+
+
+
+
PEP 2017
Assignment 12
the resistor,
as obtained
in parts
(a) and (b). (d) What fraction of
E26.23,
a 20.0-Æ
resistor
is inside
3.50 A
the energy
supplied
by the
battery is stored in the capacitor? How
100
g of pure
water that
is surrounded
does
this fraction
depend on
R? water
by
insulating
styrofoam.
If the
+ 10.0 V
R2 R 1
... atCALC
(a) Using
Eq. (26.17)
26.89
is
initially
10.0°C,
how long
will it forEthe current in a discharging capacitor,
derive an expression
take
for its temperature
to rise tofor the instantaneous power
2.00for
A P
P = i 2R dissipated in the resistor. (b) Integrate the expression
58.0°C?
to find the total energy dissipated in the resistor, and show that this
Section
Kirchhoff’s
Rules
is equal to26.2
the total
energy initially
stored in the capacitor.
26.24 .. The batteries shown in Figure E26.24
the circuit in Fig. E26.24 have
CHALLENGE PROBLEMS
negligibly small internal resist... AtheCapacitor
26.90 Find
Burglar Alarm. Figure
30.0 P26.90
ances.
current through
20.0 V
V
The
capacitance
of a capacitor
can Vbe
(a) the 30 .0-Æ resistor;
(b) the 10.0
R
affected
by dielectric
material that,
5.00 V
20
.0-Æ resistor;
(c) the 10.0-V
although not inside the capacitor, is near
battery.
C
E
enough. toInthethe
capacitor
be polarized by
26.25
circuit toshown
Figure E26.25
theFig.
fringing
electric
field
existsinnear
a
A
in
E26.25
find (a)
thethat
current
resistor
28.0 V
charged
effect
usually of
R
R;
(b) thecapacitor.
resistanceThis
R; (c)
the isunknown
emf
+
the
order
of
picofarads
(pF),
but
it
can
be
used
with
appropriate
E. (d) If the circuit is broken at point x, what
electronic
circuitry
to detect
4.00surA
is
the current
in resistor
R? a change in the dielectricE material
+ might be the
rounding
the capacitor.
a dielectric material
. Find
26.26
the emfs ESuch
1 and E2 in the cirx 6.00 V
human
and the and
effect
described
above might be used in the
cuit
of body,
Fig. E26.26,
find
the potential
6.00 A
design of aofburglar
Consider
difference
point balarm.
relative
to pointthe
a. simplified circuit shown in
Fig. P26.90. The voltage source has emf E = 10003.00
V, Vand the
capacitor
has capacitance
The electronic circuitry
C = 10.0 pF.
878
CHAPTER
26 Direct-Current
Circuits
for detecting
the current, represented as an ammeter in the diaFigure
E26.26
gram, has negligible resistance and is capable of detecting a cur(b) the
drop
the
resistor;
(c) the
the200
capac20.0
1.00
rent
thatvoltage
persists
at aacross
level
ofVat
leastV1.00
for at on
least
mAcharge
ms
6.00 V
itor; (d)
current through
the +resistor?
A long
afterburthe
after
the the
capacitance
has changed
abruptly(e)
from
C totime
C¿. The
circuit
is completed
(after
timeifconstants)
what are
the val1.00 A
glar
alarm
is designed
to bemany
activated
the
changes
by
1.00
V capacitance
E1
4.00
V
ues of(a)
theDetermine
quantities the
in parts
(a)–(d)?
+
10%.
charge
on
the
capacitor
when
it is
10.0-pF
a
b
.
26.41 charged.
A capacitor
is charged
to a potential
12.0 V and
is then
fully
(b)
capacitor
is fully ofcharged
before
the
2.00IfA the1.00
V anE2internal
connected
a voltmeter
having
3.40capaciMÆ.
V taken forofthe
intruder
is to
detected,
assuming
that
timeresistance
+ the2.00
After to
a time
of 4.00
s theisvoltmeter
readsto3.0
What are
(a) the
tance
change
by 10%
short enough
beV.ignored,
derive
an
capacitance
(b) the the
timecurrent
constant
of thethe
circuit?
equation
thatand
expresses
through
resistor R as a func26.42of .theAtime
capacitor
is connected
through(c)
a 0.895-MÆ
12.4-mF
tion
t since
the capacitance
has changed.
Determine
. Inathe
circuit shown in difference
Fig. E26.27,
current
in
26.27
resistor
constant
of find
60.0(a)
V. the
(a)the
Compute
the
rangeto of
values ofpotential
the resistance R that
will
meet
design
the
resistor;
(b)
the
unknown
emfs
and
E
(c)
theconnecresist3.00-Æ
E
;
1
2
the
charge
on
the
capacitor
at
the
following
times
after
the
specifications of the burglar alarm. What happens if R is too small?
ance
R.
that three
currents
given.
tionslarge?
areNote
made:
5.0will
s,
10.0
s,are
20.0
s, and
s. (b)
Compute
Too
(Hint:0,You
not be
able to
solve100.0
this part
analytically
the
charging
currents
at
the
same
instants.
(c)
Graph
the
results
of
but must use numerical methods. Express R as a logarithmic funcpartsof(a)R and
for t quantities.
between 0 Use
and a20trial
s. value of R and calculate
tion
plus(b)
known
.. expression
26.43the
CP In the acircuit
shownContinue
in
from
new value.
to do
this until the input
Figure
E26.43
Fig.
E26.43
both
capacitors
are
iniand output values of R agree to within three significant figures.)
tially charged
to 45.0Network.
V. (a) How
... An Infinite
S
26.91
longshown
after closing
the switch
S willFigure
15.0P26.91
As
in Fig. P26.91,
a net+
+ 20.0
50.0 V
the potential
across
capacitor
– mF
mF –
work
of resistors
of each
resistances
R1 c R1
R1
a
be
reduced
to
10.0
V,
and
(b)
what
R1 and R2 extends to infinity
will be the
current
at thatthat
time?
30.0
R2
R2 V R2 and
toward
right. Prove
the
so on
.the
26.44resistance
A resistor
capacitor
total
the ainfinite
RT ofand
b
are
connected
in
series
to
an
emf
source.
The
time
constant
for
the
d
R1
R1
R1
network is equal to
circuit is 0.870 s. (a) A second capacitor, identical to!x
the first, is
RTin= series.
R1 + What
2R12is+ the
2R1time
R2 constant for this new circuit? (b)
added
In
the
original
circuit
a second
capacitor,
identical
to the
first, to
is
(Hint: Since the network
is infinite,
the resistance
of the
network
connected
parallel
with
thealso
firstequal
capacitor.
is the time conthe
right ofinpoints
c and
d is
to RT.What
)
stant for
new circuit?
...thisSuppose
26.92
a resistor R lies along Figure P26.92
. Anofemf
26.45edge
E = in120
source
each
a cube
(12 with
resistors
all)V, a resistor with R =
b
80.0 connections
Æ, and a capacitor
C = Find
4.00 the
mF are connected in series.
with
at the with
corners.
As the capacitor
charges,
when the
equivalent
resistance
between
twocurrent
diago-in the resistor is 0.900 A,
what is
the magnitude
on each
nally
opposite
corners of
of the
the charge
cube (points
a plate of the capacitor?
. A 1.50-mF
26.46
a 12.0-Æ resisand
b in Fig.
P26.92).capacitor is charging through
a current when the
tor using
10.0-V
battery. What
willand
be the
... aBIO
26.93
Attenuator
Chains
1
1
capacitor The
has acquired
it be
4 of its maximum
4 of the
Axons.
infinite network
of resistorscharge?
shownWill
in Fig.
P26.91
is
maximum current?
26.47 .. CP In the circuit shown in Figure E26.47
+
2
electric
at its
if it carries
a 1.75-A
current?
(c) If we cut circuit.
these
varied
current
G is4
.field
bulb
what isuntil
the the
power
dissi-inisthe
Rresistors
25.65
is the ofcurrent
through
theare
body?
(b)
What
thegalvanometer
power
CALC
A midpoint
material
resistivity
r mm
is carries
4 removed,
Figure
25.13 . the
A rod
14-gauge
copper
wire
ofofdiameter
1.628
a half?
form a cable
thezero;
same
length
asperson’s
a then
single
strand,
(b) coninto
two
75.0-cm
halves,
what is
the
resistance
of each
. A
the
bridge
isthe
said
toofbeaand
balanced.
(a)safe
Show
that under
pated
in
each
of
the
remaining
bulbs?
25.80
lightning
bolt
strikes
one
end
steel
lightning
rod,
dissipated
in
his
body?
(c)
If
power
supply
is
to
be
made
by
formed
into
a
solid,
truncated
cone
of
height
h
P25.65
current of
12.5..mA.
(a) Whattoisthe
the potential
difference
across
a copper
nected end to end this
to form
a wirethe
120unknown
times asresistance
long as a issingle
25.77
According
Electrical
Code,
condition
given
Figure
E26.16
Which
light
bulb(s)
glow
brighter
as
a ms.
producing
current
burst
that
laststhe
for
Theby
rodX = MP>N.
65
increasing a(e)
its15,000-A
internal
resistance,
what
should
internal
resistance
and radii
and r2(b)
at What
eitherU.S.
endNational
(Fig.
P25.65).
r1 wire?
2.00-m length
of the
would
the potential
difference
r1 strand.
Exercises
845
wire
used
for
interior
wiring
of
houses,
hotels,
office
buildings,
(This
method
permits
very
high
precision
comparing
resistors.)
result
of
removing
Which
bulb(s)
?
R
is
2.0
m
long
and
1.8
cm
in
diameter,
and
its glow
other
isAconbe for the maximum current in4 the above situation
toinend
be4.00
1.00
mA
(a)be
Calculate
the were
resistance
ofinstead
the cone
betweenbut all else
in846
part (a)
if
the
wire
silver
of
copper,
6.00 V
CHAPTER
25plants
Current,
Resistance,
andno
Electromotive
Force
and
industrial
is(Hint:
permitted
to carry
more
than a specified
(b) ground
If thebyDiscuss
galvanometer
Gareshows
zero deflection
when
less
brightly?
why
there
differnected
to
the
35
m
of
8.0-mm-diameter
copper
wire.
or
less?
the
two
flat
end
faces.
