Current and Resistance

Transcription

1 7 Current and esstance Clcker Questons Queston N. Descrpton: Developng an understandng of resstance and resstvty. Queston n ohmc conductor s carryng a current. The cross-sectonal area of the wre changes from one end of the wre to the other. Whch of the followng quanttes vary along the wre?. The resstvty B. The current C. The current densty D. The electrc feld. only. B only. C only 4. D only 5. and B only 6. C and D only 7., B, C, and D 8. None of the above Commentary urpose: To explore and relate concepts nvolved n electrcal conducton. Dscusson: Snce charge s conserved, all the current that enters one end of the wre must ext the other, so the current through any cross-secton perpendcular to the wre s axs must be the same. Thus, quantty B does not vary. 86

2 Current and esstance 87 The resstvty of a conductor s a property of the materal tself (lke densty), and (unlke the resstance) does not depend upon the geometry of the conductor. So, quantty does not vary. The current through the wre does not vary, but as the wre expands the current spreads out over a larger area (or s channeled through a smaller area, dependng on the drecton), so the current densty (quantty C) wll vary. The mcroscopc form of Ohm s law s J E ρ, where J s the current densty at a pont n a conductor, ρ s ts resstvty, and E s the electrc feld at that pont. ccordng to ths, f the current densty vares and the resstvty does not, then the electrc feld must also vary. So, quantty D wll vary. Thus, the best answer s (6). Key onts: Conservaton of charge requres that the current through a wre s the same through any cross-secton of the wre, even f the wre s sze or shape changes. Current densty s current per unt area travelng through a conductor at a pont; ntegratng the current densty over an entre cross-secton yelds the total current. Electrcal resstvty s an nherent property of a materal, and does not depend on the amount or shape of the materal. The current densty at any pont n a conductor s proportonal to (and n the same drecton as) the electrc feld at that pont. For Instructors Only Students often fnd t easer to understand concepts n relaton to other concepts. Ths queston provdes a good context for sortng out the dstncton between current and current densty, the dstncton between resstance and resstvty, and the relatonshp between current densty and electrc feld. Students are more lkely to be comfortable wth the frst three quanttes than they are to understand the role of the electrc feld, snce current flow s usually related to voltage dfferences. Queston N. Descrpton: nkng resstvty, resstance, and resstor geometry. Queston Whch object below has the lowest resstance? ll three have length and are made out of the same materal. rea of face wthout hole rea of hole

3 88 Chapter 7. #. #. # 4. Both # and # have the same, and thers s less than the resstance of #. Commentary urpose: To relate resstance to the geometry of the resstor. Dscusson: conductor s resstance s a measure of how dffcult t s for electrcal current to flow through t. In general, the resstance of a wde wre s lower than the resstance of a narrow wre, much lke water flows more easly through a wde ppe than through a narrow ppe. For a conductor of length made out of a unform materal of resstvty ρ and havng a constant cross-sectonal area of (.e., t doesn t get wder or narrower along the drecton current s flowng), resstance s gven by ρ. Objects () and () both have the same length, materal, and cross-sectonal area, so they have the same resstance. Object () has a cross-secton that ncreases from to, so ts resstance must be less than an object wth constant cross-secton but less than one wth constant cross-secton. Thus, the best answer s (). Key onts: unform conductor s resstance depends on ts length, cross-sectonal area, and materal. The shape of the cross-secton doesn t matter. n object wth non-unform cross-sectonal area wll have a resstance greater than you would fnd f you just used ts largest cross-secton, and less than you would fnd f you just used ts smallest cross-secton. For Instructors Only You can analyze stuaton () n terms of the seres resstance of successve slces, each a dsk of ncreasng dameter. nalyzng () and () n terms of a large number of narrow wre-lke strps n parallel can help convnce students that the cross-sectonal shape doesn t matter, only ts area. naloges wth water (or perhaps molasses) flowng through ppes can help students form an ntutve understandng of how resstance depends on shape, as well as of the rules for parallel and seres resstances. QUICK QUIZZES. (d). Negatve charges movng n one drecton are equvalent to postve charges movng n the opposte drecton. Thus, Ia, Ib, Ic, and Id are equvalent to the movement of 5,, 4, and charges respectvely, gvng Id < Ib < Ic < Ia... Under steady-state condtons, the current s the same n all parts of the wre. Thus, the drft velocty, gven by v d I nq, s nversely proportonal to the cross-sectonal area.. (c), (d). Nether crcut (a) nor crcut apples a dfference n potental across the bulb. Crcut (a) has both lead wres connected to the same battery termnal. Crcut has a low resstance path (a short ) between the two battery termnals as well as between the bulb termnals. 4.. The slope of the lne tangent to the curve at a pont s the recprocal of the resstance at that pont. Note that as ncreases, the slope (and hence ) ncreases. Thus, the resstance decreases.

