N2-1. The Voltage Source. V = ε ri. The Current Source

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1 DC Cicuit nalysis The simplest cicuits to undestand and analyze ae those that cay diect cuent (DC). n this note we continue ou study of DC cicuits with the topics of DC voltage and cuent souces, the idea of an equivalent cicuit, and vaious techniques of analyzing DC cicuits. oltage and Cuent Souces We have aleady encounteed an example of an enegy souce (battey) in Note 01. ny enegy souce supplies a voltage and a cuent, of couse. t is useful, howeve, to descibe a souce specifically as a voltage o cuent souce if we place a special meaning on the tems. t is also useful to distinguish between ideal voltage and cuent souces and thei eal (that is, eal wold) countepats. The oltage Souce y an ideal voltage souce (Figue 2-1) we shall mean an enegy souce that is capable of deliveing a constant voltage acoss a load egadless of the cuent dawn by the load. Figue 2-1. Cicuit symbols fo an ideal voltage souce. load esisto ae at the same potential. n this state the souce behaves electically as if it wee shoted by a connecting wie. Though it might seem tivial at this stage the idea of a zeoed voltage souce will, in fact, pove useful in ou study of Thévenin equivalent cicuits late in this note. ε ε = ε (a) (b) Figue 2-2. eyond a cetain cuent a eal voltage souce (a) delives a voltage = ε acoss a load (appoximately), whee ε is the souce s intinsic emf and its intenal esistance. We have aleady seen in Note 01 that a cell o a battey is a moe-o-less ideal voltage souce so long as the cuent dawn fom it is small. s the cuent dawn ises, howeve, beyond a cetain point, the cell begins to behave non-ideally, that is, the voltage appeaing acoss its teminals deceases. This behavio is attibuted (at least appoximately) to the existence of an effective intenal esistance in seies with the battey s emf. Thus in these notes (whee we deem it impotant) we shall epesent a eal voltage souce by the elements dawn in Figue 2-2a. n ideal voltage souce would be one whose intenal esistance is zeo. can be measued expeimentally. The esults of a method fo finding of a eal voltage souce ae sketched in Figue 2-2b. (We shall take up this topic again in Example Poblem 2-8.) f the teminal voltage is plotted vs the output cuent beyond the point whee the souce behaves non-ideally, then the value of is (1) times the slope of the gaph. Even a dead o a zeoed voltage souce is of some small inteest. n Figue 2-3 a dead voltage souce is shown connected to a load esisto. Since the souce is dead then both sides of the souce and the 0 Figue 2-3. f the output of a voltage souce is zeo then the souce can be eplaced by a shoting wie with no change occuing in its electical behavio. The Cuent Souce cuent souce is anothe special case of an enegy souce. y ideal cuent souce (Figue 2-) we shall mean an enegy souce that is capable of deliveing a constant cuent to a load egadless of the voltage developed acoss the load. Figue 2-. Cicuit Symbols fo an ideal cuent souce. N2-1

2 eal cuent souce is much less common than a eal voltage souce. eal cuent souce can be made fom a voltage souce and a lage seies esisto o it can be constucted fom a cicuit employing a tansisto o an C chip. f you encounte a eal cuent souce in this couse at all it will likely be in the fom of a special powe supply. s you might expect the behavio of a eal cuent souce is not exactly ideal eithe. The cuent deliveed by a eal cuent souce deceases as the voltage developed acoss the load inceases beyond a cetain point. This means that a eal cuent souce behaves like an ideal souce element in paallel with an intenal esistance (Figue 2-5). n ideal cuent souce would be one whose intenal esistance is infinite. 0 R i = 0 R i Example Poblem 2-1 n Example of an ntenal Resistance f a D cell develops 1.5 acoss its teminals when unloaded but only 1.3 when connected to a 100 Ω load, what is the intenal esistance of the D cell? The fact that the D cell s teminal voltage dops fom 1.5 (effectively the cell s emf) to 1.3 when the load esisto is connected means that the cell has an intenal esistance. The cicuit is shown in Figue 2-7. ε =1.5 R =100 Ω Figue 2-5. eal cuent souce behaves (appoximately) like an ideal cuent souce with an intenal paallel esistance. Note finally fom Figue 2-6 that a cuent souce with an output of zeo can be eplaced by an open cicuit with no change in electical behavio. Zeoed cuent and voltage souces ae in a sense complementay. Figue 2-7. eal D cell connected to a load esisto. We can solve this poblem a couple of ways. When R is connected the cuent flowing must be = Δ R R 1.3 () = 100 (Ω) = The voltage dop acoss is = = Theefoe = Δ = 0.2 () () =15. Ω. Figue 2-6. zeoed cuent souce is one which can be eplaced by an open cicuit with no change in electical behavio. Let us conside a chemical cell as a eal souce element. The D cell has an intenal esistance of 15. Ω. Cicuit analysis is essentially the activity of solving fo all unknown cicuit paametes. The unknowns might be the cuents that flow in the vaious banches of the cicuit, the voltages that exist between the vaious nodes of the cicuit, the effective esistances between the vaious nodes and/o combinations of these. Kichoff s ules ae useful in solving fo these unknowns. n the next section we conside the statements of Kichoff s ules and apply them to a numbe of examples. N2-2

