8.022 (E&M) Lecture 9
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1 8.0 (E&M) Lecure 9 Topics: circuis Thevenin s heorem Las ime Elecromoive force: How does a baery work and is inernal resisance How o solve simple circuis: Kirchhoff s firs rule: a any node, sum of he currens in = sum of he currens ou (conservaion of charge a nodes) Kirchhoff s second rule: around any closed loops, he sum of EMF and poenial drops is 0 (elecrosaic field is conservaive) Power dissipaed by a resisor: P = I = I G. Sciolla MIT 8.0 Lecure 9
2 apaciors in circuis A new way of looking a problems: Unil now: charges a res or consan currens When capaciors presen: currens vary over ime s onsider he following siuaion: A capacior wih charge Q0 Æ 0 =Q 0 / A resisor in series conneced by swich s Wha happens when swich s is closed? G. Sciolla MIT 8.0 Lecure 9 3 Discharging capaciors: qualiaive Before swich s is closed: Difference in poenial beween plaes: 0 No curren circulaing in he circui (open) I s Afer swich s is closed: Difference in poenial beween capacior plaes will induce curren I As I flows, charge difference on capacior decreases Æ decreases Æ I decreases over ime G. Sciolla MIT 8.0 Lecure 9 4
3 Discharging capaciors: quaniaive Apply second Kirchhoff s law: EMF supplied by capacior : =Q/ NB: his is rue a any momen in ime Æ Q() Æ () olage drop on he resisor: I Q I 0 = No useful in his form since I=I(Q) I=dQ/d ( sign because is losing charge) Q 0 dq d = Easy inegral yields o exponenial decay of he charge: Q () = Qe 0 G. Sciolla MIT 8.0 Lecure 9 5 How o inegrae circuis To solve Q dq = 0, rewrie as: dq = d d Q Inegrae boh sides: Q () Q 0 dq Q Q () ln = Q 0 = Q () = Qe 0 0 d NB: τ= is called decay consan of he circui G. Sciolla MIT 8.0 Lecure 9 6 3
4 Soluion of circui Soluion: Q () = Qe 0 Exponenial decay of charge sored in capacior = of he circu Afer a ime, he charge decreased by /e w.r.. original value τ is called decay consan i Unis of : cgs: []= savol s /esu; []=esu/savol Æ []=s SI: []=/A; []=/; A=/s Æ []=s Derive he curren: dq d Q 0 I () = = Q 0 e = e d d Same exponenial decay as for Q() G. Sciolla MIT 8.0 Lecure 9 7 harging capaciors Now 3 elemens in circui: EMF, capacior and resisor apacior sars uncharged I s Wha happens when swich s is closed? When s is closed, curren wil l suddenly flow and will charge As charges, E opposie o EMF builds up and slows down curren I() sops when reaches G. Sciolla MIT 8.0 Lecure 9 8 4
5 harging capacior: solve he circui Solve using Kirchhoff s second law: I()=dQ/d NB: because he capacior is now charging! I s Q I = 0 dq Q Firs order differenial equaion = 0 d Soluion: Q () = e G. Sciolla MIT 8.0 Lecure 9 9 Deails of inegraion dq Q dq ( Q ) To solve = 0, rewrie as: = d d Seing: Q'= Q dq ' d = Q ' Inegraing beween =0 and : Q= Q () dq ' = d Q ( ) Q () = Q = 0 Q ' ln = = e = 0 Q () = e G. Sciolla MIT 8.0 Lecure 9 0 5
6 Graphical soluion Q() Q 0 () I() / e / e / e / Q () = e () = Q()/ = e dq () I () = = e d G. Sciolla MIT 8.0 Lecure 9 Imporan commens Soluion of circui: ( ) = e ; I ( ) = e Are Kirchhoff s laws valid a any momen in ime? Q I = e e = 0 Asympoic behavior of he capacior: A =0: I=/ as if were a shor circui A =infiniy, I=0 as if were an open circui OK! onclusion: no need o solve he differenial equaion! Soluion is an exponenial wih ime consan Asympoic behavior of gives iniial/final values for () and I() G. Sciolla MIT 8.0 Lecure 9 6
