ragsdale (zdr82) HW6 ditmire (58335) 1 the direction of the current in the figure. Using the lower circuit in the figure, we get

Transcription

1 rgsdle (zdr8) HW6 dtmre (58335) Ths prnt-out should hve 5 questons Multple-choce questons my contnue on the next column or pge fnd ll choces efore nswerng 00 (prt of ) 00 ponts The currents re flowng n the drecton ndcted y the rrows A negtve current denotes flow opposte to the drecton of the rrow Assume the tteres hve zero nternl resstnce 54 V 73 Ω I I 6 Ω 99 V Fnd the current through the 73 Ω resstor the 54 V ttery t the top of the crcut orrect nswer: 4643 A Let : 73 Ω, R 6 Ω, 54 V, 99 V I I the drecton of the current n the fgure Usng the lower crcut n the fgure, we get so + I R 0 () I R 99 V 6 Ω 0996 A Then, for the upper crcut I R I 0 (3) + I 0 I + 54 V + 99 V 73 Ω 4643 A Alternte Soluton: Usng the outsde loop + I 0 (4) I + 00 (prt of ) 00 ponts Fnd the current through the 6 Ω resstor n the center of the crcut orrect nswer: 0996 A From q () I R 99 V 6 Ω 0996 A I R I 3 At nodes, we hve I I I 3 0 () Py ttenton to the sgn of the ttery A 003 (prt of 3) 00 ponts r 9r 5 3 9r r 4 D I B

2 rgsdle (zdr8) HW6 dtmre (58335) Fnd the rto, where I s the current enterng levng the ttery Hnt: Apply the Krchhoff s lw to the loop ADA correct A Let : r, R 9 r, R 3 R 9 r, r R 5 D 3 R 3 Bsc oncept: D rcut Soluton: Bsed on Krchhoff s lw, the equton for the loop ADA s gven y + R 0 4 I B R 9 r r (prt of 3) 00 ponts Fnd the mgntude of the current 5 whch flows from to D 5 4 I 5 5 I I correct 4 5 I I I I I I I 9 I + 9 +, therefore I, 9 I Followng smlr nlyss, one fnds tht 4 9 3, so tht 3 I 4 9 I

3 rgsdle (zdr8) HW6 dtmre (58335) 3 Note: The juncton equton t D s , or 3 I 9 I 7 I 5 7 I 005 (prt 3 of 3) 00 ponts Fnd the resstnce R AB R AB 4 7 r R AB 5 r 3 R AB 80 3 r 4 R AB 6 3 r 5 R AB 84 3 r 6 R AB 4 r 7 R AB 36 r correct 8 R AB 63 8 r 9 R AB 48 r 0 R AB 35 6 r By nspecton, the followng crcut s equvlent to the orgnl crcut R 3 A R R AB R + R ( r) (9 r) r + 9 r B keywords: 36 r ponts 84 V 4 V 53 Ω 43 Ω 9 Ω Fnd the current through the 9 Ω lowerrght resstor orrect nswer: A B F r r R Let : 84 V, 4 V, r 53 Ω, r 43 Ω, R 9 Ω I From the juncton rule, I + Applyng Krchhoff s loop rule, we otn two equtons: A D r + I R () r + I R (I ) r + I R r + I (R + r ), ()

4 rgsdle (zdr8) HW6 dtmre (58335) 4 Multplyng q () y r, q () y r, Addng, r r r + r I R r r r + I r (R + r ) r + r I [r R + r (R + r )] r + r I r R + r (R + r ) (84 V) (43 Ω) + (4 V) (53 Ω) (43 Ω) (9 Ω) + (53 Ω) (9 Ω + 43 Ω) A ponts Four dentcl lght uls re connected ether n seres (crcut A), or n prllel-seres comnton (crcut B), to constnt voltge ttery wth neglgle nternl resstnce, s shown rcut A rcut B Assumng the ttery hs no nternl resstnce the resstnce of the uls s temperture ndependent, wht s the rto of the totl power consumed y( crcut A) to tht PA,Totl consumed y crcut B; e,? P A P A 8,Totl 3 P A 4 correct 4 P A 5 P A 6 6 P A 7 P A 6 8 P A 8 9 P A 4 In crcut A, the equvlent resstnce s R A 4 R, so the electrc current through ech ul s A V 4 R the power of ech ul s P A I R ( V 4 R ) R V 6 R Thus the totl power consumed y ll four uls n crcut A s P A,Totl 4 P A V 4 R In crcut B, the equvlent resstnce s R B R + R R R B R, so the electrc current through ech ul s B V R the power of ech ul s ( ) V I R R V R Thus the totl power consumed y ll four uls n crcut B s,totl 4 V R

