13. One way of expressing the power dissipated by a resistor is P = ( V)

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1 Current and esstance 9. One way of expressng the power dsspated by a resstor s ( ). Thus, f the potental dfference across the resstor s doubled, the power wll be ncreased by a factor of 4, to a value of 6 W, makng (d) the correct choce. NSWES TO EEN NUMBEED CONCETU QUESTONS. n the electrostatc case n whch charges are statonary, the nternal electrc feld must be zero. nonzero feld would produce a current (by nteractng wth the free electrons n the conductor), whch would volate the condton of statc equlbrum. n ths chapter we deal wth conductors that carry current, a non-electrostatc stuaton. The current arses because of a potental dfference appled between the ends of the conductor, whch produces an nternal electrc feld. 4. The number of cars would correspond to charge Q. The rate of flow of cars past a pont would correspond to current. 6. The 5 W bulb has the hgher resstance. Because, and both operate from, the bulb dsspatng the least power has the hgher resstance. The W bulb carres more current, because the current s proportonal to the power ratng of the bulb. 8. n electrcal shock occurs when your body serves as a conductor between two ponts havng a dfference n potental. The concept behnd the admonton s to avod smultaneously touchng ponts that are at dfferent potentals.. The knob s connected to a varable resstor. s you ncrease the magntude of the resstance n the crcut, the current s reduced, and the bulb dms.. The ampltude of atomc vbratons ncreases wth temperature, thereby scatterng electrons more effcently. OBEM SOUTONS 7. The charge that moves past the cross secton s Q ( t), and the number of electrons s Q t n ( ) e e ( 8. C s ) (. mn ) 6. s mn. 9.6 C electrons The negatvely charged electrons move n the drecton opposte to the conventonal current flow. 7. (a) From Example 7. n the textbook, the densty of charge carrers (electrons) n a copper wre s n electrons m. Wth πr and q e, the drft speed of electrons n ths wre s v d 7. Cs nq ne πr m. 6 C π. 5 m ( ) ( ) ms The drft speed s smaller because more electrons are beng conducted. To create the same current, therefore, the drft speed need not be as great.

2 9 Chapter 7 7. The perod of the electron n ts orbt s T πr v, and the current represented by the orbtng electron s Q e v e t T πr 6 9 (. 9 ms ). 6 C π 5. 9 m 5. Cs 5. m 7.4 f N s the number of protons, each wth charge e, that ht the target n tme t, the average current n the beam s Q t Ne t, gvng N ( t) 6 5 Cs. s 9 e 6. Cproton 8. 6 protons 7.5 (a) The carrer densty s determned by the physcal characterstcs of the wre, not the current n the wre. Hence, n s unaffected. The drft velocty of the electrons s v d nq. Thus, the drft velocty s doubled when the current s doubled. 7.6 The mass of a sngle gold atom s m M 97 gmol 7. 5 g 7. kg 6. atoms mol atom N The number of atoms deposted, and hence the number of ons movng to the negatve electrode, s m 5. kg n m 7. kg atom Thus, the current n the cell s Q ne t t 9 ( 9. 9 ). 6 C.78 h 6 s h 7.7 The drft speed of electrons n the lne s v d nq ne ( ) 4 v d ( m ) 6. 9 C π. m The tme to travel the length of the -km lne s then m yr t 4 7 v 4. m s.56 s 7 yr d m d ( π ) 4, or 4. m s

