Download Solutions to Homework problems Chapters 18-20.

7.
REASONING AND SOLUTION
Using Equation 16.1, we find that
Hz) = 18.19 X 10-2 m
A= vif= (343 m/s)/(418S.6
(16.1 )
I
--14. REASONING
The length L of the string is one of the factors that affects the speed of a wave
,!!a.:reling_.9nit, in .~o.faL_~~!h.espe~~~deIJen~~g
t~ m~ss_p~2:lnjt length _m{L a_~~ordingto
-.
. speed
v = ~ mlL'
F (Equation 16.2). The other factor affecting the speed is the tension F. The
f
is not directly given here. However, the frequency and the wavelength Il are given, and the
speed is related to them according to v = fll (Equation 16.1). Substituting Equation 16.1 into
Equation 16.2 will give us an equation that can be solved for the length 1.
SOLUTION
Substituting Equation 16.1 into Equation 16.2 gives
V=fA=)
mlL
F
Solving for the length L, we find that
f2A2m
(260 Hz)2(0.60 m)2(S.OX10-3 kg)
L =--=----------=
F
180N
I~
__ml
0.68
--
24. REASONING AND SOLUTION We find from the graph on the left that A = 0.060 m 0.020 m = 0.040 m and A = 0.010 m. From the graph on the right we find that T= 0.30 s0.10 s = 0.20 s. Then, f= 1/(0.20 s) = S.O Hz. Substituting these into Equation 16.3 we get
y
=A sin ( 21Cf t _ 2~ x )
and
I
y=(0.010m)sin(101Ct-S01CX)
I
26. REASONING
The speed ~f a wave on the string is given by Equation 16.2 as v ~ ~ m/L
F ,
where F is the tension in the string and m/L is the mass per unit length (or linear density) of
the string. The wavelength A is the speed of the wave divided by its frequency
(Equation 16.1).
f
SOLUTION
a. The speed of the wave on the string is
rI=
v= V(m/L)
1_15_N_=142m/S
'J0.85kg/m-·-
I
b. The wavelength is
c. The amplitude of the wave is A = 3.6 cm = 3.6 x 10-2 m. Since the wave is moving along
the -x direction, the mathematical expression for the wave is given by Equation 16.4 as
Substituting in the numbers for A,f, and A, we have
---_.
= I(
sin[( 75s-l)t
- 3.6x
---. 10-2 m)
'..
27. I SSMI REASONING
+ (18 m-I) x JI
According to Equation 16.2, the tension F in the string is given by
F = v2(m/ L). Since v = Af from Equation 16.1, the expression for F can be written
(1)
where the quantity m / L is the linear density of the string. In order to use Equation (1), we
must first obtain values for and A; these quantities can be found by analyzing the
expression for the displacement of a string particle.
f
SOLUTION
The displacement is given by y = (0.021 m)sin(25t-
2.0x).
Inspection of this
equation and comparison with Equation 16.3, y = A sin (2 nft _ 2 A
nx) , gives
A
.
,.1,=2n mor
f=~Hz
or
2n
2.0
2n f = 25 rad/s
2n=20m-1
Substituting these valuesfand
A
into Equation (1) gives
6.
REASONING
The two speakers are vibrating exactly out of phase.
conditions for constructive and destructive interference are opposite
when the speakers vibrate in phase, as they do in Example 1 in the text.
sources vibrating exactly out of phase, a difference in path lengths that
number (1, 2, 3, ... ) of wavelengths leads to destructive interference;
This means that the
of those that apply
Thus, for two wave
is zero or an integer
a difference in path
2 2 2 ...) of wavelengths leads to constructive
lengths that is a half-integer number (2.,12.,22.,
interference. First, we will determine the wavelength being produced by the speakers.
Then, we will determine the difference in path lengths between the speakers and the
observer and compare the differences to the wavelength in order to decide which type of
interference occurs.
