Name: Total Points: Multiple choice questions [120 points]

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1 Name: Toal Poins: (Las) (Firs) Muliple choice quesions [1 poins] Answer all of he following quesions. Read each quesion carefully. Fill he correc bubble on your scanron shee. Each correc answer is worh 4 poins. Each quesion has exacly one correc answer. Quesions 1 hrough 6 all refer o he same problem. Consider he circui below, consising of a baery, swich, wire, and wo oher circui elemens, denoed by boxes 1 and. The graphs below represen possible measuremens of he poenial difference, V =V a -V b, across circui elemen versus ime. Time = (verical line) denoes he ime he swich is moved. The horizonal line denoes V =. For each of he following cases, choose he appropriae graph for he poenial difference measuremen across elemen. S1 S 1 a b V A B C D E

2 Name: Toal Poins: (Las) (Firs) Quesions 1-3: The swich has been in posiion S for a long ime. A ime =, he swich is moved o posiion S1. 1 Box 1 = resisor, Box = capacior A. graph A B. graph B charge of he capacior VC = Vba ( 1 e ) C. graph C D. graph D E. graph E Box 1 = inducor, Box = resisor A. graph A B. graph B VR = IR = Vba ( 1 e ) C. graph C D. graph D E. graph E 3 Box 1 = capacior, Box = resisor A. graph AVR = IR = Vbae. For <, V R = since here is no curren. B. graph B C. graph C D. graph D E. graph E Quesions 4-6: The swich has been in posiion S1 for a long ime. A ime =, he swich is moved o posiion S. 4 Box 1 = resisor, Box = capacior A. graph A B. graph B C. graph C D. graph D E. graph E discharge of he capacior V = V e. For <, V C =V ba 5 Box 1 = inducor, Box = resisor A. graph A B. graph B C. graph C D. graph D E. graph E The inducor opposes he change of he curren. I keeps he curren running for a lile while: V R C ba ba = IR = V e. For <, V R =V ba

3 Name: Toal Poins: (Las) (Firs) 6 Box 1 = capacior, Box = resisor A. graph A B. graph B C. graph C D. graph D The curren flows in he opposie direcion (he capacior is being discharged). VR = IR = V bae. For <, he capacior is fully charged and here is no curren. E. graph E Quesions 7 hrough 9 all refer o he same problem. Consider he circui below. I conains a baery wih emf V b = 1. V, a swich S, a resisor wih R = 1 Ω, and an inducor L wih negligible resisance. A volmeer is in parallel wih he inducor (assume ha he volmeer has an infinie inernal resisance). The swich is closed a ime = s. 3 A ime = 5. 1 s afer he swich is closed, he volmeer reads V=3.68 V. S V b =1V R=1Ω L V 3 7 Wha will he volmeer read a ime 1 = 1. 1 s? A V B V V = V e b Thus, V ( ) = V e C V D. 6.1 V E. 6.3 V 1 b 1 = V e b = V b V ( Vb ) 3.68 = Wha are he unis for he value of he inducance A. esla B. weber C. henry D. ohms E. farad 3

4 Name: Toal Poins: (Las) (Firs) 9 Wha is he magniude of he value of he inducance (in he above unis)? A..5 Use V = V b e and B C D.. 1 E = L R, hus L = R V ( ln V b ) 4

5 Name: Toal Poins: (Las) (Firs) Quesions 1 hrough 19 all refer o he same problem. The picure below shows wo curren-carrying wires. A long wire parallel o he y-axis crosses he x-axis a x = 1. cm and carries an upward curren I 1 =. A. A square loop of side a = 1. cm lies in he xy-plane, cenered a he origin. I carries a counerclockwise curren I = 3. A. The +z direcion poins ou of he page. y 1 cm I 1 =. A I = 3. A 1 cm x 1 cm Quesions 1-1: Wha is he magneic field vecor (direcion, unis and magniude) along he righ segmen of he loop (a x = 5. cm) due o he verical wire? 1 direcion? A. +x r r µ Idl rˆ C. +z Use he righ hand rule and db = 4π r D. -x E. -z dl r is along +y, rˆ along -x 11 unis? A. ampere B. weber C. vol D. farad E. esla 1 magniude (in he above unis)? 5

6 Name: Toal Poins: (Las) (Firs) A. B. C. D. E. 6 Apply Ampere's law: Take as a closed pah a circle in he xz plane 8. 1 of radius 5cm and cenered a x=1cm, z= cm. π rb = µ I1 Thus 7 4π 1 B = π Quesions 13-15:Wha is he force (direcion, unis and magniude) on he righ segmen of he loop (he porion wih x = 5. cm, 5. cm < y < 5. cm) due o is ineracion wih he verical wire? 13 direcion? 14 unis? r r r A. +x F = I l B and I l r is along +y, B r along +z C. +z D. -x E. -z A. newon B. newon/meer C. esla D. vol.meer E. vol/meer 15 magniude (in he above unis)? A. B. C. D. E r F r r = I l B, F = The direcion of he ne force on he enire square loop is A. +x From 13, he force on he wire a x=5cm is along +x. The force on he wire a x=-5cm is along x and is less han he force on he wire a x=+5cm, 6