5 rgsdle (zdr8) HW6 dtmre (58335) 5 P A,Totl,Totl P A 4 Ohm s Lw There re two rules for ddng up resstnces If the resstnces re n seres, then ponts onsder the comnton of resstors shown n the fgure 45 Ω 5 Ω 43 Ω 96 Ω 78 Ω Ω 37 Ω Wht s the resstnce etween pont pont? orrect nswer: 8686 Ω Let s redrw the fgure R Bsc oncepts: R3 Let : 45 Ω, R 5 Ω, R 3 43 Ω, 96 Ω, R 5 78 Ω, R 6 Ω, R 7 37 Ω quvlent resstnce R 6 R 5 R7 R seres + R + R R n If the resstnces re prllel, then R prllel + R + R R n Soluton: The key to complex rrngements of resstors lke ths s to splt the prolem up nto smller prts where ether ll the resstors re n seres, or ll of them re n prllel It s eser to vsulze the prolem f you redrw the crcut ech tme you dd them R R 367 R 5 Step : The three resstors on the rght re ll n seres, so R 367 R 3 + R 6 + R 7 (43 Ω) + ( Ω) + (37 Ω) 0 Ω R R 3675 Step : R 5 R 367 re connected prllel, so R 3675 ( + ) R 5 R 367 R 5 R 367 R 5 + R 367 (78 Ω) (0 Ω) 8 Ω 44 Ω

6 rgsdle (zdr8) HW6 dtmre (58335) 6 c d e R 3675 Step 3: R R 3675 re n seres, so R 3675 R + R 3675 (5 Ω) + (44 Ω) 59 Ω Step 4: R 3675 re prllel, so R ( + ) R 3675 R R 3675 (96 Ω) (59 Ω) 55 Ω Ω R Step 5: Fnlly, R re n seres, so the equvlent resstnce of the crcut s 4 V R eq + R Ω Ω 8686 Ω 009 (prt of ) 00 ponts 45 Ω 6 Ω 9 Ω 84 Ω 07 V Wht s the current through 84 Ω ottomrght resstor? orrect nswer: 55 A R 3 h g Let : 6 Ω, R 84 Ω, R 3 45 Ω, 9 Ω, 4 V, 07 V R Bsc oncepts: Krchhoff s Lws V 0 round loop I 0 t crcut juncton Soluton: A key smplfcton s to relze tht R 3 re connected n prllel cn e comned mmedtely, efore pplyng Krchoff s rules c d e I I R34 h g I34 R The comned resstnce s gven y R 34 R 3 + R 3 + R 3 R 34 R 3 R 3 + (45 Ω) (9 Ω) 45 Ω + 9 Ω 897 Ω Ths now gves loop equtons juncton equton The loop equtons re V I 34 R 34 I 0 V + I 34 R 34 I R 0 f f

7 rgsdle (zdr8) HW6 dtmre (58335) 7 the juncton equton s I 34 + I I Susttutng I 34 I I n to the loop equtons, we hve ( + R 34 ) I R 34 I V R 34 I + (R + R 34 ) I V Rewrtng multplyng y fctors, we hve ( + R 34 ) R 34 I R 34 I R 34 V ( +R 34 ) R 34 I +( +R 34 ) (R +R 34 ) I ( + R 34 ) V Sutrctng, we hve [( + R 34 ) (R + R 34 ) R 34 ] I R 34 V + ( + R 34 ) V, for The voltge cross s just so the power s V 4 V 34 R 34 I 34 P V 4 ( V) 9 Ω W 0 (prt of ) 00 ponts The swtch S hs een n poston for long perod of tme R 3 D ( + R 34 ) (R + R 34 ) R 34 (6 Ω Ω) (84 Ω Ω) (897 Ω) 0879 Ω, R S we hve I R 34 V D + ( + R 34 ) V D (897 Ω)(4 V) 0879 Ω (6 Ω Ω)(07 V) Ω 55 A 00 (prt of ) 00 ponts Wht s the power dsspted n 9 Ω rghtcentered resstor? orrect nswer: W In smlr wy we cn solve for the current through R 34 gvng R V V I 34 R + R 34 ( + R ) (84 Ω)(4 V) (6 Ω)(07 V) (6 Ω)(84 Ω) + (897 Ω)(6 Ω + 84 Ω) A When the swtch s moved to poston, fnd the chrcterstc tme constnt τ R τ R 3 τ ( + R ) 4 τ + R 5 τ ( + R ) 6 τ R R 7 τ 8 τ 9 τ R 0 τ ( + R ) correct

