i I (I + i) 3/27/2006 Circuits ( F.Robilliard) 1

Transcription

1 4V I 2V (I + ) V 1 2 4Ω 6Ω 3Ω 3/27/2006 Crcuts ( F.obllard) 1

2 Introducton: Electrcal crcuts are ubqutous n the modern world, and t s dffcult to oerstate ther mportance. They range from smple drect current (DC) crcuts, through alternatng current (AC) crcuts, up to the more complex crcuts of electroncs, communcatons and computng. In all crcuts, partcles carryng electrcal charge moe around a closed path the crcut loop. We wll begn by defnng the basc physcal quanttes needed to descrbe crcuts. These quanttes nclude charge, current, potental dfference, resstance, and emf. Then we wll use these quanttes to analyse DC crcuts, n whch the charge carrers are usually electrons, and the electrons trael around the crcut n one drecton only. Although we wll consder only the smplest class of crcut, the prncples and methods we wll use are fundamental to all crcuts. 3/27/2006 Crcuts ( F.obllard) 2

3 Charge: We begn by defnng some essental crcut quanttes. In the smple model, atoms are composed of three types of partcle - protons, neutrons, and electrons. Protons and neutrons are confned to a compact central regon of the atom called the nucleus. The electrons are dstrbuted n the space surroundng ths nucleus. Charge s a fundamental quantum property possessed by protons and electrons. There are two types of charge - poste and negate. Protons carry a poste charge. Electrons carry a negate charge. The magntude of the charge on the proton s exactly equal n magntude to that on the electron. All protons and electrons n the unerse hae precsely the same magntude of charge. Neutrons carry zero net charge. Charge s, theoretcally, the most fundamental of the electrcal quanttes. SI unt of charge s the Coulomb C 1 C s the total charge on 6.24x10 18 electrons 3/27/2006 Crcuts ( F.obllard) 3

4 Current: Current s the tme-rate of flow of charge through a desgnated surface. dq dt dq n dt where: s the current flow, when charge-carrers coney a total charge of dq through a desgnated surface, n a tme nteral of dt. In materals, protons are generally confned to the nucleus, and cannot moe. Howeer, the outer electrons of atoms, n metals, and some other materals, are loosely held, and can readly moe from one atom to another, thereby consttutng an electrc current. Free electrons can also moe through a acuum, as n the electron beam of a cathode-ray tube Current s proportonal to the number of electrons that pass through a desgnated cross-secton of the conductng medum, per second. 3/27/2006 Crcuts ( F.obllard) 4

5 Ampere: SI unt of current s the Ampere A A current of 1 A flows, when 1 C of charge passes through a gen cross-secton per second 6.24x10 18 electrons per second. Although current s usually a flow of negate electrons, the flow of any charged speces, such as protons or free charged atoms ( ons ) wll consttute a current. By conenton, we deelop the theory of crcuts on the assumpton that the charge-carrers are poste. Such a current s called a conentonal current. 3/27/2006 Crcuts ( F.obllard) 5

6 Defnton of the Ampere: Although charge s the more fundamental quantty, n a theoretcal sense, the fundamental expermental unt of electrcal quanttes, n the SI system, s chosen to be the Ampere. The reason for ths has to do wth practcaltes of the expermental defnton of the quantty. When a current flows, t generates a magnetc feld, whose strength depends on the magntude of the current. Thus, the current can be defned n terms of ts assocated magnetc feld. It turns out, that current can be more accurately defned by ts magnetc feld, than charge can be defned by any known expermental means. We use the magnetc force of attracton between two long, thn, straght current-carryng conductors. For fxed separaton d, the magnetc force, F, ncreases wth the current,. If the precse separaton of the wres, d 1 m, and the magnetc F F d force between the wres, F 2 x 10-7 N, per metre length of the wres, then the current,, s defned to be precsely 1 A. 3/27/2006 Crcuts ( F.obllard) 6

