Circuit Variables. Unit: volt (V = J/C)

Transcription

1 Crcut Varables Scentfc nestgaton of statc electrcty was done n late 700 s and Coulomb s credted wth most of the dscoeres. He found that electrc charges hae two attrbutes: amount and polarty. There are two type of charges opposte charges attract and smlar polarty ones repel each other. Charge polarty s ndcated by poste and negate sgns because poste and negate charges cancel each other when brought together. As a results, the electrc charge can be descrbed by an algebrac number, q, wth unts of Coulomb (C). Because opposte charges attract each other, energy s expanded to separate them from each other. Ths energy s stored n the electrc feld between the two reseror of separated charges and s recoered when the charges are allowed to come together. The stored energy per unt charge s called the oltage or potental dfference between the two reseror of charges: = dw dq Unt: olt (V = J/C) Note that we need two reseror of charges. So oltage s between two ponts. We also use and sgns to ndcate the drecton for measurng. From defnton of oltage aboe, w s energy needed to moe a poste charge from reseror to reseror. One can defne a reference pont for measurng oltages (typcally shown as ground). The oltage between any pont and ths reference pont s call the potental of that pont. It s always assumed that s at the pont and the sgn s at the reference pont. Therefore, there s no need to ndcate and sgns for potental. Voltage between two ponts s the dfference between the potental of the two ponts (see fgure). Voltage between two charge reserors s analogous to heght dfference between two flud reseror and the same way, the potental of each pont s analogous to ts eleaton compared to some reference (e.g., sea leel). Potental Eleaton = V V = V V = V V h h h = h - - h 0 0 Sea Leel MAE40 Notes, Wnter 00

2 If we connect the charge reserors, electrc charges flow from one to other. The rate of the charge flow through a specfc area s called the electrc current: = dq dt Unt: Ampere (A = C/s) wth the current flowng n the drecton of the charge flow (t means that a poste current s assocated wth the flow of poste charge). In prncple, electrc charges generate an electrc feld and moton of the charged partcles (current) generates a magnetc feld. Ths electromagnetc feld nteracts wth all charges and affect them. The behaor of such a system s descrbed by Maxwell s equaton. Soluton of Maxwell s equatons, howeer, s dffcult and not needed expect for some cases (propagaton of electromagnetc wae and lght, antennas, etc.) In most releant engneerng cases the problem can be greatly smplfed by notng that electrc charges preferentally flow through a conductor (or a semconductor) as opposed to acuum, ar, or any nsulator. In ths case, the system can be descrbed as a crcut contanng crcut elements and connectng deal wres. Crcut theory s the scentfc dscplne that descrbes behaor of crcuts be and s bult upon the followng assumptons:. All of the electromagnetc phenomena occurs nsde each crcut element. They communcate wth the outsde world only through the oltage across and current that partcular element.. Crcut elements are connected to each other wth deal wres that do not mpede flow of charge. They can be stretched (makng them longer or shorter for example) wthout any effect on the crcut. 3. Net amount of charge cannot be accumulated n any crcut element or any locaton n the crcut. If a net charge of q enters a crcut element, the same amount of charge should leae the element. Ths means: (a) a crcut element should hae at least two termnals, (b) Because current traels through the system at a good fracton of speed of lght, we can safely assume that the total current enterng a crcut element s exactly equal to the current leang that element at any nstant tme. A note about current flow n a conductor. Consder a seres of beads on a wre. If another bead s added to the left end of wre, all of the bead moe one step to the rght and one bead wll fall off the rght end. Smlarly, when an electron enters one end of a wre, electrons n the system moe slghtly and another electron wll leae the other end of the wre. As such, whle electrons do not moe rapdly n a conductor, the current n the wre propagates at a good fracton of speed lght. Crcut Element Ideal Wre MAE40 Notes, Wnter 00

3 In the context of crcut defnton aboe, the mportant physcal quanttes are current and oltages (and electrc power). These are the crcut arables that form the bass for communcaton between crcut elements. The alue of oltage between two ponts,, s ncomplete unless the and sgns are assgned to the two ponts. Smlarly, the alue of current s ncomplete unless a reference current drecton s assgned. Howeer, as and are algebrac numbers, the assgned poston of and for oltage and drecton for the current s arbtrary the sgn of and flps accordng to ths choce. Internal of each crcut element mpose a relatonshp between the current flowng n the element and the oltage across that element. Ths relatonshp s called the element law or - characterstcs. Whle the choce of reference drectons for current through and oltage across an element s arbtrary, the element law or - characterstcs of an element depends on the reference drectons. To see ths pont, note that there exsts four possble choces for oltage and current drectons n a two-termnal crcut element: Passe Sgn Conenton Acte Sgn Conenton In two left cases, the current drecton s marked such that t flows from to sgns. Ths case s called the passe sgn conenton. In the two rght cases, current flows from to sgns. Ths case s referred to as acte sgn conenton. Obously, the - characterstc wll be dfferent n each case as the sgn of current or oltage s swtched. To resole ths confuson, we follow ths conenton:. - characterstcs of two-termnal element are wrtten assumng passe sgn conenton.. Use passe sgn conenton when markng current and oltages n a crcut as much as possble. So, n markng references for oltages and currents n the crcut, t s best to arbtrarly choose ether the oltage or current reference drecton and then use passe sgn conenton MAE40 Notes, Wnter 00 3

