CHAPTER
Current E1ectrici1y
page 598
Practice Problems
22.1
For all problems, assume that the battery voltage and
lamp resistances are constant, no matter what current
is present.
Current and Circuits
pages 591—600
594
The current through a Iightbulb connected
across the terminals of a 125-V outlet is
0,50 A. At what rate does the bulb convert
electric energy to light? (Assume 100 percent
efficiency.)
P= IV= (0.50 A)(125 V)
2
=
63 J/s
=
63W
A car battery causes a current of 2.0 A
through a lamp and produces 12 V across
it. What is the power used by the lamp?
P= IV= (2.0 A)(12 V)
=
24W
3. What is the current through a 75-W light-
bulb that is connected to a 125-V outlet?
P= IV
P
V
75W
125V
4. The current through the starter motor of a
car is 210 A. If the battery maintains 12 V
across the motor, how much electric energy
is delivered to the starter in 10.0 s?
P= IVand E= Pt
Thus, E = IVt = (210 A)(12 V)(100 s)
=
6. An automobile panel lamp with a resistance
of 33 II is placed across a 12-V battery.
What is the current through the circuit?
12V
V
036A
R
33fl
7. A motor with an operating resistance of 32 11
is connected to a voltage source. The current
in the circuit is 3.8 A. What is the voltage of
the source?
V = IR
=
(3.8 A)(32 1)
=
2V
1.2X10
A of current when
8. A sensor uses 2,OX 10 4
it is operated by a 3 0 V battery. What is the
resistance of tlic sensor circuit?
—
I
—
3.OY
fl
4
—15X10
2,OX10A
—
9. A lamp draws a current of 0.50 A when it is
connected to a 120-V source.
a.. What is the resistance of the lamp?
R=
b
f1
2
=2.4X10
=
What is the power consumption of the
lamp?
P = IV = (0.50 A)(120 V) = 6.OX 101 W
4J
2.5X10
1O A 75-W lamp is connected to 125 V.
5. A flashlight bulb is rated at 0.90 W. If the
lightbulb drops 3.0 V, how much current
goes through it?
P=1V
,
P
V
a.. What is the current through the lamp?
P
V
75W
125V
060A
b.. What is the resistance of the lamp?
0.90W
3.OV
Physics: Principles ond Problems
R==
I
O.60A
Solutions Manual
445
Chapter 22 continued
1 1 A resistor is added to the lamp in the previ
ous problem to reduce the current to half of
its original value.
a. What is the potential difference across
the lamp?
The new value of the current is
0.60 A
V
b.
0.30 A
=
2 II)
(0.30 A)(2.1 X10
IF?
=
1 V
6.3X10
53f1
F?
=
=
85mA
How much resistance was added to the
circuit?
4.5V
0.085A
V
I
14. Add a voltmeter to measure the potential
difference across the resistors in problems 12
and 13 and repeat the problems.
Both circuits will take the following form.
The total resistance of the circuit
is now
Rtotat
125V
V
0.3
=
A
1±
=
2 fi
4.2X10
Therefore,
Rres
Riamp
=
Rtota
=
4.2X 102 8
—
—
28
2.1 x10
2.1Xl0 fi
C.
Flow much power is now dissipated in
the lamp?
P
=
IV
=
(0.30 A)(6.3X 101 V)
=
19 W
Because the ammeter resistance is
assumed zero, the voltmeter readings
will be 60.0 V for Practice Problem 12 and
4.5 V for Practice Problem 13.
15. Draw a circuit using a battery a lamp, a
potentiometer to adjust the lamp’s brightness, and an on-off switch.
Lamp
page 600
12.—j [)raw a circuit diagram to include a 60.0-V
hatterv an ammeter, and a resistance of
12.5 f in series. Indicate the ammeter read
ing and the direction of the current.
c_i
Battery
Switch
4.80 A
12.Sfl
..60OV
Potentiometer
16. Repeat the previous problem, adding an
ammeter and a voltmeter across the lamp.
T
V
F?
60.OV
12.58
480A
13, Draw a series-circuit diagram showing a
4.5-V battery, a resistor, and an ammeter
that reads 85 mA. Determine the resistance
and label the resistor. Choose a direction
for the conventional current and indicate
the positive terminal of the battery.
446
Solution5 Manual
Ph vsic: Principles and Problems
Chapter 22 contrnued
Section Review
P.
=
1
vrn
=
(12 V)
112 f
2
22.1
2
P
=
V/R
=
(12 V)
/9.0
2
Current and Circuits
pages 591—600
page 600
17. Schematic I)raw a schematic diagram of a
circuit that contains a battery and a liglit
bulb, Make sure the lightbulb will light in
this circuit.
18.
1
I
Resistance Joc states that be arise I?
V/I,
if he inc reases the \ oltage, tlft resistanc e
will increase. Is Joe curl cc t Fxplain.