Imagine
slicing
the
Section 25.4 Electromotive Force and Circuits
were themaximum
same? amount of current. The table below shows the maxi25.0 V
and
what
is
the
M
=
850.0
Æ,
N
=
15.00
Æ,
P
=
33.48
Æ,
.
ent
effects
on
different
bulbs.
(a)
Find
the
potential
difference
between
the
top
of
the
steel
rod
25.71
cone
into
very
many
thin
disks,
and
calculate
the
BIO
The
average
bulk
resistivity
of
the
human
body
.
(a) the.current
in thethe
circuit
(magnitude and direction) and (b) the
E25.31
25.31 .. An
ideal6.50
voltmeter
V isdiameter
con- Figure
25.14
A wire
m long
of 2.05
mm
has awith
resist25.28
Consider
circuit
8.00 V
mum
current
for with
several
common
sizes
of wire
varnished
unknown
resistance
X?the
Figure
E25.28
max such
h equal
26.16
Consider
the
circuit
shown
and
the
lower
end
of
the copper
wire
during
thein5.0
current
burst.
volume
(in
liters)
of Ione
gasoline
athe
total
heat
of
combustion
resistance
of
disk.)
(b)
Show
that
your
(apart
from
surface
resistance
of
skin)
is about
The
Æ # m.
terminal
voltage
Vab
of
the.termi16.0-V
battery.
PROBLEMS
nected
a 2.0resistor
and ahas
battery
Æ What
Æ.
ance
of to
0.0290
material
is
wire
most
likely
made
of?
shown
in
Fig.
E25.28.
The
. A 2.36-mF
cambric
insulation.
“wire
gauge”
is. V
a 5.0
standard
0.5
V
26.82
capacitor
thatand
is the
initially
uncharged
is conFig.
The
current
Find
totalE26.16.
energy
deposited
in can
thethrough
by approxithe
cur-20.0 V
to theemf
energy
obtained
intungsten
part
(a)? filament
(See
Section
density
ofusednaltovoltage
result
agrees
Eq.The
(25.10)
when
conducting
path
between
the hands
be
represented
r15.0
rcm
..
1 =17.6;
2 the
rrod
24.0
V wire
with
5.0cylindrical
V
and with
internal
resistance
25.15
of electrical
thethe
24.0-V
battery
A
long
+ with a
.(b)
25.55
An
conductor
designed
toin carry
large
currents
Problems
847source with
Figure
3
describe
the
diameter
ofawires.
Notewith
thatconcenthe
larger
the
diameter
of E25.38
nected
in
series
with
aA,5.86-Æ
and an
emf
.
+resistor
resistor
is
4.00
the
direction
6.00-Æ
rent
burst.
gasoline
is
)
(c)
If
generator
an
average
electri900
kg>m
.
25.66
mately
as
a
cylinder
1.6
m
long
and
0.10
m
in
diameter.
The
skin
CALC
The
region
between
two
shown
Fig.isE25.31.
(a) What
0.5 Æ as of
21.2
V. What cross
(a)section
the interdiameter
1.00inmm
to be the
used
in a machine for which the tem- r2 ishas
a circular
2.50
mm
diameter
andresistance.
is 14.0
mÆ
the
wire,
the 0.45
smaller
wire
and
negligible
(a)
Just+ after the
E
= be
120What
V
. areshown.
are
the
currents
through
25.81
12.0-V
battery
has
anVin
resistance
0.24
power
output
kW
connected
battery,
much
resistance
can
made
negligible
by internal
soaking
theofhands
in and
salt
tric
conducting
spheres
withgauge.
radiito120°C
atheand
b is
4.00
Ainternal
1.6 V
16.0
iscalthe
current
in ofthe
resistor?
Æ istemperature
resistance
r ofAthe
battery
and
perature
will
range
from2.0room
to how
120°C.
It nal
2 up
long.
The
resistance
between
its
ends
is
(a)
What
is
the
0.104
Æ.
#
connection
is
made,
what
are
(i)
the
rate
at
which
electrical
+
and
awater.
capacity
of
Exercise
25.47).
The
is E energy
50.0
A 20.0-Æ
hresistance
time
will
be
required
for
it toAaluminum
charge
thecup
battery
fully?
(a)for25.0-Æ
What
the
between
the
hands
the skin
filled
with
a gauge
conducting
with
resistivity
resistance
. voltage
An
with
mass
0.130
contains
1.2
Æ
equation
iniscircuit
terms
of(see
and
What
is theifbattery
tempera¢R
¢rresistors?
¢L.) (e)
Vkg
(b)
What
is
terminal
of
the
Imax
Wire
(cm)
1A2
(b)
the resistance
of
the
will
carry
a the
current
of
12.5
atmaterial
allDiameter
temperatures
(consult
Tables
resistivity
of thebyR
material?
(b)
Ifthe
the
electric-field
magnitude
in(a)
the
aadissipated
b the
.
is
being
in
resistor;
(ii)
the
rate
at
which
.
26.17
In
circuit
shown
in
Fig.
charged
passing
10-A
current
through
it
for
5.0
h.
What
is the electri2.0
V
25.48
In
the
circuit
analyzed
in
Example
25.8
the
4.0resistor
Æ
4.00
A
resistance
is
negligible?
(b)
What
potential
difference
between
the
(a)
Show
that
the
resistance
between
the
spheres
is
given
by
r.
E26.17
kg(a)ofis
water.
Thebeheating
is placed
in in
the
waterresistor?
and ture coefficient of resistance a for the mercury
defined
R column, asFigure
battery?
(c)
What
the reading
on maximum
theelement
14
0.163 electric
18
25.1
and0.200
25.2).
What
will
the
field
this
conductor
isterminal
what
isvoltage
the in
total
current?
(c)
Ifincreasing;
the
mate1.285.0
V>m,
cal
energy
stored
theacross
capacitor
is100
(iii)that
the electrical
Vvoltage
9.0 V(b)
E26.17,
the
the
2.00-Æ
the
during
charging?
What
total
electrical
is
replaced
by
a
8.0resistor,
as
in
Example
25.9.
(a)
Calculate
the
Æ
hands
is
needed
for
a
lethal
shock
current
of
mA?
(Note
the
electrical
energy
dissipated
in
the
resistance
of
the
heating
elein
Eq.
(25.12)?
How
does
this
value
compare
with
the
temperature
.
28
voltmeter?
Explain
your
answers.
12
0.205
25
filament, and (b) what will be its resistance with that field? 25.29
A copper
transmission
rial hasenergy
free electrons
per
cubic
meter,
find
average
8.5
* is10supplied
E (i), (ii),
8.0
Vsource?
1.4
V
power
output
ofV.
the
How
dothe
the
answers
to parts
r
1
1
resistor
is
12.0
What
are
the
emf
of
the
to
the
battery
during
charging?
(c)
What
electrirate
of
conversion
of
chemical
energy
to
electrical
energy
in
the
batyour
result
shows
that
small
potential
differences
produce
danger.
ment
all
goes
into
the
cup
and
water.
The
element
itself
has
very
coefficient
of
resistivity?
Is
the
effect
of
the
change
in
length
+
25.32
shown in R
30
(c)
What The
will circuit
be the10maximum
potential
cable
kmunder
long and
10.0 cm in diameter
carries a current of 125 A.
= 0.259
adrop- over
b the full length
drift 100
speed
the
conditions
of+the
part
(b).
and
(iii)
compare?
(b)
Answer
the
same
questions
as
in
part
(a) at a
Figure
E25.32
battery
and
the
current
through
the
6.00-Æ
cal
energy
is
dissipated
in
internal
resistance
during
charging?
tery.
How
does
your
answer
compare
to
the
result
calculated
in
ous
whendrop
the skin
is damp.)
(c) With
the current
4p0.326
smallcontains
mass. How
itatakebfor the temperature
of(a)
theWhat
important?
Fig.
bat- time does
8 twomuch
40
of
theE25.32
filament?
isAcurrents
the
potential
across
the cable?
(b)inHow
muchis in part
..
25.56
plastic
tubetime
25.0
m
long
and
3.00 cm
diameter
long
after
thein
connection
is made.
(c) Answer
theVsame
quesoelectrical energy
o
resistor?
.
(d)
The
battery
is
now
completely
discharged
through
a
resistor,
Example
25.8?
(b)
Calculate
the
rate
of
dissipation
(b),
what
power
is
dissipated
the
body?
cup
and
water
to
rise
from
to
?
(The
change
of
the
34.5
25.68
21.2
C
C
(a)
What
is
the
poten..
1.00
2.00
V
Figure
P25.68
teries,
each
with
an
emf
and
6
0.412
60
25.16
electrical
energy
issolution,
dissipated
as Figure
A ductile metal wire has resistance1.6
R.VWhat
16.0will
V be the
dipped
into
awith
silver
depositing
a layer
of when
silver
0.100
mm on
E25.30
.
tions
asin
in
part
(a)electric
at
the
instant
the
charge
the6.00
capacitor
is
.
V
(b)
Derive
an
expression
for
the
current
density
as
a
function
of
26.18
A
Three-Way
Light
Bulb.
A
again
a
constant
current
of
10
A.
What
is
the
external
circuit
in
the
internal
resistance
of
the
battery.
How
does
your
answer
com+
25.72
A
typical
cost
for
power
is
$0.120
per
kilowattresistance
of
the
Nichrome
due
to
its
temperature
change
can
be
tial
difference
the
circuit
V
an internalofresistance,
andterms
two of R if it0.462
5 in
65
ad outer
thermal
energy
hour?
resistance
this wire
is stretched to three times
Section
25.5 every
Energy
anditslight
Power
inthe
Electric
Circuits
thick uniformly
over
surface
of
tube.
Ifisbrightness
this
coated
4.00
V time.
0.50
V
one-half
final
value.
a (c) Use
Vab between
radius,
in
of the
potential
difference
the spheres.
three-way
bulb
hasenergy
three
resistance?
(e)the
What
total
supplied
totube
theWhat
exterpare
toneglected.)
theFind
result
calculated
inthat
Example
25.8?
theb results
hour.
