Physics 111 Fall 2007 Electric Currents and DC Circuits

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1 Physics 111 Fall 007 Elecric Currens and DC Circuis 1 Wha is he average curren when all he sodium channels on a 100 µm pach of muscle membrane open ogeher for 1 ms? Assume a densiy of 0 sodium channels per µm of surface and a flow rae of 1000 ions per ms hrough each channel The average curren is given by 0channels 1000ions channel ion Q 1µ m I 1 10 s 19 C 100µ m A 08nA The immediae cause of many deahs is venricular fibrillaion, an uncoordinaed quivering of he hear as opposed o proper beaing An elecric shock o he ches can cause momenary paralysis of he hear muscle, afer which he hear will someimes sar organized beaing again A defibrillaor is a device ha applies a srong elecric shock o he ches over a ime inerval of a few milliseconds The device conains a capacior of several microfarads, charged o several housand vols Elecrodes called paddles, abou 8 cm across and coaed wih conducing pase, are held agains he ches on boh sides of he hear Their handles are insulaed o preven injury o he operaor, who calls Clear! and pushes a buon on one paddle o discharge he capacior hrough he paien s ches Assume ha an energy of 00 J is o be delivered from a 00-µF capacior a To wha poenial difference mus i be charged? Consider he following, a defibrillaor conneced o a µf capacior and a 47 kω resisor in a RC circui The circuiry in his sysem applies 000 V o he RC circui o charge i b Wha is he ime consan of his circui? c Wha is he maximum charge on he capacior?

2 d Wha is he maximum curren in he circui during he charging process? e Wha are he charge and curren as funcions of ime? f How much energy is sored in he capacior when i is fully charged? a The energy sored in his sysem is given fromu 1 C V so ha he poenial difference is V U C 00 ( J) CV V 6 b The ime consan is given as τ RC Ω 10 F 1 s c The maximum charge on he capacior is he produc of he capaciance and he maximum volage, which is 6 Q CV 10 F 000V 0 160C max max d The maximum curren is given from Ohm s Law Vmax 000V Imax 0 106A R Ω e The charge and currens are hus given 0 s 160C 1 e and () 1 s I 0 106A 1 e Q 1 as () f The maximum amoun of energy sored when he capacior is fully charged 1 Qmax 1 is Emax CVmax QmaxVmax 400J C An immersible heaer coil is o be designed o hea an insulaed conainer wih 4 liers of disilled waer from 0 o o 0 o C in less han 0 minues a How much energy mus be inpu o hea he waer o his emperaure? b To hea he waer, wha minimum power mus be supplied? c If a 1 V power supply is o be used, wha minimum curren mus flow in he heaing coil? d Wha mus be he oal resisance of he heaing coil? Is his a maximum or minimum resisance o hea he waer in 0 minues or less?

3 Firs we need o conver liers ino kilograms The conversion is ha one lier of waer has a mass of l kg Therefore we have 4kg of disilled waer ha we need o hea a The amoun of hea energy needed o hea he waer is 1000cal 4186J o Q mwaercwaer T 4 kg 0 C 0 10 J 1kg 1cal b The minimum power ha mus be supplied is Energy Inpu 0 10 J P 78 9W ime 1800s 789W c The minimum curren ha mus flow is found from P IV I A 1V 789W d The resisance needed is P I R R 0 Ω For he minimum curren ( A) his represens a maximum resisance 4 Analyze he circui shown below o find he currens flowing hrough and he power generaed in each resisor 6 V 1 kω 4 kω 1 kω kω For he 1 kω and kω resisors, we can add hem since hey are in series o ge he equivalen resisance of ha branch o be 4 kω Nex, he wo 4 kω resisors are in parallel and he equivalen resisance for hese wo are given as he reciprocal of he sum of he reciprocals of he individual resisances This gives an equivalen resisance of kω Lasly his leaves a baery and wo resisors in series, namely 1 kω and kω These las wo resisors are in series and he equivalen resisance is kω Therefore we have a kω equivalen circui resisance and by Ohm s Law he curren V V developed by he baery is I 6 circui A 1 7mA R 00Ω eq Now working backwards hrough he circui, his oal curren has o flow hrough he 1 kω resisor and he power dissipaed across his resisor is P I R 17mA 1kΩ 4 ( ) mw 1 kω 1kΩ 4