No, resistance depends on the device,
When V increases, so wiN I.
Resistance oti want to measure the i esis
tance of a long piece of wire. Show how you
would construct a circuit with a hatte, a
volt metei an am metei, and the wire to he
tested to make the ineasti cement Specify
what von would measure and hox you
would compute the resistance.
19.
4.0 W increase
21.
Energy A
circuit converts
E
when it is operated
I
\ 3.0mm
=
4J
4.4X10
.
I’riii IJ)!(’’ 111(1 PI)J)I(’i1)
I Ii.
The potential energy of the charges
decreases as they flow through the resis
tor. This decrease in potential energy is
used to produce heat in the resistor.
Practice Problems
22.2
Using Electric Energy
pages 601—605
page 603
23. A I 5- electric heater operates on a I 20—V
outlet.
is
the
c.
rn i
cut thi ough the heatci?
120V
15
I low muc h energy
in 30.0 s
=
Rt
2
I
=
4J
29X10
=
80A
is
used b the heater
(15 f)(30.0 s)
2
(8.0 A)
I low inu h thermal enei gy
in this time?
is
I iheratecl
2.9X 10 J, because all electric energy
is converted to thermal energy.
24. A 39-a
battery.
resistor is
a. What
V
F?
‘:
for
1(60.0 mm)
/
=
E
1 i(
J)J
of
22. Critical Thinking We say that power is
dissipated” in a resistor ‘lb dissipate is to
use, to waste, or to squander. What is “used”
when charge flows through a resistor?
b.
Power A circuit has 1 2
of resistance and
is connected to a 1 2 V battery I)etermine
the change in power if the resistance
decreases to () Q
J
/ 2 2X10
3J\
V
R
20,
2.2X 3fl3
energy whc n it is operated fbr ‘3.0 mm.
Deterin inc the amoti ii t of energy it will
a. What
Measure the current through the wire
and the potential difference across it.
Divide the potential difference by the
current to obtain the wire resistance.
16W
=
_
2
zS.P=P
=
1
16W—12W=4,OW
P
convert
4’
12 W
=
is
connected across a 45-V
the current in the circuit?
45V
39
12A
Sllu1 I,)11,s
\‘
1(111 11(11
‘147
Chapter 22 continued
take to charge the battery using a current of
73 A? Assume that the charging voltage is
the same as the discharging voltage.
(t3)IVt
Ecnarge
=
(1 3)(55 A)(1 2 V)(1 O h)
=
858 Wh
858Wh
(75 A)(12 V)
IV
3L Rework the previous problem by assuming
that the battery requires the application of
14 V when it is recharging.
(t3)IVt
Echarge
=
(1 3)(55 A)(1 2 V)(1 O Ii)
858 Wh
t
—
E
IV
—
B5BWh
(75 A)(14 V)
h
—
Section Review
Using Electric Energy
pages 601—605
22.2
page 605
32 Energy A car engine drives a generator,
which produces and stores electric charge in
the car’s battery The headlamps use the elec
tric charge stored in the car batteiy List the
forms of energy in these three operations.
Mechanical energy from the engine
converted to electric energy in the gen
erator; electric energy stored as chemi
cal energy in the battery; chemical
energy converted to electric energy in
the battery and distributed to the headlamps; electric energy converted to
light and thermal energy in headlamps
P
1
2
P
1R
2
V
=
IR
2
1
v
/R
2
)
1
(O.5V
=
2
1
V
=
0.25
35. Efficiency Fvaluate the impact of research
to improve power transmission lines on
society and the environment.
Research to improve power transmis
sion lines would benefit society in cost
of electricity. Also, if less power was
lost during transmission, less coal and
other power-producing resources
would have to be used, which would
improve the quality of our environment.
36. Voltage Why would an electric range and
an electric hot-water heater be connected to
a 240-V circuit rather than a 120-V circuit?
For the same power, at twice the volt
age, the current would be halved. The
R loss in the circuit wiring would be
2
I
dramatically reduced because it is pro
portional to the square of the current.
37. Critical Thinking When demand for elec
tric power is high, power companies some
times reduce the voltage, thereby producing
a “brown out” \‘hat is being saved?
Power, not energy; most devices will
have to run longer.
Chapter Assessment
Concept Mapping
page 610
38. Complete the concept mrp using the fol
lowing terms: watt, current, reszsta?ice
3 Resistance A hair dryer operating from
120 V has two settings, hot and warm. In
which setting is the resistance likely to be
smaller? Why?
Hot draws more power, P = 1V so the
fixed voltage current is larger. Because
I = VIR the resistance is smaller,
34. Power Determine the power change in a
circuit if the applied voltage is decreased by
one-half.
Physi
5:
Pnnciples and Problems
Solutions Manual
449