(a)
Some
people
leave
their
porch
light
on
all
the
9.00
V is
of
Fig.idealized
P25.68?
(b)
What
iselectrical
the
.
c
b
resistors.
(a)terms
the
A
4 current
0.519
its
original
length,
assuming
the density
and resistivity
of85theof 25.30
An
ammeter
is
.
+
25.39
Light
Bulbs.
rating
of ahigh)
light
bulb
(such
as filaments.
a in
isthe
then nal
connected
across
aThe
battery,
what
willabe
the
current?
.12.0-V
26.83
Apower
resistor
and
resistor
connected
in
224-Æ
589-Æ
(c)and
Show
that
the result
inresistance
part
(a) reduces
to of
Eq.the
(25.10)
settings
(low,
medium,
but
only
two
(a) A par..
resistor?
(f)to
What
total
energy
isand
dissipated
the
parts
(b)change
toresistor
calculate
the
net
output
thetoyearly
cost
keep
a4.00-V
75-W
bulband
burning
day
night?
(b)are
Sup25.62
terminal
voltage
of
the
A
with
is
connected
tobattery.
a The
battery
that
5.0
Vpower
9.0
Vwhen
in
the (a)
circuit
(magnitude
and
material
do not
when
the
wire
is Rstretched?
(Hint:
connected
battery
as
shown
in electrical
100-W
bulb)
theticular
power
itat dissipates
when
connected
across
a120-V
25.57 ..
Onais
your
first
day
work
as an
electrical
technician,
you
2.00
10.0
series
across
aemf
90.0-V
line.
(a) V
What
isV(b)
the
voltage
across
each
(a)
What
considerations
determine
the
maximum
current-carrying
L
=
b
a
separation
between
the
spheres
is
small.
three-way
light
bulb
connected
across
a
line
can
disinternal
resistance?
(g)
Why
are
the
answers
to
parts
and
(e)
not
8.0
V
1.4
V
How
does
your
result
compare
to
the
electrical
power
dissipated
in
pose
your
refrigerator
uses
400
W
of
power
when
it’s
running,
and
r
=
0.40
Æ
battery?
(c)
A
battery
with
has
emf
12.0
V
and
internal
resistance
.
For
what
two
–
direction);
(b) the
amount
of metal
does terminal
not change, so stretching out+ the wire will Fig.
E25.30.
Find
(a) thethe
reading
of
120-V
potential
difference.
What
isW,
theper
resistance
of
(a)
a piece
100-W
are
to
determine
resistance
meter
of
a +long
of two dresistor
0V
6.0W.
...
resistor?
(b)
voltmeter
connected
across
the the
reads
224-Æ
capacity
of The
household
wiring?
(b)
A total
of
4200
W
power
to asked
a W?
25.67
temperature
ofresistor
resistance
in is
Eq.
sipate
W, A
120
180
Describe
filaments
are
the
same?
the 8.0resistor
calculated
fordissipated
thiscoefficient
circuit
Example
25.9?
Æ
it runs
hours
a60day.
What
isorthe
yearly
cost
of how
operating
10.30
V 8and
internal
resistance
will
the
power
ininthe
becof80.0
voltage
theRas
16.0-V
batVcross-sectional
affect
itsvalues
area.)
the
ammeter,
(b)
the
current
through
8.00
Vyour
0.50
V
ab of of
bulb
and
(b)
a
60-W
bulb?
(c)
How
much
current
does
each
bulb
wire.
The
company
you
work
for
is
poorly
equipped.
You
find
a the
a
23.8
V.
Find
the
voltmeter
resistance.
(c)
Find
reading
of the
.
be
supplied
through
the
wires
of
a
house
to
the
household
electri..
a in can
(25.12)
equals
the
temperature
coefficient
of
resistivity
Eq. 0.50
arranged
in the
the25.81
bulb,with
and charge
calculate
the
resistance
of each
filament.
..
25.82
Repeat
Problem
and
discharge
+ currents
25.49
A
25.0bulb
is
connected
across
the
terminals
of
Æ
refrigerator?
is
inserted
in
cir25.63
CP
BIO
Struck
by
Lightning.
Lightning
strikes
Æ
.
tery;
(c)
the
potential
differ25.17 In household wiring, copper wire 2.05 mm in diameter is the
4.00Æ
resistor,
(c)and
the an
terminal
draw
in normal
battery,
a 30
voltmeter,
ammeter,
but
no meter
for the
directly
same
voltmeter
iffilament
it iswith
connected
across
resistor.
(d)How
The
589-Æ
cal
appliances.
thehigh
potential
difference
across
group
appli..use?
(25.6)
only
if If
the
coefficient
of
thermal
expansion
is
small.
(b)
Suppose
the
with
the
higher
resistance
burns
of
charging
and
discharging
times
now
be 1.7
h out.
12.0-V
battery
having
of
What
percent3.50
Æ
8.00
Vwill
25.73
A
12.6-V
battery
negligible
internal
resistance
atA.
d,The
with
its car
negative
40aof
ms.
currents
as
asainternal
25,000
A
that
last
forthe
about
If A
a .cuit
ence
of Find
point
a with
respect
to
point
c. resistance.
(d)
Using
25.20
as
V
voltage
of
battery.
often
used.
the
resistance
of
24.0-m
length
of Fig.
this
wire.
acinvolve
25.40
Ifthe
aresistance
“75-W”
bulb
(see
Problem
25.39)
isthe
connected
across
measuring
(an
ohmmeter).
You
put
leads
from
4.00
V
readings
on
this
voltmeter
are
lower
than
the
“true”
voltages
(that is,
ances
is
120
V,
determine
the
gauge
of
the
thinnest
permissible
cylindrical
column
of
mercury
is
in
a
vertical
glass
tube.
At
20°C,
much
power
will
the
bulb
dissipate
on
each
ofthe
the
three
brightness
rather
than
5.0toh.a What
differences
in of
performance
do you
see?
age of person
the power
of thebybattery
dissipated
across
the properties,
internal
is connected
series
combination
a 3.2- Æ resistor
that
obeys
connected
to the
negis potential
struck
a bolt
lightning
these
the terminal
model,
graph
the
rises
andisofdrops
in thiswith
circuit.
avoltmeter
220-V potential
difference
(as
used
in Europe),
how
much
across
the
terminals
of is
the
battery,
and
the meter
reads
without
voltmeter
present).
Would
it (low,
be
possible
tobut
design
a volt..lawCP
wire
thatwill
can
be
used.
(c) Suppose
wire
used
in this
house
is
of
the
length
ofpass
the
mercury
column
isWe
12.0
cm.
The
diameter
the
settings?
What
will
be
the
medium,
or high)
on
25.83
Consider
the
cir-that
resistance
and
hence
isSnot
available
to thethe
bulb?
Ohm’s
and
a the
thermistor
doesbrightness
Ohm’s
.current
ative
terminal
of
the
8.00-V
battery.
What
isnot
theobey
difference
oflaw
potenthrough
his
body.
shall
assume
that
his of
mass
25.33
When
switch
in
Fig.
E25.33
power
itcut
dissipate?
12.6 V.does
Youshown
off
aFig.
20.0-m
length
of wire
and
connect
it to2the
meter
that
gave
readings
higher
than
the
“true”
voltages?
Explain.
the
gauge
found
in
part
(b)
and
has
total
length
42.0
m.
At
what
.
Figure
P25.83
each
setting?
(c)
Repeat
part
(b)
for
the
situation
in
which
the
filamercury
column
is
1.6
mm
and
doesn’t
change
with
temperature
cuit
in
P25.83.
The
25.50
An
idealized
voltmeter
is
connected
across
the
terminals
instead
has
a
current–voltage
relationship
with
,
V
=
aI
+
bI
a
=
75 voltmeter
kg, that heVisof
wet
all, heFigure
is in a E25.33
rainstorm) and therefore .tial Vbc between the terminals of the 4.00-V battery now?
is open,isthe
the(after
battery
25.41
European
Light
In with
Europe
standard
voltage
.inBulb.
battery,
with
an ammeter
series
with
it toRmeasure
theout.
in to the plates of a
26.84
resistor
iscurrent
connected
=thecurrent
850
Æ
rate
isresistance
energy
dissipated
inÆ
theappliance
wires?
(d)
house
is (which
built
in
ment
with
the
lower
resistance
burns
because
glass
small
coefficient
of
The coefemf
has
interof a 3.08
15.0-V
battery,
and
a 75.0is The
also
connected
What
is the
through
3.2- Æ
3.8 source
Æ .and
b potential
=negligible
1.3AÆ>A.
has
ofa1.0
and
that
histhermal
body
isexpansion.
all
water
isa 25.69
kÆ,
The
difference
across
the
terminals
ofthe
a battery
reads
V.a When
thehas
switch
is
closed,
Rcut
2 United
in
homes
isThe
220ammeter
V26.19
instead
of
the
120A.
V
used
in
the
States.
the
wire.
reads
7.00
You
then
off
a= 40.0-m
..
charged
capacitor
with
capacitance
Just
before the
C battery
4.62
mF.
community
where
thevoltmeter
consumer
costmercury
ofapproximation).
electric
energy
isTable
$0.11
per
Vgiven
Working
Late!
You
are
working
late
in your
ficient
of
volume
expansion
of
the
is
in
17.2,
nal
resistance.
The
resistors
across
its
terminals.
If
the
reads
11.3
V:
(a)
how
much
resistor?
reasonable
for
a
rough,
but
plausible,
(a)
By
how
is
8.40
V
when
there
is
a
current
of
1.50
A
in
the
from
the electronics
the voltmeter reading drops to 2.97 V,
Therefore
a
“100-W”
European
bulb
would
be
intended
for
use
length
of
wire
and
connect
it
to
the
battery,
again
with
the
ammeter
connection
is
made,
the
charge
on
the
capacitor
is
6.90
kilowatt-hour.
If
the
house
were
built
with
wire
of
the
next
larger
..
shop
and
find
that
you
need
various
resistors
for
a
project.
