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FUNDAMENTALS
OF CIRCUITS
31.1. Solve:
of lead from
a
mechanical pencil is
From Table 30.1, the resistivity of carbon is
p
=
3.5
x
R
m. From Equation 31.3, the resistance
p~
p~
(3.5
x
lo'
R
m)(0.06 m)
A
m'
n(0.35~10' m)'
R===
=
5.5
R
31.2. Solve:
length
L
of a wire with a crosssectional area
A
and having a resistance
R
is
(a)
From Table 30.1, the resistivity of aluminum is
p
=
2.8
x
lo'
R
m. From Equation 31.3, the
(IO
x
lo4 m)z(lOOO
R)
L==
=
3.57 m
2.8
x lo4
R
m
P
(b)
The number of turns is the length of the wire divided by the circumference of one
turn.
Thus,
3.57 m
2n(1.5 x
10"
m)
=
379
31.3. Solve:
The current
I
in a wire when a potential difference is applied to the ends of the wire can be
obtained from
Ohm's
law:
I
=
AV/R.
Using
R
=
pL/A,
(3.0
V)lc(0.40
x
m)'
(1.5~
lo4
R
m)(0.50 m)
=
2.0
A
where we have used nichrome's resistivity from Table 30.1.
31.4. Solve:
The potential difference between the ends of a copper wire that carries a current can
be
obtained
from Ohm's law and the relationship
R
=
p
L/A
.
Finding the resistivity of copper from Table 30.1,
L
A
~(0.5
x
m)'
(3.0 A)(1.7
x
lo*
C2
m)(0.20 m)
AV=fR=Ip=
=
13 mV
31.5. Solve:
diameter copper wire. That is,
We need an aluminum wire whose resistance and length are the same as that of a 0.50mm
=9
r.
=
ErCu
=
2.8
x
\1(0.25
R
m
mm)
=
0.32
mm
1.7
x
IO'
R
m
We need a 0.64mmdiameter aluminum wire.
31.6. Solve:
The slope of the
I
versus
AV
graph, according to
Ohm's
law, is the resistance
R.
From the graph in
Figure
Ex3
I .6, the
slope
of the material is
311
~.x.
..._.
..I
.
"...
._,
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Chapter 31
31.7. Solve:
9vf

i;.
.
From the circuit in Figure Ex37.1, we see that 50
R
and 100
R
resistors are connected in series across the battery.
Another resistor of 75
R
is also connected across the battery.
31.8. Solve:
that resistor is connected to resistor and a capacitor in parallel.
In Figure Ex31.8, the positive terminal of the battery
is
connected to a resistor. The other end of
31.9.
Model:
Assume that the connecting wires are ideal.
Visualize:
Solve:
I,
=
10 V/5
SZ
=
2
A downward. Using Kirchhoff's junction law, we see that
Please refer to Figure Ex3 1.9.
The current in the
2
R
resistor
is
=
6
V/2
R
=
3 A to the left. The current in the
5
R
resistor is
I
=
i
I2
=
3
A
I
2
A
=
5
A
This current flows toward the junction, that is, downward.
31.10. Model:
The batteries and the connecting wires are ideal.
Visualize:
Solve:
(a)
Choose the current
I
to be in the clockwise direction. If
I ends up being a positive number, then the
current really does flow in
this
direction.
If
I
is negative, the current really flows counterclockwise. There are no
junctions,
so
I
is the same for all elements in the circuit. With the 9 V battery being labeled
1
and the 6 V battery
being labeled
2.
Kirchhoff
s
loop law is
XAY =A%at, +AVK
+4V,,,
=+&,
lR&2
=O
Please refer to Figure Ex3 1.10.
Note the signs: Potential is gained in battery I, but potential is lost both in the resistor and in battery
2.
Because
I is
positive, we can say that
I
=
0.100 A flows from left to right through the resistor.
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 Spring '08
 Medvedev

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