4 This 17 ma curren has o flow hrough he remainder of he circui which has an equivalen resisance of kω The poenial drop across his resisor is given from Ohm s lawv I R 17mA kω V However, his branch of he circui, kω kω 4 expanding he circui back o is original sae, has a 4 kω resisor Thus, he curren hrough his resisor is given by using Ohm s Law again, knowing he poenial drop 4V across he resisor Thus we findv4 kω I4k ΩR I 4kΩ 0 8mAand he power 4kΩ dissipaed across he 4 kω resisor is P kω I kωr ( 08mA) 4kΩ 89mW 4 4 Finally, we know ha 17 ma of curren was developed by he circui and ha 08 ma flowed ou of he juncion hrough he 4 kω resisor Thus, 08 ma mus flow o he boom branch o conserve charge This gives he curren hrough he 1 kω and kω resisors as 08 ma and he power dissipaed across each as P I R 08mA 1kΩ 1 and P ( ) mw ( 08mA) kω 1 mw 1 kω 1kΩ 1 kω I kωr 8 respecively As a las check, he power developed by he circui is given as P I R 17mA kω 10 Since energy has o be ( ) mw circui circui eq 1 conserved in he circui we can add all of he energy (per uni ime) ha was dissipaed across he circui and we find 1019 mw, wih he discrepancy due o rounding errors RC ime consans can be easily esimaed by measuring he ime (known as he halfime) for he capacior volage o decrease o half of some arbirary saring value when discharging hrough a resisor Furher, we know ha he volage across he capacior will vary asv() V e o RC Show how a single measuremen of he halfime can be used o deermine he RC ime consan (Hin: subsiue V() V o /) V RC RC o o V RC o 1 () V e V e ln() RC ln Thus measuring he ime i akes for he capacior o discharge o have of i s iniial poenial, we see ha we need only one measuremen o calcuale he ime consan

5 6 A simple RC series circui has a 100 µf capacior a If he ime consan is 0 s, wha is he value of he resisor? b Suppose ha a second idenical resisor is insered in series wih he firs Wha is he new ime consan of he circui? c Suppose he second idenical resisor is placed in parallel wih he firs resisor, sill conneced o he capacior Wha is he new ime consan in his case? d Suppose ha we use he single resisor from par a, bu now a second idenical capacior is conneced in series wih he firs Wha is he new ime consan of he circui? e Suppose ha we use he single resisor from par a, bu now a second idenical capacior is conneced in parallel wih he firs Wha is he new ime consan of he circui? τ 0s a The resisance is given fromτ RC R 10 Ω 6 C F b The equivalen resisance for hese wo resisors in series is he sum of he individual resisances, or 1x10 6 Ω and he new ime consan is 6 6 τ R series C 1 10 Ω F s series 100 c The equivalen resisance for hese wo resisors in parallel is he reciprocal of he sum of he reciprocals of he individual resisances, or x10 Ω and he new ime consan 6 is τ R parallel C 10 Ω F s parallel d The equivalen capaciance for hese wo capaciors in series is he reciprocal of he sum of he reciprocals of he individual capaciances, or 0x10 - F and he new ime consan is τ 0 10 Ω 0 10 F s series RC series e The equivalen resisance for hese wo capaciors in parallel is he sum of he individual capaciances, or 00x10-6 Ω and he new ime consan is 6 τ 0 10 Ω F s parallel RC parallel 100