7 Potental Dfference: To cause a current to flow n a conductor, we set up an electrc feld through the conductor. The electrc forces whch then act on the free electrons of the conductor, cause those electrons to accelerate. As a consequence of ther moton, the electrons collde wth the atoms of the conductor nelastcally, thereby loosng the energy they ganed from the electrc feld, to the atoms. Ths causes the conductor atoms to brate. We percee ths braton of the atoms as heat, whch s transferred to the surroundngs of the conductor, and lost to the system of electrons and conductor atoms. In summary, the electrons gan energy from the electrc feld, and lose t through collsons wth conductor atoms. Work s done by the feld, to moe the electrons. The potental dfference ( PD ) between two ponts n a crcut, s defned as the work that has to be done per unt charge, to moe that charge from one pont to the other. SI unt: Work/Charge J/C Volt V There s a PD of 1 V between two ponts, f 1 J of work must be done, to moe 1 C of charge, from the frst pont, to the second. 3/27/2006 Crcuts ( F.obllard) 7

8 + A B - conductor Ohm s Law: If we send a current,, through a conductor, AB, a potental dfference (whch s a potental loss),, wll occur between A & B. If we ncrease, then wll also ncrease. Say we make an expermental plot of the alues of, aganst. It turns out, that the shape of the - cure depends on the nature of the conductor. For some conductors, we wll get a lnear relatonshp. For other conductors, the relatonshp wll be non-lnear. Ohmc conductor most metals, carbon. Non-Ohmc conductor semconductor dode, plasma tubes 3/27/2006 Crcuts ( F.obllard) 8

9 esstance: For Ohmc conductors, the potental across the conductor s proportonal to the current flowng through t, thus Ohm s Law. where s a constant of proportonalty that depends on the composton and geometry of the partcular conductor, and s called the resstance of the conductor. Note: Ohm s law apples to a partcular conductor. SI unt for : V/A Ohm Ω ( kω 10 3 Ω; MΩ 10 6 Ω ). Crcut Symbol Fxed : Ω or Ω Varable : Ω or Ω 3/27/2006 Crcuts ( F.obllard) 9

10 A Smple Pcture: Ohm s law s a property of some materals. It s not really a law of Physcs. We can get a smple pcture of what s happenng, as follows. When an electrc feld s appled along a conductor, free electrons wll moe (drft) along that conductor, collde wth conductor atoms, and suffer energy losses n those collsons. The gan n potental energy of electrons from the feld, wll equal the losses n collsons. As the feld s ncreased, the potental losses are ncreased n proporton, by more energetc collsons, occurrng more frequently. Because of the ncrease n the acceleratng feld, the aerage drft elocty of electrons, between collsons also ncreases. Therefore the electrons moe more rapdly along the conductor, between collsons, whch consttutes an oerall ncrease n current. In ohmc conductors (such as X, Y, Z, n the fgure), an ncrease n electrc feld strength, whch corresponds to an ncrease n potental dfference across the conductor, s proportonal to the ncrease n oerall effecte drft elocty of electrons (e. to the ncrease n current through the conductor). 3/27/2006 Crcuts ( F.obllard) 10 X Y Z

11 esstors n Combnaton: There are only two dfferent ways to connect two resstors, and r, n a crcut n e seres e parallel e n out r In a seres connecton, an nput electron, e, has NO choce. If t goes through, t must go through r. It can do nothng else. e out r In a parallel connecton, an nput electron, e, has an excluse-or choce. It can go through ether or r, but not both. And t has no other choce. Example: A C B A s not n seres wth B, snce an electron could go through A then C. B s n parallel wth C. 3/27/2006 Crcuts ( F.obllard) 11

12 Seres: Consder two resstors, and r, connected n seres. It s mportant to be able to fnd the total equalent resstance, ρ, of the combnaton. equalent resstance. r ρ r current through both & r. PD across both & r. For unt charge passng through and r, by conseraton of energy - PD across alone r PD across r alone Energy lost n ρ thus ρ Energy Energy lost n + lost n r + + r...by Ohm's law + r r In seres, resstors add drectly. 3/27/2006 Crcuts ( F.obllard) 12

