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251
C
URRENT
,
R
ESISTANCE
,
AND
E
LECTROMOTIVE
F
ORCE
25.2.
IDENTIFY:
/
IQ
t
=
.
Use
d
I nqvA
=
to calculate the drift velocity
d
.
v
SET UP:
28
3
5.8 10 m .
n
−
=×
19
1.60 10
C
q
−
.
EXECUTE:
(a)
2
420 C
8.75 10
A.
80(60 s)
Q
I
t
−
==
= ×
(b)
d
.
=
This gives
2
6
d
28
19
3
2
8.75 10
A
1.78 10
m s.
(5.8 10 )(1.60 10
C)( (1.3 10
m) )
I
v
nqA
π
−
−
−−
×
=
×
××
×
EVALUATE:
d
v
is smaller than in Example 25.1, because
I
is smaller in this problem.
25.8.
IDENTIFY:
/
t
=
.
Positive charge flowing in one direction is equivalent to negative charge flowing in the
opposite direction, so the two currents due to Cl
−
and
+
Na are in the same direction and add.
SET UP:
+
Na and Cl
−
each have magnitude of charge
qe
= +
EXECUTE:
(a)
16
16
19
total
Cl
Na
(
)
(3.92 10
2.68 10 )(1.60 10
C)
0.0106 C.
Qn
n
e
−
=+
=
×+ ×
×
=
Then
total
0.0106 C
0.0106A
10.6 mA.
1.00 s
Q
I
t
=
=
(b)
Current flows, by convention, in the direction of positive charge. Thus, current flows with Na
+
toward the
negative electrode.
EVALUATE:
The Cl
−
ions have negative charge and move in the direction opposite to the conventional current
direction.
25.12.
IDENTIFY:
E
J
ρ
=
, where
/
J IA
=
.
The drift velocity is given by
d
.
=
SET UP:
For copper,
8
1.72 10
m
−
=×Ω
⋅
.
28
3
8.5 10 / m .
n
EXECUTE:
(a)
52
32
3.6 A
6.81 10 A/m .
(2.3 10
m)
I
J
A
−
×
(b)
85
2
(1.72 10
m)(6.81 10 A/m )
0.012 V m.
EJ
−
×
Ω
⋅
×
=
(c)
The time to travel the wire’s length
l
is
28
3
19
3
2
4
d
(4.0 m)(8.5 10
m )(1.6 10
C)(2.3 10
m)
8.0 10 s.
3.6 A
ln q A
l
t
vI
×
=
1333 min
22 hrs!
t
=≈
EVALUATE:
The currents propagate very quickly along the wire but the individual electrons travel very slowly.
25.16.
IDENTIFY:
Apply
L
R
A
=
and solve for
L
.
SET UP:
2
/4
AD
=
, where
0.462 mm
D
=
.
EXECUTE:
8
(1.00
)(
4)(0.462 10
m)
9.75 m.
1.72 10
m
RA
L
−
−
Ω×
=
×Ω
⋅
EVALUATE:
The resistance is proportional to the length of the wire.
25.36.
IDENTIFY:
The sum of the potential changes around the circuit loop is zero.
Potential decreases by
IR
when
going through a resistor in the direction of the current and increases by
E
when passing through an emf in the
direction from the
−
to + terminal.
SET UP:
The current is counterclockwise, because the 16 V battery determines the direction of current flow.
25
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Chapter 25
EXECUTE:
16.0 V 8.0 V
(1.6
5.0
1.4
9.0
)
0
I
+−
−
Ω
+
Ω
+
Ω
+
Ω
=
16.0 V 8.0 V
0.47 A
1.6
5.0
1.4
9.0
I
−
==
Ω+
Ω
(b)
16.0 V
(1.6
)
ba
VI
V
Ω
=
, so
16.0 V
(1.6
)(0.47 A)
15.2 V.
ab a
b
VVV
−
−
Ω
=
(c)
8.0 V
(1.4
5.0
)
ca
V
++Ω
+
Ω
=
so
(5.0
)(0.47 A)
(1.4
)(0.47 A) 8.0 V
11.0 V.
ac
V
=
=
(d)
The graph is sketched in Figure 25.36.
EVALUATE:
(0.47 A)(9.0
)
4.2 V.
cb
V
=Ω
=
The potential at point
b
is 15.2 V below the potential at point
a
and
the potential at point
c
is 11.0 V below the potential at point
a
, so the potential of point
c
is
15.2 V 11.0 V
4.2 V
−=
above the potential of point
b
.
Figure 25.36
25.47.
IDENTIFY
and
SET UP:
By definition
.
P
p
LA
=
Use
,
and
PV
IEV
L
I J
A
=
to rewrite this expression in terms
of the specified variables.
EXECUTE:
(a)
E
is related to
V
and
J
is related to
I,
so use
P = VI.
This gives
VI
p
LA
=
and
so
E
JpE
J
LA
=
(b)
J
is related to
I
and
ρ
is related to
R,
so use
2
.
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 Spring '11
 Zhu

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