7 Time consan of circui (E9) Simple circui wih EMF = 3 =.3 F =.7 Ω Quesions: Wha are and I? erify ha ime consan is () 3 e / I () = EMF e = 5. s s = ( /e )=.9 when =5. If formula is correc EM G. Sciolla MIT 8.0 Lecure 9 3 erify ime consan (E8) circui wih EMF = squared 5 pulses ariable iniially = 0.3 µf ariable iniially = 400 Ω = 00 Ω Display on scope and I( ) erify ha ime consan is () 5 e / I AG () 0mA e / EMF G A G. Sciolla MIT 8.0 Lecure 9 4 7
8 erify ime consan (E8) circui wih EMF = squared 5 pulses ariable iniially = 0.3 µf ariable iniially = 400 Ω = 00 Ω EMF G Assuming τ= Wha happens when we double? τ = == τ 0 Æ (I AG ) raises (falls) wice as fas A How should we change o have he same effec? ==( ) Æ : 400 Æ 900 Ω G. Sciolla MIT 8.0 Lecure 9 5 More complicaed circuis Wha if he circui is more han jus a series of and? onsider he following circui: s alculae Q() on he capacior Soluion: Kirckhoff s laws will solve i: TEDIOUS! Use Thevenin s Theorem G. Sciolla MIT 8.0 Lecure 9 6 8
9 Thevenin equivalence Thevenin s heorem: Any combinaion of resisors and EMFs wih erminals can be replaced wih a series of a baery O and a resisor T where O is he open circui volage T = O /Ishor where Ishor is he curren going hrough he shored erminals or T = eq wih all he EMF shored In our case: A T O B T () = O e Once he circui is reduced, he soluion is known: Q G. Sciolla MIT 8.0 Lecure 9 7 Thevenin s demonsraion Prove ha O is he open circui volage ( ) = O ) Æ T () ( ) Since Q exp( = O exp T So O is he asympoic for he capacior Since for Æ infiniy, Æopen circui: O = of he open circui A B O T Prove ha T = O /I shor wih I shor = curren hrough shored erminals There is only one curren going hrough he reduced circui A =0, behaves like a shor Æ A =0 Ishor= O / T Æ T = O /I shor G. Sciolla MIT 8.0 Lecure 9 8 9
10 Solve he acual problem alculae O and T = O /I shor for our problem: T O O = Q () = e Shoring is makes irrelevan in he circui: I shor = ( ) O = = Thevenin I () = e I shor NB: This is //, same resisance we would ge if we sh ored EMF! ( ) G. Sciolla MIT 8.0 Lecure 9 9 Thoughs on Thevenin The imporance of Thevenin: When we have a messy sysem or resisors and EMFs, we can reduce i o a simple EMF in series jus measuring I shor and open : Any unknown combinaion of s and EMFs O T areful: Thevenin works only when he elemens in he box follow Ohm s law, i.e. linear relaion beween and I G. Sciolla MIT 8.0 Lecure 9 0 0
11 Oscillaing circui (E3) circui wih: EMF = k = 0. µf =.5 MΩ Fluorescen ligh in parallel wih capacior EMF ( FL <<< when curren flows; ~infinie oherwise) Why is ligh flashing a ν~ Hz? Iniially he capacior will sar charging (no curren hrough he lamp) When >cerain value ~ k Æ curren flows hrough fluorescen ligh discharging he capacior very quickly The process will sar again ν~/τ=/=4 Hz G. Sciolla MIT 8.0 Lecure 9 Oscillaing circui (E3) circui wih: EMF = k = 0. µf =.5 MΩ Fluorescen ligh in parallel wih capacior ( FL <<< when curren flows; ~infinie oherwise) EMF NB: charging and discharging ime consans are very differen! harging: fluorescen ligh is ~ open circui: τ charge = Discharge: fluorescen ligh has a (very small) resisance FL Thevenin: T =// FL ~ FL τ = T ~ FL << discharge G. Sciolla MIT 8.0 Lecure 9
12 Noron s heorem Any combinaion of resisors and EMFs wih erminals can be replaced wih a parallel of a curren generaor I N and a resisor T where T is he equivalen resisance of he circui wih all he EMF shored and all he curren sources open (same as Thevenin!) I N = O / T I N T T = // = /( ) O I N = = = T // G. Sciolla MIT 8.0 Lecure 9 3 Summary and Oulook Today: circuis Thevenin s heorem Nex ime: Magneism emember: don miss office hours Bring your problems and le s find soluions ogeher! G. Sciolla MIT 8.0 Lecure 9 4
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