HW_7_solutions - Version Dfli — Homework 7 —...

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Unformatted text preview: Version Dfli — Homework 7 — Hofinianr] — [55745) i This print—out should have 12 questions. Multiple—choice questions Ina},r continue on the next column or page — find all choices before answering. Internal Resistance [I4 fl'fl'l 19.0 points The emf of a batter}.r is E = 11 "v' . When the batteryr delivers a current of [1.9 A to a load. the potential difference between the terminals of the battery is 9 "'v' volts. Find the internal resistance of the battery. Correct answer: 2.22222 fl. Explanation: Given: 5 = 11V. WW 2 9 V. and I = [LB :1. The potential difierence across the internal resistance is E — Wand. so the internal resis— tance is given bv 43—Hour! 3" = -——— I llV-QV {3.9 A 2.22222 fl . Four Series Copper Wires ["32 10.0 points Four copper wires of equal length are con— nected in series. Their cross-sectional areas are 1.5 crng. 2.2 cmE. 2.? crng. and E: cmg. If a voltage of 7'3 V is applied to the ar— rangement, determine the voltage across the 2.2 cm2 wire. Correct answer: 19.15158 V. Explanation: Let : :11 = 1.5 c1112. A2 = 2.2 c1112, A3 = 2.? cm‘v’. A4 : 5 ctnfl. and trim: 2 T3 V. The resistance of the copper wire is propor— tional to the length of the wire and inversely proportional to the ertrss sectional area of the wire. Since the lengths are the san’le1 and the current i is the same for all resistors connected in series1 taking 1" out —._ etc, we have A i_ We! _E rl+r2+r3+n rg'i where Via: is the difierence in voltage across all for wires and ler is the difference in voltage across the second wire only. Solving for V2._ we have 7'2 L2 _ r1+ T2 + 1'3 + “Ir-1.er 1 Ar =a+ifi+im' A1 :12 A3 A4 1 = 2.2 cm2 1 l. l l. lbemz + 22cm2 2'r'crn2 +5Cm2 x [73 it] = Four Resistors JMS nos 10.!) points Four resistors are connected as shown in the figure. at“ Find the resistance between points a and t. CoIIEct answer: 35.9911 fl. Version [1131 - Homework T — Hofimann — {EEHEJ 2 Explanation: = 14.11194 Q + 52 ii = [56.1094 ii . 3:223 3 E.- The parallel connection of R123 and R4 LEE : R1 = 21 [1‘ gives the equivalent resistance Hg = is it, {11325211. #2 1 +L:R4+R123 H4 2 To ft, and Rob R123 R4 Rl'23 R4 ii 1' is = 97 V. Rab = L“ . _ . H123 + H4 DhfllhlflWlSVZIR. [fifimmrflnmu] A. good rule of thumb is to eliminate jtlnc— = tions connected by zero resistanoe. fifi‘mm H + ‘9 fl = 35.9911 11. or combining the above steps, the equivalent m e R3 resistance is n. b ( R] R2 + Rn) “- R. d H1 + H: t Hob = R1 R2 — R R. £1 + R2 + :i + i The parallel connection of R1 and R2 gives (21 Hi {43 it) + 32 H {W m the equivalent resistance _h 2] 11 + 43 ft ' 1 1 1 R2 + R. _ [:21 it) [43 o) n _=_ _= —— rest so H12 R|+Hg HIE: 2152+43fl+1 +r RI RE — as ILiriii o R . = _....__—._— — II I: n 1 . 12 RI + RE :1 _ {21 ii} {43 ii] _ 21 it + 43 fl Holt SF sortev 24 : 14.109411. [1114 111.11 points H12 Rs Find the equivalent resistance of the circuit. shown in the figure. fl 1: 1-1.5 i'i 3.511 The series connection of R12 and Hp, gives the equivalent resistance Risa = R12 + Rn 1|Iriersion {ltll — Homework T" — Hoffman“ — (SET-'15) 3 Correct answer: 31.54154 ll. Explanation: Let: R1 = 1351, 3-2 = l3 fl . R15 = 14.5 H. R4 = 3.5 fl. and Hg 2 14 ll. R34 = R3 + H4 2 {1-1.5 Q] + (3.551] = 18 fl 1 1 '1 3.. = __ __ m (Re i 334) 1 1 1 = [111111) + [11111)] =?.5-1839 11. so Rm = 31 + H.234 + Rs = {13 n) + [rsesssnj + (14.11] = 34.554114 Resistor Combination Ill BEE lilfl points The following diagram shone part of an elec— trical eirenit. '26 ll .