PHY1 Elecriciy Topic 7 (Lecures 1 & 11) Elecric Circuis n his opic, we will cover: 1) Elecromoive Force (EMF) ) Series and parallel resisor combinaions 3) Kirchhoff s rules for circuis 4) Time dependence in circuis eading from Young & Freedman: For his opic, read secions 5.4 & 5.5, he inroducion o chaper 6 and secions 6.1 o 6. & 6.4. nroducion Charges will always move sponaneously o a posiion of lower poenial energy. (Posiive charges mover o lower poenial, negaive charges o a poin a more posiive elecrosaic poenial.) A curren will herefore no coninue o flow in a circui on is own some device mus be presen o raise he charges o a higher poenial energy again. We can see he same hing if we consider conservaion of energy. Curren flowing in a circui which possesses resisance will dissipae elecrical power, and his power mus be supplied by some exernal agency. A hird way of expressing his is ha in any closed loop or circui, he poenial difference in raversing he circui and reurning o he same poin mus be zero; since posiive charge drops in poenial as i flows along a wire, some means mus exis o raise i in poenial for i o ge back o is sar poin. The device required o complee he circui is said o be a source of elecromoive force (emf), and may be a baery or an elecromagneic generaor which conver chemical or mechanical energy respecively ino increased elecrosaic poenial energy. Elecromoive Force The diagram alongside shows a simple circui of a baery conneced via wo wires o a resisor. The arrows show convenional curren direcion he direcion aken by a hypoheical posiive carrier paricle. When he carrier eners he negaive erminal of he baery, i has is poenial raised o a fixed posiive value in order o reach he posiive erminal. As i moves hrough he wire and resisor, i falls o a lower poenial, is elecrosaic energy being convered o hermal energy due o collisions wih he laice. reurns o he negaive erminal a he same poenial as i sared. + The baery mus do work o move charges o he erminals agains he repulsion of he charges already presen here. n his case, chemical energy is convered ino elecrical poenial energy. The emf of he device is defined as he work done per uni charge moved from negaive o posiive erminal. has he unis of Vol (which, remember, is 1 Joule per Coulomb). should be clear from he unis ha he emf is no really a force bu he energy per charge, as is poenial! For his reason i is common o use he abbreviaion emf in preference o he words elecromoive force. PHY1 Elecriciy 41
nernal esisance and Terminal Volage An ideal source of emf would simply raise charges o he required poenial. eal devices, such as baeries and generaors, also have some inernal resisance, r. The erminal volage, V T, available o he exernal circui, hen depends on he curren flowing. The poenial difference across he inernal resisance mus be r, so he erminal volage is V T = r. [1] r V T The exernal volage is also given by V T =. We can furher express his as = r + [] So = + r esisors in Series and Parallel We previously examined circuis wih wo or more capaciors conneced ogeher, and saw ha hey behaved like a single capacior, wih a capaciance which we could calculae from he individual capaciance values. We will now see ha a similar siuaion exiss for combinaions of resisors, which can again be expressed as an effecive overall resisance. esisors in Series Two resisors in series will have a common curren flowing hrough hem. The volage across 1 is given by V 1 = 1. Similarly V =. The oal volage across he combinaions is jus V = V 1 + V = ( 1 + ). Defining he overall resisance V in he usual way, = = 1 +. For N resisors in series, = 1 + + + N. [4] (n a series combinaion, he oal resisance is always greaer han any individual resisance.) esisors in Parallel When wo resisors are in parallel, i is he volage across hem which is common, and he curren splis o flow hrough he resisors separaely. We V V herefore have 1 = and =, wih he oal curren being given by 1 = 1 +. Defining he overall resisance as before as For N resisors in parallel, 1 1 + 1 1 = = +. V 1 V =, we have 1 1 1 1 1 1 = + + +. [5] 1 N (n a parallel combinaion, he oal resisance is always less han he smalles individual resisance.) [3] PHY1 Elecriciy 4
Kirchhoff s ules Some circuis are more complicaed han series or parallel connecions. For hese, i can be helpful o use Kirchhoff s juncion and loop rules. Kirchhoff s juncion rule is a saemen of he conservaion of charge. The algebraic sum of he currens enering and leaving a juncion is zero. 1 3 4 Σ =. [6] n he figure alongside, his means 1 + 3 4 =. Charge is neiher creaed nor desroyed a he juncion, and i does no accumulae here. Kirchhoff s loop rule is a saemen of he conservaion of energy. The algebraic sum of he changes in poenial around any closed loop is zero. Σ V =. [7] Thus when any charge passes in a closed loop around a circui, i reurns o he same poenial. When charge passes from he negaive o he posiive erminal of a source of emf, is poenial is raised. When i passes hrough a resisance, i falls in poenial. An illusraion of he use of Kirchhoff s rules is given in he examples a he end of his opic. Circuis Circuis conaining only componens of consan resisance and emf will have a seady curren flowing. When a capacior is included in he circui, he curren is likely o change wih ime. This is because he volage across he capacior depends on he charge i holds, and i akes ime for a flowing curren o gradually charge or discharge he capacior. (f a capacior were conneced wih ideal wires, of zero resisance, o an ideal emf, wih no inernal resisance, hen i would be charged insanly.) Capacior Discharge Consider he circui alongside, consising of a capacior and resisor in parallel conneced across an ideal baery wih emf. As long as he + + swich is closed, he poenial difference across boh capacior and C resisor is, and he charge on he capacior is = C. The swich is hen opened a ime =. The capacior hen discharges hrough he resisor, and he curren is jus he rae of discharge, d =. [8] d The poenial difference, V, is V = =. C d Using [8], = C d earranging and inegraing, we have So 1 d d = ln = or = e. [9] PHY1 Elecriciy 43
