phys122 practice - Name[please print last first total...

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Unformatted text preview: Name [please print}: .. .... .. ................-.. last first total points It. [25 points} Two vertical straight and tier].r long parallel wires are separated by H = fiiicm. The parailel electric current: through each of them are i. = IDA and 1'; = 2|] it respectively. {Fe = ala- tt lfl'TTmfi‘tJ ii] [iii paints] I’II‘I electron [e = —l.fi it 10'“ C] has an upward eel-Deity 1'} = 1.5 3-: Ifli‘mfe along a line at exactly half—distance between the two currents. Find the net farce {magnitude and directitrn} acting on the electrern. “—5: “ah-lg 4—11 aila =5— 3. . as . .... i... lemma» ' it \Th e- t: IL_ It. 'L E I.._II ngG} ewe: ee ;7%-:( :'[email protected] hi I? paint] Bummer new a line 111 1:112:13“ at the 2twe-enrreflte and perpeardicuiarti them. Determineall paints film; this line where the net magnetic field finishes. Me. 125,. «km, whit: him» flue ‘hib (fitted-5. : FD :1} iii! a t2 '2‘- 1'1. '9'. 1 1 )1—-— :1": if: "-:"- £3 2...: tEH-CI-I... 4.3. 'c. St} E} {Sp-pints] Sketch the magfi‘citic fijdasafnfietih'net'distance alert; the line-described in h}. namely the line perpendicular to the two currents and' III the plane fer-1mm! by the two currents. Use the diagram below, where the current a" in at the arigin and the current 1': is at D. Please note that this part is somewhat challenging. Name {phase print]: III-I'I'I-II'III'II'HII'Htin-IFIIFl-IH-Hl I-I-III I-IIIIIIIIII-IlI-Illtlltfl-eie-n'... §"|.||-H ............. last that total points 1. {35 Flt-L] A equnre mil with N = It!!! turns and the ice of each side «at the met. = (1.501113: aeedhrenexternal uniform magneticfiedperpendimlartethe ane of the tail. [U ' errn field mean: that the field. has the same magnitude and dire:- tinn throught the epace at any given time} Even though the magnetic field is uniform. it! intensity varies III-Pith time aeeerdi tn the fattening relationsh: Em“) = Ifleinflt]. wheretieineecnndeend Bin Tesla. henetreeietaneeet'thecei i; R = 1911. e] [1e tn] Find the nregnetie that due to the external field thmngh the coil as e fitnctien e tirne. GE?” 2 IHLE'EwkEl-D e 2'30 51 34:) mm in} {1D [rte] Find the electric current meme in the enii. Let“ 2%): ilafirgfle -5}"3mC5JF') meafi- W e] 15 tat] The In ‘tnde eithe n] etit fidcl created. by it straight segment at wire of ienglth Eon each for the square afihe neuter ofthre equate N-tnrn nail is _ Haiti} Haiti — m 1Ill’tlere pa, = 4! :n: Iii-7 T Infi-L Determine the intensity ofthe net magnetic field at the center at" the ceii an a funetinn of time in terms of the given quantities and specify its directien. mmlusepfinl}: SflLuTIDMS flan) {fim} 2. [35 points mm] at. £133} MWIEMMEY. + interlaced aura? 3' lhjg'lhflJ-flflsfl. n. wuhnw: B. [H915] ChargeZismuflddimulybclnw clgargcYum Ducfihfim A. awayfiumml’uitwin part i. In II: 59m prufidud. draw arrows to infirm the uppmfimart fireman {If the In: cinch-ital fame amt-:11 “charge? “durum: 2. Hill: acumen-i=3! [clue is arm for dfltfi' charge. 31:1: 34:) Explicit]; [H0 cm mammal.) .I'lfllfll: fl“: ' ' "ca .l-U'lll I'MIIL'I .1. {35 pts total} A resistor R = ltl-EI ohms. and an ideal inductor L are connected in series to a T«voltage supply. 1’. A plot of the current. in this circuit {[1}, as a lunction ofltimc after the power supply is connected is shown below. 