6 7 A 100 µf capacior wired in a simple series RC circui is iniially charged o 10 µc and hen discharged hrough a 10 kω resisor a Wha is he ime consan of he circui? b Wha is he iniial curren ha flows? c How much charge is lef on he capacior afer 1 ime consan? d Wha is he curren afer 1 ime consan? e How much charge is lef on he capacior afer ime consans have elapsed and wha curren is flowing hen? a The ime consan is he produc of he resisance and he capaciance, or 10s b The iniial curren ha flows is found from Ohm s Law 6 Q Q C 6 V C IR I A 10µ A C RC 10µ C c Afer 1 ime consan we have Q( ) Qo e µ C 68µ C e d The curren ha flows afer one ime consan is 10µ A I( ) Io e µ A 68µ A e e Afer ime consans we have s 10µ C Q( s) Qo e µ C 00µ C and he curren ha is flowing is e s 10µ A I ( s) Io e µ A 00µ A e 8 Find he reading ha each of he (ideal) meers would have in he following circui 10 V A V1 1 kω kω V 10 kω A1 1 kω kω For he 10 kω and kω resisors, we can add hem since hey are in series o ge he equivalen resisance of ha branch o be 6 kω

7 In he middle branch we have a kω resisor in series wih a 10 kω resisor The equivalen resisance of his branch is he sum of he individual resisors, or 1 kω Nex, we have he lower wo branches in parallel Thus he 6 kω resisor is in parallel wih he 1 kω resisor and he equivalen resisance for hese wo are given as he reciprocal of he sum of he reciprocals of he individual resisances This gives an equivalen resisance of 411 kω Lasly his leaves a baery and wo resisors in series, namely 411 kω and 1 kω These las wo resisors are in series and he equivalen resisance is11 kω Therefore we have a 11 kω equivalen circui resisance and by Ohm s Law he curren developed by he baery and passes hrough ammeer A is V V I 10 circui A 0mA R 110Ω eq This curren passes hrough he 1 kw resisor, so he poenial drop across i as displayed by volmeer V 1 is given by Ohm s LawV k IR 0mA 1kΩ V 1 Ω The poenial drop across he boom wo branches mus be he same and is 8V (10V V by Kirchoff s rule for poenial drops) The curren ha flows hrough he middle branch is given by Ohm s Law wih an V 8V equivalen resisance of 1 kω and ha curren is I 0 6mA R 1kΩ By he juncion rule, ma of curren flows ino he juncion of he middle and boom branch, and 06 ma flows hrough he middle branch so a curren of 18mA mus flow hrough he boom branch as well as hrough ammeer A 1 Lasly we need he poenial drop across he 10 kω resisor in he middle branch The curren in his branch is 06 ma and by Ohm s Law V k IR 06mA 10kΩ 6 V, which is wha would be displayed on volmeer 10 Ω V 9 Two 100 m 14 gauge wires (16 mm diameers), one of copper and one of aluminum, are soldered ogeher and he 00 m wire is hen conneced o a 6 V dc power supply wih unlimied curren a How much curren flows in he wire? b Wha is he poenial across each 100 m secion of wire? c How much power is developed in each secion of wire? d How much oal power was delivered by he baery? Does his equal he power dissipaed by he circui? If no, where did he exra energy go?

8 a To deermine he curren ha flows in he wire we need he equivalen resisance for heses wo secions of wire Le s model hem as wo resisors in series The equivalen resisance is he sum of he individual resisances ha we now need o calculae For he LCu 8 100m copper secion we have RCu ρ Cu Ωm 081Ω ACu π ( m) while for he aluminum LAl 8 100m secion RAl ρ Al 8 10 Ωm 14Ω Thus he AAl π m ( ) equivalen resisance is R eq R Cu + R Al 16Ω Thus he curren ha flows hrough V 6V boh secions I 78A R 16Ω eq b The poenial difference across each secion is given for copper, V IR 78A 081Ω V and V Cu Cu 7 Al IRAl 78A 14Ω 7V for aluminum Noe, ha he sum of he poenial drops across each resisor sums o 6V as i should c The power dissipaed in he copper secion is P I R 78A 081Ω 6, while for aluminum ( ) W ( 78A) 14Ω 10 W Cu Cu PAl I RAl 4 d The power developed by he baery is P I R ( 78A) 16Ω 16 7W, which is equivalen o he power dissipaed across each secion of resisor 10 The Earh s amosphere is able o ac as a capacior, wih one plae he ground and he oher he clouds and in beween he plaes an air gap Air, however, is no a perfec insulaor and can be made o conduc, so ha he separaion of charges from he cloud o ground can be bridged Such an even is called a lighning srike For his quesion, we ll model he amosphere as a spherical capacior and ry o calculae he number of lighning srikes ha happen every day Firs, le s assume ha he clouds are disribued around he enire Earh a a disance of 000 m above he ground of area 4pR Earh, where R Earh 6400km a Wha is he resisance of he air gap? Nex we need o calculae he capaciance of he Earh-cloud capacior Here we will Q use he fac ha he capaciance C and we ll calculae V Assuming ha we V

