**Unformatted text preview:**

familiari (nf3795) – DC circuits – yeazell – (58395) 1This print-out should have 16 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 (part 1 of 2) 10.0 pointsYou can obtain only four 20 Ω resistor s fromthe stockroom.How can you achieve a resistance of 50 Ωunder these circumstances?1. 1 in series with 3 in parallel2. 2 in series with 2 in parallel correct3. 4 in parallel4. 2 in series5. 3 in parallel6. 4 in series7. None of these8. 2 in parallel9. 3 in seriesExplanation:Let : R = 20 Ω .Placing two resistors of the same si ze in par-allel halves the original individual resistance:1Rnew=1R+1R=2RThus we needReq= R + R +R2Two resistors in series with two parallel resis-tors:Req= 20 Ω + 20 Ω +20 Ω2=50 Ω .002 (part 2 of 2) 10.0 pointsWhat can you do if you need a 5 Ω resistor?1. 2 in series with 2 in parallel2. None of these3. 3 in series4. 2 in parallel5. 3 in parallel6. 2 in series7. 1 in series with 3 in parallel8. 4 in parallel correct9. 4 in seriesExplanation:Placing 4 resistors of the same value inparallel will quarter the or iginal individualresistance:1Req=1R+1R+1R+1R=4R.Four parallel resistors:Req=R4=20 Ω4=5 Ω .003 10.0 pointsAfter a 6.67 Ω resistor is connected acrossa battery with a 0.23 Ω internal resistance,the electric potential between the physicalbattery terminals is 12 V.What is the rated emf of the battery?Correct answer: 12.4138 V.Explanation:Let : R = 6.67 Ω ,r = 0.23 Ω , andV = 12 V .The current drawn by the ex ternal resistoris given byfamiliari (nf3795) – DC circuits – yeazell – (58395) 2I =VR=12 V6.67 Ω= 1.7991 A .The output voltag e is reduced by the inter-nal resistance of the battery byV = E − I r ,so the electromotive force isE = V + I r= 12 V + (1.7991 A) (0.23 Ω)=12.4138 V .004 10.0 pointsFour resistors are connected as shown in thefigure.21 Ω53 Ω86 Ω91 V47 ΩS1abcdFind the resistance between points a and b.Correct answer: 37.8222 Ω.Explanation:R1R3R4EBR2S1abcdLet : R1= 21 Ω ,R2= 47 Ω ,R3= 53 Ω ,R4= 86 Ω , andEB= 91 V .Ohm’s law is V = I R .A good rule of thumb is to eliminate junc-tions connected by zero resistance.R2R3R1R4abcdThe parallel connection of R1and R2givesthe equivalent resistance1R12=1R1+1R2=R2+ R1R1R2R12=R1R2R1+ R2=(21 Ω) (47 Ω)21 Ω + 47 Ω= 14.5147 Ω .R12R3R4abThe series connection of R12and R3givesthe equivalent resistanceR123= R12+ R3= 14.5147 Ω + 53 Ω= 67.5147 Ω .R123R4abfamiliari (nf3795) – DC circuits – yeazell – (58395) 3The parallel connection of R123and R4gives the equivalent resistance1Rab=1R123+1R4=R4+ R123R123R4Rab=R123R4R123+ R4=(67.5147 Ω) (86 Ω)67.5147 Ω + 86 Ω= 37.8222 Ω .or combining the above steps, the equivalentresistance isRab=R1R2R1+ R2+ R3R4R1R2R1+ R2+ R3+ R4=(21 Ω) (47 Ω)21 Ω + 47 Ω+ 53 Ω(86 Ω)(21 Ω) (47 Ω)21 Ω + 47 Ω+ 53 Ω + 86 Ω=37.8222 Ω .005 (part 1 of 2) 10.0 pointsTwo identical light bulbs A and B are con-nected in series to a constant voltage source.Suppose a wire is connected across bulb B asshown.EA BBulb A1. will burn more brightly. correct2. will burn less brightly.3. will burn as brightly as before.4. will go out.Explanation:Because the wire is added in parallel tolight bulb B, the equivalent resistance nowbecomes just the resistance of light bulb A.Since the resistance dropped, the current hadto increase which means that light bulb A willget brighter.006 (part 2 of 2) 10.0 pointsand bulb B1. will burn as brightly as before.2. will burn more brightly.3. will burn less brightly.4. will go out. correctExplanation:Since the wire is the path of zero resistance,all the current wil l flow through the wire andnone will flow through light bulb B. With nocurrent to light the light bulb, bulb B will goout.007 10.0 pointsConsider resistors R1and R2connected inseriesER1R2and in parallelER1R2to a source of emf E that has no internalresistance.How does the power dissipated by the resis-tors in these two cases compare?1. It is greater for the series connection.2. It is different for each connection, but onefamiliari (nf3795) – DC circuits – yeazell – (58395) 4must know the values of R1and R2to knowwhich is greater.3. It is greater for the parallel connection.correct4. It is the same for both connections5. It is different for each connection, but onemust know the values of E to know which isgreater.Explanation:The power dissipated by the resistors isP =E2Req.The equivalent resistance for a series con-nection isRs= R1+ R2.The equival ent resista nce for a parallel con-nection isRp=R1R2R1+ R2.Regardless of the values of R1and R2, Rp<Rs, so more power is dissipated in the parallelconnection.008 (part 1 of 5) 10.0 pointsAssume the battery is ideal (it has no in-ternal resistance) and connecting wires haveno resistance. Unlike most real bulbs, the re-sistances of the bulbs in the questions belowdo not change as the current through themchanges. Three identical bulbs are in the cir-cuit as shown below in the figure. (The switchS is initially closed.)EABCSWhich of the following correctly ranks thebulbs in brightness?1. Bulb A is the brightest, B is next bright-est, and C is the least brightest.2. None of these is correct.3. Bulb B and C are equally bright, andeach is brighter than A.4. Bulb A is the brightest, and B and C areequally bright. correct5. All bulbs are equally bright.Explanation:Bulb B and C are connected parallel, theyhave the same potential difference, t hus theyare equally bright. IA= IB+ IC, so A thebrighter than B and C.009 (part 2 of 5) 10.0 pointsWhich of the following correctly ranks thecurrent flowing through the bulbs?1. B ul b B and C have the same current, andeach has more current than A.2. B ul b A has the largest current, and C hasthe smallest current.3. Bulb A has the largest current, and B andC have the same current. correct4. None of these is correct.5. All bulbs have the same current flowingthrough them.Explanation:As mentioned in t he last part, B and Chave the same potential difference and currentI =VR, while the current of A is the sumof that of B and C.010 (part 3 of 5) 10.0 pointsWhich of the following correctly ranks thepotential difference across these bulbs?1. The potential difference is l a r gest acrossfamiliari (nf3795) – DC circuits – yeazell – (58395) 5A, and smallest across C.2. None of these is correct.3. All bulbs have the same potential acrossthem.4. Bulb B and C

View Full Document