7. REASONING AND SOLUTION Using Equation 16.1, we find that

A= vif= (343 m/s)/(418S.6 Hz) = 18.19 X10-2 m I (16.1 )

---

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

-. v = ~ F (Equation 16.2). The other factor affecting the speed is the tension F. The speedmlL' .

is not directly given here. However, the frequency f and the wavelength Il are given, and thespeed is related to them according to v = fll (Equation 16.1). Substituting Equation 16.1 intoEquation 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=) FmlL

Solving for the length L, we find that

f2A2m (260 Hz)2(0.60 m)2(S.OX10-3 kg) I~__L =--=----------= 0.68 mlF 180N --

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 ~ ~ F ,m/L

where F is the tension in the string and m/L is the mass per unit length (or linear density) ofthe string. The wavelength A is the speed of the wave divided by its frequency f(Equation 16.1).

SOLUTIONa. The speed of the wave on the string is

v= rI= 1_15_N_=142m/S IV(m/L) 'J0.85kg/m-·-

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 alongthe -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( 3.6x 10-2 m) sin[( 75s-l)t + (18 m-I) x JI- ---.---_. ' ..

27. I SSMI REASONING 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), wemust first obtain values for f and A; these quantities can be found by analyzing theexpression for the displacement of a string particle.

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 nx) , givesA

2n f = 25 rad/sorf=~Hz

2nand2n=20m-1

or,.1,=2n m

A .

2.0

Substituting these valuesfand A into Equation (1) gives

6. REASONING The two speakers are vibrating exactly out of phase. This means that theconditions for constructive and destructive interference are opposite of those that applywhen the speakers vibrate in phase, as they do in Example 1 in the text. Thus, for two wavesources vibrating exactly out of phase, a difference in path lengths that is zero or an integernumber (1, 2, 3, ... ) of wavelengths leads to destructive interference; a difference in path

lengths that is a half-integer number (2.,12.,22., ...) of wavelengths leads to constructive2 2 2

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 theobserver and compare the differences to the wavelength in order to decide which type ofinterference occurs.

SOL UTION According to Equation 16.1, the wavelength A is related to the speed v andfrequency.Lof the sound as follows:

~ ••••- .••-. ".r' _._ •

----- "-'- _..." .•.••r.

A=~- 343 m/sf - 429 Hz =0.800 m

Since. ABC in Figure 17.7 is a right triangle, the Pythagorean theorem applies and thedifference !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 is I destructive I (the

speakers vibrate out of phase).

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 theinterference occurring at C is constructive or destructive. If the listener hears no sound,destructive interference occurs, so

n = 1, 3, 5, ... (1)

SOLUTION Since v = Af , according to Equation 16.1, the wavelength of the tone is

A=~= 343m/sf 68.6Hz =5.00m

Speaker B will be closest to Speaker A when n = 1 in Equation (1) above, so

nA 5.00 m

d2 =-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 illThen

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:

.h = v1(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

tension F in the cello string can be found by using Equation 16.2, v = .J F I(m I L) .

SOLUTION 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 thelength 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

v of these waves is given by v = ~ F (Equation 16.2), where F is the tension in the stringm/L

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 visuallyinspecting the standing wave pattern. The frequency of the waves is related to the speed ofthe waves and their wavelength byi= v/1 (Equation 16.1).

SOLUTION

a. The speed of the waves is

-~ F =J 280N =1180m/slv-:- m/ L 8.5xl0-3kg/m

b. Two loops of any standing wave comprise onewavelength. Since the string is 1.8 m long and consistsof three loops (see the drawing), the wavelength is

1=t(1.8 m)=ll.2 ml?

c. The frequency of the waves is

i=~= 180 m/s -1150 Hzl1 1.2 m

1.8 m

3;>

35. I SSMI REASONING 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 tubeopen 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 the450-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

L~n(_V J~3[ 343m/s ]~IO.57 rnl4fn 4( 450 Hz)

10. REASONING The drawing at the rightshows the set-up. The force on the +q ~0.50 m~+q +2qcharge at the origin due to the other +qcharge 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 theleft, since each is repulsive. The sum ofthe 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~

(0.50 m)2~

Force due to +qcharge at x=+O.50 m

+ k~ql

( d)2~Force due to +2q

charge at x=+d

= 2 klqllql(0.50 m)2

vTwice the force due to

+q charge atx=+O.50 ill

Rearranging this result and solving for d give

kl2qllql_ klqllql

(d)2 - (0.50m)2

or or d =±0.7l m

We reject the negative root, because a negative value for d would locate the +2q charge tcthe 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·

•• .I

i1. SOLUTIONa. 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:

q2

: +y --.' ..

~ ! ~_3_0_~n n +x---_?~~Q~~?~~Q~--_.

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 theforces.