7 Name: Toal Poins: (Las) (Firs) since he wire is farher away. The forces on he wires a y=+5cm and y=- 5cm cancel ou. C. +z D. x E. no direcion (F=) 17 The direcion of he ne orque on he enire square loop is A. +z C. z D. y E. no direcion (=) All of he forces acing on he loop go hrough he cener of he loop. Quesions 18-19: The square loop is now roaed so ha i lies in he yz-plane, cenered a he origin. I carries a counerclockwise curren I = 3. A as viewed from he verical wire. The wo verical segmens (a z = ± 5. cm) are equidisan from he verical wire (which is sill a x = 1. cm, z = ). y I 1 =. A 1 cm 1 cm I = 3. A x z 1 cm 18 The direcion of he ne force on he enire square loop is A. +x C. +z No force on he wires a y=+5cm and 5cm. For he wires a z=+5cm and 5cm, see drawing on he righ. F B z B F Top view: y ou of he page 7 x

8 Name: Toal Poins: (Las) (Firs) D. x E. no direcion (F=) 19 The direcion of he ne orque on he enire square loop is A. +z C. z D. y (see drawing above) E. no direcion (=) 8

9 Name: Toal Poins: (Las) (Firs) Quesions hrough 3 all refer o he same problem. Consider a square wire loop of side a lying in he xy plane, cenered on he origin, as shown below. The +z direcion poins ou of he page in his drawing. y B r a a x Quesions -5: A uniform magneic field r B = Bzˆ fills he region (field lines direced ou of he page). The magniude of B r is B=B. Wha is he magniude of he ne flux hrough he loop? A. B Φ=B a aking he area vecor along +z a B. 4aB C. B a D. E. π B a π Ba 4 Quesions -5:The magneic field magniude is decreased linearly from B o in a ime, i.e. B = B 1 Quesions 1-: A ime =, wha is he induced emf (unis and magniude) around he loop? 1 unis? A. newon B. joule C. esla D. vol E. ampere 9

10 Name: Toal Poins: (Las) (Firs) magniude? A. B a Ba B. dφ d db d Φ = Ba, emf = = a = (independen of ) Ba C. πba D. E. No enough informaion 3 The direcion of he curren in he op segmen of he loop (a y = +a /) a ime B a = is A. +x (clockwise in he loop) B. x (counerclockwise in he loop) To compue he flux, we chose he area vecor along +z (ou of he page). The posiive direcion around he loop is counerclockwise. Since he emf is posiive, he curren flows in a counerclockwise direcion. C. no curren in he loop 4 The direcion of he ne force on he square loop a ime A. +x C. +z D. -x E. No direcion (F=) (see drawing) = B r is F F I F 5 The direcion of he ne orque on he square loop a ime = is F A. +x C. +z D. y E. No direcion (=) All of he forces go hrough he cener of he loop 1

11 Name: Toal Poins: (Las) (Firs) Quesions 6-3: The magneic field is reurned o is original value (B=B ). In a ime 1, he angle θ beween he loop and he xy plane is increased from o 6, i.e. o θ ( ) = 6, and hen held a θ = 6. In he op view of he square loop below, +y 1 poins ou of he page. x θ r B = B zˆ n z 6 A ime = 1 (θ = 3 ), he magniude of he flux hrough he loop is A. increasing B. decreasing Φ=B a cosθ and cosθ decreases as θ goes from o 6. (ake he area vecor, n, as indicaed on he figure) C. no changing 7 The direcion of he curren in he op segmen of he loop ( a y = +a /) a ime 1 = (θ = 3 ) is (referring o he figure above) A. owards he op righ corner of he page dφ B. owards he boom lef corner of he page emf = >. The posiive d direcion around he loop is se by he choice of he area vecor defined in 6. C. no curren in he loop 8 The direcion of he ne force on he square loop a ime A. +x C. +z = 1 (θ = 3 ) is 11

12 Name: Toal Poins: (Las) (Firs) D. x E. no direcion (F=) Since B is uniform, he forces on any wo opposie sides of he square loop cancel ou (same B, same lengh, opposie direcions for he curren). 9 The direcion of he ne orque on he square loop a ime A. +x = µ B, µ is along n, B is along z C. +z D. y E. no direcion (=) 3 The magniude of he induced emf in he loop is larges a = 1 (θ = 3 ) is Α. θ= Β. θ=3 C. θ=45 dθ π D. θ=6 emf = Ba sin θ = Ba sin θ.for θ beween and 6, sinθ is d 3 larges a θ=6. E. all angles beween and 6 (he emf is consan) 1 1