n oher words, he charge on he capacior decays away exponenially. We can also consider how he volage and curren change. Since V = and =, [9] also gives us V = e. C C f we differeniae [9] wih respec o ime, we obain e e = e = = Noe ha all hree quaniies, charge, volage and curren, die away wih he same ime consan τ =. When =, he charge 1 (for example) has decayed o e 37% of is iniial value. We can also define he half-life T ½ of he charge, ha is he ime required for i o fall o is original value. Equaion [8] hen gives us 1 = = e T1 Taking he naural logarihm, his gives T1 = ln =.693τ. Charging a Capacior Now consider he circui illusraed, where he capacior and resisor are in series. niially he capacior is uncharged, hen he swich is closed a ime =. We can employ he loop rule [7], passing clockwise round he circui, o obain C =. C + (emember ha he poenial falls as we follow he curren hrough he resisor or capacior.) The same resul can also be obained by considering ha he sum of he poenial differences across he resisor and he capacior mus add up o he emf from he baery. The curren now is charging he capacior, so we have d =. d Therefore d C = = =. Here we are defining d C C C = C as he final, fully charged value for he charge on C. 1 d d = ln ( ) = ln 1 = e = C ( ) 1.5.37 T ½ = e [1] τ PHY1 Elecriciy 44
We see ha we have he same ime consan, τ =, bu in his case his gives he ime required for he curren o rise o 63% of is final value. Once again, we can also examine he volage across he capacior: V = = ( 1 e ) = ( 1 e C C ). And finally he curren (hrough eiher resisor or capacior) is given by d = = e = e = e. d.63 τ Noe ha here is he iniial curren, and ha he funcional form is he same as i was for he case of discharging he capacior. PHY1 Elecriciy 45
Puing Wha You Have Learn no Pracice uesion 7.1 A baery has an emf of 1.6 V and an inernal resisance of.15ω. is conneced o a load resisance of 3.Ω. (a) Wha is he erminal volage of he baery? (b) How much power is (i) delivered o he load resisor; (ii) dissipaed inside he baery; (iii) produced in oal by he baery? Soluion (a) The erminal volage depends on he curren, so we firs work ha ou. (b) (i) The exernal power is (ii) The inernal power is P 1.6 =.58 A + r = 3.15 =. V = r = 1.6.58.15 = 1.5 V. ex.58 3..774 W 774 mw = = = =. P in r.58.15 39 mw = = =. (iii) The oal power is Po = = 1.6.58 = 813mW, which is he sum of he answers o pars (i) and (ii), as i should be! uesion 7. 1 r 1 1 r B 3 r3 1 3 The circui above consiss of 3 differen imperfec baeries conneced o wo equal resisors. Find he currens 1, and 3 leaving he baeries, and he poenial difference from A o B, V AB. Take 1 = 6 V, r 1 = 1Ω, = 1 V, r = Ω, 3 = 1 V, r 3 = 3Ω and 1 = = Ω. Soluion Applying Kirchhoff s juncion rule a B, 1 + + 3 = 3 = 1 (1) (Noe ha he currens 1,, 3 canno all be flowing in he direcions indicaed owards poin B. We do no know which flow in he oher direcion, bu his will become apparen from he calculaed values being negaive.) Applying Kirchhoff s loop rule for he lef hand circui, in clockwise direcion, r + r = ( ) A 1 1 1 1 1 1 1 r1 + 1 + r = () Applying Kirchhoff s loop rule for he righ hand circui, in clockwise direcion, r + + r = (3) 3 3 3 3 1 + + 3 3 = Subsiue (1) in (3) r ( )( r ) ( r ) ( r r ) (4) 1 + 3 + + 3 3 = PHY1 Elecriciy 46
Now subsiue in he componen values given o give simulaneous equaions we can solve: () 6 1 1+ 1 = 1 + = 4 (5) 1 (4) 1 1 3 + 5 1 = 3 5 = (6) 1 55 46 = 1 + 4 5 (5) + (6) ( ) 1 1 =.18 A Subsiue in (5) 4 + 1 (.18) = =.89 A Subsiue in (1) 3 = 1 =.93A VB VB = r = 1.89 = 9.8 V. uesion 7.3 The circui alongside shows a 1 µf capacior conneced in parallel wih a 15Ω resisor and a 3. V baery. The swich, which was iniially closed, is opened a ime =. Wha is he curren flowing hrough he resisor immediaely afer he swich is opened? Afer how long will he curren have dropped o 1% of his value? Wha is he energy dissipaed in he resisor during he complee discharge? Soluion niially, he volage across boh capacior and resisor is, so he curren is The curren dies away according o = 3.. A = 15 =. + + C f = e C. =.1, hen = ln.1 C ln1 C = 6 = C ln1 = 1 1 15 ln1 =.69s. akes.69 s for he curren o decay o 1% of is iniial value. The answer o he final par of he quesion mus jus be ha he energy sored on he capacior is 1 dissipaed in he resisor, i.e. U = C. However, we can demonsrae his by inegraing he insananeous power. We have P =, and = e = e U = P d = e C d C = e C 6 C C 3 1 1 = = = = 4 4.5 1 J PHY1 Elecriciy 47
The oal energy dissipaed in he resisor is.45 mj. Problems from Young & Freedman for Topic 7: Try o do exercises 5.8 o 5.38, 5.46 o 5.53, 5.61 o 5.84, 6.1 o 6.34, 6.4 o 6.51 and 6.55 o 6.78. The laer problems in each chaper are more challenging. (Numerical answers o odd-numbered quesions are available a he back of he book.) PHY1 Elecriciy 48