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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 I 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 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 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 I