4 Current and esstance From Ohm s law, I V 6.. Ω. 6.. Consder the expresson for resstance: ρ ρ. Doublng all lnear dmensons π r ncreases the numerator of ths expresson by a factor of, but ncreases the denomnator by a factor of 4. Thus, the net result s that the resstance wll be reduced to one-half of ts orgnal value. 7. (a). The resstance of the shorter wre s half that of the longer wre. The power dsspated, ( ) (and hence the rate of heatng), wll be greater for the shorter wre. Consderaton of the expresson I mght ntally lead one to thnk that the reverse would be true. However, one must realze that the currents wll not be the same n the two wres. 8.. Ia Ib > Ic Id > Ie I f. Charges consttutng the current I a leave the postve termnal of the battery and then splt to flow through the two bulbs; thus, Ia Ic + Ie. Because the potental dfference s the same across the two bulbs and because the power delvered to a devce s I( V), the 6-W bulb wth the hgher power ratng must carry the greater current, meanng that I > I. Because charge does not accumulate n the bulbs, all the charge flowng nto a bulb c e from the left has to flow out on the rght; consequently, Ic leavng the bulbs recombne to form the current back nto the battery, I f Id Ib Id and Ie I f. The two currents (a). The power dsspated by a resstor may be expressed as I, where I s the current carred by the resstor of resstance. Snce resstors connected n seres carry the same current, the resstor havng the largest resstance wll dsspate the most power.. (c). Increasng the dameter of a wre ncreases the cross-sectonal area. Thus, the cross-sectonal area of s greater than that of B, and from ρ, we see that < B. Snce the power dsspated n a resstance may be expressed as ( ), the wre havng the smallest resstance dsspates the most power for a gven potental dfference. NSWES TO MUTIE CHOICE QUESTIONS. The average current n a conductor s the charge passng a gven pont per unt tme, or I Q t e n t, so the number of electrons passng ths pont per second s n I 6. Cs. 9 electron s 9 t e.6 C electron makng (c) the correct choce.. The drft velocty of charge carrers n a conductor s gven by v d I nq. Wres and B carry the same current I. lso, they are made of the same materal, so the densty and charge of the charge carrers (n and q) are the same for the two wres. Snce the cross-sectonal area s πr, and r rb, then wre has a cross-secton 4 tmes that of B. Ths makes v vb 4 and the correct choce s (e). B. Snce B and rb r, we fnd that B ρ rb ( r r ) ρ ρ π, and the π π 4 correct response for ths queston s choce (d).

5 9 Chapter 7 4. The resstance of a conductor havng length and a crcular cross-secton s ρ ρ ρ r r. The value of π π r for each of the three wres s: for wre ; r 4r for wre ; and for wre. Thus, wre has the smallest resstance and choce (c) s 9r the correct answer. 5. The kwh s kwh kw h W 6 s J s 6 s. 6 6 J. Thus, the combnaton of choce has unts of energy. The other combnatons have unts of: charge per unt tme, or current for (a); energy per unt tme, or power for (c); current tme, or charge for (d); and power dvded by tme for (e). None of these are unts of energy, and choce s the only correct response to the queston. 6. When the potental dfference across the devce s V, the current s so the resstance s I V Ω and (a) s the correct choce. 7. When the potental dfference across the devce s V, the current s.5 so the resstance s I V.5. Ω and s the correct choce. 8. The temperature varaton of resstance s gven by + α T T, where s the resstance of the conductor at the reference temperature T, usually. C. Usng the gven resstance at T 9. C, the temperature coeffcent of resstvty for ths conductng materal s found to be α. 55 Ω C T T 9. C. C. Ω. The resstance of ths conductor at T. C, where wll be T T. C (. C ) 4. C + ( 4 C ) ( C). Ω Ω so s the correct choce. 9. The power consumpton of the set s ( ) I ( V )( 5) Thus, the energy used n 8. h of operaton s E t of cost ( 4. kwh )( 8. cents kwh ) 9cents. The correct choce s (c)... W. kw.. kw 8. h 4. kwh, at a cost. The current through the resstor s I. V. Ω., and the charge passng through n a s nterval s Q I t (. Cs )( s ). C. Thus, (c) s the correct choce.. esstors n a parallel combnaton all have the same potental dfference across them. Thus, from Ohm s law, I V, the resstor wth the smallest resstance carres the hghest current. Choce (a) s the correct response.. esstors n a seres combnaton all carry the same current. Thus, from Ohm s law, I, the resstor wth the hghest resstance has the greatest voltage drop across t. Choce (c) s the correct response.

6 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 EVEN NUMBEED CONCETU QUESTIONS. In 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. In 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 V, 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 SOUTIONS 7. The charge that moves past the cross secton s Q I( t), and the number of electrons s Q I 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 I I 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.