8 rgsdle (zdr8) HW6 dtmre (58335) 8 In chrgng n R crcut, the chrcterstc tme constnt s gven y τ R, where n ths prolem R s the equvlent resstnce, or R + R 0 (prt of ) 00 ponts 37 MΩ MΩ 5 MΩ µf 5 V S S hs een left t poston for long tme It s then swtched from to t t 0 Determne the energy dsspted through the resstor R lone from t 0 to t orrect nswer: µj seres they shre common current, I The correspondng power consumptons y R R 3 re respectvely P I R P 3 I R 3 Ths shows the correctness of the frcton, e P /(P + P 3 ) R /(R + R 3 ) Alternte soluton: More formlly, notng tht the ntl current s I 0, the totl energy dsspted R + R 3 y R s U R 0 0 I(t) R dt I 0 R e t/[(r+r3)] dt ( ) [ R (R ] + R 3 ) R + R 3 e t/[(r+r3)] R R + R 3 ( MΩ) ( MΩ) + (37 MΩ) µj 0 ( µf) (5 V) Let : 5 V, 5 MΩ, R MΩ, R 3 37 MΩ, µf The totl energy dsspted ponts In the fgure elow the ttery hs n emf of 3 V n nternl resstnce of Ω Assume there s stedy current flowng n the crcut 3 V Ω U R+R3 dssp 9 Ω 6 Ω Snce R R 3 re n seres, the energy R dsspted y R s only frcton of R + R 3 the totl energy: ( ) ( ) U R dssp R + R 3 We oserve tht power consumpton consderton provdes n ndependent check on the frcton used Snce the two resstors re n 4 µf Fnd the chrge on the 4 µf cpctor orrect nswer: 95 µ Let : 9 Ω,

9 rgsdle (zdr8) HW6 dtmre (58335) 9 R 6 Ω, r n Ω, V 3 V, 4 µf The equvlent resstnce of the three resstors n seres s R eq + R + r n (9 Ω) + (6 Ω) + ( Ω) 6 Ω, so the current n the crcut s I V R eq, the voltge cross R s V I R R V R eq (6 Ω) (3 V) (6 Ω) 4875 V Snce R re prllel, the potentl dfference cross ech s the sme Hence the chrge on the cpctor s Q V (4 µf) (4875 V) 95 µ 04 (prt of ) 00 ponts The swtch hs een open for long perod of tme R V Immedtely fter the swtch s closed, the current suppled y the ttery s I 0 V ( + R ) R S I 0 V R 3 I 0 V correct 4 I 0 5 I 0 0 V + R Before the swtch s closed, there s no chrge on the cpctor, so the voltge s zero cross the cpctor t ths tme Becuse t s not possle to chnge the chrge on the cpctor lke step functon (or the current should e nfntely lrge), mmedtely fter the swtch s closed, the voltge cross the cpctor ( R ) s stll zero Therefore, the voltge cross s V ; e, thnk of the cpctor s eng short-crcut for ths nstnt of tme So the current suppled y the ttery, whch s the sme s the current gong through, s I 0 V 05 (prt of ) 00 ponts A long tme fter the swtch hs een closed, the current I suppled y the ttery s I V R I V ( + R ) R 3 I 0 4 I 5 I V V + R correct After long tme, the cpctor hs een chrged remned stle Tht mens the current gong through s the sme s the current gong through R ; e, thnk of the cpctor s eng open-crcut for ths tme So we cn wrte down the equton V I + I R,