‘l'i' El A 71111 B . Find the equivalent resistance R“, between points A and B of the resistor network. 1. RAB = 13 11 2. ft“; = 42 it correct 3. R“, = 11' n 4. R” s s? n s. RAE, = 74 n s. RAE = .13 n ’3'. RAB = 45 fl 3. RMg = 53 fl 9. RH = at} it 113. R” : €139 Explanation: R5 R: A é. B Let: R1=35H1 H2=3Tfl1 33:12.”. R1=ti35L R5=26§L Rfi=29fl. and 37:?851. Start from the rightrhanrl sirle in determining the equivalent resistances. Step 1: H1. [132. and Rs are in series. so 3123 = 31+ Rs + Rs —_ :1511+3Tfl+ 1:211 = 84 fl . Step 2: Hg; is now parallel with R4. so 1 __ 1 _i_fi’1+R123 Hl'fl-l 3123 I R4 lites H4 Version ["31 — Homework 7 - Hoffinann — [fifiTalfi] -1 Rurqu R123 + R4 [at n} [as n) 2 at n + as it = fill: 51. R1234 = Step 3: R1234 is in series with R5 and Hg, so R3=R12M+R5+Rfi =3fifl+2fifl+2lifl =91fl. Step 4: Finally R2i is parallel with Ry, so 1 _ l +1_R1+R5 Esq—R5 R»?— Rer _ are? R“? * at + H7 m {91 nurse] _ tnti+rsri = 429. AP EM 1993 MC 7!] flflfi 1|].l] points The following diagram shows at closed elec— trical circuit. The nmmeter in the center of the resistive network reads zero amperee. Find the electric resistance Rx. 1.Rx=14n. 2.Rx=iflfl. s.Rx=sn. 4.Rx=lflfl. 5.1L =sn. {LRx '—ti£l. T.Rx =11 fl. S.Rx=22fl. 9. RT = 24 fl . correct 10. HI = if: Q. Explanation: Let: 31:3!!! 32:31}! and Hg =95}. If the anirneter reads zero I_.. = G A, the two ends of the amineter should he an equipoten- tial: Vufl = $4. This means that the potential drop from .J’: or y to each side of the amineter through either the upper it or lower t5 part of the circuit must he the same. II"(Jr—M:=In-Ir1‘r1 {1} i;a,—ts=neiLi {2} Vii — Vy = It! R2 {3} Ve-Vy= “R: {4} Setting the potential across the ammeter equal l—‘L = 1”} (Le, Eq. 1 = 2 and Eq. 3 =. 4], we have In £1 = If R3 and In Hg = £31133, Luzfl'izfl n at n2 Version [FIJI — Homework 7 — Hoffinann — {5131745} 5 Ra 33 RI _ R1 _WQHMH _ so =2ML Current in Circuit flflT (part 1 of 2] HM] points Consider the circuit shown below. ‘What equation does the loop DCFED yield"? 1. £2 -!- igrg -r it? = L'] 2.51—tgr2miR=D 3.52—t-2r2+ifl=fl a—o+am—ne=n 5. —£2 —t?1"g - 2iR=fl 6. —£'; -t21|"2 ~—tR= I] 7. £2 — tg’r'g — if! = I] correct 8. —E.'g — igrg + 2151?. = I] Q.£g+t3rg+iR=fl 1"]. —51 + tzfr‘z — Hi = fl Explanation: Recall that Kirchhol'f's loop rule Sl-ill'lLi-i that the sum of the potential differences across all the eiernents around a closed circuit loop is zero. If a resistor is traversed in the direction of the current. the change in potential is — IR. if an emf source is traversed from the — to + terminals1 the change in potential is +5. Apply the opposite sign for traversing the elements in the opposite direction. Thus by inspection, : £2 — “Ii-R —t21"2 = [1 {ME {part 2 of 2} 113.0 points Let 51: £2 = E. and r1: r3 = r. where 5 = II Vandr=5fl AlsoletR= 1.851. Find the current i. Hint: From symmetry. one expects i] = 112. Correct answer: 2.55314 A. Explanation: Let: £1:£2=£=11V, rl=r2=r=fifl1 and R=1.8fl. E] = £2 and r1 = r2. This implies that t; = i2. (Why? Look at the loops DCFED and aeeea. Notice the identit}r of the two loop equations.) Hence the junction rule yields t1+t2=2i2=t _ i 32:— 2 Substituting this into the loop equation DCFED. . r Eg—tR—Erg=[l Solving for i yields Charging RC Circuit I11 DDS 10.0 points Version flfll — Homework ? — Hofimsnn - {56?45] I Thus the corresponding potential across R; i E .r Discharging RC Circuit {11 DIG 19.9 points Let Rl=R2=Rs=R- Initiallyr the switch is in position "b" and the capacitor is uncharged. At time t = U the switch is moved to position “:1”. Determine the potential difference across the resistor R1 at r = R1 (3. Find the characteristic time of the circuit when the switch is in position ‘13". 