13 Parallel: Two resstors, and r, connected n parallel, hae an equalent sngle resstor ρ. equalent resstance. r r ρ Input current,, splts nto two currents, (through ) and r (through r). PD across the parallel combnaton. By conseraton of charge - thus gng r...by Ohm's r 1 r Law In parallel, resstors add by recprocals. 3/27/2006 Crcuts ( F.obllard) 13

14 Example: For the followng crcut, fnd the total equalent resstance between A and B (all resstances n ohm). A Seres: r B 1.0 edraw: A parallel 1 r r B edraw: edraw: A Seres: r A 5.0 B B Answer: 5.0 ohm 3/27/2006 Crcuts ( F.obllard) 14

15 Electromote Force (emf): Electrons lose energy as they pass through the resste elements n a crcut. Ths energy must be replaced, for the electron current to contnue. Ths energy s suppled by a power supply (PS), such as a cell, battery, generator, or electronc power supply. The emf, ε, of a PS s the energy suppled per unt charge that passes through the PS. SI Unt for emf: (work)/(charge) J/C Volt V Note: A PS s an energy source, whereas a resstor s an energy snk, for charge carrers n a crcut When charge-carrers pass through a PS, they generally encounter resstance along the nternal path wthn the PS tself. Ths nternal resstance,, can be regarded as beng n seres wth the PS. s the energy lost per unt charge as t flows through the PS. 3/27/2006 Crcuts ( F.obllard) 15

16 ε + - or Crcut Symbol for emf : ε + - ε emf of the PS nternal resstance of the PS. emf s n Seres & Parallel: ε ε 2 ε where: ε ε 1 + ε In seres, both the emf s and the nternal resstances are addte. For the parallel connecton, we wll only consder the case where both emf s are the same, and ther nternal resstances are neglgble. ε ε ε + - where: ε ε 1 0 3/27/2006 Crcuts ( F.obllard) 16

17 A a Swtches, Fuses & Earths: Sngle pole swtch: A s connected, or dsconnected (swtched) to a. A B a b Double pole, double throw, swtch: A s swtched to a, and B to b, smultaneously. A a b c otary swtch: A s swtched to one only of a, b, or c. A a Fuse, safety swtch or crcut breaker: A s dsconnected from a, f the current exceeds a set alue. A or AE Earth: A s connected to earth (zero potental) 3/27/2006 Crcuts ( F.obllard) 17

18 The Smple Crcut: By use of the rules for addng resstances, and emf s, many crcuts can be reduced to a sngle emf, n seres wth a sngle resstance. + - ε Inddual emf s of the crcut hae been lumped nto a sngle emf, ε, wth a sngle seres nternal resstance,. Inddual resstances of the crcut hae been lumped nto a sngle equalent external resstance,, n seres wth the emf. the current drawn from the emf the current through the external resstance the oltage across the termnals of the emf the oltage across the external resstance,. Such crcuts are called smple crcuts, and can be analysed usng essentally the crcut methods that we hae deeloped so far. Some crcuts consst of a network of emf s and resstors, whch cannot be reduced to ths smple form. These are called complex crcuts, and requre more general methods for ther analyss. 3/27/2006 Crcuts ( F.obllard) 18

19 Basc equaton of the Smple Crcut: ε + - energy gen per unt charge to charge - carrers as they pass through the emf Ths equaton s the applcaton of the conseraton of energy to the smple crcut We assume that the crcut has reached a steady-state condton. We apply the conseraton of energy to the charge-carrers as they go round the crcut. energy lost per unt charge by charge - carrers n the external and nternal resstances + + usng Ohm's Law Ths equaton, together wth Ohms Law, and the rules for combnng resstors, and emf s, are suffcent to analyse the smple crcut. 3/27/2006 Crcuts ( F.obllard) 19

20 Potental around a Smple Crcut - ε + Say we earth the negate termnal of the cell, whch wll set ts potental to zero. We then plot the potental of ponts around the crcut, mong clockwse from ths earthed pont. - ε + ε (ε ) (ε ) 0 Potental rses, from zero, to ε, n the cell, drops by ( ) n the nternal cell resstance, and fnally falls, by a further ( ), back to zero n the external resstance. 3/27/2006 Crcuts ( F.obllard) 20