2 n no: one cos I“! I11 I11! tit-ate} a} {It} puts} 1t'tfrite an expression for Ht} in tormsut' the circuit parameters V. R, L. ifs}: .E/{.£‘ «"7 '1': 9f: - as = V —- .F Ire; Ti [it—fir ._ Mia. 15- ‘L/g I={fi*e'9 flat-I» ‘fw I-’{ = zit-r5 = 'fihffi’iai‘: flglmis mid“: L“: 45.13; I" '3') {I0 3313] The inductor L, is approximately one of those Values {henry]_ 1" 5'9? 5'. Circlethehestenswer. F'- I?" gflt=ffllp‘a?$= fifH so 12 s @ s c] [5 pts} The supply voltage V, is appl'flfimfltfllf oneof these values [in volts]. Cll‘ClE thfl heet answer. :zfld‘r’ If: llfflfiafl fl'. iii son ms to so :1} [.5 pts} fine of these combinations of parameters is the slope of the graph of fit] at t = E}. Circle the best answer. REL we REV we LIV sis in”; e] [5 pts] Justify your answer to part II] and use the measured slope to find the supply # values- H at: -cfl‘fii- Jail- ilk- Name {place print}: ................................. . .............................................. lact- iiret tutti pciiltl 4. [35 pm} Consider an ideal LG circuit a: menu. Suppme that G' = 1 HF and L =1Handthatatnmeinitieltimet= fl thereisnccurrcntin the circuit and the chargeunthecapacitcrplatcs is Q = 1 MC. 3.] [ll] [its] Find the perinci of the electromagnetic oscillations in this circuit. —- 2: t CPL L "‘ :3 1"” {a "T;- antihi- 6.131135%, 1:] [Ill]I lite-l Find the maximum value ci the currmt in the circuit. . QUIET :Q 1 1mg; :- mQui-nac = '1'}: 555—? c] pte. Replacetheca. Ichwitht'eracapacitnreC = [L5 pFandC: = LittFfiunflILHdtfith thein uctcraeehwm below. Determinithe period cfccciiiation cfthecurrentthrough the inductorL _‘~_. 2 .L «i ' _. s F -—.- C‘. c C” C"\ c133}; C‘I L ‘ f: “ C-‘ T: 2?. LCE, :: 3.3V“? Se; in f?- -cl} 1113.} Reconcctthe two capacitance shaminthe diagram below. Determine thencnperiodclccciilaticncltbemrrcntthmghtheinductcrh c“ —. gel: 2)} T192; ‘EL-Cex =- Effluent} Sec. d—_-_.—n—l—|II Hm: (plus: print]: EDLLFI'ICINS Paints inn first: 5. [30 minis] The: main] bar: {1. 2. and 1] I: marknd all mifluuniihcrmdasm Thcfulln'WingubsnnI-atiuns A. A mmadalqraatudtnt: icndlfllmmmdih - annd 131mm and 3A B I! ' flflwhmflfl nun In: In: It'yuu do and; ham: lnfligh inform-lion in answer any of the qufiiuls huh-w. :xplai: wit].r mi. 1%.. i. [Spa] laharlamngnatmanuamgnfiindimuhar? Explain. marmmmmmm Wfltmmthen WEN ii. ['J'pu] Iahariamagmuraaumagnaiindirmbafl Explain. Burl mmmmmmwmmflmndflm. Wamzlmmiasfi thwmmthemmlWMHmndzdmdfl mamflcmiaa. Jammmmmmmmmm hurl mnnntheamhat flumthemumefludhmm B. [MFG] flumdufafiamthmcmlhmcnd4misuhmadmamm3fiandlfl- Mmflmmhnmdfl. imp-act ifbmughtmrlfi? Explain. A HWJAWZB, Hilkhamaftheanmem Wilma” Wanamflm mtaranumu mw.i‘fltis anmmmmmi “flaws! m w” a“? n m a $23 {Since mm, so #3 Human: m #2.! since mmmmmmmmams .mmrom, 5'4 #Emidatflnctflmthismamm. m {2. [3.11:] The diam-am Willow! a bar magnet am an a fixed pin-.15: rod II'III has negative chi-yo. ‘nwmagnatisfmammmubmtapim throughfluunlar.and1hamd isin Iinn lith 1h: pivot. point. Duaa th: magic: route? if an. indicate aha dimalicn in m which it walnut-11:. Elplain. WMofJMiHMEWiInflm, mun mmnashameflcmfi. There rag. themaetmflmm flhemnetmuid inrizgnithfvoflhve' mm tatfihotium nth! Mamgmthetm mgmfifinat immammmnsnfiam Wham.) 3H“? {El-I] Name {please print]: .. I'I'F'lllllllfllihiil-IIIHII'I'IIIll III Il'IIII-H liillii-iI-IIIII-I-II Ilill-I III-IIIIIIi-ll-Iu ........ ti. 3!] points] Consider the folletving two electric :ireuits With R1 = III, R; = an, 33.: andtheeanfafthebatbuy5=fifl 1 E1; ill-1' 22a; : ng-IL 1 L 32*“; ‘3 V: i. R: rail-=— Tl" = E.“- b) [15 pinball-law should one connect the the: m with] the me battery so as to have the mmmum electric ELIE!“ through the battery? Yen can choose in we we. twn er all three resistors- TILL “unwi- fitted-{SM a»... weiglw I; i .- '2. Qt! we: \Us Li tum-I‘wtiqi mkrmHVE-filf mat. _ T 1T+T+I 15i— E g M ._ "'l": ‘1: .1- / R Li "'3' .L‘ 9-11]. Name [please print]: ................................... . last _ first tetal points i. [Iii] points] r'tn insulated metallic sphere el' t'tttlius e: carries a net pesitit'e charge q. l+ .Jt metallic shell. concentric with the sphere. has an inner ratiius it anti an cutter radius e {e c: it e: e] and carries a zero net charge. Let us cienete with c the distance item the center ei the sphere to an arbitrary paint The electric Ftehi vanishes for D E r s: e. a} [lit peints] Draw the electric iieltl lines+ I!) [ll] peints] The spherical shell has an induced charge —Q en the inner surface and an induced charge Q on the enter surface. Using Gauss law determine 0. [9% 5?. amegiqh. 5m: {ti n“. 56kg“? lactic; $50 M m {tell its. ”this-Law 1,56%:0 2 Lit—Q =3 Q: c] El'fl peintsl Determine the magnitude cf the electric field as a functien sf dist ce 1* EC!) and draw it. Vii-ink Eat] ‘1 test. sfi~£tt1=fi Bet-ace. EL: I. it. Name [please print}: (last) ififii] Tetal Points:— {15 ts tctall An electric field line diagram cf the eletiiric field in a certain regien cl' space is shewn at right. [fipts] fin the diagram. draw an arrnw to indicate the directien cf the eiectriceiferce that would be erected an a small positive test charge at paint: it. B, and C'. litheeleetrical farce is acre at any paint. state se esplicitly. [Gpts] lIEttttumpare the nitudes cf the eiectricci farce en the small posit tie test charge at the fellawing pairs cf paints: Fer parts i and ii. the magnitude ef the electrical :l'crce en the small punitive teat charge will he directl nienel the magnitude ut' the electri: filial the paints in questinn. i. Paints at. and i5“. Explain. Ele- E. because the electric field line density near paint it is greater than tltttt near paint .ti'. ii. Paints ti and C. Explain. E, -c ELIMH'II-Sfi the electric field line density near paint it is less titan that near paint C. [l3pte] n. sntttll test charge tunfl was ariginally used in detetntine the electric field as represented an the diagram. a. retell test charge 4- 24'. is new used te measure the electric field. i. is the magnitude of the eiectricfield at int d as measured h the new teat c e greater titan. c uni in. at less than that measured by the cal-Final test charge".I Heinlein. berg q The magnitude cf the electric field as measured by any smell teat charge is defined as the magnitude of the electric farce etterted an the test charge divided by the arnuuat cf charge. in this case. the magnitude cf electrical t'ctce an the teat charge will duuhle. but an will the atnuunt el' charge. 