251
C
URRENT
,
R
ESISTANCE
,
AND
E
LECTROMOTIVE
F
ORCE
25.2.
I
DENTIFY
:
/
I
Q t
=
.
Use
d
I
n q v A
=
to calculate the drift velocity
d
.
v
S
ET
U
P
:
28
3
5.8
10
m
.
n
−
=
×
19
1.60
10
C
q
−
=
×
.
E
XECUTE
:
(a)
2
420 C
8.75
10
A.
80(60 s)
Q
I
t
−
=
=
=
×
(b)
d
.
I
n q v A
=
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
−
−
−
−
×
=
=
=
×
×
×
×
E
VALUATE
:
d
v
is smaller than in Example 25.1, because
I
is smaller in this problem.
25.8.
I
DENTIFY
:
/
I
Q 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.
S
ET
U
P
:
+
Na
and
Cl
−
each have magnitude of charge
q
e
= +
E
XECUTE
:
(a)
16
16
19
total
Cl
Na
(
)
(3.92
10
2.68
10 )(1.60
10
C)
0.0106 C.
Q
n
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.
E
VALUATE
:
The Cl
−
ions have negative charge and move in the direction opposite to the conventional current
direction.
25.12.
I
DENTIFY
:
E
J
ρ
=
, where
/
J
I
A
=
.
The drift velocity is given by
d
.
I
n q v A
=
S
ET
U
P
:
For copper,
8
1.72
10
m
ρ
−
=
×
Ω⋅
.
28
3
8.5
10
/m .
n
=
×
E
XECUTE
:
(a)
5
2
3
2
3.6 A
6.81 10 A/m .
(2.3 10
m)
I
J
A
−
=
=
=
×
×
(b)
8
5
2
(1.72
10
m)(6.81
10 A/m )
0.012 V m.
E
J
ρ
−
=
=
×
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
v
I
−
−
×
×
×
=
=
=
=
×
1333 min
22 hrs!
t
=
≈
E
VALUATE
:
The currents propagate very quickly along the wire but the individual electrons travel very slowly.
25.16.
I
DENTIFY
:
Apply
L
R
A
ρ
=
and solve for
L
.
S
ET
U
P
:
2
/4
A
D
π
=
, where
0.462 mm
D
=
.
E
XECUTE
:
3
2
8
(1.00
)(
4)(0.462
10
m)
9.75 m.
1.72
10
m
RA
L
π
ρ
−
−
Ω
×
=
=
=
×
Ω⋅
E
VALUATE
:
The resistance is proportional to the length of the wire.
25.36.
I
DENTIFY
:
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.
S
ET
U
P
:
The current is counterclockwise, because the 16 V battery determines the direction of current flow.
25
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252
Chapter 25
E
XECUTE
:
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 +
Ω +
Ω
(b)
16.0 V
(1.6
)
b
a
V
I
V
+
−
Ω =
, so
16.0 V
(1.6
)(0.47 A)
15.2 V.
a
b
ab
V
V
V
−
=
=
−
Ω
=
(c)
8.0 V
(1.4
5.0
)
c
a
V
I
V
+
+
Ω +
Ω =
so
(5.0
)(0.47 A)
(1.4
)(0.47 A)
8.0 V
11.0 V.
ac
V
=
Ω
+
Ω
+
=
(d)
The graph is sketched in Figure 25.36.
E
VALUATE
:
(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.
I
DENTIFY
and
S
ET
U
P
:
By definition
.
P
p
LA
=
Use
,
and
P
VI
E
VL
I
JA
=
=
=
to rewrite this expression in terms
of the specified variables.
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 Spring '11
 Zhu
 Resistor, Electrical resistance, Ω, Series and parallel circuits, loop rule

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