21 Solng Smple Crcut Problems: Steps: The general pattern s - 1. use the rules for seres and parallel to fnd the total external resstance 2. use ε + to fnd the total current,, drawn from the emf 3. use Ohm to fnd partcular oltages, and currents 3/27/2006 Crcuts ( F.obllard) 21

22 ε10v Example: For Fg.1, fnd () total external resstance,. () total current,, drawn from the PS. 2.6Ω () oltage,, across the 6 Ω. () current, 6, n the 6 Ω. 4 6 Fg.1 4 4Ω 6 6Ω ε10v Ω 2.4Ω Fg.2: Equalent Crcut () Total resstance, : (parallel resstors) thus () oltage, : 6 () Total current, : thus Applyng Ohm s law to the external resstance of the equalent crcut (Fg.2) 2.0 x V () current, 6 : Applyng Ohm s law to the 6Ω resstance of the orgnal crcut (Fg.1) 6 / 6 4.8/6 0.8V 2 A 3/27/2006 Crcuts ( F.obllard) 22

23 + - ε Measurement of ε and for a PS: nterceptε grad- ( expermental pont ) The emf, and nternal resstance of a PS cannot be measured drectly, usng a multmeter. The followng ges a precse method for ther measurement. Connect a arable external resstor,, across the termnals of a PS. If we decrease, from ts maxmum alue, the current,, wll ncrease, but the PD,, across, wll smultaneously decrease. Plottng the alues of aganst the correspondng alues of, ges a straght lne. ε + + thus: - + ε negate gradent ntercept Ths has the form y m x + c (a straght lne) By measurng the -ntercept and gradent of the plot, we can fnd ε and. 3/27/2006 Crcuts ( F.obllard) 23

24 Power In Crcut Elements: So far, we hae consdered the total amounts of energy per unt charge, beng gen to, or lost by, charge-carrers, as they pass through the emf s and resstors of a smple crcut. Ths takes no account of the tme-rate of energy gan or loss power. dq P dt Let P the power dsspated n a resstor, when t conducts a current,, wth an assocated PD of. Say a charge dq passes through the resstor,, n a tme dt. From the defnton of PD, the energy, dw, lost by the charge dq, as t passes through the resstor s - dw dq snce Ddng both sdes by dt we get (snce dw dt Thus P P dw dt dq dt...(1) and 3/27/2006 Crcuts ( F.obllard) 24 dq dt ) dw dq Power oltage x current

25 dq P Power Dsspated n esstance: P.(1) dt By use of Ohm s law, we get two other useful ersons of (1). From Ohm and (2) (1) : (3) (1) :...(2)...(3) 2 P 2 P Summary: The power, P, dsspated n a resstor,, whch has a current,, flowng through t, wth a PD of, across t, s gen by - P 2 3/27/2006 Crcuts ( F.obllard) 25 An example: In power transmsson lnes, the, s kept small, to mnmse power losses 2. 2

26 Example: Fnd the current drawn by a 40 W lamp, runnng on a 240 V supply. P40W 240 V Example: thus P P A Fnd: () the power dsspated n the external 8 ohm resstor of the followng crcut () the total power suppled, by the emf, to the crcut. 1 6 ε20v + - 2Ω 8Ω () Frst fnd the current: thus Thus Power n 8 ohm W () Total power suppled 2 A power lost n both and 2 ( + ) 2 2 (8 +2) W 3/27/2006 Crcuts ( F.obllard) 26

27 Krchoff s ules: If a crcut cannot be reduced to an equalent smple crcut (a sngle emf n seres wth a sngle external resstance), then the methods we hae used so far wll fal. For such complex crcuts, we need a more general method. Krchoff s ules ge such a general method, and allow for the soluton of any DC crcut. Furthermore, Krchoff ges the bases for the analyss of more general AC, and electronc crcuts Krchoff realsed that any network of crcut elements conssts of two basc elements: 1. Node a pont n a crcut where 3 or more conductng paths meet. 2. Loop any closed conductng path n a crcut. There are two Krchoff rules. The frst rule deals wth nodes; the second rule deals wth loops. Krchoff assumes that the crcut s n a steady state. 3/27/2006 Crcuts ( F.obllard) 27