Force

F

1_

y component

+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

30.00

-- -.------ ,--.----.- ...- ~

IS. REASONING The unknown charges canbe determined using Coulomb's law toexpress the electrostatic force that eachunknown charge exerts on the 4.00 fiC

charge. In applying this law, we will usethe fact that the net force points downwardin the drawing. This tells us that theunknown charges are both negative andhave the same magnitude, as can beunderstood with the help of the free-bodydiagram for the 4.00 fiC charge that is qA

shown at the right. The diagram showsthe 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 unknowncharges have the same magnitude) will the resultant force point vertically downward. Thisoccurs because the horizontal components of the forces cancel, one pointing to the right andthe other to the left (see the diagram). The vertical components reinforce to give theobserved 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! (4.00XlO-6 C)\qBIF=k-----=k-----r2 r2

The magnitude of the net force acting on the 4.00 fiC charge, then, is the sum of themagnitudes of the two vertical components F cos 30.00 shown in the free-body diagram:

(4.00XlO-6 C)lqA\ (4.00XlO-6 C)lqBILF=k 2 cos30.0o+k - cos30.0°r r

Solving for the magnitude of the charge gives

IqAI= (LF)r2

(405 N)( 0.0200 ill)2 6=---------------= 2.60xlO- . C

2(S.99Xl09 N.m2/C2)(4.00XlO-6 C)cos30.0o

Thus, we have qA = qB = 1-2.60XlO-6 cl.- --~----_ .._._ .._.--~._.__ .-------------------

29. REASONING AND SOLUTION

a. In order for the field to be zero, the point cannot be between the two charges. Instead, itmust be located on the line between the two charges on the side of the positive charge andaway from the negative charge. If x is the distance from the positive charge to the point inquestion, then the negative charge is at a distance (3.0 m + x) meters from this point. Forthe field to be zero here we have

or Iq-I _lq+1

(3.0 m+x)2 -~

Solving for the ratio of the charge magnitudes gives

~ _ 16.0 ,LiC_ (3.0 m+x)2

Iq+l- 4.0,LiC - x2or 4.0= (3.0 m+x)2

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 AppendixCA) 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 spotbetween 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 Ithe 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 = klql/ r2, to find the magnitude of the

electric field due to each point charge.

b. The drawing shows the electric field produced by the charges q I and q2 at point B, whichis located at x = +6.0 cm .

•E2

Since q} is positive, the electric field points away from it. At point B, the electric field

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 fieldisE=+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 N.m2/C2)(21X10-6 C)=-----------+-------(3.0 X10-2 m)2 (9.0 x 10-2 m)

=] -6.2x107 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 C) (8.99X109 N.m2/C2)(21X10-6 C)= 2 + 2

(3.0X10-2 m) (3.0xlO-2 m)

67. ISSMI REASONING Consider the drawing

at the right. It is given that the charges qA'

ql' and q2 are each positive. Therefore, the FA2

charges qI and q2 each exert a repulsive

force on the charge qA' As the drawing

shows, these forces have magnitudes FAl

(vertically downward) and FA2 (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.

------4d

d

q2 = +3.0 ~C

SOLUTION The basis for our solution is the fact that the horizontal component of FAD

must cancel out the horizontal force FA2' 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 qA and q2 is 4d. The horizontal component of FAD is

FAU cos e , which must be equal to FA2' so that we have

klqAllqul cose = klqAllq2!

( 4d)2 + d2 (4d)2

.••... -- .~ ~ ..~--.--'-

or kdcose=&l17 16

The drawing in the REASONING, reveals that cos e = (4 d) I J( 4d)2 + d2 = 4 I JU .Therefore, we find that

~( ~ )=~17 JU 16or

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 byEquation 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 - EPE B

F = EPE A -EPEB = 9.0x10-4 J = !4.5X10-3 N Is 0.20 m -----

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=~= 4.5xlO-3 Nqo 1.5xlO-6 C = !3.0X103 N/C I

The direction is the same as that of the force on the positive charge, namelyI from A toward B I.

.--'~_.,~---...._---~..,. _.-.-----=------

22. REASONING AND SOLUTION The figure at the rightshows two identical charges, q, fixed to diagonallyopposite corners of a square. The potential at corner A iscaused by the presence of the two charges. It is given by r

Since both charges are the same distance from corner B, thisis 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 byEquation 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.

+7.0q

d

-S.Oq

+3.0q +S.Oq

d

-d

•

Id

pd

•d•-3.0q

+7.0q

SOLUTION Starting at the upper left corner of the rectangle, we proceed clockwise andadd up the six contributions to the total electric potential at P (see the drawing):

k(+7.0q) k(+3.0q) k(+S.Oq) k(+7.0q) k(-3.0q) k(-S.Oq)

V = ~d2 +(~r + ~ +~d2 +(~r +~d2 +(~r + ~ +~d2 +(~r_ k(+14.0q)

-F(%)Substituting q = 9.0 X 10-6 C and d= 0.13 m gives

18. REASONING 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 , where q can be either a positive or negativer

quantity, depending on the nature of the charge. We will apply this expression to obtain thepotential created at the empty corner by each of the three charges fixed to the square. Thetotal potential at the empty corner is the sum of these three contributions. Setting this sumequal to zero will allow us to obtain the unknown charge.