SOL UTION According to Equation 16.1, the wavelength A is related to the speed v and
frequency.Lof the sound as follows:
~
••••- .••-.
".r'
_._
•
-----
A=~-
"-'-
_..." .•.••r.
343 m/s
f - 429 Hz =0.800
m
Since. ABC in Figure 17.7 is a right triangle, the Pythagorean theorem applies and the
difference !1d in the path lengths is given by
We will now apply this expression for parts (a) and (b).
a. When dBC = 1.15 m, we have
Since 1.60 m = 2 (0.800 m) = 2A, it follows that the interference
speakers vibrate out of phase).
is
I destructive I
(the
b. When dBC = 2.00 m, we have
!1d = ~d~
+d~c -dBC
= )(2.50 m)2 +(2.00 m)2 -2.00 m = 1.20 m
Since 1.20 m = 1.5 (0.800 m) = (1~) A, it follows that the interference is I constructive
I
(the
speakers vibrate out of phase).
7.
ISSMllwwwl REASONING
The geometry of the positions of the loudspeakers and the
listener is shown in the following drawing.
c
The listener at C will hear either a loud sound or no sound, depending upon whether the
interference occurring at C is constructive or destructive. If the listener hears no sound,
destructive interference occurs, so
(1)
n = 1, 3, 5, ...
SOLUTION
Since v = Af , according to Equation 16.1, the wavelength of the tone is
A=~=
f
343m/s
68.6Hz =5.00m
Speaker B will be closest to Speaker A when n = 1 in Equation (1) above, so
nA
d2
5.00 m
=-2-+dl =-2-+l.00m=3.50m
From the figure above we have that,
Xl
= (1.00 m) cos 60.0° = 0.500 m
y = (1.00 m) sin 60.0° = 0.866
ill
Then
or
Therefore, the closest that speaker A can be to speaker B so that the listener hears no sound
IS
Xl
+ x2 = 0.500 m + 3.39 m = 13.89 m I·
23. I SSMI REASONING
The fundamental frequency .h is given by Equation 17.3 with n = 1:
= v 1(2L). Since values for .h and L are given in the problem statement, we can use this
expression to find the speed of the waves on the cello string. Once the speed is known, the
.h
tension F in the cello string can be found by using Equation 16.2, v =
SOLUTION
.J
F I(m I L) .
Combining Equations 17.3 and 16.2 yields
2L.h = ~ m~ L
24. REASONING
The frequencies In of the standing waves on a string fixed at both ends are
given by Equation 17.3 as
fn = n (2:),
where n is an integer that specifies the harmonic
number, v is the speed of the traveling waves that make up the standing waves, and L is the
length of the string. For the second harmonic, n = 2.
SOLUTION
The frequency
h of the second
harmonic is
-----...--
29.
REASONING
A standing wave is composed of two oppositely traveling waves. The speed
F
v of these waves is given by v = ~ m/L
(Equation 16.2), where F is the tension in the string
and m/L is its linear density (mass per unit length). Both F and m/L are given in the
statement of the problem. The wavelength 1 of the waves can be obtained by visually
inspecting the standing wave pattern. The frequency of the waves is related to the speed of
the waves and their wavelength by = v/1 (Equation 16.1).
i
SOLUTION
a. The speed of the waves is
-~
v-:-
m/F L
=J 8.5xl0-3kg/m
280N
=1180m/sl
1.8 m
b. Two loops of any standing wave comprise one
wavelength. Since the string is 1.8 m long and consists
of three loops (see the drawing), the wavelength is
1=t(1.8
m)=ll.2
ml
?
3;>
c. The frequency of the waves is
i=~=
1
35.
I SSMI
REASONING
180 m/s -1150 Hzl
1.2 m
AND SOLUTION
The distance between one node and an adjacent
antinode is A/4. Thus, we must first determine the wavelength of the standing wave. A tube
open at only one end can develop standing waves only at the odd harmonic frequencies.