'l'l'rerel'ute. the magnitude cf the electric field at paint A as measured by the new test charge is equal In that as measured by the criginei test charge. it. Is the magnitude at the electric-at ecscrted en the new test ch et int it renter than. e eat re. ar icsr than that exerted an the erfghtca] teat charge“.ll Explain. “7E“ PD 3 q The magnitude ut' the eiecuicel farce en a test charge at paint it is equal tn the prcdttct at" the magnitude cf the electric field at paint rt with the quantityr cl' charge an the test charge. Since the magnitude cf the electric field at paint d is the same fer hcth test charger. then the magnitude at the electrical feree en the new test charge will he greater titan that en the nrlgirtel lest charge since the new test charge has there charge. "W' -me Heme I’Mfi [print Lee! rel Tetnl Fee'trte J. hflj‘fl' pie] A negatively charged emhee red I: hteught eleee te {hut rte-t teuehlng} the hell than uncharged eleetreeeeee. While the red in eleee he the hell yett mementeritr teeth the hell with your halter. Then rte-tr mature the ember rertl. I'm- away. an the diegreree heIew eketeh the tigne et’ the ehergee en the pane et' the eleetreeeepe including the hell* widget. and leaf hath he'l'ere and ether teudting the hell. Sheer the peeitielr til“ the leaf fer the "after ten I1 else. Den the eleetreeeepe he“: I. eet cheese et the end? "In. whet he the eign? Er 1‘ Explain. IIIIIIIIIIII-Ifllllli .l-IIIIIIIIIIIIIIIIII IEIIIIIIIIIIII-lll-l Infllllll'lllllllllll Infill-IIIIIIIIIIIIII IIIH-II-II-IIIIIIII- III-III-IIIIIII-Ill- III-IIIIIIIIIIIIIII- IfiFII-IfiF-IHIIIIII-l .LJIIIILJIIiLAIII-II IllII-IE-IIIIIIIIII- IEFIIIIEFIIIIIIIIIII ILJI-IILJIIIIIIIIII- Ill-IIIIIIIIIIIIIII- 'II-II-IIIIIIIIIIII III-IIIIIIIIIIII-II “E“ hHN pate] rt lee; untrflm'th' ehergee red with tetel thee-1e H} 1: theme there in ere-u eettiee. The fem tttte te thie red en e eherp 9 ie eheten fer t'l'fl- difl'etertt Ie-eet'tem et' 9. The magnitude-e et‘ the Few: are 3 "unite” Ind L5 'unite"+ Drew efetee Vector replicating the fame an e {the hlee'lt that} at the Intention shown due ttt ttm identieal ehergett rede. eeeh with chug-e +11. Drew the 1teeter eeeefultr he the left herrtl diagram there. What ii the munitudettfthe fume {in ““11in? 31’5“: Ernie; It' the red ett right were “meted weeld the meinitude er the three due Ie the left reel ineteeee ee deeteeee eeremeitttheflme? Ertpleirt. fill. ”‘PIM f'fih fimflhfi ##1## iIL-I_ {hr-tr. {Inf-refining ‘33.: fine. due ft: (4%" rte-J ML? We? fl-e'fig. dial-t rte-rd themed-fl:- Name {please print}: ...................................................................................... lest - firat total points 1. E30 paints] Three identical capacitors {1'1 = G; = C: = C are connected as shown in the figure below. The Battery pmficlee a. knnwrl voltage V. «9+6: I a]: [II] pcinta] Determine the equivalent capacitance between the paint: it and ‘f in terms of G, which il assumed tn be a knnwn quantity. Ci Ml. C3, TH- tamiicik l. 1 II _-" i1 1‘!— C’L'b LL fibfifib Wfit‘ Cal, {£11. C: It. It?" -a 3f:- _ 53. (3:1 - EC. h} [ll] points] Determine the charge: nn all the pla e5 capacitor-a when the switch .5' is clccecl in terms cf 15' and 'Ir', assumed tn he kncwn quantitiec. Q: Cent-U: 4%51‘] ff” c] [ll]I points] New replace the capacitor C; with a wire and leave all the ether elements of the circuit unchanged. Determine the equivalent capacitance between K and '1’ cf this new circuit and the charges on the platen cf 01 and 0:. Cl in- m‘ QWEci-‘tmmt cue ‘Efixn. 0-941 W ne’fidt C5. C1 fl Nmmmllrinl}: EflLUTmN Palm: Ida: fin: .[15puints] Mwlargatlinahaclaafuhargc 4w 4'" have uait'arm charge dmsilias 4a.; and 45;... 