3 Kichoff s Rules Conside the multibanch cicuit dawn in Figue 2-8. Suppose that all the cicuit paametes ae known except fo the cuents 1, 2, and 3. You cannot solve fo these cuents by means of esisto eductions because of the existence of the voltage souces in the two banches of the cicuit. (Convince youself of this.) To solve this poblem you need to apply Kichoff s ules. good fist step in poblems of this kind is to numbe the nodes and to edaw the cicuit to make it as ecognizeable as possible (Figue 2-8b). You can numbe the nodes any way you like. n symbols, Note 02 Δ i = 0 aound a closed loop 1 [2-1] i Kichoff s Cuent Rule Kichoff s cuent ule (abbeviated KCR) is commonly stated in these wods: The sum of the cuents flowing into a node is zeo. n symbols, n = 0 into a node 2 [2-2] n Ω Note the phase into a node. The cuents indicated in Figues 2 ae shown flowing into node 2. We assumed these diections and they ae completely abitay. f in the subsequent analysis we find that the numeical value of a cuent is negative then the actual diection of the cuent is opposite to the diection assumed. 5 2 Figue 2-8a. n example of a multibanch cicuit Ω Figue 2-8b. Figue 2-8a edawn to make it moe ecognizeable NOTE: We ae using hee the tems banch, loop and node. banch is a physical section of a cicuit compising wie, voltage souce, esistos, and so on. node is a point whee two o moe banches meet. loop is an imaginay closed path that we tavese in the task of calculating potential changes. The path defined by the nodes in Figue 2-8 is a loop. The path maked by the nodes 2- is a banch (and of couse the nodes ae numbeed). Kichoff s oltage Rule Kichoff s voltage ule (abbeviated KR) is commonly stated in these wods: The sum of voltage changes aound a closed loop, o path, is zeo. 3 Example Poblem 2-2 pplying Kichoff s Rules Solve fo the cuents 1, 2 and 3 in Figues 2-8 using Kichoff s ules. To apply Kichoff s ules coectly we must obseve the signs of the vaious voltage changes and cuent diections consistently. pplying KR aound the loop shown in Figue 2-8b defined by the nodes , we get = 0. [2-3] (The fist voltage change we encounte is acoss the 3 battey which is ve, the change acoss the Ω esisto in the banch 1-2 is negative, and so on). pplying KR aound the loop defined by the nodes (moving clockwise), we get = 0. [2-] (The fist voltage change is that acoss the 2Ω esisto which is ve, and so foth.) pplying KCR to node 2 we get 1 t can be shown that KR aound a closed loop follows fom the fact that the electic field within the conducto foming the loop is a consevative field. 2 t can be shown that KCR is a consequence of chage consevation. N2-3