7 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 I t T πr 6 9 (. 9 ms ). 6 C π 5. 9 m 5. Cs 5. m 7.4 If N s the number of protons, each wth charge e, that ht the target n tme t, the average current n the beam s I Q t Ne t, gvng N I( 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 I 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 I t t 9 ( 9. 9 ). 6 C.78 h 6 s h I 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 I d ( π ) 4, or 4. m s

8 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 I 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 I. and cross-sectonal area 5. 6 m, the drft speed of the conducton electrons n ths wre s v d I nq. Cs (. 7 9 m )(. 6 C) m 4 ms. V 7. From Ohm s law, I Ω

9 94 Chapter 7 7. I max max 8 6 Thus, f 4. 5 Ω, ( ) max V and f Ω, ( ) 6. V max 7. The volume of the copper s V Snce V 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 Insertng 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. V 7. (a) From Ohm s law,. Ω I 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 V 7.5 (a) I 4. Ω From ρ, ρ ( m) ( Ω ) π. 4. m Ω m

10 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 V 7.7 The resstance s. 5 Ω, so the resstvty of the metal s I 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) I I max mn 4 cm cm cm 5. cm 5. m V m ( 6. )( 8. ) 8. 8 ρ ( ) 8 7. Ω m mn max mn V m ( 6. )( 5. ) 7. 8 ρ ( ) 8 7. Ω m max 7.9 The volume of materal, V 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, I 5. 6 Ω 5 5 V ubber-soled shoes and rubber gloves can ncrease the resstance to current and help reduce the lkelhood of a serous shock.

11 96 Chapter 7 7. If 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 I, where ρ. Combnng these relatons, we obtan E I I( ) ρ or E ρ( I ) 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, I I ff, so the current n ntarctca s I f I T T I 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. If 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

12 Current and esstance t 8 C, I + α T T ( Ω ) +. 5 or I. 6 6 m 5. V ( C )( 8 C C) 7.8 If 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. V and the current wll be I. Ω. t. C, + α T T ( (. ) + 9. ( C) Ω ). C. C. Ω V Thus, the current s I 9. V. Ω The resstance of the heatng element when at ts operatng temperature s V 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

13 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, V 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 I( V) I V. W 8.. V, so the current must be. From Ohm s law, the resstance s I 8. V 4. 4 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 I( V) 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

14 Current and esstance The power requred s I( V) 7.6 (a) The power loss n the lne s V. W loss I 7. Ω km 6 km 5. W 5 MW The total power transmtted s 8 nput ( ) I 7 V 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 ( ) I ( V )(. ) 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 ( ) I ( V )(.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 V 49 W 4. 4 Ω

15 Chapter (a) t the operatng temperature, ( ) I ( V )(.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. V ( ) V I.5, and V ( ) V I.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. I I 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 -V battery s I t 55 h. Thus, the stored energy s Energy stored t ( ) I t ( V ) 55 h 66 W h. 66 kwh cost kwh $. kwh $ cents ( ) 5 I 75 V.. 5 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

16 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 V V 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 V 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

17 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. In 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. V (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 I t ( 6. )( h ) ( 6. C s )( 6 s ). 6 5 C

18 Current and esstance 7.5 The energy avalable n the battery s Energy stored t ( ) I t ( ) I t V 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. V W 9. V 67. 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 V 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 I ( 5 )(. 79 Ω) 8. 9 V 89 µ V

19 4 Chapter Ohm s law gves the resstance as V I. 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 If 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

20 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) If you upgraded, the new power requrement would be: ( ) W 9 W and the requred current would be 9 W I 6. 4 > V 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 I 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 I t 8 C 5. s (a) From ( ) I, the current s I 8. W. V 667. 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

21 6 Chapter The volume of alumnum avalable s V mass densty 5 kg. 7 kg m m (a) For a cylnder whose heght equals the dameter, the volume s d d V d π π 4 4 ( m ) V 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, V, so the length of an edge s 5 ( V) 4. 6 m. 4 9 m The resstance between opposte faces s 8 ρ ρ ρ 8. Ω m m m Ω 7 Ω 5. V 7.6 The current n the wre s I 5. Ω Then, from v d I nq, the densty of free electrons s I 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 I the maxmum safe current s I max 6. W Ω, Thus, the system s not adequately protected by a 4. fuse.

22 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 V mass 5. g m V densty 7.86 g cm 6 cm m, the cross-sectonal area of the wre s V Ω 7 MΩ (a) ρ ρ ρ From, the length of the wre s gven by V V V ρ (. Ω ). 6 ( m ) Ω m 9. m π d The cross-sectonal area of the wre s 4 6 4V m d π π 9. m V. Thus, the dameter s 4 9. m.9 mm 7.65 The power the beam delvers to the target s (. )( ). 6 5 I 4 V 5 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

23 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 V. 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 [] If 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 I V t and the power that s delvered to the cleaner s delvered I 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

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