1J54 e 1 1 . T I — 2.25sJ os,ego i s. or. 8.3—9 2‘ T _ REG 1 _ 3_ = — 4.osse 1 T (R14-jofl 5' 153—05: 4. r = {R1 + R210 correct 5 -2 5 T — _-.w...%.__.m 6'1 E ‘ ‘(s1+s2)o 1 rec” . = F 5 T RIC 3.28e'flfi 7.T==ngc 9.1.EEE_L5 3.T=i..r‘R] It}. 115 E's—“'5 correct 9_ T = RI R? C Explanation: ' 'When 3 is at a. the loop equation is given 10- T = R1 C by Expianation: 5—i—4Uh+flfi=fl C In charging E111 R C circuit, the characteris— . . tic time constant is iven be This gwcs g ” s c s o T=RC1 2 —R R 9W : E'JHI‘.‘ '3‘ . . . . 1+ '2 hR where In thls problern R Is the equivalent At t = RC“. resistance. or i i=fif'! R=m+nt Version {lfll »- Homework 'r' - Hofi'inann - [55745] T Four Resistors and Capacitor 011 [part 1 of 2) 10.0 points The circuit has been connected as shown in the figure for a "long" time. 'What is the magnitude of the electric po— tential across the capacitor? Correct answer: 10 ‘2". Explanation: Let: 31:45“. H3=52£L "After a long time” implies that the capac- itor C is full}r charged. so it acts as an open circuit with no current flowing to it. The equivalent circuit is ___.t it ag=e1+e2=45n+52o=1mio and Ra=R3+Ri=2fl+23fl=25flt 55 s 25 v I = --— = = [-1.2 A. d * 221 men 5 an s 25 ‘v' r = —=—.——=1.5. '4 ab 25 52 Heroes R1. and across R3 53: IbR3=lillili|II2fl1|= 2V. Since 51 and 53 are "measured" from the same point ""rt"._ the potential across C must he 5o=53—51 =2V—l?V=—lflv licl =. 012 [part 2 of2) 10.1] points If the hatter}.r is disconnected. how long does it take for the 1.roltage across the capacitor to drop to a value of V[t] = where Eu is the e initial voltage across the capacitor? Correct answer: 3'3"] as. Explanation: 1|fifith the battery removed. the circuit is where R§=R1+Rg=rlfiil+2fl=fiflil at=eg+ad=...2o+25o=75n Versifln EH11 — Humewnrk T — anfma and an the time constant is T '=" RWC = {3UflH13pF} =39fl ,ns. The capacitor discharges according to g: —m Qt: fl firm =l E}. c .t 1 —— = In (—j = —lne T I? t= THHE} = -I[3‘Elfl ,ns] (—1] =Eflflfln5. -—l __ r1, +IZERH- in Mamie! ,. 59 IL]: 1 2. £1... LIE-{1.1L QIHZL _ 11.1,” it") a. a": 1-; 47—” +5123. Mt in.-.$t.rifir! 1 at”. = _..E_.+ 9.}- any; {—fl'fl .. WEN-1 arm E" {1+ “1!. J“: FH‘U'LLI . {1:— _ Qum- ‘ flrup‘f‘; 1 LL’ “Ah-£33..) “ma — 11'” r 3* EL“ “571.... / [1ij : WC‘M ____________H________ . flr r; ‘13. r L eq— 29:2,... _ _ _ , I11 4 11M .:._ Attila? _ _ n? f Fich— . EH1. _ Inn.- REL—5 ., _ .. 2'1;+{' : ‘2! w . [a $..¢.~a..i'_._fl,__ — = (might: m + ea —— 1&- '-r _ 9-, +..-’21,+!3r._'5_ E h Jam“, '- .Hni'.H-QJS. 1-11-1- H .41. flu = “>41- ._ Nut-+1.? M's --.,=::~:F é} )- 7 This _£K_;Fa_ wfi_5__JuM [*5 EA”. L111? _.= LL'rLflL- t 3 r1. . - @HCC‘; ( 1: nag/fl." )__I_%. ‘flg_'.-.'1:'T.fi_‘“’>._ ... «fix—sf. .. .__.... 39.-.? T “49.! a” r -..i.-r‘~ £361.55 at 2.11 ‘1 El Eu» :_._(_HSJ)(‘?_{) 7—: 3D -- . mm _. MW 3% rid LC‘SGJUJMQ‘T) .s._ f f: =.. Afiwflfi) . ...
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HW_7_solutions - Version Dfli — Homework 7 —...

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