3 Current and esstance ssumng that, on average, each alumnum atom contrbutes three electrons, the densty of charge carrers s three tmes the number of atoms per cubc meter. Ths s densty N n mass per atom ρ M N ρ, M 6 6. mol or n. 7 g cm cm m gmol m The drft speed of the electrons n the wre s then v d ne 5. Cs 9 (.8 m ) 6. 9 C ( 4. ) 5 4. m s 6 m 7.9 (a) Usng the perodc table on the nsde back cover of the textbook, we fnd M Fe g mol ( g mol )( kg g) kg mol From Table 9., the densty of ron s ρ Fe 786. kg m, so the molar densty s ρfe 786. ( molar densty) Fe M Fe kg m 4. kg mol 5 mol m (c) The densty of ron atoms s N densty of atoms ( molar densty) 6. atoms mol mol atoms.. 8 m 5 m (d) Wth two conducton electrons per ron atom, the densty of charge carrers s n ( charge carrers atom)( densty of atoms) electrons atoms atom m 7. 9 electrons m (e) Wth a current of. and cross-sectonal area 5. 6 m, the drft speed of the conducton electrons n ths wre s v d nq. Cs (. 7 9 m )(. 6 C) m 4 ms. 7. From Ohm s law, Ω

4 94 Chapter 7 7. max max 8 6 Thus, f 4. 5 Ω, ( ) max and f Ω, ( ) 6. max 7. The volume of the copper s Snce m densty. kg 8.9 kg m. 7 m, ths gves. 7 m. [] (a) From ρ, we fnd that 8 ρ 7. Ω m ( 4. 8 m).5 Ω nsertng ths expresson for nto Equaton [] gves m. m, whch yelds 8. m πd. 7 m From Equaton [],, or 4 d m 4. m π π ( 8. m) 4. 8 m. 8 mm. 7. (a) From Ohm s law,. Ω 9.5 Usng ρ and data from Table 7., the requred length s found to be πr ρ ρ (. Ω) π (. 79 m) 5 8 Ω m 7. m 8 ρ ρ 47. Ω m 5m 7.4 πd 4. Ω π. 4 m 7.5 (a) 4. Ω From ρ, ρ ( m) ( Ω ) π. 4. m Ω m

5 Current and esstance Usng ρ Cu and data from Table 7., we have ρcu ρ π l Cu π, whch reduces to r rl r r Cu ρ ρ Cu and yelds r r Cu ρ ρ Cu 8 8. Ω m 8 7. Ω m The resstance s. 5 Ω, so the resstvty of the metal s 6. ( πd 4) (. 5 Ω) π. m ρ 4 5. m Thus, the metal s seen to be slver Ω m 7.8 Wth dfferent orentatons of the block, three dfferent values of the rato are possble. These are: cm cm 4 cm 8 cm cm cm 4 cm cm 8. m,,. m and (a) max mn 4 cm cm cm 5. cm 5. m m ( 6. )( 8. ) 8. 8 ρ ( ) 8 7. Ω m mn max mn m ( 6. )( 5. ) 7. 8 ρ ( ) 8 7. Ω m max 7.9 The volume of materal, π r, n the wre s constant. Thus, as the wre s stretched to decrease ts radus, the length ncreases such that πr πr ( f ) f ( ) gvng f r r r r ( 4. ) 6 5. f The new resstance s then f f f 6 ρ ρ ρ ρ π 6 4 r π r π 4 r f f Ω 56 Ω ( ) 7. (a) From Ohm s law, 5. 6 Ω 5 5 ubber-soled shoes and rubber gloves can ncrease the resstance to current and help reduce the lkelhood of a serous shock.