But
alas, (a)
all
its
resistivity
at
20°C
is
given
in
Table
25.1,
and
its
temperature
have
resistances
=
6.00
Æ
R
power
is
being
dissipated
by
the
appliance,
and
(b)
what
is
the
25.74
A
cylindrical
copper
cable
1.50
km
long
is
connected
1
degrees
Celsius
this lightning bolt
negative
to the positive
terminal.
When the
current(a)isIf3.50
in the mC.
and themany
ammeter
A reads
1.65would
A. Find
r Eincrease the temwith
a 220-V
potential
difference
(see
Problem
25.40).
youAhow
ininis
series
toRmeasure
the
The
ammeter
reads
4.20
Even
C What
Estored
What
is
the
initially
in
theRA.
capacitor?
(b)
is the
diameter
than
that
found
ingiven
part
(b),
what
would
savings
you
have
is
aenergy
big
box
of 10 .0-Æ
resistors.
you
can make
coefficient
of
isof
in Table
25.2.
Atthe
20°C,
what
1Show
and
.current.
The
capacitor
=220.0-V
4.00
Æ
+ be
internal
resistance
the
across
potential
difference.
(a)
What
should
be
its What
diam2 a
of
75resistivity
kg battery?
of
water?
(b) Given
that
the (a)
internal
body
temreverse
direction,
the
potential
difference
becomes
10.20
V.
(a)
the
emf,perature
the internal
resistance
the
bring
“100-W”
European
bulb
home
to the
States,just
what
though
the equipment
you
available
to United
you
isresistor
limited,
your
electrical
power
dissipated
the
the
costs
in
one
Assume
thetemperature
appliances
E25.51
.electricity
of
following
equivalent
resistances
by
a after
combination
of your
the
between
the
ends
of Figure
thethat
mercury
column?are
(b)kept
The a is
has
Cthe
=have
9.00
mF
25.51 perature
In resistance
the
circuit
in37°C,
Fig.year?
E25.51,
eter capacitance
so thateach
it resistance
produces
heat
at.battery?
a rate in
of(b)
75.0
W?is(b)
What
theconnection
about
would
person’s
actually
of (b)
the
What
the
emf
ofis the
battery,
and
theis
circuit
resistance
R. the
should
bethe
its internal
U.S.of
rating?
How
much
current
will
the
boss assures
you
its
high
quality:
The
ammeter
has
very
small
ispower
made?
(c)
What
is
theÆ,
electrical
power
dissipated
in
on
for
an
average
of
12
hours
a
day.
resistors:
(a)
(b)
(c)
(d)
10
.0-Æ
35
1
.0
Æ,
3
.33
Æ,
7 .5theÆ.resistor
mercury
column
is
heated
to
60°C.
What
is
the
change
in
its
resisR
When
the
capacitor
is
fully
electric
field
inside
the
cable
under
these
conditions?
find
(a)
the
rate
of
conversion
of
internal
that
much?
Why
not?
What would 1.0
happen
first?
battery?
Assumeincrease
that the
two
meters
are
ideal,
12.0
V
V
100-W
European
draw
normal
use
inresistance.
the United
States?
A fluorescent
resistance,
and thebulb
very
large
What
..(c)In
a
d coeffi. Ininhas
atvoltmeter
the
instant
when
the
energy
stored
capacitor
has
decreased
25.78
Compact
Compact
26.20
the
circuit
shown
in
Fig.
tivity?
What
is
theFluorescent
change
in
length?
Explain
..
charged,
magnitude
of
the
charge
on
its
plates
istheQis
=the
36.0
mC
+ why
25.75 ....the
A
Nonideal
Ammeter.
Unlike
theinidealized
amme(chemical)
energy
to
energy
S
theelectrical
Bohr
model
of its
theBulbs.
hydrogen
atom,
thethe
electron
BIO
A person
with
body
resistance
between
his
hands
of.
so
they 25.64
don’t
affect
the
circuit.
.25.70
E26.20
25.42
Aofbattery-powered
global
positioning
system
(GPS)
resistance
1 meter
of wire?
toemf
half
the
value
calculated
inispart
(a)? Figure
bulbs
are
much
more
efficient
at the
producing
light
than
are
ordinary
15expansion,
E26.20,
the
rate
at
which
dissiR
cient
of
volume
rather
than
the
coefficient
of
linear
Calculate
the
.
E
ter
described
in
Section
25.4,
any
real
ammeter
has
a
nonzero
within
the
battery;
(b)
the
rate
of
dissipa1
.
makes
around
nucleus.
What
is
the
average
10
kÆ
accidentally
grasps
the
terminals
of
a
14-kV
power
supply.
rev>s
6.0
*
10
25.34 In the circuit of Fig. E25.32,
operating
onlength
9.0 .V
draws
aisenergy
current
ofiswelding
0.13
A.
How
much
25.58
A 2.0-mm
of
wire
made
by
the
end
of a battery
26.85
A
capacitor
that
uncharged
is10.0
connected
in
.. internal
incandescent
bulbs.
initially
cost
much (d)
more, butisthey
lastreceiver
far .(a)
pating
electrical
is
20.0
W.RP25.84.
expansion,
determines
change
inelectron?
length.
25.84
CP
Consider
the
circuit
ininitially
Fig.
The
resistance.
(a)
Anresistance
ammeter
resistance
connected
inV 10.0 V
tion 5.0ofcurrent
electrical
inThey
thethe
battery;
at aenergy
point
on
the
orbit
of replaced
the
If the
ofwith
theshown
power
supply
Æ, what
Aisis2000
the
is removed
and
by a What
resistor the
of change
Æ resistor
electrical
energy
does
it
consume
during
1.5
h?
120-cm-long
silver
wire
to
the
end
of
an
80-cm-long
copper
wire.
series
with
a
resistor
and
an
emf
source
with
E 2.00
=resist110
V and neglonger
and
use
much
less
electricity.
According
to
one
study
of
(a)
Find
and
What
is(b)
the
R1negligible
Ra2battery
. (b)
in
itsofresistance?
(Hint:
Since
the
percentage
in r and L is
. dissipation
has
emf
60.0
V
and
internal
resistance.
,
=
R
Æ
series
with
a
resistor
R
and
of
emf
and
internal
E
(c) the25.65
rate
of this
electrical
20.0
V
2
the
current
through
the
person’s
body?
What
is
the
power
CALCR.AWhen
material
resistivity
ischanges
10.0
V
10.0
V
Figure
unknown
resistance
isofdone,
an idealrvoltmeter
con. Consider
25.43
a resistor
with
length
L,Find
uniform
cross-sectional
Each
of3.00
wire
is, and
0.60
mm
in
diameter.
The
wire
is atcircuit
roomattained
c(25.10)
ligible
internal
resistance.
Just
the
is completed, the
these
bulbs,
aa compact
bulbit that
produces
as much
light
a 100-W
bderive
of
the
(c)
the after
curarethe
small,
you
may
find
helpful
from
Eq.as
an piece
. After
capacitors
have
Cance
mF
Cmeasured
= 6.00
mF
The
current
by
the
ammeter
is Ito
Find
the safe
current
energyacross
in
external
resistor.
1 = r.
2 battery?
inemf
his
body?
(c)
If the
power
supply
be
made
by V 5.0 V
formed
into
solid,
truncated
cone
of
height
h Vcurrent
A.-5
nected
the
points
b and
c reads
1.9
V.to
Find
(a) 5.0
the
in area
P25.65
Water
A, dissipated
and
uniform
resistivity
that
isgiven
carrying
a6.5
current
with
unirboth
temperature,
so the
resistivities
are
asR
inis
Table
25.1.
A potencurrent
through
the
resistor
The
time5.0
A.
*
10
incandescent
bulb
uses
only
23
W
of
power.
The
compact
bulb
..
rent
through
and
the
10.0-Æ
their
final
charges,
the
charge
on
is
.
(a)
What
isconstant for
C
Q
=
18.0
mC
25.52
typical
flashlight
2 what
1 should
1
increasing
its internal
resistance,
the internal
resistance
andAand
radii
andsmall
(Fig. P25.65).
r1 the
r2 at either
the
circuit
(b)
resistance
R. end
(c) Graph
the potential rises
r1
form
current
density
J.
Use
Eq.
(25.18)
to
find
the
electrical
power
tial
difference
of
5.0
V
is
maintained
between
the
ends
of
the
the
circuit
is
What
are
the
resistance
of
the
resistor
5.2
s.
lasts
10,000
hours,
on
the
average,
and
costs
$11.00,
whereas
the
resistor.
Calculate
total
elec- R1to
30.0
V 5.0and
V the
thefor
final
charge
on C(d)
? (b)
What
isthe
the
resistance
? be 1.00
contains
batteries,
having
of between
1.5 V, connected in
2current
be
the
maximum
in
the
above
situation
mA
(a)two
Calculate
theeach
resistance
ofan
theemf
cone
and
drops
in
this
circuit
(see
Fig.
25.20).
dissipated
per unitwire.
volume,
p. Express
your result
incopper
terms of
(a) +E
2.0-m composite
(a)power
What
isthe
thecapacitor?
current
in
the
section?
capacitance
ofconsumption
incandescent
bulb
costs
only
$0.75,
but
lasts
just
750
hours.
The
trical
in
all
the
series
with
a
bulb
having
resistance
(a)
If
the
internal
resist17
Æ.
.
or less?
endshown
faces. (Hint:
the battery is and J; (b)
25.35 the
Intwo
the flat
circuit
in Fig.Imagine
E25.32,slicing
the 16.0-V
...
J and
(c) Einand
r; 26.86
(b) WhatS is
the. current
ther.silver
section?
the magniAn R-C
circuit (c)
hasWhat
a timeisconstant
RC. (a) If the circuit is
study
assumed
that
electricity
costs
$0.080
perdelivered
kilowatt-hour
ance ofcone
the
batteries
negligible,
what
is
to theand that .Figure
25.71 Electric
very is
many
thin
andpower
calculate
BIO The
average
bulk
resistivity
of Sthe pulses
human body
removed
andinto
reinserted
with
the disks,
opposite
polarity,
sothe
that its negaP25.84
25.44
Eels.
Electric
eels
generate
electric
tude of EBIO
in the
copper?