28 Krchoff ule 1: ule 1 (K1): Conseraton of charge at a node. The sum of currents nto a node the sum of currents out of that node 1 2 n out 2A 6A 4 N 3 1A N 5A In a steady-state crcut, there s no gan, nor loss, of charge at a node. If we don t know the physcal drecton of a current, we assume a drecton, and proceed consstently on that bass. If the numercal soluton for that current s poste, we guessed correctly; f the soluton s negate, the current drecton s the reerse to that assumed. 3/27/2006 Crcuts ( F.obllard) 28

29 Krchoff ule 2: ule 2 (K2): Conseraton of energy around a loop. Algebrac sum of around a loop emf's loop algebrac sum of potental drops across the conductors around the same loop ( ) ( r) loop We assume that each pont n the loop has a fxed potental (s n steady state). If a poste test charge traerses a smple crcut loop, ts potental wll ncrease n the emf s, and decrease n the resstances. When t has completed the loop, ts potental must return to the alue t had at ts startng pont. The gans must equal the losses! Howeer, n complex crcuts, the stuaton s a lttle more noled than ths. Note: that we are talkng here about the change n a scalar quantty, the potental of a test charge. An ncrease s a poste change, a decrease s a negate change. Ths s nothng to do wth drecton n the ector sense. 3/27/2006 Crcuts ( F.obllard) 29

30 K2 Sgn Conenton: In a smple crcut, test charges, +dq, can always be taken around the loop n the drecton of the loop current, so that the test charge wll gan potental n emf s, and lose potental n resstances. Ths s llustrated n Fg.1 +dq ε Fg.1 Howeer, n a complex crcut, we may be forced to moe our test charge through emf s and resstors, n drectons opposte to those of Fg.1. In ths case, the test charge wll lose potental n emf s, and gan potental n resstors, the opposte of Fg.1. To keep correct an account of the potental, we need a sgn conenton The drectons of Fg.1 are taken as poste, and the reerse drectons, as negate. + drecton of () s the drecton of the current,. + drecton of ε s from the negate termnal to the poste ε 3/27/2006 Crcuts ( F.obllard) 30 +

31 A e15v D I2A 5Ω + r4ω 5A B E5V C K2 Example: Consder a loop ABCD, whch s part of some network. Let us wrte K2 for ths loop. K2 : loop ( ) ( r) loop Traerse the loop clockwse (n drecton ABCD). We wll call ths the poste loop drecton ( ndcated by the blue crcular arrow). The current drectons are shown (by the maue arrows); the poste drectons for the two emf s are ndcated (by the red arrows). K2 Algebracally: -E + e -I + r K2 Numercally: (2)(5) + (5)(4) thus If the drecton of a current, or an emf s n the poste loop drecton, t s wrtten as poste; f opposte to the loop drecton t s negate. 3/27/2006 Crcuts ( F.obllard) 31

32 Steps n Network Problems: In the usual stuaton, we are gen the emf s, and must fnd all currents. 1. Label drectons of known emf s wth an arrow, on the crcut dagram. 2. Assume drectons for unknown currents and label on the dagram.. 3. For each crcut loop, choose a poste loop drectons (clockwse, or antclockwse), and label t. 4. Apply K1 to nodes (ths can be done on the dagram, n smple cases). 5. Apply K2 to loops. 6. Sole node and loop equatons for unknowns. 7. Interpret your numercal answers as physcal drectons on a crcut dagram. Note: If we get a negate soluton for a current ( or emf, f unknown ) we guessed the wrong drecton at the start of the problem the physcal drecton s the reerse of that assumed! 3/27/2006 Crcuts ( F.obllard) 32