9 Q have a spherical charge disribuionv k, so ha V is he difference in poenial r beween he lower plae (he Earh s surface) and he upper plae (he clouds) and ha in a ypical day, x10 C of charge is spread over he surface of he Earh b Wha is V? c Wha is he capaciance of he Earh-cloud capacior? Since he accumulaed charge will dissipae hrough he air, we have a simple RC circui d Wha is he ime consan for his discharge ha is spread over he whole surface of he earh? Experimenally i is found ha for each lighning srike abou C of negaive charge is delivered o he ground e Wha is he number of lighning for his amoun of charge? f Approximaely how long would i ake he Earh-cloud capacior o discharge o 0% of is iniial amoun? So o answer he quesion of how many lighning srikes per day, we know ha we ge par e number of srokes in par f amoun of ime g How many lighning srokes per day? a The resisance of he air-gap is given 10 m ρ 91Ω A 4π L 1 as R ( 10 Ωm) 6 ( m) b Assuming ha he Earh-cloud sysem can be modeled as wo spherical charge disribuions separaed by a fixed disance, we find he difference in poenial o be V V lower V upper k Q r lower Q r upper 9 Nm 9 10 C 1 10 C r earh r Q 10 C c The capaciance is hereforec 1 0F V V earh V 000m + d The ime consan is he produc of he resisance and capaciance of he circui Thus he ime consan isτ RC 91 Ω 10F 98 s

10 e For each lighning srike, C of charge is delivered Thus he number of lighning srikes is he oal charge divided by C per srike, or 0,000 srikes f To calculae he ime, we use he equaion for a discharging capacior RC max 98 s Q() Q e 000 e 17s 048hr g So we ge 0,000 srikes every 048hrs, or 41,667 srikes per hour Thus in 4 hours we ge abou 1,000,000 srikes This value, even if only approximae, is no far from he observed number of abou 1 million per day!! 11 The suden engineer of a campus radio saion wishes o verify he effeciveness of he lighning rod on he anenna mas as shown below The unknown resisance R x is beween poins C and E Poin E is a rue ground bu is inaccessible for direc measuremen because his sraum is several meers below he Earh s surface Two idenical rods are driven ino he ground a A and B, inroducing an unknown resisance R y The procedure is as follows Measure resisance R 1 beween poins A and B, hen connec A and B wih a heavy conducing wire and measure resisance R beween poins A and C (a) Derive an equaion for R x in erms of he observable resisances, R 1 and R (b) A saisfacory ground resisance would be R x < 00 Ω Is he grounding of he saion adequae if measuremens give R 1 10 Ω and R 600 Ω? a c R 1 R y R R y x b (a) For he firs measuremen, he equivalen circui is as shown in Figure 1 Figure 1 R ab R 1 R y + R y R y a R c so R y 1 R 1 (1) R y R y R x Figure

11 For he second measuremen, he equivalen circui is shown in Figure Thus, R ac R 1 R y + R x () Subsiue (1) ino () o obain: R 1 1 R 1 + R x, or R x R 1 4 R 1 (b) If R 1 10 Ω and R 600 Ω, hen R x 7 Ω The anenna is inadequaely grounded since his exceeds he limi of 00 Ω

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