19-1
C
URRENT
, R
ESISTANCE
,
AND
D
IRECT
-C
URRENT
C
IRCUITS
19
Answers to Multiple-Choice Problems
1.
B
2.
B
3.
D
4.
B, D
5.
A
6.
A
7.
C
8.
A
9.
C
10.
C
11.
C
12.
C
13.
C
14.
A
15.
E
Solutions to Problems
19.1.
Set Up:
An electron has charge of magnitude
The current tells us how much
charge flows in a given time. We can find the number of electrons that correspond to a certain amount of charge.
Solve: (a)
(b)
(c)
Reflect:
The amount of charge associated with typical currents is quite large. In Chapter 17 we found that the force
between objects that have net charge 1.0 C is immense. The same amount of charge enters and leaves any part of the
circuit each second and each circuit element remains neutral.
19.2.
Set Up:
Solve:
19.3.
Set Up:
The number of ions that enter gives the charge that enters the axon in the specified time.
Solve:
19.4.
Set Up:
Negative charge moving from
A
to
B
is equivalent to an equal magnitude of positive charge going from
B
to
A.
The current direction is the direction of flow of positive charge.
Solve:
The total positive charge moving from
B
to
A
is
Positive charge flows from
B
to
A
so the current is in this direction.
19.5.
Set Up:
gives the amount of charge that passes a point in the wire in time
From
we can find
the number of electrons, since an electron has charge
Solve:
Reflect:
Typical currents correspond to the flow of a large number of electrons.
1
5.50
3
10
2
3
C
/
s
2
1
1
1.60
3
10
2
19
C
/
electron
2
5
3.44
3
10
16
electrons
/
s
2
e
.
D
Q
D
t
.
D
Q
5
I
D
t
I
5
D
Q
D
t
5
1.85 C
30 s
5
62 mA.
1.85 C.
D
Q
5
1
5.11
3
10
18
1
2
3
3.24
3
10
18
421
1.60
3
10
2
19
C
2
5
I
5
D
Q
D
t
.
I
5
D
Q
D
t
5
9.0
3
10
2
8
C
10
3
10
2
3
s
5
9.0
m
A
D
Q
5
1
5.6
3
10
11
ions
21
1.60
3
10
2
19
C
/
ion
2
5
9.0
3
10
2
8
C
I
5
D
Q
D
t
.
D
Q
5
I
D
t
5
1
25,000 A
21
40
3
10
2
6
s
2
5
1.0 C
I
5
D
Q
D
t
D
t
5
D
Q
I
5
7.50 C
1.50 A
5
5.00 s
D
Q
5
I
D
t
5
1
1.50 A
21
300 s
2
5
450 C
1.50 A
5
1
1.50 C
/
s
2
1
1
1.60
3
10
2
19
C
/
electron
2
5
9.38
3
10
18
electrons
/
s
e
5
1.60
3
10
2
19
C.
1A
5
1 C
/
s.
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19.6.
Set Up:
For copper, Table 19.1 gives that
and for silver,
Solve: (a)
(b)
so
19.7.
Set Up:
From Table 19.1, copper has
Solve:
19.8.
Set Up:
For copper,
Solve:
19.9.
Set Up:
The length of the wire in the spring is the circumference
of each coil times the number
of coils.
Solve:
Reflect:
The value of
we calculated is about a factor of 100 times larger than
for copper. The metal of the spring
is not a very good conductor.
19.10.
Set Up:
For aluminum,
For copper,
Solve:
so
19.11.
Set Up:
For aluminum,
For copper,
Solve: (a)
(b)
19.12.
Set Up:
Solve:
R
new
5
r
L
new
p
1
d
new
/
2
2
2
5
r
3
L
p
1
2
d
/
2
2
2
5
3
4
r
L
p
1
d
/
2
2
2
5
3
4
R
d
new
5
2
d
L
new
5
3
L
;
R
5
r
L
A
5
r
L
p
1
d
/
2
2
2
.
L
5
R
p
r
2
r
c
5
1
2.00
3
10
2
4
V
2
p
1
0.750
3
10
2
3
m
2
2
1.72
3
10
2
8
V
5
0.0205 m
5
2.05 cm
R
5
r
c
L
p
r
2
.
R
5
r
al
L
A
5
1
2.63
3
10
2
8
V
#
m
21
3.80 m
2
1
1.00
3
10
2
2
m
21
5.00
3
10
2
2
m
2
5
2.00
3
10
2
4
V
r
c
5
1.72
3
10
2
8
V
#
m.
r
al
5
2.63
3
10
2
8
V
#
m.
R
5
r
L
A
.
d
c
5
d
al
Å
r
c
r
al
5
1
3.26 mm
2
Å
1.72
3
10
2
8
V
#
m
2.63
3
10
2
8
V
#
m
5
2.64 mm.
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- Spring '09
- RODRIGUEZ
- Physics, Charge, Current, Resistance, Resistor, Electrical resistance, Series and parallel circuits, Direct-Current Circuits
-
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