4 to mark the other arable. Note that t may not be possble or practcal to mark eery element usng passe sgn conenton. In ths case, A good rule s to mark eery element except oltage and current sources usng passe sgn conenton. Electrc power produced or absorbed n an element: Power: P = dw dt Unt: Watt (W=J/s). Snce = dw/dq and = dq/dt (both total derates), then: P = dw dt P = = dw dq dq dt Usng the defnton of the oltage, one fnds that f we use passe sgn conenton: P < 0 P > 0 Element s producng power Element s absorbng power Obously, f we use acte sgn conenton: P < 0 P > 0 Element s absorbng power Element s producng power Example: Fnd the power absorbed or suppled by elements A and B f =5Aand = 0 V. For element A, - P = = 5 0 = 3, 000 W = 3 kw A B Snce element A reference drectons follow passe sgn conenton and P>, element A s absorbng 3 kw of power. For element B, P = = 5 0 = 3, 000 W = 3 kw Snce element B reference drectons follow acte sgn conenton and P >,element B s supplyng 3 kw of power. MAE40 Notes, Wnter 00 4

5 Defntons KIRCHHOFF LAWS A crcut conssts of crcut elements attached to each other wth deal wres or connectors. Node: A node s a pont n the crcut that s connected to two or more crcut elements wth deal wres. Loop: A loop s a closed path n a crcut through at least two crcut elements whch return to startng node wthout passng through any node twce. Loop Node Node Note that a node should not be taken lterally as a geometrc pont on a specfc crcut dagram as the deal wres connectng the elements can be moed and stretched. For example, any pont on the deal wres connected to the four elements n the bottom of the fgure s a node (as hghlghted n the fgure). All these ponts represent the same node. Krchhoff Current Law (KCL): KCL follows from our assumpton that no net charge can be accumulated at any pont n the crcut (or n a node): Sum of currents enterng a node s equal to sum of currents leang a node. Alternately, as a current leang a node can be replaced wth a current enterng a node wth opposte sgn, KCL can be stated as: Krchhoff Current Law (KCL): or Algebrac sum of currents enterng a node s zero. Algebrac sum of currents leang a node s zero. I use the last erson of KCL n these notes as ths erson s also used n the node-oltage method. How to apply KCL. You should hae dentfed the nodes and mark currents and ther reference drectons on t.. Draw a closed path around the node and mark a startng a pont. 3. Go around the path n the clockwse drecton. Wheneer you pass a wre, wrte down the current wth a sgn f extng and a sgn f enterng. 4. Stop when you are back to the startng pont (and wrte = 0). MAE40 Notes, Wnter 00 5

6 Example: Wrte KCL for the marked node. 3 4 =0 3 4 Example: Fnd () (3) () = 0 = 4 A 3 Krchhoff Voltage Law (KVL): KVL follows from the defnton of the oltage between ponts. Voltage s defned as the amount energy to moe unt charge from one pont to another. Zero net energy should be expanded f a charge q s moed around a closed loop and s returned back to ts orgnal poston,.e., Σ W = qσ =0 Σ =0 Krchhoff Voltage Law (KVL): Algebrac sum of all oltages around any closed path n a crcut s zero. or Sum of oltage drops around a loop - Sum of oltage rses around a loop = 0. How to apply KVL. You should hae marked oltages and ther reference drectons on t.. Draw the loop and mark a startng a pont. 3. Go around the path n the clockwse drecton. Wheneer you go oer a crcut element, wrte down the oltage across element wth a sgn f enterng the termnal of an element and a sgn f enterng the termnal of an element. 4. Stop when you are back to the startng pont (and wrte = 0). MAE40 Notes, Wnter 00 6

7 Example: Wrte KVL for the marked loop. b _ b c a =0 a c KVLs and KCLs are constrants on crcut arables whch arse because of the crcut arrangement (attachment of connecton wres). In addton, nternal of each crcut element mpose a relatonshp between the current flowng n the element and the oltage across that element (element Laws or - characterstcs). Combnaton of these two set of constrants results n a unque set of alues for the crcut arables (currents and oltages). In a crcut wth E elements, there are E crcut arables ( and for each element). We need E equatons to fnd these crcut arables. If the crcut has N nodes, we can wrte N KCL equatons (KCL on the N th node s exactly the sum of KCL on the other N nodes). We can also wrte E N ndependent KVLs and E- characterstc equatons: No. of KCL equatons: N No. of KCV equatons: E N No. of - characterstcs: E Total E If the - characterstc s a lnear relatonshp between and, the element s a lnear element. If a crcut s made of lnear element, the resultng set of E equatons n E arables for a lnear algebrac set of equatons. In ths course, we only use lnear elements, thus, the term lnear crcut theory. MAE40 Notes, Wnter 00 7