10 rgsdle (zdr8) HW6 dtmre (58335) 0 whch gves the current I s V I + R R t + R Ω + 3 Ω 4 Ω 06 (prt of ) 00 ponts The crcut hs een connected s shown n the fgure for long tme 48 V Ω Ω µf 3 Ω 47 Ω Wht s the mgntude of the electrc potentl cross the cpctor? orrect nswer: V S R R 3 + Ω + 47 Ω 48 Ω, Across, I t R t 48 V 4 Ω A I R 48 V 48 Ω A I t ( A) ( Ω) V cross R 3 3 I R 3 ( A) ( Ω) V Snce 3 re mesured from the sme pont, the potentl cross must e so Let : Ω, R 3 Ω, R 3 Ω, 47 Ω, µf 0 5 F I I t R 3 t I t R I S 3 V V V V 07 (prt of ) 00 ponts If the ttery s dsconnected, how long does t tke for the voltge cross the cpctor to drop to vlue of V (t) 0 e, where 0 s the ntl voltge cross the cpctor? orrect nswer: 0 µs Wth the ttery removed, the crcut s After long tme mples tht the cpctor s fully chrged, so t cts s n open crcut wth no current flowng to t The equvlent crcut s l I l R 3 R Ir r I t It Il I r R R 3 Req Ieq I I

11 rgsdle (zdr8) HW6 dtmre (58335) where R l + R 3 Ω + Ω Ω, R r R + 3 Ω + 47 Ω 60 Ω ( R eq + ) R l R r ( Ω + 60 Ω 0 Ω, so the tme constnt s ) τ R eq (0 Ω) ( µf) 0 µs The cpctor dschrges ccordng to Q t e t/τ Q 0 V (t) e t/τ 0 e t ( ) τ ln lne e t τ (lne) (0 µs) ( ) 0 µs ponts A negtvely chrged prtcle movng t 45 ngles to oth the x-xs y-xs enters mgnetc feld (pontng out of of the pge), s shown n the fgure elow q v Wht s the ntl drecton of deflecton? F (+ĵ + ˆk ) F ) (+ˆk + î 3 F ( ĵ + î) 4 F 0 ; no deflecton 5 F ) ( ˆk + î 6 F (+ĵ + î) correct 7 F ( ĵ î) 8 F ( ĵ + ˆk ) 9 F ( ĵ ˆk ) 0 F ) ( ˆk î Bsc oncepts: Mgnetc Force on hrged Prtcle: F q v B Rght-h rule for cross-products F F F ; e, unt vector n the F drecton Soluton: The force s F q v B B ( ) ˆk, v v ( î + ĵ), z Fgure: î s n the x-drecton, ĵ s n the y-drecton, ˆk s n the z-drecton y x q < 0, therefore, F q v B [ ( )] q v B ( î + ĵ) ˆk q v B (+ĵ + î) F (+ĵ + î)

12 rgsdle (zdr8) HW6 dtmre (58335) Ths s the seventh of eght versons of the prolem ponts A negtvely chrged prtcle movng t 45 ngles to oth the z-xs x-xs enters mgnetc feld (pontng towrds the ottom of the pge), s shown n the fgure elow x q v Fgure: î s n the x-drecton, ĵ s n the y-drecton, ˆk s n the z-drecton Wht s the ntl drecton of deflecton? F ) ( ˆk + î F (+ĵ ˆk ) 3 F (+ĵ î) 4 F 0 ; no deflecton 5 F ĵ 6 F +ĵ correct 7 F (+ĵ + ˆk ) 8 F ( ĵ + ˆk ) 9 F ) (+ˆk î 0 F ( ĵ + î) Bsc oncepts: Mgnetc Force on y z hrged Prtcle: F q v B Rght-h rule for cross-products F F F ; e, unt vector n the F drecton Soluton: The force s F q v B B ( î), v ( ) v +ˆk + î, q < 0, therefore, F q v B [( ) ] q v B +ˆk + î ( î) q F +ĵ v B (+ĵ) Ths s the thrd of eght versons of the prolem 00 (prt of 6) 00 ponts A devce ( source ) emts unch of chrged ons (prtcles) wth rnge of veloctes (see fgure) Some of these ons pss through the left slt enter Regon I n whch there s vertcl unform electrc feld (n the ĵ drecton) 03 T unform mgnetc feld (lgned wth the ±ˆk-drecton) s shown n the fgure y the shded re q m 5 cm +300 V Regon I 3 cm z Regon of Mgnetc Feld 03 T y Regon II Fgure: î s n the drecton +x (to the rght), ĵ s n the drecton +y (up the pge), ˆk s n the drecton +z (out of the pge) x