(d) What
the
magnitude
of E
the sildischarging,
howislong
will
it take for
itsinstored
energy
the
bulbs
are
on
for
4.0
h
per
day.
(a)
What
is
the
total
cost
(includ# m. Theto be reduced
h delivbulb?
(b)
If the
batteries
for
5.0a.h,(b)
what
the
total
resistance
of one
such
disk.)
Show
thatcurrent
yourenergy
(apart
from
surface
resistance
of
the
skin)
is
about
5.0
Æ
tive
terminal
is now
nextlast
to point
Find
(a)is the
in the
cir- along
their
skinisthat
beofused
to stun
an between
enemy
they of
come
ver? (e)
What
the
potential
difference
ends
the
tocan
its initial
value?
(b) If when
itthe
is charging,
how long will it take
1>e
ing
the
price
of
the
bulbs)
to
run
each
bulb
for
3.0
years?
(b)
How
R
ered
to
the
bulb?
(c)
The
resistance
of
real
batteries
increases
as
1
result agrees
Eq. (25.10)
when
conducting
path
between
the hands
canpulses
be represented
r1 = rvoltage
cuit (magnitude
andwith
direction);
(b) the
terminal
Vba of the into
2.
contact
with
it.for
Tests
these
be up toapproxisilver
section
of wire?
the have
storedshown
energythat
to reach
its maximum
value?
1>e of can
much
do
you save
over
3.0between
years
iftwo
youconcenusenegligible,
a compact
fluorescent
. CALC
they run
down.
If
initial
internal
resistance
what
25.66
mately
as a cylinder
1.6
m long
andeven
0.10 larger).
m in diameter.
The skin
The
region
16.0-V
battery;
(c)the
the
potential
difference
point a with
Vacis of
.
.
500
V
and
produce
currents
of
80
mA
(or
A
typical
25.59
A
3.00-m
length
of
copper
wire
at
20°C
has
a
r
26.87
Strictly
speaking,
Eq.
(26.16)
implies that an infinite
bulbconducting
instead
of an
incandescent
bulb?
(c)
What
is the
the power
resistance
of resistance can be made negligible by soaking the1.20-m2
is
the
combined
internal
resistance
of
both
batteries
when
hands
in
salt
tric
spheres
with
radii
a
and
b
is
respect to point c. (d) Graph the potential rises and drops in this pulse
R2 much
C2 are
Cdischarge
E1.60
for with
10 ms.
Whatof
power
how
deliv- completely.
1energy
long lasts
section
diameter
mm
and
a to1.80-m-long
amount
time and
is
required
a section
capacitor
Yet
a “100-W”
bulb?
it(Assume
actually uses
to the bulb
has
decreased
to half
its (Remember,
initial
value?
that only
the 23 W water. (a) What
is
the
resistance
between
the
hands
if
the
skin
filled
with
afluorescent
conducting
material
with
resistivity
circuit
(see
Fig.
25.20).
ered
the unfortunate
enemy
with
assuming
abe
steady
withtodiameter
0.80for
mm.
There
is a single
currentapulse,
of
2.5
mAmay
in the
1.60practical
purposes,
capacitor
considered to be fully
of (a)
power
operates
acrossActually,
120
V.) the
resistance
ofShow
theand
bulb
isthe
constant.
will
change
some.r.
that
resistance
between
spheres
given bycurrent?resistance is negligible? (b) What potential difference between the
25.36
The
following
measurements
were itmade
on a is
Thyrite
mm-diameter section.
(a) What
0.80-mmdischarged
afterisa the
finitecurrent
lengthinof the
time.
To be specific, consider a
SmA? (Note that
what when the current through the filament changes, because this
is needed for
lethal Failure.
shock current
of 100
resistor:
.hands
25.45
BIO
Treatment
of aHeart
A connected
heart
diameter
section?
(b) What
iscapacitance
the magnitude
ofdefibrillator
in a the
E Sto
capacitor
with
C
resistor R to be fully
r
1
1
changes the temperature of the filament and hence the resistivity of
resultthe
shows
small
potential
differences
produce
is1.60-mm-diameter
used your
to enable
heartthat
to(c)
start
beating
ifqit
has stopped.
This
isdangerR =
a - b4.00
section?
What
is the
magnitude
of E
in the
discharged
if
its
charge
differs
from
zero
by
no more than the
I
1A2
0.50
1.00
2.00
the filament wire.)
ous
currents
when
the
skin
is
damp.)
(c)
With
the
current
in part
4p a
b
done
by
passing
a
large
current
of
12
A
through
the
body
at
25
V
0.80-mm-diameter
section?
(d)
What
is
the
potential
difference
charge
of
one
electron.
(a)
Calculate
the
time
required
to reach this
25.53 . A
“540-W”
electric
heater
is
designed
to
operate
from
(b),
what
power
is
dissipated
in
the
body?
Vab 1V2 2.55
3.11
3.77
4.58
for
a verythe
short
time,
usually
about
3.0mF,
(a)
What
power
does
between
ends
of
the
3.00-m
length
ofms.
wire?
state
if
and
C
=
0.920
R
=
670
kÆ,
=
7.00
Q
0
.
(b) Derive
an is
expression
for the
current
density
function of .25.72 A typical cost for electric power is $0.120 per kilowatt- mC. How
120-V lines.
(a) What
its resistance?
(b) What
current
doesasitadraw?
the
defibrillator
totime
theDensity
body,
and
(b)this?
how (b)
much
is Q , is the time
25.60
Criticaldeliver
Current
in Superconductors.
One
many
constants
is
Forenergy
a given
(a)
aterms
function
ofto
I. (b)
Thyrite
obey
Ohm’s
law?
V
Vabdoes
radius,
in
of the
potential
between
the spheres. hour. (a) Some
(c) Graph
If the
line
voltage
drops
110Does
V, difference
what
power
the heater
ab as
people leave their porch light on all the time. What0 is
transferred?
problem with some
of the to
newer
required
reachhigh-temperature
this state alwayssuperconducthe same number of time conHow
tell?
(c)
Graph
theinresistance
as
a(25.10)
function
R Actually,
= Vabto>IEq.
(c)you
Show
that
result
(a) reduces
when the .the yearly cost
take?can
(Assume
that
thethe
resistance
ispart
constant.
it will
change
to
keep aof75-W
bulb burning
day and
night?
(b) Sup25.46
Consider
theenough
circuit
Fig. density
E25.32.
(a)practical
What
the
total
tors is getting
a large
current
for
use
withstants,
independent
of the values
of Cisand
R?
Why or why not?
of
I. separation
L = b in- temperature.)
a between the(d)
spheres
small.coils are
because
of the change
The is
heater
pose
your
refrigerator
usesis400
W of power
when
it’sÆrunning,
and
.
rate
at
which
electrical
energy
dissipated
in
the
5.0and
out
causing
the
resistance
to
reappear.
The
maximum
current
26.88
CALC
The
current
in
a
charging
capacitor
is given by Eq.
.
...theThe
25.37
The
following
measurements
of current
andresistance
potential
difa in Eq. it runs 8 hours a day. What is the yearly cost of operating your
25.67
temperature
coefficient
of
metallic,
so that
resistance
of the heater
decreases
with decreas9.0resistors?
(b)
What
is
the
power
output
of
the
16.0-V
batÆ
density
for
which
the
material
will
remain
a
superconductor
is
(26.13).
(a)
The
instantaneous
power
supplied
by
the
battery is Ei.
ference
were made
athe
resistor
constructed
of Nichrome
wire: ais in Eq. refrigerator?
(25.12)
equals
temperature
coefficient
resistivity
ing temperature.
Ifonthe
change
of resistance
withoftemperature
tery?
Atcritical
what rate
is electrical
energy
being
converted
to other
called(c)the
current
density
of
the
material.
In
1987,
IBM
Integrate
this
to
find
the
total
energy
supplied
by
the
battery. (b)
(25.6)
if the
of2.00
thermal
expansion
is small. A 25.73 .. A 12.6-V car battery with negligible internal resistance
taken into
the coefficient
electrical
consumed
Iaccount,
1A2only will
0.50
1.00 power
4.00by the heater
forms
in the
battery?
(d) Show
that the
power
output
of
the is i 2R. Integrate
research
labs8.0-V
had produced
thin
filmspower
with
critical
current
densities
The
instantaneous
dissipated
in the
resistor
cylindrical
of mercury
is in ain
vertical
glass
tube. At 20°C, is connected
be larger
or smallercolumn
than what
you calculated
part (c)?
Explain.
to
a
series
combination
of
a
3.2resistor
that
obeys
Æ
2 the overall rate of dissipation of electrical
16.0-V
equals
of 1.0 *battery
much
current
could
an 18-gauge
10 5 A>cm
. (a)
to How
find the
total
energy
dissipated
in thewire
resistor. (c) Find the
3.88column
7.76
15.52
theVlength
the mercury
is 12.0 cm.
The diameter of the Ohm’s lawthis
ab 1V2 of1.94
and
a thermistor that does not obey Ohm’s law but
energy
in the rest
offinal
theSection
circuit.
Section
25.6
Theory
of
Metallic
Conduction
(see Example
25.1
in
25.1)
ofinthis
material
carry
and
stillthat this equals the
energy
stored
the
capacitor,
and
show
2
mercury column is 1.6 mm and doesn’t change with
temperature ..
instead
a current–voltage
relationship
with a =
= those
aI + used
bI , in
25.47
The has
capacity
of
a(b)storage
battery,
suchV
as
.. VPure
(a)
Graph
a function of I. (b)
Does Nichrome
obey
Ohm’s
remain superconducting?
Researchers
are
trying
to
energy
supplied
by the
battery
lessdevelop
the energy dissipated in
ab as silicon
1.0expansion.
* 10 16
25.54
approximately
free
because
glass has contains
a small coefficient
of thermal
The coef- 3.8 Æ and total
Whatinisampere-hours
the current through
b =systems,
1.3 Æ>A.
6 # h2.the
automobile
electrical
is
rated
A2. 3.2- Æ
1A
law?
How
can
you
tell?