Thus, for a tube of length L producing sound at the third harmonic (n = 3), L = 3(A/4).
Therefore, the wavelength of the standing wave is
1=1L=1(1.5
m)=2.0
m
and the distance between one node and the adjacent antinode is A/4 = I 0.50 m I.
39.
I SSM I
REASONING
The natural frequencies of a tube open at only one end are given by
Equation 17.5 as in = n( 4~), where n is any odd integer (n = 1,3,5, ... ), v is the speed of
sound, and L is the length of the tube. We can use this relation to find the value for n for the
450-Hz sound and to determine the length of the pipe.
SOLUTION
a. The frequency in of the 450-Hz sound is given by 450 Hz = n( 4~)'
Likewise, the
frequency of the next 'higher harmonic is 750 Hz = (n + 2) ( 4~ ), because n is an odd
integer and this means that the value of n for the next higher harmonic must be n + 2.
Taking the ratio of these two relations gives
"' '" A TT
(
n + 2)(
AVT
I __
,')
Solving this equation for n gives n = [].
b. 'Solving the equation 450 Hz = n( 4VL) for L and using n = 3, we find that the length of
the tube is
450m/s
Hz) ]~IO.57
4fn J~3[ 4(343
L~n(_V
rnl
10. REASONING
The drawing at the right
~0.50
m~+q
+2q
shows the set-up. The force on the +q
charge at the origin due to the other +q
charge is given by Coulomb's
law
+q I ••
d
(Equation 18.1), as is the force due to the
+2q charge. These two forces point to the
left, since each is repulsive. The sum of
the two is twice the force on the +q charge at the origin due to the other +q charge alone.
SOLUTION
Applying Coulomb's law, we have
k~
k~ql
+
(0.50 m)2
~
Force due to +q
charge at x=+O.50 m
=
( d)2
klqllql
(0.50 m)2
Force due to +2q
charge at x=+d
Twice the force due to
+q charge atx=+O.50 ill
~
2
v
Rearranging this result and solving for d give
kl2qllql_
(d)2
klqllql
-
or
or
d =±0.7l
m
(0.50m)2
•• .I
We reject the negative root, because a negative value for d would locate the +2q charge tc
the left of the origin. Then, the two forces acting on the charge at the origin would hav~
different directions, contrary to the statement of the problem. Therefore, the +2q charge i~
located at a position of I x = +0.71 m I·
i1.
q2
':
SOLUTION
a. The magnitude F12 of the force exerted on q1
by q2 is given by Coulomb's law, Equation 18.1,
where the distance is specified in the drawing:
+y
.. --.
n +x
~---_?~~Q~~?~~Q~--_.
!
~_3_0_~n
q1
1.30ci
_.•
q3
Since the magnitudes of the charges and the distances are the same, the magnitude of F 13is
the same as the magnitude ofF12, or F13 = 0.213 N. From the drawing it can be seen that the
x-components of the two forces cancel, so we need only to calculate the y components of the
forces.
Force
y component
F
+F12
sin 23.0° = +(0.213 N) sin 23.0° = +0.0832 N
+F13
sin 23.0° = +(0.213 N) sin 23.0° = +0.0832 N
Fy=+0.166N
1_
--
-.------
,--.----.-
-~
...
IS. REASONING The unknown charges can
be determined using Coulomb's law to
express the electrostatic force that each
30.00
unknown charge exerts on the 4.00 fiC
charge. In applying this law, we will use
the fact that the net force points downward
in the drawing.
This tells us that the
unknown charges are both negative and
have the same magnitude, as can be
understood with the help of the free-body
qA
diagram for the 4.00 fiC charge that is
shown at the right. The diagram shows
the attractive force F from each negative charge directed along the lines between the charges.
Only when each force has the same magnitude (which is the case when both unknown
charges have the same magnitude) will the resultant force point vertically downward. This
occurs because the horizontal components of the forces cancel, one pointing to the right and
the other to the left (see the diagram). The vertical components reinforce to give the
observed downward net force.