8 Lnear Crcut Elements Resstor V = / R R - characterstc: = R Resstance R, Unt Ohm (Ω) Conductance G = /R, Unt Semens (S) P = = R P>0 : Resstor always absorbs power Independent Voltage Source (IVS) s - characterstc: = s for any s = s Independent Current Source (ICS) s s = s - characterstc: = s for any Some Obseratons:. - characterstcs (element laws) of lnear crcut element are lnes n - plane.. - characterstcs of IVS and ICS are ndependent of sgn conenton (e.g., equally correct for both passe and acte sgn conentons). 3. Resstors always absorb power. IVS and ICS can ether absorb or supply power. MAE40 Notes, Wnter 00 8

9 Short Crcut - characterstc: =0forany Note that a short crcut element s a specal case of a resstor (wth R =0)orandeal oltage source (wth s =0). Open Crcut - characterstc: =0forany Note that an open crcut element s a specal case of a resstor (wth R )orandeal current source (wth s =0). Swtch - characterstc: Swtch open: Open crcut = 0forany Swtch closed: Short crcut = 0forany MAE40 Notes, Wnter 00 9

10 Procedure Crcut Analyss usngkvl and KCL. Note how you can calculate problem unknown (e.g., power dsspaton n an element) from the crcut arables.. Go through the crcut n an orderly fashon (e.g., from left to rght, top to bottom). Take each element and a) Identfy nodes (termnals of each element should be connected to a node.) b) Assgn oltages and currents and ther reference drectons to each element. Use passe sgn conenton. c) Identfy crcut arables. Ths wll tell you how many equatons you hae to wrte. d) Wrte down - characterstcs equaton. 3. If you hae N nodes, wrte N KCLs. 4. Calculate no. of KVLs you need: N KV L =E (E) (N ) = E N (where E s no. of elements). Note that as some element laws (e.g., IVS, ICS) result n tral equatons for some crcut arables, t s better to count the crcut unknowns from step aboe and use N KV L = N unknowns N element laws N Nodes. 5. Choose N KV L loops and wrte KVLs. Choose loops that go through the smallest number of elements. 6. Sole the system of equatons and fnd crcut arables. 7. Check your soluton by applyng your soluton to some KVL and KCL (specally KVL on loops you hae not consdered). 8. Fnd problem unknown, f any, from the crcut arables. Note: It s not always possble or practcal to assgn passe sgn conenton to all elements. If so, note that - characterstcs of IVS and ICS are ndependent of sgn conenton. You do NOTneed to use passe sgn conenton for these elements. But be careful f you need to fnd power suppled or absorbed by IVS or ICS. Note: You can readly reduce the number of equatons to sole by half f you substtute - characterstcs equatons n KVLs and KCLs to get a set of E equatons n ether oltages or currents (or a combnaton of both). In fact, after you are fluent n usng the aboe procedure, you only need to mark ether the current or oltage for an element and use the element law to drectly wrte the other crcut arable on the crcut dagram. Then, wrte only KCLs and KVLs and sole. MAE40 Notes, Wnter 00 0

11 Example: Fnd. Followng the procedure, we frst mark the crcut arables ( unknowns: 0,,, 3, 4, 0,,, 3, 4, 5 ) and dentfy the nodes, and wrte the - characterstcs equatons: 0 = 5 0 = V 5 Ω 5Ω 70Ω 0 0Ω 8Ω = 5 3 = = 8 4 Note that we hae 6 elements so we hae crcut arables. Howeer, the IVS element law specfes the oltage across ts termnal, so we hae crcut arables whch hae to be found. The crcut has four nodes. Snce we need to wrte KCL only n N nodes, we choose not wrte KCL n the bottom node. the KCLs are: 0 5 = 0 3 = = 0 80 V 5 5 Ω Ω 5Ω 0 0Ω Ω We need to wrte N KV L = 5 3 = 3 KVLs. Choosng 3 loops wth smallest number elements, we get: 0 3 = 0 (80) = = 0 Aboe are eleen equatons n eleen unknowns that can be soled. The numbers of equatons to be soled can be haled by usng - characterstcs equatons to substtute n KVLs and KCLs. For example, f we substtute for oltages from - characterstcs equatons n KVLs, we get 6 equatons n 6 unknown currents: 0 5 = = 0 3 = = = = 0 MAE40 Notes, Wnter 00

12 Note: As can be seen, een relately small crcuts results n a large number equatons to be soled. Seeral technques for reducng the number of equaton to be soled are ntroduced next. MAE40 Notes, Wnter 00