Solutionsto Bulb questions Note: We did some basic circuits with bulbs in fact three main onesI can think ofI have summarized our results below.For the final exam, you must have an understanding of the brightness of bulbs in these three circuits (for like and unlike bulbs) as well as answers to questions 6 and 7.Questions 8 onwards are a little complicated and will not be on the final exam (simply because we did not discuss such circuits in class).I have the answers below for these questions for the sake of completeness. Basic Bulb circuits Series circuit If identicalbulbs(same resistance, rated for the same watts) are in series, they willbe the same brightness since the current througheach bulband the voltageacross each bulb willbe the same.So inthe circuit below,ifthe bulbsare both roundor bothlong,they willbe equallybright. If youhave two different bulbsinseries (e.g. a roundbulband a longbulb)the one withthe greater resistance (the longbulb)willhave the greater voltage across it.The same current flows throughboth the bulbsand since brightnessdepends on power output(voltage x current), the longbulbwillbe brighter.Parallel Circuit If two identicalbulbs(roundor long)are inparallel,they willhave the same brightness( same voltage across each and same current througheach bulb). If the two bulbsare not the same, then bothbulbswillhave the same voltagebut greater current willflow throughthe bulbwith the smallerresistance.So ifa roundbulband a longbulbare inparallel,then the roundbulbis brighterthan the longbulb. Combination of series and parallel ClearlyA must be brighterthan B and C which are equallybright. Usingthe current modelwe couldsay that A gets the entire battery current but B and C onlyget halfthe current. Usingthe voltagemodel,we can say that because B and C combineto form a smallerresistance there is a smallervoltagedrop across the parallelcombinationthan bulbA.So usingthe voltagemodel,we end up withthe same ranking:A > B=C. But what happens if we unscrew C (or B)? Then the current modeltellsus that B shouldget brightersince itgets allthe battery current (justlikeA) and the brightness of A should not change since it is upstream. What didwe actuallyobserve inclass? When we unscrew C (or B), B gets dimmer and so does A! The onlyway to understand thisisto use the voltagemodel. When we unscrew C thus removingitfrom the circuit,the resistance of B is more than the parallelcombinationof B and C.The greater resistance reduces the current throughthe battery.Lesser current means less brightnessfor A.Since B isinseries withA, ithas the same brightnessas A (lesser than when C was includedinthe circuit).6)LightBulbs Does a light bulbobey Ohms Law?Explain. As youknow by now,lightbulbsdo NOT obey Ohms Law.How do we know this?Rememberwe measured the current througha bulbwithdifferent voltagevalues appliedacross it?The plotof voltage vs. current is NOT a straightlinefor a bulb. Thismay nothave been obvioustoyou from yourdata unless youtooka lotof data. Why doesnt a bulbobey Ohms Law?A bulbfilamentisdesignedto carry a large current.The bulb filamentgets very hot and glowsmuch likered-hot coal ina coal fire.When the filamentgets hot its resistance (which is the inverse of the slope of the graph) increases.Thus the resistance of the bulb changes as itstays litwhich means the slope of the graph does not stay constant. 7) What determines the brightness of a light bulb?I s it the current through the bulb, the voltage across the bulb or both? Both.If youconnect a longbulband a roundbulbinseries, the longbulbglowsbut the roundbulbdoes not.Both of them have the same current flowingthroughthem(they are inseries) but the longbulbhas a greater voltage across it because it has more resistance (even thoughV= IR isnot obeyed exactly,voltage dropdoes depend onthe resistance).If you connect a roundbulband a longbulbinparallel,however, it is the roundbulbthat isbrighterbecause now the voltage across bothbulbsis the same but the current is larger throughthe roundbulb(smallerresistance). 8)P =V x I but assumingbulbsobey Ohms law (well,approximately,at lower currents) I = RV .Combiningthese two relations,we have P = RV) (2. At theirnormaloperatingtemperatures (before the filamentgets very hot),from 2VR P , the bulbs displayresistances 22120 V240 60 WVR P, and 2120 V190 75 W ,and 2120 V72 200 W . The nominal60 W lamphas greatest resistance. When they are connected inseries, they allcarry the same smallcurrent. Here the highest-resistance bulbglows mostbrightly(likethe longbulbwhen connected in series withthe roundbulb)and the one withlowest resistance is faintest. This isjustthe reverse of their order of intensityif they were connected inparallel,as they are designed to be. 9) Note:Something like this will not be on the final exam since we did not do circuits with switches and shorting wires.However the voltage across the bulbs in series is still relevant for the final exam (and useful to know). Suppose the battery voltage12 V and each lamp has2R . Before the switch isclosed the current is 12 V2 A6 . The potentialdifference across each lamp is 2 A 24 V . The power of each lamp is 2 A 4 V 8 W , totaling24 W for the circuit.Closingthe switch makes the switch and the wires connected to ita zero-resistance branch. Allof the current throughA and B willgo throughthe switch and (b) lampC goes out,withzero voltageacross it. Withless totalresistance, the (c) current inthe battery 12 V3 A4 becomes larger than before and (a) lampsA and B get brighter.(d) The voltageacross each of A and B is 3 A 26 V ,larger than before. Each converts power 3 A 6 V 18 W , totaling 36 W,which is(e) an increase. 10)Again nothing like this will be on the final but this is good to know. Remember bulbbrightnessdepends on current and voltage.The bulbsare identical(thisis what we will assume since we were not toldotherwise).Then bulbsthat are inseries shouldhave the same current and the same voltage drop. So 1 = 2 since each of them have a voltagewhich ishalf the battery voltageand the same current.3 = 6 and 4 = 5 (same reason as above) However the parallelcombinationof 4 and 5 mustbe inseries with3 and 6.While3 and 6 have the same voltageacross them (same resistance), the parallelcombination4 and 5 willhave less voltageacross them (smallerresistance) and 4 and 5 willalsohave lesser current (since the current splitsat the junction). Since the totalvoltage across DG is now splitamongmore bulbs,3 and 6 willhave less voltageacross them compared to 1 and 2 butmore voltageacross them than 4 and 5. So the finalrankingis: 1 = 2 > 3 = 6 > 4 = 5 11) B1 >B4>B2 =B3>B5 = B6 It actuallyhelps to replace allthe bulbsby resistors to analyze thiscircuit(keepinginmindthat the measured currents and voltages willNOT be quitethe same as that predicted byOhms Law).Thismay be hard to understand withour limitedexposure,but withidenticalbulbs,we can rank them based on voltagealone. Clearly,B1 is the brightest.Now B2 and B3 together offer a smallerresistance than B4, B5 and B6.So the latter grouphas a greater voltage dropacross the parallelbranches.StillB5 and B6 get onlyhalfof this voltageacross each which isless than the voltagedrop across B2 and B3.B4, of course gets allof it,so is brighterthan B2 and B3. 12) (Circuiton next page) Assume the switch is open (as shown below so there isa break inthe wire).State which bulbsor combinationof bulbsare inseries and whichones are inparallel. C, D and B are in paralleland the combinationisinseries withA. b) Rank the bulbsaccording to brightnessand explainyour reasoning. A gets allthe battery current so shouldbe brightest. B, C and D are inparalleland get the same current and the same voltageacross them. So the rankingisA> B=C=D. c) Now assume the switch is closed.State which bulbsare in series and whichones are inparallel. Closingthe switch willmake C and D goout (the current willtake the easy path throughthe switch and bypass C and D.No current throughthe bulbmeans the bulbwillnot glow). Then A and B are inseries. d) Rank the bulbsaccording to brightnessand explainyour reasoning. A = B (series) C and D do not light!