SOLUTION Since we know from the REASONING that the unknown charges have the
same magnitude, we can write Coulomb's law as follows:
(4.00XlO-6 C)lqA!
F=k-----=k-----
(4.00XlO-6 C)\qBI
r2
r2
The magnitude of the net force acting on the 4.00 fiC charge, then, is the sum of the
magnitudes of the two vertical components F cos 30.00 shown in the free-body diagram:
(4.00XlO-6 C)lqA\
LF=k
r2
(4.00XlO-6 C)lqBI
cos30.0o+k r -
cos30.0°
Solving for the magnitude of the charge gives
IqAI=
(LF)r2
(405 N)( 0.0200 ill)2
=---------------=
2(S.99Xl09 N.m2/C2)(4.00XlO-6
-
--~----_
.. _._ .._.--~._.
cl.
__ .-------------------
Thus, we have qA = qB = 1-2.60XlO-6
2.60xlO- 6. C
C)cos30.0o
29. REASONING AND SOLUTION
a. In order for the field to be zero, the point cannot be between the two charges. Instead, it
must be located on the line between the two charges on the side of the positive charge and
away from the negative charge. If x is the distance from the positive charge to the point in
question, then the negative charge is at a distance (3.0 m + x) meters from this point. For
the field to be zero here we have
Iq-I
or
_lq+1
(3.0 m+x)2 -~
Solving for the ratio of the charge magnitudes gives
~
Iq+l-
_ 16.0 ,LiC_ (3.0 m+x)2
4.0,LiC -
or
4.0= (3.0 m+x)2
x2
x2
Suppressing the units for convenience and rearranging this result gives
or
4.0x2 = 9.0 + 6.0x + x2
or
3x2 -6.0x-9.0
=0
Solving this quadratic equation for x with the aid of the quadratic formula (see Appendix
CA) shows that
x=3.0m
or x=-1.0m
We choose the positive value for x, since the negative value would locate the zero-field spot
between the two charges, where it can not be (see above). Thus, we have I x = 3.0 m I.
b. Since the field is zero at this point, the force acting on a charge at that point is 10NI.
b
?
The electric field produced by q4 at the origin points toward the charge, or along the +y
direction. The net electric field is, then, E = -£3 + E4, where E3 and E4 can be determined
.by using Equation 18.3.
SOLUTION The net electric field at the origin is
=j +3.9x106 N/C I
The plus sign indicates that I the net electric field points along the +y direction, .
31. ISSMI REASONING
a. The drawing shows the t)Vopoint charges ql and q2' Point A is located at x = 0 cm, and
point B is at x = +6.0 cm.
Since q I is positive, the electric field points away from it. At point A, the electric field E}
points to the left, in the
-x direction.
Since q2 is negative, the electric field points toward it.
At point A, the electric field E2 points to the right, in the +x direction. The net electric field
is E = -E} + E2. We can use Equation 18.3, E =
electric field due to each point charge.
klql/
to find the magnitude of the
r2,
b. The drawing shows the electric field produced by the charges q I and q2 at point B, which
is located at x = +6.0 cm .
•
E2
Since
is positive, the electric field points away from it. At point B, the electric field
q}
points to the right, in the +x direction. Since q2 is negative, the electric field points toward
it. At point B, the electric field points to the right, in the +x direction. The net electric field
isE=+EI
+E2'
SOLUTION
a. The net electric field at the origin (point A) is E = -El + E2:
-(8.99X109
N.m2/C2)(8.5X10-6
=-----------+-------
C)
(8.99X109
(3.0 X 10-2 m)2
=] -6.2x107
N.m2/C2)(21X10-6
C)
(9.0 x 10-2 m)
N/C I
The minus sign tells us that the net electric field points along the
-x axis.
b. The net electric field at x = +6.0 cm (point B) is E = El + E2:
(8.99X109
N.m2/C2)(8.5XlO-6
=
2
(3.0X10-2
m)
C)
+
(8.99X109
N.m2/C2)(21X10-6
2
(3.0xlO-2
m)
C)
67.
