circuit also varies, which causes the potential drop across
the internal resistance of the battery to vary.
S
ET
U
P
:
The largest current will occur when
R
= 0, and the smallest current will occur when
R
→
∞
. The largest
terminal voltage will occur when the current is zero (
R
→
∞
) and the smallest terminal voltage will be when the
current is a maximum (
R
= 0).
E
XECUTE
:
(a)
As
R
→
∞
,
I
→
0, so
V
ab
→
E
= 15.0 V, which is the largest reading of the voltmeter. When
R
=
0, the current is largest at (15.0 V)/(4.00
Ω
) = 3.75 A, so the smallest terminal voltage is
V
ab
=
E
– rI =
15.0 V –
(4.00
Ω
)(3.75 A) = 0.
(b)
From part (a), the maximum current is 3.75 A when
R
= 0, and the minimum current is 0.00 A when
R
→
∞
.
(c)
The graphs are sketched in Figure 25.66.
E
VALUATE
:
Increasing the resistance
R
increases the terminal voltage, but at the same time it decreases the
current in the circuit.
Figure 25.66
25.67.
I
DENTIFY
:
Apply
.
L
R
A
ρ
=
S
ET
U
P
:
For mercury at 20 C
°
,
7
9.5
10
m
ρ
−
=
×
Ω⋅
,
1
0.00088 (C
)
α
−
=
°
and
5
1
18
10
(C )
.
β
−
−
=
×
°
E
XECUTE
:
(a)
7
2
(9.5
10
m)(0.12 m)
0.057
.
(
4)(0.0016 m)
L
R
A
ρ
π
−
×
Ω⋅
=
=
=
Ω
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2516
Chapter 25
(b)
0
(
)
(1
)
T
T
ρ
ρ
α
=
+
Δ
gives
7
1
7
(60
C)
(9.5
10
m)(1
(0.00088 (C
)
)(40 C
)
9.83 10
m,
ρ
−
−
−
°
=
×
Ω⋅
+
°
=
×
Ω⋅
°
so
8
3.34
10
m.
ρ
−
Δ
=
×
Ω⋅
(c)
0
V
V
T
β
Δ
=
Δ
gives
0
(
)
A L
A
L
T
β
Δ
=
Δ
.
Therefore
5
1
4
0
(18
10
(C
)
)(0.12 m)(40 C
)
8.64
10
m
0.86 mm.
L
L
T
β
−
−
−
Δ
=
Δ
=
×
°
° =
×
=
The cross sectional area of the
mercury remains constant because the diameter of the glass tube doesn't change. All of the change in volume of the
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 Spring '06
 Buchler
 Physics, Current, Resistance, Force, Electromotive Force, Volt, Resistor

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