(c)
What
is
the
resistance
of
the
resistor
in
superconductors
with
critical
current
densities
of
1.0
*
10
A>cm
electrons
per cubic
meter.expansion
(a) Referring
to mercury
Table 25.1,
calculate
the 17.2,
ficient
of volume
of the
is given
in Table
resistor?
#
can supplywire
a current
50 A forwould
1.0 h, be
or needed
25 A for
h batterycylindrical
ohms?
What diameter
of suchofa material
to
mean free time t for silicon at room temperature. (b) Your answer 50-A
26.50 . In the circuit sho
in Fig. E26.50, C = 5.90
E = 28.0 V, and the emf has ne
gible resistance. Initially the cap
itor is uncharged and the switch
in position 1. The switch is t
moved to position 2, so that
capacitor begins to charge. (a) W
will be the charge on the capacit
long time after the switch is mo
to position 2? (b) After the switch
the charge on the capacitor is m
value of the resistance R? (c) H
to position 2 will the charge on
the final value found in part (a)?
26.51 . A capacitor with C =
shown in Fig. E26.50 with a res
source with E = 18.0 V and neg
the capacitor is uncharged and
switch is then moved to position
charge. After the switch has been
is moved back to position 1 so th
(a) Compute the charge on the
thrown from position 2 back to p
drops across the resistor and a
described in part (a). (c) Comput
tor and across the capacitor just a
tion 2 back to position 1. (d) Co
10.0 ms after the switch is thrown
Section 26.5 Power Distrib
26.52 . The heating element of
when connected to a 240-V line.
ing element? Is 12-gauge wire la
(b) What is the resistance of the
ating temperature? (c) At 11 cen
per hour to operate the dryer?
electric field.
SET UP: E = ρ J = ρ 20 [1 + α (T − T0 )]
I
π r2
EXECUTE: E = (5.25 × 10−8 Ω ⋅ m)[1 + (0.0045/C°)(120°C – 20°C)](12.5 A)/[π (0.000500 m) 2 ] = 1.21 V/m.
(Note that the resistivity at 120°C turns out to be 7.61 × 10−8 Ω ⋅ m. )
EVALUATE: This result is fairly large because tungsten has a larger resistivity than copper.
(b) IDENTIFY: Relate resistance and resistivity.
SET UP: R = ρ L /A = ρ L /π r 2
EXECUTE: R = (7.61 × 10−8 Ω ⋅ m)(0.150 m)/[π (0.000500 m) 2 ] = 0.0145 Ω
EVALUATE: Most metals have very low resistance.
(c) IDENTIFY: The potential difference is proportional to the length of wire.
SET UP: V = EL
EXECUTE: V = (1.21 V/m)(0.150 m) = 0.182 V
EVALUATE: We could also calculate V = IR = (12.5 A)(0.0145 Ω) = 0.181 V, in agreement with part (c).
25.16.
IDENTIFY: The geometry of the wire is changed, so its resistance will also change.
ρL
SET UP: R =
. Lnew = 3L. The volume of the wire remains the same when it is stretched.
A
L
A
EXECUTE: Volume = LA so LA = Lnew Anew . Anew =
A= .
3
Lnew
Rnew =
25.17.
ρ Lnew
Anew
=
ρ (3L)
A/3
=9
ρL
A
= 9 R.
EVALUATE: When the length increases the resistance increases and when the area decreases the resistance
increases.
ρL
IDENTIFY: R =
.
A
SET UP: For copper, ρ = 1.72 × 10−8 Ω ⋅ m. A = π r 2 .
EXECUTE: R =
(1.72 × 10−8 Ω ⋅ m)(24.0 m)
π (1.025 × 10−3 m)2
= 0.125 Ω
EVALUATE: The resistance is proportional to the length of the piece of wire.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15.2 V − 11.0 V = 4.2 V above the potential of point b.
Figure 25.32
25.33.
IDENTIFY: The voltmeter reads the potential difference Vab between the terminals of the battery.
SET UP: open circuit I = 0. The circuit is sketched in Figure 25.33a.
EXECUTE: Vab = ε = 3.08 V
Figure 25.33a
SET UP: switch closed The circuit is sketched in Figure 25.33b.
EXECUTE: Vab = ε − Ir = 2.97 V
r=
ε − 2.97 V
I
3.08 V − 2.97 V
r=
= 0.067 Ω
1.65 A
Figure 25.33b
Vab 2.97 V
=
= 1.80 Ω.
I
1.65 A
EVALUATE: When current flows through the battery there is a voltage drop across its internal resistance
and its terminal voltage V is less than its emf.
IDENTIFY: The sum of the potential changes around the loop is zero.
SET UP: The voltmeter reads the IR voltage across the 9.0-Ω resistor. The current in the circuit is
counterclockwise because the 16-V battery determines the direction of the current flow.
EXECUTE: (a) Vbc = 1.9 V gives I = Vbc / Rbc = 1.9 V/9.0 Ω = 0.21 A.
And Vab = IR so R =
25.34.
(b) 16.0 V − 8.0 V = (1.6 Ω + 9.0 Ω + 1.4 Ω + R )(0.21 A) and R =
5.48 V
= 26.1 Ω.
0.21 A
(c) The graph is sketched in Figure 25.34.
EVALUATE: In Exercise 25.32 the current is 0.47 A. When the 5.0-Ω resistor is replaced by the 26.1-Ω
resistor the current decreases to 0.21 A.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
25-14
Chapter 25
(a) The rate of conversion of chemical energy to electrical energy in the emf of the battery is
P = ε I = (12.0 V)(2.00 A) = 24.0 W.
(b) The rate of dissipation of electrical energy in the internal resistance of the battery is
P = I 2 r = (2.00 A) 2 (1.0 Ω) = 4.0 W.
(c) The rate of dissipation of electrical energy in the external resistor
R is P = I 2 R = (2.00 A) 2 (5.0 Ω) = 20.0 W.
25.52.
EVALUATE: The rate of production of electrical energy in the circuit is 24.0 W. The total rate of
consumption of electrical energy in the circuit is 4.00 W + 20.0 W = 24.0 W. Equal rate of production and
consumption of electrical energy are required by energy conservation.
IDENTIFY: The power delivered to the bulb is I 2 R. Energy = Pt.
SET UP: The circuit is sketched in Figure 25.52. rtotal is the combined internal resistance of both batteries.
EXECUTE: (a) rtotal = 0. The sum of the potential changes around the circuit is zero, so
1.5 V + 1.5 V − I (17 Ω) = 0. I = 0.1765 A. P = I 2 R = (0.1765 A) 2 (17 Ω) = 0.530 W. This is also
(3.0 V)(0.1765 A).
(b) Energy = (0.530 W)(5.0 h)(3600 s/h) = 9540 J
P
0.265 W
0.530 W
=
= 0.125 A.
= 0.265 W. P = I 2 R so I =
R
17 Ω
2
The sum of the potential changes around the circuit is zero, so 1.5 V + 1.5 V − IR − Irtotal = 0.
(c) P =
3.0 V − (0.125 A)(17 Ω)
= 7.0 Ω.
0.125 A
EVALUATE: When the power to the bulb has decreased to half its initial value, the total internal resistance
of the two batteries is nearly half the resistance of the bulb. Compared to a single battery, using two
identical batteries in series doubles the emf but also doubles the total internal resistance.
rtotal =
Figure 25.52
25.53.
V2
= VI . V = IR.
R
SET UP: The heater consumes 540 W when V = 120 V. Energy = Pt.
IDENTIFY: P = I 2 R =
V2
V 2 (120 V) 2
so R =
=
= 26.7 Ω
P
540 W
R
P 540 W
(b) P = VI so I = =
= 4.50 A
V 120 V
EXECUTE: (a) P =
V 2 (110 V) 2
=
= 453 W. P is smaller by a factor of (110/120)2 .
R
26.7 Ω
EVALUATE: (d) With the lower line voltage the current will decrease and the operating temperature will
decrease. R will be less than 26.7 Ω and the power consumed will be greater than the value calculated in
part (c).
(c) Assuming that R remains 26.7 Ω, P =
SET UP: For mercury at 20°C, ρ = 9.5 × 10−7 Ω ⋅ m, α = 0.00088 (C°) −1 and β = 18 × 10−5 (C°)−1.
EXECUTE: (a) R =
ρL
A
=
(9.5 × 10−7 Ω ⋅ m)(0.12 m)
(π /4)(0.0016 m) 2
= 0.057 Ω.
(b) ρ (T ) = ρ0 (1 + α∆T ) gives
ρ (60° C) = (9.5 × 10−7 Ω ⋅ m)(1 + (0.00088 (C°) −1 )(40 C°) = 9.83 × 10−7 Ω ⋅ m, so ∆ρ = 3.34 × 10−8 Ω ⋅ m.
(c) ∆V = βV0∆T gives A∆L = A( β L0∆T ). Therefore
∆L = β L0∆T = (18 × 10−5 (C°) −1)(0.12 m)(40 C°) = 8.64 × 10−4 m = 0.86 mm. The cross-sectional area of
the mercury remains constant because the diameter of the glass tube doesn’t change. All of the change in
volume of the mercury must be accommodated by a change in length of the mercury column.
ρL
L∆ρ ρ∆L
+
(d) R =
gives ∆R =
.
A
A
A
(3.34 × 10−8 Ω ⋅ m)(0.12 m)
(95 × 10−8 Ω ⋅ m)(0.86 × 10−3 m)
= 2.40 × 10−3 Ω.
(π /4)(0.0016 m)
(π /4)(0.0016 m) 2
EVALUATE: (e) From Eq. (25.12),
⎞
1 ⎛ R
1 ⎛ (0.057 Ω + 2.40 × 10−3 Ω) ⎞
− 1⎟ =
− 1⎟ = 1.1 × 10−3 (C°) −1.
α=
⎜
⎜
0.057 Ω
∆T ⎝ R0
⎠
⎠ 40 C° ⎝
∆R =
25.68.
2
+
This value is 25% greater than the temperature coefficient of resistivity and the length increase is
important.
IDENTIFY: Consider the potential changes around the circuit. For a complete loop the sum of the potential
changes is zero.
SET UP: There is a potential drop of IR when you pass through a resistor in the direction of the current.