I SSMI REASONING Consider the drawing
at the right.
-- -- --
It is given that the charges qA'
ql' and q2 are each positive.
Therefore, the
d
F A2
4d
charges qI and q2 each exert a repulsive
q2
= +3.0 ~C
force on the charge qA' As the drawing
shows, these forces have magnitudes F Al
(vertically downward) and F A2 (horizontally
to the left). The unknown charge placed at
the empty corner of the rectangle is qu' and it exerts a force on qA that has a magnitude
FAU' In order that the net force acting on qA point in the vertical direction, the horizontal
component of F AU must cancel out the horizontal force F A2' Therefore, FAD must point as
shown in the drawing, which means that it is an attractive force and qu must be negative,
since qA is positive.
SOLUTION
The basis for our solution is the fact that the horizontal component of FAD
must cancel out the horizontal force F A2' The magnitudes of these forces can be expressed
using Coulomb's law F = k!qllq'll r2 , where r is the distance between the charges q and q'.
Thus, we have
and
where we have used the fact that the distance between the charges qA and qu is the diagonal
of the rectangle, which is ~( 4d)2 + d2 according to the Pythagorean theorem, and the fact
that the distance between the charges q A and q2 is 4d. The horizontal component of FAD is
FAU
cos e , which must be equal to F A2' so that we have
cose = klqAllq2!
klqAllqul
.••...
( 4d)2
-- .~ ~ ..
+ d2
~--
kdcose=&l
17
(4d)2
16
.--'-
The drawing in the REASONING,
Therefore, we find that
17 ~
JU )=~
16
~(
or
reveals
that
cos e = (4 d) I J( 4d)2 + d2 = 4 I JU
or
As discussed in the REASONING, the algebraic sign of the charge qu is I negative
I.
.
cll9
'G
4.
REASONING
Equation 19.1 indicates that the work done by the electric force as the
particle moves from point A to point B is WAB = EPEA - EPEB. For motion through a
distance s along the line of action of a constant force of magnitude F, the work is given by
Equation 6.1 as either +Fs (if the force and the displacement have the same direction) or -Fs
(if the force and the displacement have opposite directions). Here, EPEA - EPEB is given to
be positive, so we can conclude that the work is WAB = +Fs and that the force points in the
direction of the motion from point A to point B. The electric field is given by Equation 18.2
as E = F/qo' where qo is the charge.
SOLUTION
a. Using Equation 19.1 and the fact that WAB = +Fs, we find
WAB
= +Fs = EPE A
F = EPE A -EPEB
s
-
EPE B
= 9.0x10-4
N I
0.20 m J = !4.5X10-3
-----
As discussed in the reasoning, the direction of the force is I from A toward B
I.
b. From Equation 18.2, we find that the electric field has a magnitude of
E=~=
qo
4.5xlO-3 N
1.5xlO-6 C = !3.0X103 N/C I
The direction is the same as that of the force
I from A toward B I.
.-
,~---_---~
-'~_.
....
..,.
on the positive
charge,
namely
_.-.-----=------
22. REASONING AND SOLUTION
The figure at the right
shows two identical charges, q, fixed to diagonally
opposite corners of a square. The potential at corner A is
caused by the presence of the two charges. It is given by
r
Since both charges are the same distance from corner B, this
is equal to the potential at corner B as well. If a third charge,
q3' is placed at the center of the square, the potential at corner A (as well as corner B)
becomes
16.
REASONING The electric 'potential at a distance r from a point charge q is given by
Equation 19.6 as V = kq / r. The total electric potential at location P due to the six point
charges is the algebraic sum of the individual potentials.
