CHAPTER
20
ELECTRIC CIRCUITS
CONCEPTUAL QUESTIONS
____________________________________________________________________________________________
1.
REASONING AND SOLUTION
a. When
S
1
is set to position
A
, current can flow from the generator only along the path that
requires
S
2
to be set to position
A
.
Hence, the light will be on when
S
2
is in
position
A
.
b. When
S
1
is set to position
B
, current can flow from the generator only along the path that
requires
S
2
to be set to position
B
.
Hence, the light will be on when
S
2
is in
position
B
.
______________________________________________________________________________
2.
REASONING AND SOLUTION
When an incandescent light bulb is turned on, the
tungsten filament becomes white hot.
Since the voltage is constant, the power delivered to
the
light
bulb
is
given
by
Equation
20.6c:
.
From
Equation 20.5,
P
=
V
2
/
R
0
0
[1
(
)]
R
R
T
T
α
=
+
−
, where
a
is the temperature coefficient of resistivity and is a positive
number. Thus, as the filament temperature increases, the resistance of the wire increases,
and as the filament heats up, the power delivered to the bulb decreases.
____________________________________________________________________________________________
3.
SSM
REASONING AND SOLUTION
Two materials have different resistivities.
Two
wires of the same length are made, one from each of the materials. The resistance of each
wire is given by Equation 20.3:
/
R
L
A
ρ
=
, where
ρ
is the resistivity of the wire material,
and
L
and
A
are, respectively, the length and crosssectional area of the wire. Even when the
wires have the same length, they may have the same resistance, if the crosssectional areas
of the wires are chosen so that the ratio
/
L
A
ρ
is the same for each.
____________________________________________________________________________________________
4.
REASONING AND SOLUTION
The resistance of a wire is given by Equation 20.3:
/
R
L
A
ρ
=
, where
ρ
is the resistivity of the wire material,
L
is the length of the wire, and
A
is its crosssectional area.
Since the crosssectional area is proportional to the square of the
diameter, a doubling of the diameter causes the crosssectional area to be increased four
fold.
From Equation 20.3, we see that doubling both the diameter and length causes the
resistance of the wire to be reduced by a factor of 2.
____________________________________________________________________________________________
5.
REASONING AND SOLUTION
One electrical appliance operates with a voltage of
120 V, while another operates with a voltage of 240 V.
The power used by either appliance
is given by Equation 20.6c:
.
Without knowing the resistance
R
of each
appliance, no conclusion can be reached as to which appliance, if either, uses more power.
P
=
V
2
/
R
____________________________________________________________________________________________
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Chapter 20
Conceptual Questions
1031
6.
REASONING AND SOLUTION
Two light bulbs are designed for use at 120 V and are
rated at 75 W and 150 W.
The power used by either bulb is given by Equation 20.6c:
.
We see from Equation 20.6c that, at constant voltage, the power used by a bulb
is inversely proportional to the resistance of the filament.
Therefore, the filament resistance
is greater for the 75W bulb.
P
=
V
2
/
R
____________________________________________________________________________________________
7.
REASONING AND SOLUTION
Rather than state how many watts of power an appliance
uses, appliance instructions often give statements such as "10 A, 120 V" instead.
This
statement gives the current used by the appliance for a specific voltage.
It provides all the
information necessary to determine the power usage of the appliance from Equation 20.6a:
P
=
IV
.
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 Spring '08
 holland
 Current, Resistance, Light, Resistor, ohm, Electrical resistance

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