33 Example: 4V I 2V (I + ) 3V 1 2 4Ω 6Ω 3Ω Fnd the magntudes and drectons of the currents delered by the three cells of the network llustrated. Label unknown currents n 4V & 2V cells. (We need to guess these drectons.) Use K1 to label thrd current. Label poste drectons for emf s. Take clockwse as poste for both loops. Wrte loop equatons (K2) for both loops: loop ( ) ( r) loop Loop 1: +4-3 ( I + ) 4 + () I + 10 (1) Loop 2: (I) 3 - ( I + ) I - 4. (2) 3/27/2006 Crcuts ( F.obllard) 33

34 1 4 I + 10 (1) 1-7 I (2) Example: 4V 0.20 A 0.26 A I 2V (1)x2: 2 8 I (3) (2)x5: 5-35I (4) (3) + (4): 7-27I I - 7/ A (5) (5)->(1): 1-28/ /27 55/27 hence A (I + ) 0.06 A 3V 4Ω The physcal drectons for I & need to be nterpreted by labellng on the crcut dagram. Snce I s negate, ts physcal drecton (red arrow) wll be opposte to the assumed drecton (purple arrow). Snce s poste, ts physcal drecton wll be n the same drecton as that guessed (purple arrow). 3/27/2006 Crcuts ( F.obllard) 34 6Ω 3Ω The thrd current can be found by - (I + ) ( ) -0.06A Ths s also labelled.

35 Capactance: So far, we hae dscussed DC networks, whch contan only emf s and resstances. Crcuts n general, may contan other crcut elements, among whch are capactors, nductors, dodes, transstors, gates, and a huge array of ntegrated crcut chps for arous functons. Here we wll look at capactance, and capactors. Electrcal capactance s a property of a system. It s essentally a measure of the system s ablty to store energy n an electrc feld. Consder a system of two bodes, A & B. +q (-q) A Say we transfer charge carrers, carryng a total charge of (+q), from B to A. Ths wll ge A a poste charge of (+q), and leae a resdual negate charge of (-q) on B. Because of the separaton of charge, an electrc feld, and consequent potental dfference,, wll exst between A and B. As more charge s transferred to A, the PD,, between A and B wll ncrease, n proporton to the total charge, q, transferred. The two bodes, A & B consttute a capactor. The gradent of the -q lne s used to defne the capactance of ths capactor. q B 3/27/2006 Crcuts ( F.obllard) 35

36 Defnton of Capactance: The capactance, C, of a capactor, s defned as the rato of the charge, q, on ether conductor to the magntude of the potental dfference,, between them +q A C q (-q) B The capactance of a capactor s a measure of ts ablty to store charge, and electrcal potental energy. Capactance wll depend on the geometry, sze and composton of a system. Capactance s numercally equal to the charge, q, that must be moed (from B to A) to rase the potental dfference,, by one olt. SI Unt: Coulomb/Volt C/V Farad F (Common unts used are µf, nf and pf) 3/27/2006 Crcuts ( F.obllard) 36

37 Parallel Plated Capactor: Ths s a common geometry for a capactor, and conssts of two parallel conductng plates separated by an nsulator. A d To charge the capactor, an emf s connected across the plates. Ths causes a charge of (+q) to be transferred from one plate to the other, producng a potental dfference of across the plates. crcut symbol ε + (+q) -q +q The capactance, C, of a parallel plated capactor depends on the plate area, A, and on plate separaton, d, and s gen by - q C 0 A d where: A area of each plate d separaton of the plates ε 0 constant permttty of acuum x C 2 N -1 m -2 (We wll dere ths result n the next secton - electrc felds.) 3/27/2006 Crcuts ( F.obllard) 37

38 A d Example: Fnd the capactance of a parallel plate capactor wth plates 10 cm square, separated by 1 mm, n ar. C 0 A d ( x10 ) 8.85x pf x F Note: To make the capactance, C, large, we can ncrease the plate area, A, or reduce the plate separaton, d. There are practcal lmtatons on the sze of the plates, and when d s reduced below a certan separaton, short crcutng, or arcng, wll take place between the plates. Thus practcal capactors n electroncs are typcally no larger than µf. 3/27/2006 Crcuts ( F.obllard) 38