8.0 V − 4.0 V
EXECUTE: (a) I =
= 0.167 A. Vd + 8.00 V − I (0.50 Ω + 8.00 Ω) = Va , so
24.0 Ω
Vad = 8.00 V − (0.167 A)(8.50 Ω) = 6.58 V.
(b) The terminal voltage is Vbc = Vb − Vc . Vc + 4.00 V + I (0.50 Ω) = Vb and
Vbc = +4.00 V + (0.167 A)(0.50 Ω) = +4.08 V.
(c) Adding another battery at point d in the opposite sense to the 8.0-V battery produces a counterclockwise
10.3 V − 8.0 V + 4.0 V
current with magnitude I =
= 0.257 A. Then Vc + 4.00 V − I (0.50 Ω) = Vb and
24.5 Ω
Vbc = 4.00 V − (0.257 A) (0.50 Ω) = 3.87 V.
25.69.
EVALUATE: When current enters the battery at its negative terminal, as in part (c), the terminal voltage is
less than its emf. When current enters the battery at the positive terminal, as in part (b), the terminal
voltage is greater than its emf.
IDENTIFY: In each case write the terminal voltage in terms of ε , I and r. Since I is known, this gives two
equations in the two unknowns ε and r.
SET UP: The battery with the 1.50-A current is sketched in Figure 25.69a.
Vab = 8.4 V
Vab = ε − Ir
ε − (1.50 A)r = 8.4 V
Figure 25.69a
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25.73.
⎟
1 yr
⎝ 3 ⎠ ⎝⎜
⎠ ⎝ 3.60 × 10 J ⎠
EVALUATE: Electric lights can be a substantial part of the cost of electricity in the home if they are left on
for a long time!
IDENTIFY: Set the sum of the potential rises and drops around the circuit equal to zero and solve for I.
SET UP: The circuit is sketched in Figure 25.73.
ε − IR − V = 0
ε − IR − α I − β I 2 = 0
β I 2 + (R + α )I − ε = 0
EXECUTE:
Figure 25.73
25.74.
The quadratic formula gives I = (1/2 β ) ⎡ −( R + α ) ± ( R + α ) 2 + 4 βε ⎤
⎣⎢
⎦⎥
I must be positive, so take the + sign
I = (1/2 β ) ⎡ −( R + α ) + ( R + α ) 2 + 4 βε ⎤
⎣⎢
⎦⎥
I = −2.692 A + 4.116 A = 1.42 A
EVALUATE: For this I the voltage across the thermistor is 8.0 V. The voltage across the resistor must then
be 12.6 V − 8.0 V = 4.6 V, and this agrees with Ohm’s law for the resistor.
(a) IDENTIFY: The rate of heating (power) in the cable depends on the potential difference across the
cable and the resistance of the cable.
SET UP: The power is P = V 2 /R and the resistance is R = ρ L /A. The diameter D of the cable is twice its
V2
V2
AV 2 π r 2V 2
=
=
=
. The electric field in the cable is equal to the potential
R ( ρ L /A)
ρL
ρL
difference across its ends divided by the length of the cable: E = V /L.
EXECUTE: Solving for r and using the resistivity of copper gives
radius. P =
r=
Pρ L
πV
2
=
(50.0 W)(1.72 × 10−8 Ω ⋅ m)(1500 m)
π (220.0 V)2
= 9.21 × 10−5 m. D = 2r = 0.184 mm
(b) SET UP: E = V /L
EXECUTE: E = (220 V)/(1500 m) = 0.147 V/m
25.75.
EVALUATE: This would be an extremely thin (and hence fragile) cable.
IDENTIFY: The ammeter acts as a resistance in the circuit loop. Set the sum of the potential rises and
drops around the circuit equal to zero.
(a) SET UP: The circuit with the ammeter is sketched in Figure 25.75a.
EXECUTE: I A =
ε
r + R + RA
ε = I A ( r + R + RA )
Figure 25.75a
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(b) E = Pt = IVt = (30 A) (19.2 V) (1.7) (3600 s) = 3.53 × 10 J.
(c) Ediss = Pdisst = I 2 Rt = (30 A)2 (0.24 Ω) (1.7) (3600 s) = 1.32 × 106 J.
(d) Discharged at 30 A: I =
ε
r+R
gives R =
ε − Ir = 12.0 V − (30 A)(0.24 Ω) = 0.16 Ω.
I
30 A
(e) E = Pt = I 2 Rt = (30 A)2 (0.16 Ω) (1.7) (3600 s) = 8.81× 105 J.
(f) Since the current through the internal resistance is the same as before, there is the same energy
dissipated as in (c): Ediss = 1.32 × 106 J.
25.83.
EVALUATE: (g) Again, part of the energy originally supplied was stored in the battery and part was lost in
the internal resistance. So the stored energy was less than what was supplied during charging. Then when
discharging, even more energy is lost in the internal resistance, and what is left is dissipated over the
external resistor. This time, at a higher current, much more energy is lost in the internal resistance. Slow
charging and discharging is more energy efficient.
IDENTIFY: No current flows through the capacitor when it is fully charged.
SET UP: With the capacitor fully charged, I =
EXECUTE: VC =
. VR = IR and VC = Q /C.
VR
Q 36.0 µ C
4.00 V
= 0.667 A.
=
= 4.00 V. VR1 = VC = 4.00 V and I = 1 =
R1 6.00 Ω
C 9.00 µ F
VR2 = IR2 = (0.667 A)(4.00 Ω) = 2.668 V.
25.84.
ε
R1 + R2
ε = VR1 + VR2
= 4.00 V + 2.668 V = 6.67 V.
EVALUATE: When a capacitor is fully charged, it acts like an open circuit and prevents any current from
flowing though it.
IDENTIFY: No current flows to the capacitors when they are fully charged.
SET UP: VR = RI and VC = Q /C.
EXECUTE: (a) VC1 =
Q1 18.0 µ C
=
= 6.00 V. VC2 = VC1 = 6.00 V.
C1 3.00 µ F
Q2 = C2VC2 = (6.00 µ F)(6.00 V) = 36.0 µ C.
(b) No current flows to the capacitors when they are fully charged, so
VR2 = VC1 = 6.00 V. I =
R2
=
6.00 V
= 3.00 A.
2.00 Ω
60.0 V − 6.00 V
= 18.0 Ω.
3.00 A
EVALUATE: When a capacitor is fully charged, it acts like an open circuit and prevents any current from
flowing though it.
IDENTIFY and SET UP: Follow the steps specified in the problem.
q a
EXECUTE: (a) ∑ F = ma = q E gives
= .
m E
q aL
=
(b) If the electric field is constant, Vbc = EL and
.
m Vbc
R=
25.85.
ε − IR2
VR2
ε = IR1 + IR2 .
I
=
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Direct-Current Circuits
28.0 V − 18.0 V
= 5.00 Ω
2.00 A
(c) Apply Kirchhoff’s loop rule to loop (2): −(6.00 A)(3.00 Ω) − (4.00 A)(6.00 Ω) + ε = 0
ε = 18.0 V + 24.0 V = 42.0 V
EVALUATE: Can check that the loop rule is satisfied for loop (3), as a check of our work:
28.0 V − ε + (4.00 A)(6.00 Ω) − (2.00 A) R = 0
28.0 V − 42.0 V + 24.0 V − (2.00 A)(5.00 Ω) = 0
52.0 V = 42.0 V + 10.0 V
52.0 V = 52.0 V, so the loop rule is satisfied for this loop.
(d) IDENTIFY: If the circuit is broken at point x there can be no current in the 6.00-Ω resistor. There is
now only a single current path and we can apply the loop rule to this path.
SET UP: The circuit is sketched in Figure 26.25b.
R=
26-11
(b) Apply the loop rule to loop (3): +10.0 V − (20.0 Ω) I 2 − 5.00 V = 0 and I 2 = 0.250 A.
(c) I1 + I 2 = 0.333 A + 0.250 A = 0.583 A
26.25.
EVALUATE: For loop (2) we get
+5.00 V + I 2 (20.0 Ω) − I1(30.0 Ω) = 5.00 V + (0.250 A)(20.0 Ω) − (0.333 A)(30.0 Ω) =
5.00 V + 5.00 V − 10.0 V = 0, so that with the currents we have calculated the loop rule is satisfied for this
third loop.
IDENTIFY: Apply Kirchhoff’s point rule at point a to find the current through R. Apply Kirchhoff’s loop
rule to loops (1) and (2) shown in Figure 26.25a to calculate R and ε . Travel around each loop in the
direction shown.
(a) SET UP:
26-12
Figure 26.25a
EXECUTE
: +28
.0 V − (3Inc.
.00All
Ω)rights
I − (5reserved.
.00 Ω) I This
= 0 material is protected under all copyright laws as they currently exist.
© Copyright
2012 Pearson
Education,
No portion of this
28.material
0 V may be reproduced, in any form or by any means, without permission in writing from the publisher.
EXECUTE: Apply Kirchhoff’s point rule to point a: ∑ I = 0 so I + 4.00 A − 6.00 A = 0
I = 2.00 A (in the direction shown in the diagram).
(b) Apply Kirchhoff’s loop rule to loop (1): − (6.00 A)(3.00 Ω) − (2.00 A) R + 28.0 V = 0
−18.0 V − (2.00 Ω) R + 28.0 V = 0
28.0 V − 18.0 V
= 5.00 Ω
2.00 A
(c) Apply Kirchhoff’s loop rule to loop (2): −(6.00 A)(3.00 Ω) − (4.00 A)(6.00 Ω) + ε = 0
ε = 18.0 V + 24.0 V = 42.0 V
EVALUATE: Can check that the loop rule is satisfied for loop (3), as a check of our work:
28.0 V − ε + (4.00 A)(6.00 Ω) − (2.00 A) R = 0
28.0 V − 42.0 V + 24.0 V − (2.00 A)(5.00 Ω) = 0
52.0 V = 42.0 V + 10.0 V
52.0 V = 52.0 V, so the loop rule is satisfied for this loop.
(d) IDENTIFY: If the circuit is broken at point x there can be no current in the 6.00-Ω resistor. There is
now only a single current path and we can apply the loop rule to this path.