6 96 Chapter 7 7. f a conductor of length has a unform electrc feld E mantaned wthn t, the potental dfference between the ends of the conductor s E. But, from Ohm s law, the relaton between the potental dfference across a conductor and the current through t s, where ρ. Combnng these relatons, we obtan E ( ) ρ or E ρ( ) ρj 7. Usng + α T T wth 6. Ω at T (from Table 7. n the textbook), the resstance at T 4. C s ( 6. Ω ) + 8. ( C ) ( 4. C. C) 8. C and α slver. ( C ) 6. Ω 7. From Ohm s law, ff, so the current n ntarctca s f T T f + α + α ( Tf T ). 9 C (. 58. C. C ) + 9. ( C) ( 88. C. C) (a) Gven: lumnum wre wth α. C (see Table 7. n textbook), and. Ω at T. C. f 46. Ω at temperature T, solvng + α ( T T ) gves the fnal temperature as ( ) T T Ω. Ω. C α 9. C 58 C The expanson of the cross-sectonal area contrbutes slghtly more than the expanson of the length of the wre, so the answer would be slghtly reduced For tungsten, the temperature coeffcent of resstvty s α. C. Thus, f 5 Ω at T C, and 6 Ω at the operatng temperature of the flament, solvng + α ( T T ) for the operatng temperature gves ( ) T T Ω 5 Ω C. C α 45 C For alumnum, the resstvty at room temperature s ρ 8. Ω m and the temperature coeffcent of resstvty s α l 9. ( C). Thus, f at some temperature, the alumnum has a resstvty of 8 8 ρ ( ρ ) 7. Ω m 5. Ω m Cu solvng ρ ρ + αl T T for that temperature gves 8 5. Ω m T T + 8 ρρ 8. Ω m C +. α l 9. C C

7 Current and esstance t 8 C, + α T T ( Ω ) +. 5 or. 6 6 m 5. ( C )( 8 C C) 7.8 f 4. Ω at T C and 4. 4 Ω at T 9. C, then + α T T gves the temperature coeffcent of resstvty of the materal makng up ths wre as α T T ( 4. Ω) ( 9. C C). 7.9 (a) The resstance at. C s ρ 4. 4 Ω 4. Ω 8 (. 7 Ω m )( 4. 5 m) π ( 5. m). Ω C 9. and the current wll be. Ω. t. C, + α T T ( (. ) + 9. ( C) Ω ). C. C. Ω Thus, the current s 9.. Ω The resstance of the heatng element when at ts operatng temperature s 5 W. 7 Ω ρ From + α ( T T ) + T T α ρ + T T α, the cross-sectonal area s ( 8 5 Ω m )( 4. m) + (. 4 ( C) )( C. C).7 Ω m

8 98 Chapter 7 7. (a) From ρ, the ntal resstance of the mercury s ρ 7 ( 9. 4 Ω m )(. m) π. m 4. Ω Snce the volume of mercury s constant, f f gves the fnal crosssectonal area as f ( f ). Thus, the fnal resstance s gven by ρ f ρ f f. The fractonal change n the resstance s then f 7. The resstance at. C s f f ρ f ρ f or a.8% ncrease. + α T T ( ) +.9 C. Ω C. C 7 Ω Solvng + α T T for T gves the temperature of the meltng potassum as T T + α 5. 8 Ω 7 Ω. C + 9. ( C ) 7 Ω 6. C 7. (a) The power consumed by the devce s ( ). W 8.., so the current must be. From Ohm s law, the resstance s Ω 7.4 (a) The energy used by a -W bulb n 4 h s E t ( W )( 4 h ) (. kw )( 4 h ). 4 kwh and the cost of ths energy, at a rate of $. per klowatt-hour s cost E rate ( 4. kwh)( $. kwh) $ 9. The energy used by the oven n 5. h s E t ( ) t (. C s )( J C) kw Js and the cost of ths energy, at a rate of $. per klowatt-hour s cost E rate ( kwh)( $. kwh) $ h kwh

9 Current and esstance The power requred s ( ) 7.6 (a) The power loss n the lne s W loss 7. Ω km 6 km 5. W 5 MW The total power transmtted s 8 nput ( ) 7 7. W 7 MW Thus, the fracton of the total transmtted power represented by the lne losses s loss 5 MW fracton loss. 7 or 7.% 7 MW nput 7.7 The energy requred to brng the water to the bolng pont s E mc( T) (. 5 kg ) 4 86 J kg C ( C. C ) 6. 5 J The power nput by the heatng element s nput ( ) ( )(. ) 4 W 4 J s Therefore, the tme requred s 5 E 6. J t s mn 67 4 Js 6 s. mn nput 7.8 (a) E t ( 9 W )( h ) ( 9 J s )( 6 s ). 5 J The power consumpton of the color set s ( ) ( )(.5 ) W Therefore, the tme requred to consume the energy found n (a) s 5 E. J t. Js s mn 6 s 8 mn 7.9 The energy nput requred s E mc( T) (.5 kg ) 4 86 J kg C ( 5. C. C ) 5. 5 J and, f ths s to be added n t. mn 6 s, the power nput needed s 5 E 5. J 49 W t 6 s The power nput to the heater may be expressed as ( ), so the needed resstance s 49 W 4. 4 Ω