Id
d
d
+7.0q
+S.Oq
+7.0q
+3.0q
-
d -3.0q
•
d
•
p
d
-S.Oq
SOLUTION
Starting at the upper left corner of the rectangle, we proceed clockwise and
add up the six contributions to the total electric potential at P (see the drawing):
k(+7.0q)
V = ~d2
_
k(+3.0q)
k(+S.Oq)
k(+7.0q)
k(-3.0q)
k(-S.Oq)
+(~r + ~ +~d2 +(~r +~d2 +(~r + ~ +~d2 +(~r
k(+14.0q)
-F(%)
Substituting q = 9.0
18.
REASONING
X
10-6 C and
d= 0.13 m gives
The potential V created by a point charge q at a spot that is located at a
distance r is given by Equation 19.6 as V = kq
r , where q can be either a positive or negative
quantity, depending on the nature of the charge. We will apply this expression to obtain the
potential created at the empty corner by each of the three charges fixed to the square. The
total potential at the empty corner is the sum of these three contributions. Setting this sum
equal to zero will allow us to obtain the unknown charge.
and
SOLUTION
The drawing at the right shows the three
charges fixed to the corners of the square. The length of
each side of the square is denoted by L. Note that the
distance between the unknown charge q and the empty
corner is L. Note also that the distance between one of the
1.8-,uC charges and the empty corner is r = L, but that the
distance between the other 1.8-,uC charge and the empty
corner is r = -J L2 + L2 = hL, according to the Pythagorean'
theorem.
L
+1.8 uC •.
L
L
+1 8,liC ••
L
q
Using Equation 19.6 to express the potential created by the unknown charge q and by each
of the 1.8-,uC charges, we find that the total potential at the empty corner is
23.
I SSMI
REASONING
-Initially, the three charges are infinitely far apart. We will proceed
as in Example 8 by adding charges to the triangle, one at a time, and determining the electric
potential energy at each step. According to Equation 19.3, the electric potential energy EPE
is the product of the charge q and the electric potential V at the spot where the charge is
placed, EPE = q V. The total electric potential energy of the group is the sum of the energies
of each step in assembling the group.
SOLUTION Let the corners of the triangle be numbered clockwise as 1,2 and 3, starting
with the top corner. When the first charge (ql = 8.00 )lC) is placed at a corner 1, the charge
has no electric potential energy, EPE1 = O. This is because the electric potential VI produced
by the other two charges at corner 1 is zero, since they are infinitely far away.
Once the 8.00-!-.lCcharge is in place, the electric potential V2 that it creates at corner 2 is
where r21 = 5.00 m is the distance between corners 1 and 2, and
ql
= 8.00 !-.lC. When the
20.0-!-.lCcharge is placed at corner 2, its electric potential energy EPE2 is
~ ( 20.0 X 10-6
l
l
5.00 m)_( 8_.0_0_X_1
4-( 8_.9_9_x1_0_9N_. m_2_/C_2
0_-6_C_J= 0.288 J
The electric potential V3 at the remaining empty corner is the sum of the potentials due to
the two charges that are already in place on corners 1 and 2:
where ql = 8.00 !-.lC,r31 = 3.00 m, q2 = 20.0 !-.lC,and r32 = 4.00 m. When the third charge
(q3
= -15.0 !-.lC)is placed at corner 3, its electric potential energy EPE3 is
=(-IS.OXIO-6
C)(8.99XI09
N.m2/C'J(
3.00m
8.00xI0-6
C + 20.04.00m
X 10-6 C - -1.034J
J-
The electric potential energy of the entire array is given by
EPE = EPEI + EPE2 + EPE3 = 0 + 0.288 J + (-1.034 J) = 1-0.746 J I
32. REASONING AND SOLUTION
Let point A be on the x-axis where the potential is
515 V. Let point B be on the x-axis where the potential is 495 V. From Equation 19.7a, the
electric field is
E
= __~_V = __VB~-_V=A
!1s
!1s
495V-515V
= - -2(6.0Xl0-3
= -1.7x103V/m
m)
The magnitude of the electric field is 11.7 x 103 Vim I. Since the electric field is negative, it
points to the lIen I, from the high toward the low potential.