4W’"-"‘*‘£“ reap-actively. Thamflaaumpmllalmaach "* amaraadaapammhyamflldiflanaad. A: n." ‘. :nlargamant of lb: nanlfl raglan of me plates is f" '- shuwnmigm. PaintstdRam-unqual , .:*—"'—T4;.~é1—~T-.-w*‘f distances m m: lafl sheet. 1 I . :P I. E F: A. i. [data] In d1: diagram above. draw arrows to “a '1 It show 15-.- dimcfian arm: clmric field at ‘1‘ I tach fifth: filllfl' paints. a f (Mmmdwmmm. MM Huh-..“ mm" are find fly amupmfim) E. [GI-ta] Raul: 1h: magnitndas uflhc alumni: field at puinls .P dining]: 1": Explain. Til the fafi' mm Hr: right afbafll shuts. flu! amid-awn: due fr: III: Iberia have the man lid-action. The alumni-rude ofi‘hair mrdr m Ia Ihenfm mar at Paid Tainan alpaim R lands were may have ayaaira firearm Sim .rhafidd'ga “Handy cmshfl dine: notdrpmdmdidmm Ilia magnitude nffirfifdmflhrqadfllfialafpdma cadmium Pisa'q'mimi". iii. [finial lathe clam-it: 1301th at paintPgr-m mm. lan- adamequdmmat alpn'mtR? Explain. Illawmdmfiyrhmm-kfieflmapnaflmpahrchWfimmmfimMIPmfi: shahpmflhwafiumfiwemdflspbnmmmamdmm. I'm-workmansh- charmmwflfiomMahatmpmfliamgamaburbmaamflerabmfuumfw. Illa kahlharahupaaflfll. Bde-—Wfqfllbapuflnfhfdw'mnce WaMum-fl Elmira-Macaw»: La. mpmidmpamflmbekurhnmmwfi B. Thcmoslmammmphmdhymmfingplaflnflchlfimthammulthugeas thealmlilmdauaa. Then-gamusitymthalcfl surfldadhhalaflplauis-flob. i. [3pm] Find Illa charge din-nail}.r an Illa right all-face affine lafl platc. Explain. Siam: the drag: dammit: ca barks-affine: qrdae kfi Wamlaifupm-dapffiacwlflmflymm I right snaffle must be rid}. ii. [61:15] Find I'll: charge dmily an 111: left SUI-fan: of the right plate. (Hm.- liaa Ihe Gaussian mrlida ham i1 Illa diagram.) Explain. Mihffifimfierifluanqufflwfimmim ard'mamlflafldcmdmmflfiafiflrlmflndfidfl. mfiurh-aaghhamfl Ingram-rum. mdmmbadlnmlmadcfiwgr. RWMkafimfix-ad the righplflflfllfllhfltfieh+flp mm :2} ll'""'.F Ill-III- I'll“ :3. a) [13? pts} A. 24 1|lur'izult power supply is enuneeted in the circuit as she-tun below. 11551 t "2 I; E H E Q 24 V i] Threugh which resister is the current the smallest? L2. ohms ii]I What is the ratio of Current threugh R1 to eurreut through R. '3 Iflfl: 2... least-t: RrI::-Jr‘l 3-5:: 4:2. iii} What is the ratio of current through R; to eurreat through R, r ' 3 rara= 3ft Items-xi «Elf-I; 3:1: “‘E-E-‘E‘J-fr iv] 1What is the ratislul’ veltage acre-rs R1 ta voltage across R; ‘3 are = _. _L_/ res-t a It =41; M 4‘? =r’JI. that :— fi:-El,rf-§fi 15’1"?- hmrpe} Ir: -JI-Jtr. J. Era-L ” - .1." "ti- emf a at 3 7r”? What is the resistance of an idea] voltmeter? gym, file-I What is the resistance til an ideal ammeter'i' In the diagram below the resisters R. are ldentieal. An ideal voltmeter is eeuneeted tn terminals a and B in the diagram helm and the veltage measured is 5 units. In a separate measurement an ideal ammeter is cenueeted ts terminals A and B in the diagram below and the current measured is $25 amperea. What is the ernl 5‘, at the hasten:r and what is the value of R? Slaw haw you dhtaiu recur answers. R '- fi Vin-3': “rt-T: Ham: {please print]: Hr. mm. 1m ' first "£351 points ‘1. [21} points] The resistances of the misturs in the circuit heiuw have the flue: R1=3fl,Rg=fiflandfla=liflq Theemf'snflhetmhatteriamfizglr’aud 5&3 In .L-5 '11. E. . . ll ‘ C3, ‘ '11 E: R 1.. gm 3.