10 Chapter (a) t the operatng temperature, ( ) ( )(.5 ) 84 W From + α ( T T ), the temperature T s gven by T T + α The resstances are gven by Ohm s law as. ( ).5, and ( ).8 Therefore, the operatng temperature s T. C + (. 5) (. 8) 4 (. ( C) ) C 7.4 The resstance per unt length of the cable s. Wm. 5 Ω m From ρ, the resstance per unt length s also gven by ρ. Hence, ρ the cross-sectonal area s πr, and the requred radus s r 8 ρ ( ) 7. Ω m 5 π π. Ω m. 6 m 6. cm 7.4 (a) The ratng of the - battery s t 55 h. Thus, the stored energy s Energy stored t ( ) t ( ) 55 h 66 W h. 66 kwh cost kwh $. kwh $ cents ( ) W 5 6 W 5 µ W 7.44 (a) E t ( 4. W)( 4. d )( 4. h d ). 4 4 Wh. 4 kwh cost E rate ( kwh)( kwh). 4 $. $. 6 E t (.97 kw )(. mn )( h 6 mn ) kwh cost E ( rate) 4.85 kwh $. kwh $ cents (c) E t ( 5. kw)( 4. mn )( h 6 mn ). 47 kwh cost E rate (.47 kwh)( kwh) 46 $. $. 4.6 cents

11 Current and esstance 7.45 The energy saved s E ( hgh low ) t ( 4 W W)( h). 9 and the monetary savngs s Wh 9. kwh savngs E rate (. 9 kwh)( $. 8 kwh) $. cents 7.46 The power requred to warm the water to C n 4. mn s Q mc T t t (. 5 kg ) ( 4 86 J kg C )( C C) ( 4. mn )( 6 s mn) 5. W The requred resstance (at C) of the heatng element s then 4 Ω 5. W so the resstance at C would be 4 Ω + T α T +.4 C C C 4 Ω We fnd the needed dmensons of a nchrome wre for ths heatng element from ρ ρ ( πd 4) 4ρ π d, where s the length of the wre and d s ts dameter. Ths gves 8 4 d 4 5 ρ Ω m ( 48. ) π π( 4 Ω) 8 m 8 Thus, any combnaton of length and dameter satsfyng the relaton d 48. m wll be sutable. typcal combnaton mght be 8 4. m and d 48. m. m 8. m 8. mm Yes, such heatng elements could easly be made from less than 5. cm of nchrome. The volume of materal requred for the typcal wre gven above s d π π ( m) (. m ) ( 4. ) 6 7 cm m The energy that must be added to the water s J E mc( T) ( kg ) C 5 C kg C and the cost s cost E rate ( 5 kwh)( kwh) $. 8 $. m kwh 5 kwh 6.6 J 4. cm