.•..
;;
•..
,---
40. REASONING AND SOLUTION The capacitance is C = qo!Vo = qlTl. The new charge q is,
therefore,
q = qoV = (5.3Xl0-Se)(9.0V)
6.0 V
= 18.0 x lo-Sel
-- ----------------------.---
-----_._._-~.-
Vo
-
._------_ .._---_.~._.-
l.
~-.---'--_._--~------~----'''''-----..-'
7.
._.' _c_~._'_
.~-,
~__~_~_
~
.....
_ --- -_. ---
ro_
,--
REASONING As discussed in Section 20.1, the voltage gives the energy per unit charge.
Thus, we can determine the energy delivered to the toaster by multiplying the voltage V by
the charge !1q that flows during a time !1t of one minute. The charge can be obtained by
solving Equation 20.1, 1= (!1q)/(!1t), since the current! can be obtained from Ohm's law.
SOLUTION Remembering that voltage is energy per unit charge, we have
Energy = V !1q
Solving Equation 20.1 for !1q gives !1q = I !1t, which can be substituted in the previous result
to give
Energy = V !1q = VI !1t
I
According to Ohm's law (Equation 20.2), the current is = VIR, which can be substituted in
the energy expression to show that
Energy = VI!1t =
v(
V)!1t = V2R!1t = (120 V)2
R
140 (60 s) = I 6.2 x 10 4 J
14. REASONING AND SOLUTION
I
Solving Equation 20.5 for ayields
,~"
43.
I SSM I
-.•..,,, •...
'/~
REASONING The equivalent series resistance Rs is the sum of the resistances of
the three resistors. The potential difference V can be determined from Ohm's law as
V = IRs.
SOLUTION
a. The equivalent resistance is
Rs
=250+450+750=114501
b. The potential difference across the three resistors is
V = IRs = (0.51 A)(l45 0) = 174 V
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53. [§sIYfj REASONING Since the resistors are connected in parallel, the voltage across each
one is the same and can be calculated from Ohm's Law (Equation 20.2: V = 1R). Once the
voltage across each resistor is known, Ohm's law can again be used to find the current in the
second resistor. The total power consumed by the parallel combination can be found
calculating the power consumed by each resistor from Equation 20.6b: P = 12 R. Then, the
total power consumed is the sum of the power consumed by each resistor.
SOLUTION Using data for the second resistor, the voltage across the resistors is equal to
V = 1R = (3.00 A)(64.0 0)= 192 V
a. The current through the 42.0-0 resistor is
1=
V _ 192 V
R -
42.0 0 = I 4.57 A
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b. The power consumed by the 42.0-0 resistor is
while the power consumed by the 64.0-0 resistor is
Therefore the total power consumed by the two resistors is 877 W + 576 W = 11450 W
58.
I .
REASONING We will approach this problem in parts. The resistors that are in series will be
combined according to Equation 20.16, and the resistors that are in parallel will be combined
according to Equation 20.17.
SOLUTION The 1.00 0, 2.00 0
and 3.00 0 resistors are in series
with an equivalent resistance of
Rs = 6.00 O.
This equivalent resistor of 6.00.0 is
in parallel with the 3.00-.0 resistor,
2.00Q
6.00Q
6.00Q
2.00 Q
6.00Q
so
1
1
-=--+Rp
6.00.0
Rp
1
3.00.0
=2.000
2.00Q
This new equivalent resistor of
2.00 .0 is in series with the 6.00-.0
resistor, so Rs' = 8.00 O.
8.00Q
".
Rs' is in parallel with the 4.00-Q
resistor, so
2.00 Q
111
-=---+--
R/
8.00 Q
4.00 Q
I
Rp
=2.67 Q
Finally, Rp' is in series with the 2.00-Q, so the total equivalent resistance is 14.67 Q I·
-------~-,.,......--------...-------------~--_ _------------•.
bo
SOLUTION Since
RI
and
R2
.---------
..•....