SOLUTION The drawing at the right shows the threecharges fixed to the corners of the square. The length of +1.8 uC •.each side of the square is denoted by L. Note that the

distance between the unknown charge q and the emptycorner is L. Note also that the distance between one of the1.8-,uC charges and the empty corner is r = L, but that the

distance between the other 1.8-,uC charge and the empty +1 8,liC ••corner is r = -J L2+ L2 = hL, according to the Pythagorean' L qtheorem.

Using Equation 19.6 to express the potential created by the unknown charge q and by eachof the 1.8-,uC charges, we find that the total potential at the empty corner is

and

L

L

L

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 electricpotential energy at each step. According to Equation 19.3, the electric potential energy EPEis the product of the charge q and the electric potential V at the spot where the charge isplaced, EPE = q V. The total electric potential energy of the group is the sum of the energiesof each step in assembling the group.

SOLUTION Let the corners of the triangle be numbered clockwise as 1,2 and 3, startingwith 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 X10-6 4-( 8_.9_9_x1_0_9N_. m_2_/C_2)_(8_.0_0_X_10_-6_C_Jl= 0.288 Jl 5.00 m

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( 8.00xI0-6 C + 20.0 X10-6 CJ-3.00m 4.00m - -1.034J

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

495V-515V = -1.7x103V/m

= - -2(6.0Xl0-3 m)

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, theelectric field is

E = __~_V = __VB~-_V=A!1s !1s

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)Vo 6.0 V = 18.0 x lo-Sel

-- ._------_ .._---_.~._.-----------------------.--- -----_._._-~.- -

l.

~-.---'--_._--~------~----'''''---.-' --.- .~-,._.' _c_~._'_ ~__~_~_ ~ ....._ --- -_. --- ro_ ,--

7. 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 bysolving 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 resultto give

Energy = V !1q = VI !1t

According to Ohm's law (Equation 20.2), the current is I = VIR, which can be substituted inthe energy expression to show that

Energy = VI!1t = v( V)!1t = V2 !1t = (120 V)2 (60 s) I 4R R 140 = 6.2 x 10 J I

14. REASONING AND SOLUTION 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 asV = IRs.

SOLUTIONa. 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 I

53. [§sIYfj REASONING Since the resistors are connected in parallel, the voltage across eachone is the same and can be calculated from Ohm's Law (Equation 20.2: V = 1R). Once thevoltage across each resistor is known, Ohm's law can again be used to find the current in thesecond resistor. The total power consumed by the parallel combination can be found

calculating the power consumed by each resistor from Equation 20.6b: P = 12R. Then, thetotal 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 VR - 42.0 0 = I 4.57 A I

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 I .

58. REASONING We will approach this problem in parts. The resistors that are in series will becombined according to Equation 20.16, and the resistors that are in parallel will be combinedaccording to Equation 20.17.

SOLUTION The 1.00 0, 2.00 0and 3.00 0 resistors are in series

with an equivalent resistance of

Rs = 6.00 O.

This equivalent resistor of 6.00.0 isin parallel with the 3.00-.0 resistor,so

1 1 1-=--+-Rp 6.00.0 3.00.0

Rp =2.000

This new equivalent resistor of2.00 .0 is in series with the 6.00-.0

resistor, so Rs' = 8.00 O.

2.00Q

2.00 Q

2.00Q

6.00Q

6.00Q

8.00Q

6.00Q

".

Rs' is in parallel with the 4.00-Q

resistor, so

111-=---+--R/ 8.00 Q 4.00 Q

IRp =2.67 Q

2.00 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 R3 is wired in parallel with the equivalent resistor R12, so the equivalent

resistance RI23 of this combination is

_1__ 1 1 1 1--+-=--+- ..RI23 R3 Rl2 48 Q 24 Q or RI23 = 16 Q

(20.17)

The resistance R4 is in series with the equivalent resistance R123, so the equivalent resistance

RAB between the points A and B is

...•.•. --......... ...'-- ~., ..- ~-'--'------"-' .~-'''''-r' _

61. I SSMI REASONING When two or more resistors are in series, the equivalent resistance is

given by Equation 20.16: Rs = RI + R2 + R3 + .... Likewise, when resistors are in

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 resistorsRp 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 resistanceof 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 parallelcombination (9.0 Q and 8.0 Q) and the series combination (4.0 Q and the 6.0 Q) are inparallel; 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-Qcombination 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.

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. The8.00-0 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 totalcurrent that flows through the 20.0-0 resistor. Similar reasoning can be used to find thecurrent 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

b . d.t:: . 1 1 1 1 d' f-'o tame lrom EquatIOn 20.17, - =-+-+-+ ... , an IS equal to 5.33 O. Therelore,Rp Ri R2 R3

the equivalent resistance of the upper branch of the circuit is

Rupper = 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 beV==(0.750A)(25.30)=19.0V. Since the lower branch is in parallel with the upperbranch, the voltage across both branches must be the same. Therefore, the current throughthe 9.00-0 resistor is, from Ohm's law,

V,19 = lower = 19.0 V

~ 9.000 = 12.11 A I