12 Chapter (a) For tungsten, Table 7. from the textbook gves the resstvty at T. C 9 K 8 as ρ 56. Ω m and the temperature coeffcent of resstvty as α 45. ( C ) 45. K. Thus, for a tungsten wre havng a radus of. mm and a length of 5. cm, the resstance at T 9 K s ρ ρ 56. π r ( m) 8 Ω ( 5. m) (. m) π 7. Ω From Stefan s law, the radated power s σ et 4, where s the area of the radatng surface. Note that snce we are computng the radated power, not the net energy ganed or lost as a result of radaton, the ambent temperature s not needed here. n the case of a wre, ths s the cylndrcal surface area π r. The temperature of the wre when t s radatng a power of 75. W must be T σ e W ( W m K ) π (. m )(. 5 m)(. ) 4 or T 45. K (c) ssumng a lnear temperature varaton of resstance, the resstance of the wre at ths temperature s + ( )(. 45 K 9 K) + α T T. 7 Ω 4. 5 K gvng 7. Ω (d) The voltage drop across the wre when t s radatng 75. W and has the resstance found n as part (c) above s gven by. 7 Ω 75. W. (e) Tungsten bulbs release very lttle of the energy consumed n the form of vsble lght, makng them neffcent sources of lght The battery s rated to delver the equvalent of 6. amperes of current (.e., 6. C/s) for hour. Ths s Q t ( 6. )( h ) ( 6. C s )( 6 s ). 6 5 C

13 Current and esstance 7.5 The energy avalable n the battery s Energy stored t ( ) t ( ) t 9 h. W h The two headlghts together consume a total power of requred to completely dscharge the battery s 6 W 7 W, so the tme Energy stored. W h t 7 W 5 h 7.5 ssumng a constant resstance, the power consumed by the devce s proportonal to the square. Thus, of the appled voltage, ( ) ( ) or 6 5. W W 7.5 The temperature varaton of resstance s descrbed by + α T T, where s the resstance at T C. Thus, f an alumnum wre α 9. ( C ) has, we have + α T C, and ts temperature must be T C+ C+ α 9. ( C) 8. C 7.5 From ( ), the total resstance needed s Thus, from 48 W ρ 8. Ω ρ, the length of wre requred s (. Ω ). 6 ( m ) Ω m. m. km 7.54 The resstance of the 4. cm length of wre between the feet s ρ 8 (. 7 Ω m )(. 4 m) π. m Ω, so the potental dfference s 6 5 ( 5 )(. 79 Ω) µ

14 4 Chapter Ohm s law gves the resstance as. From ρ, the resstvty s gven by ρ ( ). The results of these calculatons for each of the three wres are summarzed n the table below. ( m) ( Ω) ρ ( Ω m) The average value found for the resstvty s ρ Σρ Ω av m whch dffers from the value of ρ 5 8 Ω m. 5 6 Ω m gven n Table 7. by.% t temperature T, the resstance of the carbon wre s c c + α c T T, and that of the nchrome wre s n n + α n( T T ). When the wres are connected end to end, the total resstance s + ( + )( ) + + α α T T c n c n c c n n f ths s to have a constant value of. kω as the temperature changes, t s necessary that c + n. kω [] and α + α [] c c n n From equaton [],. kω, and substtutng nto Equaton [] gves c n. k Ω. 5 C +. 4 n n C Solvng ths equaton gves n 5.6 k Ω ( nchrome wre) Then, c. k Ω 5. 6 k Ω 4.4 k Ω carbon wre

15 Current and esstance (a) The total power you now use whle cookng breakfast s + 5 W. 7 kw The cost to use ths power for.5 h each day for. days s cost ( t) rate (. 7 kw ). 5 h day. $. $. 6 ( days) ( kwh) f you upgraded, the new power requrement would be: ( ) W 9 W and the requred current would be 9 W 6. 4 > No, your present crcut breaker cannot handle the upgrade (a) The charge passng through the conductor n the nterval t 5. s s represented by the area under the vs. t graph gven n Fgure Ths area conssts of two rectangles and two trangles. Thus, Q rectangle + rectangle + trangle + trangle ( ) + ( ) 5. s. 4. s. s (. s. s )( 6.. ) + ( 5. s 4. s ) 6.. Q 8 C The constant current that would pass the same charge n 5. s s Q t 8 C 5. s (a) From ( ), the current s 8. W The tme before the stored energy s depleted s 7 Estorage. J t 5. s 8. J s Thus, the dstance traveled s 4 d v t (. ms ). 5 s 5. m 5. km