} [IE [Minn] Determine the magnitudas and directimu a!" the electrical unreal: II, I; and I: flowing through the three rain-.15. Ii = Ea 4:1,: H5»! wheki’bhma flit " 35121.. "'- 1 L Ilazfijafll. = it, It =- 3%: 9" ng A- I III *1 I“ i t— 1, 3__ 1'3: ._..Pr 2. LEW-k arm—algae: I1, :- 3é-Pr --— mafia b] [1n paints] Damn: the pmver dimipated in the entire til-wit. '1?" Pp .: Tk‘fii-L I222: 1%“ = 13.251117 f} a. mm... (mm: Print}- .................................................................... -. ......... ~ - 2...:- lm first tutti. paints REL 1- [25 paints] A flu”! WIFE IMP can. flee-l3 rotate around it; sides at, nee figun: __-._';_'~‘.' hehW._The side nfthve square in L- — 511 cm lung. The mere of the ride die n = III] 3. _ L-rjrfll‘ 1infinite the ether sides, ah the and tie have negligible mate. The gravititinnal eceelerntiun' "__*' - ' is g = 9.3 mini. #11 eiettrical current r' 131' when diredion-end mepituiie is flaring through this loop. The loop is an a vertiefl homageneous magnetielield 3 =11”. The "L" loop iein equilibrium and the-aide: beand dam—frame]: angledfl = ti“ with the wtical. * IF? I . 11. \{f‘t m‘fiufiifi'htt A W wait we. alum“ it“ w new we: ‘ a m mg“ m ”a“... . 1 1| _ ._.1 W QM? 1....- nib-mum. ‘clu a. k. (EWA fig: Ila—kc. .ifleiaeumuiii. 'r . Hm {Phase prim}: MM Total Pail": M ITII'I'J' 2. [15 pain: natal] a. 1:"ng inmlmed wire is famed in:- a squirt Imp i1 I mica] h nun-mu is held hm‘izflnufllglufl mun wither.“ m Neglecllheml'5magnelic IL A. {film} Dmmmnnflmflyfiwd'ugnmshuwin flu: dhacfimflumrflmdnflhecmupanmeflcpuius. phinhow ymhmmmwmmwuyaudid. Th- m audit fun in me diurfimgrh: wmmu With thzfiuu: #15: 11‘ hand curling in list 'rcm'an affine :1:an [hr Ehmub pgiflfl III! the dirfifl'flt q'lfiefieflm the Jump: Hwfi'dd amid: B. [11pm] fiilcthewirclanpisheldingnlfma bummiulheplmuhhulmpis ught nwmccmpassandkldflmmanhuwn. i. Isthemagnimdcuflh: ‘ fieldall'hc Wmmflgmcr Imdlmnt um! rathemnmlludcul' 1h: magnetic fir.“ Ilfllnmmpmilplnh? Explain. Gunner that. hit Mp5: qfnr rpasin'an. [Mmguflcfie alight-mm: qr; magnum: f: the vumrmofflwfidn‘affiebap milk: find! of the hu- magnfl. Since then Mali-EH! an: Men-fan flair an: E: yawn than the fit Ill: loap dime. ii. anmumwnnfliempmwdhmmshmmlpmflmcdimiunfl:nun]:and of the campus nadir poinls. Explain how 3w haw It} draw I11: armw as you did. A: :I‘m'ed abate. the WWW“ lfi: localim affllr campus; is flu.- vmar nun nflhtfirlcf of III: [mp [which point: mum the hp am: page in #1: 10pm diagmmj and liltfidd affine bar Magnet {Which point: In 1h: rigid. 4:. [3pm] naminhaphnnwalhwedmmm Mtnhurizmlalm'usshm {'Ihchu mishaldinplm} Willlhnhnpsidaul'fllc Impmmaurafmrpageqdiualhmnw rmuininpla'cr.’ Explam ynlrlemmnl. Hat-brand the Ira-pail: offh- loapisghm by FI-iLX‘BL Whmlhthfitflq'fltbarmgnfl thrcfiefl. It: dimnim 1': on aflh page. mm name -" rm‘ '3 Total Points a} [3 pts} A power supply. a switch, a resistor. and an ideal capacitor are connected in a series circuit. Initially the switch is open and the capacitor is uncharged. When the switch is closed [at t = [I] 1I-Irhich of the lollowing quantities do not change suddenly at t = t}? Mark with “3' those that do not change suddenly. current through resistor :1 Imltage across resistor I :— voltage across capacitor :3 charge on capacitor K {3 nuts) Melee a sketch of the voltage across the resistor as a function of time. Ill-II...
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    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

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    Jill Tulane University ‘16, Course Hero Intern