13 Name: Toal Poins: (Las) (Firs) PROBLEM 1[4 poins] Equipoenials in a plane perpendicular o 3 charged rods are shown in he figure below. The rod on he lef has charge +q. The wo rods on he righ have charge q. The line connecing poins BCD is a field line. The poenial difference beween adjacen equipoenials is 1. V; he V= equipoenial is labeled. -q +q -q 1). [1ps] On he figure, skech he elecric field line ha passes hrough poin A (marked wih black do). Use an arrowhead o indicae direcion. The elecric field line is perpendicular o he equipoenial lines. The elecric field is direced from +q o q. ). [1ps] Find he elecric poenial a poins A, B, C and D. Explain. Coun he number of equipoenial lines from he poin o he V= equipoenial line. Use ha V=1V beween wo adjacen field lines. V decreases in he direcion of he elecric field (which is from +q o q) V A =+1V, V B =+5V, V C =+1V, V D =-1V 13

14 Name: Toal Poins: (Las) (Firs) 3). [1ps] Rank he magniude of he elecric field a poins A, B, and C. Explain. The smaller he spacing beween he equipoenial lines, he greaer he elecric field V since E = r The spacing beween he equipoenial lines is smalles a B, hen a C hen a A E A <E C <E B 4). [1ps] How much work is done by he elecric field in moving an elecron from poin A o poin D? Jusify your answer. Is he work by he elecric field posiive or negaive? Explain. D W = q( VA VD ) = e(1 ( 1)) = ev = 3. 1 The work done by he E field on he elecron is negaive. A 19 J 14

15 Name: Toal Poins: (Las) (Firs) PROBLEM [4 poins] In he circuis below, bulbs 1-5 are idenical, and he baeries are idenical and ideal. Boxes X, Y, and Z conain unknown arrangemens of bulbs. I is known ha R x =R y >R z _ X + _ Y 3 + _ Z 5 1). [1 ps] Rank he poenial difference across box X, box Y, and box Z in order of increasing absolue value. Explain your reasoning. Using Kirchhoff s rule for he 3 circuis: V ba =V X +V 1 V ba =V Y +V V ba =V Z +V 4 The oal resisance of he circui conaining X is greaer han he oal resisance of he circui conaining Y. This is because here is an exra pah provided by bulb 3 in he circui conaining Y and ha R X =R Y. Thus he curren hrough bulb 1 (=curren hrough he baery in he circui conaining X) is greaer han he curren hrough bulb (=curren hrough he baery in he circui conaining Y). I follows ha V >V 1. Then using he firs wo equaions above we find: V X >V Y The oal resisance of he circui conaining Z is less han he oal resisance of he circui conaining Y since R Z <R Y. The curren hrough bulb 4 is greaer han he curren hrough bulb. I follows ha V 4 >V. Then using o he las wo equaions above, we find V Z <V Y. To conclude: V Z <V Y <V X ). [1 ps] Is he brighness of bulb 3 greaer han, less han, or equal o he brighness of bulb 5? Explain. We have V 3 =V Y and V 5 =V Z, hus V 3 >V 5 The greaer he volage across a ligh bulb, he brigher he ligh bulb. 3 is brigher han 5. 15

16 Name: Toal Poins: (Las) (Firs) 3). Rank he brighness of bulbs 1, and 4 in order of increasing brighness. Explain. From he above discussion, V 4 <V <V 1 The greaer he volage across a ligh bulb, he brigher he ligh bulb. Thus in order of increasing brighness: 1<<4 4). [1ps] Suppose a second box idenical o box Z is conneced o he circui as shown. Deermine wheher he brighness of each of he bulbs below will increase, decrease, or remain he same. Explain. + _ 4 Z 5 Z a. Bulb 4 Since he oal resisance of he circui decreases (one exra pah), he curren hrough he baery increases. The curren hrough 4 is he same as he curren hrough he baery. 4 ges brigher. b. Bulb 5 V 4 +V 5 =V ba V 4 increases since 4 ges brigher. Thus V 5 decreases, and 5 ges dimmer. 16

CHAPTER 12 DIRECT CURRENT CIRCUITS

CHAPTER 12 DIRECT CURRENT CIRCUITS CHAPTER 12 DIRECT CURRENT CIUITS DIRECT CURRENT CIUITS 257 12.1 RESISTORS IN SERIES AND IN PARALLEL When wo resisors are conneced ogeher as shown in Figure 12.1 we said ha hey are conneced in series. As