SET UP: The circuit is sketched in Figure 26.25b.
R=
Chapter
26
Figure
26.25b
I=
26.26.
8.00 Ω
= 3.50 A
EVALUATE: Breaking the circuit at x removes the 42.0-V emf from the circuit and the current through the
3.00-Ω resistor is reduced.
IDENTIFY: Apply the loop rule and junction rule.
SET UP: The circuit diagram is given in Figure 26.26. The junction rule has been used to find the
magnitude and direction of the current in the middle branch of the circuit. There are no remaining unknown
currents.
EXECUTE: The loop rule applied to loop (1) gives:
+ 20.0V − (1.00 A)(1.00 Ω) + (1.00 A)(4.00 Ω) + (1.00 A)(1.00 Ω) − ε1 − (1.00 A)(6.00 Ω) = 0
ε1 = 20.0 V − 1.00 V + 4.00 V + 1.00 V − 6.00 V = 18.0 V. The loop rule applied to loop (2) gives:
+20.0 V − (1.00 A)(1.00 Ω) − (2.00 A)(1.00 Ω) − ε 2 − (2.00 A)(2.00 Ω) − (1.00 A)(6.00 Ω) = 0
ε 2 = 20.0 V − 1.00 V − 2.00 V − 4.00 V − 6.00 V = 7.0 V. Going from b to a along the lower branch,
Vb + (2.00 A)(2.00 Ω) + 7.0 V + (2.00 A)(1.00 Ω) = Va ⋅ Vb − Va = −13.0 V; point b is at 13.0 V lower
potential than point a.
EVALUATE: We can also calculate Vb − Va by going from b to a along the upper branch of the circuit.
Vb − (1.00 A)(6.00 Ω) + 20.0 V − (1.00 A)(1.00 Ω) = Va and Vb − Va = −13.0 V. This agrees with Vb − Va
calculated along a different path between b and a.
Figure 26.25b
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26-12
Chapter 26
EXECUTE: +28.0 V − (3.00 Ω) I − (5.00 Ω) I = 0
28.0 V
= 3.50 A
8.00 Ω
EVALUATE: Breaking the circuit at x removes the 42.0-V emf from the circuit and the current through the
3.00-Ω resistor is reduced.
IDENTIFY: Apply the loop rule and junction rule.
SET UP: The circuit diagram is given in Figure 26.26. The junction rule has been used to find the
magnitude and direction of the current in the middle branch of the circuit. There are no remaining unknown
currents.
EXECUTE: The loop rule applied to loop (1) gives:
+ 20.0V − (1.00 A)(1.00 Ω) + (1.00 A)(4.00 Ω) + (1.00 A)(1.00 Ω) − ε1 − (1.00 A)(6.00 Ω) = 0
I=
26.26.
ε1 = 20.0 V − 1.00 V + 4.00 V + 1.00 V − 6.00 V = 18.0 V. The loop rule applied to loop (2) gives:
+20.0 V − (1.00 A)(1.00 Ω) − (2.00 A)(1.00 Ω) − ε 2 − (2.00 A)(2.00 Ω) − (1.00 A)(6.00 Ω) = 0
ε 2 = 20.0 V − 1.00 V − 2.00 V − 4.00 V − 6.00 V = 7.0 V. Going from b to a along the lower branch,
Vb + (2.00 A)(2.00 Ω) + 7.0 V + (2.00 A)(1.00 Ω) = Va ⋅ Vb − Va = −13.0 V; point b is at 13.0 V lower
potential than point a.
EVALUATE: We can also calculate Vb − Va by going from b to a along the upper branch of the circuit.
Vb − (1.00 A)(6.00 Ω) + 20.0 V − (1.00 A)(1.00 Ω) = Va and Vb − Va = −13.0 V. This agrees with Vb − Va
calculated along a different path between b and a.
Figure 26.26
26.27.
IDENTIFY: Apply the junction rule at points a, b, c and d to calculate the unknown currents. Then apply
the loop rule to three loops to calculate ε1, ε 2 and R.
(a) SET UP: The circuit is sketched in Figure 26.27.
Direct-Current Circuits
C=
t
=
R ln(V /V0 )
4.00 s
⎛ 12.0 V ⎞
(3.40 × 10 Ω)ln ⎜
⎝ 3.00 V ⎟⎠
6
26-19
= 8.49 × 10−7 F
(b) τ = RC = (3.40 × 106 Ω)(8.49 × 10−7 F) = 2.89 s
EVALUATE: In most laboratory circuits, time constants are much shorter than this one.
26.42.
IDENTIFY: For a charging capacitor q (t ) = Cε (1 − e−t/τ ) and i (t ) =
ε
R
e − t/ τ .
SET UP: The time constant is RC = (0.895 × 106 Ω) (12.4 × 10−6 F) = 11.1 s.
EXECUTE: (a) At t = 0 s: q = Cε (1 − e−t/RC ) = 0.
At t = 5 s: q = Cε (1 − e −t/RC ) = (12.4 × 10−6 F)(60.0 V)(1 − e −(5.0 s)/(11.1 s) ) = 2.70 × 10−4 C.
At t = 10 s: q = Cε (1 − e−t/RC ) = (12.4 × 10−6 F)(60.0 V)(1 − e−(10.0 s)/(11.1 s) ) = 4.42 × 10−4 C.
At t = 20 s : q = Cε (1 − e−t/RC ) = (12.4 × 10−6 F)(60.0 V)(1 − e−(20.0 s)/(11.1 s) ) = 6.21 × 10−4 C.
At t = 100 s : q = Cε (1 − e −t/RC ) = (12.4 × 10−6 F)(60.0 V)(1 − e −(100 s)/(11.1 s) ) = 7.44 × 10−4 C.
(b) The current at time t is given by: i =
At t = 0 s : i =
60.0 V
8.95 × 105 Ω
60.0 V
ε
R
e−t/RC .
e−0/11.1 = 6.70 × 10−5 A.
e−5 / 11.1 = 4.27 × 10−5 A.
8.95 × 105 Ω
60.0 V
At t = 10 s : i =
e−10/11.1 = 2.72 × 10−5 A.
8.95 × 105 Ω
60.0 V
At t = 20 s : i =
e−20/11.1 = 1.11 × 10−5 A.
8.95 × 105 Ω
60.0 V
At t = 100 s : i =
e−100/11.1 = 8.20 × 10−9 A.
8.95 × 105 Ω
(c) The graphs of q (t ) and i (t ) are given in Figure 26.42a and b.
At t = 5 s : i =
EVALUATE: The charge on the capacitor increases in time as the current decreases.
Figure 26.42
26.43.
IDENTIFY: The capacitors, which are in parallel, will discharge exponentially through the resistors.
SET UP: Since V is proportional to Q, V must obey the same exponential equation as Q,
V = V0e – t /RC . The current is I = (V0 /R) e – t /RC .
EXECUTE: (a) Solve for time when the potential across each capacitor is 10.0 V:
t = − RC ln(V /V0 ) = –(80.0 Ω)(35.0 µF) ln(10/45) = 4210 µs = 4.21 ms
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
−2t /RC
0
dt = 0 2
= 0 = U 0.
∫ e
2C
RC 2 0
RC 2
EVALUATE: Increasing the energy stored in the capacitor increases current through the resistor as the
capacitor discharges.
IDENTIFY and SET UP: When C changes after the capacitor is charged, the voltage across the capacitor
changes. Current flows through the resistor until the voltage across the capacitor again equals the emf.
EXECUTE: (a) Fully charged: Q = CV = (10.0 × 10−12 F)(1000 V) = 1.00 × 10−8 C.
(b) E =
26.90.
(b) The initial current just after the capacitor is charged is I 0 =
ε − VC ′
R
=
ε
R
−
q
. This gives
RC ′
q ⎞ −t/RC ′
⎛ε
, where C ′ = 1.1C.
i (t ) = ⎜ −
⎟e
⎝ R RC ′ ⎠
(c) We need a resistance such that the current will be greater than 1 µA for longer than 200 µs. Current
equal to this value requires that at t = 200 µs, i = 1.0 × 10−6 A =
e −(2.0×10
−4 s)/R (11×10−12 F)
. This says 1.0 × 10−6 A =
1⎛
1.0 × 10−8 C ⎞
⎜⎜1000 V −
⎟
R⎝
1.1(1.0 × 10−11 F) ⎟⎠
7
1
(90.9)e − (1.8×10 Ω )/R and
R
18.3R − RlnR − 1.8 × 107 = 0. Solving for R numerically we find 7.15 × 106 Ω ≤ R ≤ 7.01 × 107 Ω.
26.91.
EVALUATE: If the resistance is too small, then the capacitor discharges too quickly, and if the resistance is
too large, the current is not large enough.
IDENTIFY: Consider one segment of the network attached to the rest of the network.
SET UP: We can re-draw the circuit as shown in Figure 26.91.
⎛ 1
1 ⎞
EXECUTE: RT = 2 R1 + ⎜
+
⎝ R2 RT ⎟⎠
−1
= 2 R1 +
R2 RT
. RT 2 − 2 R1RT − 2 R1R2 = 0.
R2 + RT
RT = R1 ± R12 + 2 R1R2 . RT > 0, so RT = R1 + R12 + 2 R1R2 .
EVALUATE: Even though there are an infinite number of resistors, the equivalent resistance of the
network is finite.
Figure 26.91
26.92.
IDENTIFY: Assume a voltage V applied between points a and b and consider the currents that flow along
each path between a and b.
SET UP: The currents are shown in Figure 26.92.
EXECUTE: Let current I enter at a and exit at b. At a there are three equivalent branches, so current is
I /3 in each. At the next junction point there are two equivalent branches so each gets current I /6. Then at
b there are three equivalent branches with current I /3 in each. The voltage drop from a to b then is
⎛ I⎞
⎛ I⎞
⎛ I⎞
V = ⎜ ⎟ R + ⎜ ⎟ R + ⎜ ⎟ R = 65 IR. This must be the same as V = IReq , so Req = 65 R.
⎝ 3⎠
⎝ 6⎠
⎝ 3⎠
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.