,
-.......--
are wired in series, the equivalent resistance Rl2 is
(20.16)
The resistor
resistance
is wired in parallel with the equivalent resistor
R3
RI23
so the equivalent
of this combination is
_1__ --+-=--+1
1
RI23
The resistance
R12,
R4
R3
1
48 Q
Rl2
1 ..
24 Q
(20.17)
or
is in series with the equivalent resistance
RI23
R123,
= 16 Q
so the equivalent resistance
R AB between the points A and B is
...•.•.
--.........
'-
...
- ~., ..-
~-'--'------"-'
61. I SSMI REASONING When two or more resistors are in series, the equivalent resistance is
Rs = RI + R2 + R3 + ....
Likewise, when resistors are in
given by Equation 20.16:
parallel, the expression to be solved to find the equivalent resistance is given by Equation
20.17: _1_ = _1 +_1_ + _1_+ .... We will successively apply these to the individual resistors
Rp
RI
R2
R3
in the figure in the text beginning with the resistors on the right side of theflgure.
SOLUTION
Since the 4.0-Q and the 6.0-Q resistors are in series, the equivalent resistance
of the combination of those two resistors is 10.0 Q. The 9.0-Q and 8.0-Q resistors are in
parallel; their equivalent resistance is 4.24 Q. The equivalent resistances of the parallel
combination (9.0 Q and 8.0 Q) and the series combination (4.0 Q and the 6.0 Q) are in
parallel; therefore, their equivalent resistance is 2.98 Q. The 2.98-Q combination is in
series with the 3.0-Q resistor, so that equivalent resistance is 5.98 Q. Finally, the 5.98-Q
combination and the 20.0-Q resistor are in parallel, so the equivalent resistance between the
points A and B is I 4.6 Q I.
.~-'''''-r'
_
65.
I SSMII
wwwl
REASONING
Since we know that the current in the 8.00-0 resistor is
0.500 A, we can use Ohm's law (V = 1R) to find the voltage across the 8.00-0 resistor. The
8.00-0 resistor and the 16.0-0 resistor are in parallel; therefore, the voltages across them
are equal. Thus, we can also use Ohm's law to find the current through the 16.0-0 resistor.
The currents that flow through the 8.00-0 and the 16.0-0 resistors combine to give the total
current that flows through the 20.0-0 resistor. Similar reasoning can be used to find the
current through the 9.00-0 resistor.
SOLUTION
a. The voltage across the 8.00-0 resistor is V8 = (0.500 A)(8.00 0) = 4.00 V . Since this is
also the voltage that is across the 16.0-0 resistor, we find that the current through the
16.0-0 resistor is 116= (4.00 V)/(16.00)=
0.250 A. Therefore, the total current that
flows through the 20.0-0 resistor is
ha
= 0.500 A + 0.250 A =1 0.750 A
I
b. The 8.00-0 and the 16.0-0 resistors are in parallel, so their equivalent resistance can be
.
1
1
. d.t::
ob tame
lrom EquatIOn
20.17, - 1 =-+-+-+
Rp
the
Rupper
equivalent
resistance
Ri
of
R2
the
1
f-'
... , an d' IS equal to 5.33 O. Therelore,
R3
upper
branch
of
the
circuit
is
= 5.33 0 + 20.00 = 25.3 0, since the 5.33-0 resistance is in series with the 20.0-Q
resistance. Using Ohm's law, we find that the voltage across the upper branch must be
V==(0.750A)(25.30)=19.0V.
Since the lower branch is in parallel with the upper
branch, the voltage across both branches must be the same. Therefore, the current through
the 9.00-0 resistor is, from Ohm's law,
V,
19 =
lower
~
= 19.0 V
9.000 = 12.11 A
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