13 rgsdle (zdr8) HW6 dtmre (58335) 3 In order for n on to pss through oth slts on strght lne, whch of the followng condtons must e true for the forces on the on? ( F s the electrc force vector F B s mgnetc force vector) F F B F F B 3 F 0, FB 0 4 F 0, FB 0 5 F F B 6 F F B 7 F F B correct 8 F F B, thus t s mpossle 9 F F B To otn strght ort, the upwrd downwrd forces need to cncel The force on chrged prtcle s F F + F B q ( + v B) For the force to e zero, we need F + F B 0, F F B Therefore, the forces re equl opposte the mgntude of forces re equl; e, F F B 0 (prt of 6) 00 ponts In whch drecton (reltve to the coordnte system shown ove) should the mgnetc feld pont n order for negtvely chrged ons to move long the pth shown y the dotted lne n the dgrm ove? or B 0 ; drecton undetermned B ˆk correct 3 B +ˆk The force due to the mgnetc feld provdes the centrpetl force tht cuses the postve ons to move n the semcrcle As the negtvely chrged on exts the regon of the electrc feld, F B q v B, so y the rght-h rule the mgnetc feld must pont out of the pge ˆk down the pge; e, ĵ ( or n the z-drecton ), snce the force F s n the drecton the vector product q q v v +î B? F B F B ĵ, î ˆk ĵ, snce F q v B F ( ĵ) [ F F q v q v B ] B [ (+î) ( ˆk )] ĵ, consequently B ˆk s correct 0 (prt 3 of 6) 00 ponts If the ons re postvely chrged, the electrc feld must e downwrd for chrge to move through Regon-I undeflected nnot e determned snce F B F Flse correct

14 3 True The ove sttement s Flse ecuse, s we see from prevous dscusson, tht the feld s downwrd, ndependent of the chrge of the em gven n the sttement of the queston 03 (prt 4 of 6) 00 ponts In Regon I, the electrc potentl etween the pltes s 300 V, the dstnce etween the pltes s 5 cm, the mgnetc feld n oth Regons I II s 03 T Wht s the speed of sngly chrged on tht psses through oth slts mkes t nto Regon II? orrect nswer: m/s Let : B 03 T, V (300 V) d (5 cm) rgsdle (zdr8) HW6 dtmre (58335) N/ Snce the electrc mgnetc forces on the on re equl, q q v B v B N/ 03 T 04 (prt 5 of 6) 00 ponts The ons tht mke t nto Regon II re oserved to e deflected downwrd then follow crculr pth wth rdus of r 03 m The chrge on ech on s 0 8 Wht s the mss of the ons? orrect nswer: kg Let : r 03 m 03 m q The on wll e deflected upwrd n m/s Regon-II wth the sme crculr rdus The rdus of crculr pth tken y chrged prtcle n mgnetc feld s gven y r mv q B m q B r v ( 0 8 ) (03 T)(03 m) m/s kg 05 (prt 6 of 6) 00 ponts An on wth the sme mss wth twce the chrge mgntude, only postvely chrged, gets through Regon-I The on wll e deflected upwrd n Regon-II wth twce the crculr rdus The on wll e deflected upwrd n Regon-II wth hlf the crculr rdus 3 The on wll e deflected downwrd n Regon-II wth hlf the crculr rdus correct 4 The on wll pss through Regon-II undeflected 6 The on wll e deflected downwrd n Regon-II wth the sme crculr rdus 7 The on wll e deflected downwrd n Regon-II wth twce the crculr rdus In Regon-II the mgnetc force F B s now F B + q ( v B) m Thus, the force s twce the mgntude ut opposte n sgn Hence, the centrpetl ccelerton v hs the opposte drecton (upwrd) r so tht the deflecton s n tht drecton mv q v B r

15 rgsdle (zdr8) HW6 dtmre (58335) 5 r mv q B Thus the rdus of the ort s hlf of wht t ws efore; e, when the rdus ws r mv q B