16 6 Chapter The volume of alumnum avalable s mass densty 5 kg. 7 kg m m (a) For a cylnder whose heght equals the dameter, the volume s d d d π π 4 4 ( m ) and the dameter s d π π The resstance between ends s then ρ ρd πd 4 5 ( ) 8 4 ρ 4. 8 Ω m π d π m For a cube,, so the length of an edge s 5 ( ) 4. 6 m. 4 9 m The resstance between opposte faces s 8 ρ ρ ρ 8. Ω m m m Ω 7 Ω The current n the wre s 5. Ω Then, from v d nq, the densty of free electrons s n de( r ) 5 v π ms. 6 Cπ 5. m ( ) ( ) or n m 7.6 Each speaker has a resstance of 4. Ω and can handle 6. W of power. From the maxmum safe current s max 6. W Ω, Thus, the system s not adequately protected by a 4. fuse.

17 Current and esstance 7 ( outer nner ) 7.6 The cross-sectonal area of the conductng materal s π r r. Thus, ( ) 5 ρ 5. Ω m 4. m π (. m) 5. m 7.64 The volume of the materal s Snce mass 5. g m densty 7.86 g cm 6 cm m, the cross-sectonal area of the wre s Ω 7 MΩ (a) ρ ρ ρ From, the length of the wre s gven by ρ (. Ω ). 6 ( m ) Ω m 9. m π d The cross-sectonal area of the wre s m d π π 9. m. Thus, the dameter s 4 9. m.9 mm 7.65 The power the beam delvers to the target s (. )( ) W The mass of coolng water that must flow through the tube each second f the rse n the water m t c T as temperature s not to exceed 5 C s found from 5 m. J s t c ( T ) 4 86 J kg C 5 C.48 kg s 7.66 (a) t temperature T, the resstance s ρ, where ρ ρ + α( ) T T, + α ( T T ), and + α ( T T ) + α T T Thus, T T T T ρ + α( ) + α ( ) + α T T + α T T + α T T + α ( T T ) contnued on next page

18 8 Chapter 7 ρ 8 ( 7. Ω m )(. m) π (. ) Then + α T T gves. 8 Ω + ( C)( C). 8 Ω Ω The more complex formula gves ( ) Ω 7 C 8. C 6 + ( 7 C)( 8. C). 48 Ω Note: Some rules for handng sgnfcant fgures have been delberately volated n ths soluton n order to llustrate the very small dfference n the results obtaned wth these two expressons Note that all potental dfferences n ths soluton have a value of. Frst, we shall do a symbolc soluton for many parts of the problem, and then enter the specfed numerc values for the cases of nterest. From the marked specfcatons on the cleaner, ts nternal resstance (assumed constant) s where 55 W Equaton [] f each of the two conductors n the extenson cord has resstance c, the total resstance n the path of the current (outsde of the power source) s t + c Equaton [] so the current whch wll exst s t and the power that s delvered to the cleaner s delvered t t 4 t Equaton [] The resstance of a copper conductor of length and dameter d s ρ ρ Cu Cu πd 4 c 4ρCu πd Thus, f c, max s the maxmum allowed value of c, the mnmum acceptable dameter of the conductor s d mn 4ρ π Cu c, max Equaton [4] contnued on next page

19 Current and esstance 9 (a) f c. 9 Ω, then from Equatons [] and [], t ( ). Ω. Ω 55 W + 8. Ω and, from Equaton [], the power delvered to the cleaner s delvered ( ) 55 W ( ) Ω ( 55 W ) 47 W f the mnmum acceptable power delvered to the cleaner s mn, then Equatons [] and [] gve the maxmum allowable total resstance as + 4 ( ) t, max c, max mn mn so ( ) c, max When mn 55 W, then c, max ( ) ( ) ( ) mn mn mn 55 W 55 W 55 W. 8 Ω Ω m 5. m and, from Equaton [4], d mn π. 8 Ω 6. mm When mn 5 W, then c, max 5 W 55 W 55 W. 7 9 Ω Ω m 5. m and d mn π. 7 9 Ω 9. mm