39 Capactances n Parallel: In crcuts, capactors are often wred together n seres or n parallel. We need to be able to compute the oerall equalent capactance of such combnatons. An equalent capactor s one that could completely replace the combnaton n a crcut. When the same potental dfference exsts across both the equalent capactance and the orgnal combnaton, both would be storng the same total charge. Consder two capactors, C 1 & C 2, n parallel. C (+q 1 ) 1 The equalent capactor s C. If the same potental dfference,, exsts (+q) C (+q 2 ) C across each, then both must store the 2 same total charge. Thus q q 1 + q 2 From defnton of capactance C C 1 + C 2 Thus C C 1 + C 2 In parallel, capactors add drectly. 3/27/2006 Crcuts ( F.obllard) 39

40 (-q) (+q) (-q) C 1 C 2 (+q) Capactors n Seres: Consder two capactors, C 1 & C 2, n seres. The equalent capactor s C. C (+q) 1 2 To charge the combnaton, we moe a charge (+q) from the left plate of C 1, to the rght plate of C 2. The central wre connectng the two capactors contans free charges. The resdual (-q) on the left plate of C 1, wll attract an equal but opposte charge, (+q), to ts rght plate, from the central wre. Smlarly, C 2 wll attract a resdual charge (-q) to ts left plate, from the central wre. Thus both the seres capactors carry the same charge, q. Howeer, because ther capactances are dfferent, they wll hae dfferent nddual PD s, 1 & 2. From conseraton of thus q C 1 C From defnton of 1 q C 1 1 C q C C 2 energy capactance Seres capactors add by recprocals. 3/27/2006 Crcuts ( F.obllard) 40

41 Example: Fnd the total capactance between A and B of nf 2.0 nf A B 0.8 nf In seres, thus thus C C 1.2 nf 5 edraw: 1.2 nf In parallel, thus C nf A 0.8 nf B Thus total capactance between A & B s 2.0 nf. A B 2.0 nf 3/27/2006 Crcuts ( F.obllard) 41

42 Energy Stored n a Capactor: When a capactor s charged, work s done n separatng charges wthn the capactor. Ths energy comes from the emf that separates the charges. If the capactor s dscharged through a resstance, ths stored energy s transferred to the resstance, where t s dsspated as heat. Say a capactance, C, has separated charge, q, and potental dfference,. (-q) C ε + (+q) We separate a small addtonal amount of charge dq. To do ths we need to do a small amount of work dw. Because the change n charge s small, the potental dfference,, wll reman essentally constant. From the defnton of potental dfference, the work done, dw, s gen by - dw dq..(1) 3/27/2006 Crcuts ( F.obllard) 42

43 Usng defnton of C q dw dq..(1) q C Energy Stored n a Capactor... q C capactance ( 2) ( 2) ( 1) : dw dq... ( 3) The total energy, E, stored n the capactor s the sum of all the elements of work, dw, done n storng the total charge, q. Total work dw E E q C q 0 q C dq...( 4) Usng the defnton of capactance C q q C ( 4) : E q 2 ( 4) : E C Summary: 2 q 1 E q 2C C The energy stored n a capactor s proportonal to the square of the charge separated, or to the square of the PD across the capactor. 2 3/27/2006 Crcuts ( F.obllard) 43

44 Example: An electrc car needs to output 20 kw of power for 3 hours, before beng recharged. If ths energy were to be stored n a 1 mf capactor, what would the PD across the capactor termnals need to be? Total energy to be stored E Power x tme (20 x 10 3 ) x (3 x 60 x 60) 216 MJ But E 1/2 C 2 thus (216 x 10 6 ) 1/2 (1 x 10-3 ) 2 whch ges 2 (216 x 10 6 ) x 10 3 x x hence 6.60 x 10 5 V 0.66 MV Ths s clearly a lethally hgh oltage, and hence mpractcal! 3/27/2006 Crcuts ( F.obllard) 44

45 4V I 2V (I + ) V 1 2 4Ω 6Ω 3Ω 3/27/2006 Crcuts ( F.obllard) 45