4 N = 0. [2-5] We can now solve the thee simultaneous equations (eqs[2-3] though [2-5]) in the usual way. The esults ae 1 = 1 9, 2 =, 3 = One of the cuents (3) tuns out to be negative. This means that 3 s tue diection is opposite to the diection we assumed. To check the consistency of these esults we can find the total potential change aound (stating fom node 1): 5 x 1 x2 3 = This is zeo as equied by KR. We now continue with thee methods that ae based on Kichoff s Rules: the methods of Loop cuents, node voltages and supeposition. The Method of Loop Cuents The method of loop cuents (also called mesh cuents) is an advance on the basic Kichoff s ules. s we have seen in the basic application of Kichoff s ules we assign a tue cuent to each banch of a cicuit and then apply KR and KCR. The solutions of the esulting simultaneous equations ae the tue cuents the cuents that ae actually flowing. ut in the method of loop cuents we assign an imaginay cuent, numbeed as usual, to each loop. n a banch of the cicuit that is shaed by a numbe of loops moe than one loop cuent will be flowing. We must find the tue cuents fom the solutions of the simultaneous equations; they ae, in geneal, the sum o the diffeence of adjacent loop cuents. Used caefully this method is supeio to the usual method in that we need apply only KR and we need solve a fewe numbe of equations. Let us conside an example. Example Poblem 2-3 The Method of Loop Cuents Solve fo the cuents in each banch of the cicuit in Figue 2-9 by the method of loop cuents. We begin by naming the thee loops 1, 2 and 3 as shown. To each of these loops we assign an imaginay cuent 1, 2 and 3 (we use a pime to emind us that these ae not in geneal the tue cuents). Fo example, the tue cuent flowing though the top 3Ω esisto is 1 2 (o the evese). Thee cuents would mean odinaily that we would need to solve thee simultaneous equations. ut one of the loop cuents we know aleady, 1 = 3, making only two equations actually necessay in this example ' ➁ 3 Ω 3 Ω Ω 2' ➂ 3' 12 Figue 2-9. n example of a multibanch cicuit to illustate the method of loop o mesh cuents. WRNNGS: Make sue that when you move aound a loop you go continuously in the same clockwise o counteclockwise diection. Take the cuent that is flowing in the diection you tavel as positive. void applying KR to the loop containing a cuent souce as the voltage acoss a cuent souce is usually not known. 3 pplying KR to loops 2 and 3 keeping the wanings in mind we get 2'3' 32'31' 22'= 0 [2-6] and 33'31' 3'2' 12 = 0. [2-7] Substituting 1 = 3 into eq[2-6] and [2-7] we can solve them in the usual way. The esults ae: 1'= 3, 2'= 51 7, 3'= Having the loop cuents, we can now solve fo the tue cuents. f we take the tue (unpimed) cuent 1 = 1 = 3, 2 as the cuent though the 2Ω esisto, 3 the cuent though the uppe 3Ω esisto, the cuent though the Ω esisto and 5 the cuent though the lowe 3Ω esisto we have 2 = 2'= 1.08, 3 = 1' 2'= 1.92, 3 cuent souce is conventionally labelled with the value of cuent it is able to sustain though itself.

5 = 3' 2'= 0.90, 5 = 1' 3'= Note that the cuent though the Ω esisto is negative; this means that it is actually flowing to the left. Method of Node oltages The method of node voltages also goes by the name of nodal analysis. The method is anothe vaiation on Kichoff s ules and has the advantage in some instances in equiing only KCR. The method consists of fist choosing an abitay (but convenient) node to seve as a zeo voltage efeence. The voltages at all the othe nodes ae then assigned and efeenced to this zeo voltage. KCR is then applied to each node and the equations esulting ae solved in the usual way. n applying this method you may wish to follow this fou step ecipe : 1 Choose a efeence node whose voltage can be set to zeo. This node is abitay though it is usually best to choose the node that is connected diectly to the negative side of the voltage souce. 2 ssign voltages 1, 2, etc. to the vaious nodes. 3 pply KCR to the vaious nodes to deive the simultaneous equations. Solve the equations. Let us conside an example. Example Poblem 2- The Method of Node oltages Solve fo the voltages indicated in the cicuit in Figue 2-10 by the method of node voltages Ω 1 Ω 2 Ω 3 6 Ω Figue multi-banch cicuit fo nodal analysis. This method lends itself well to compute adaptation. t is used in the pogam Spice and othes. Note 02 Following the ecipe we obtain these equations fo nodes 2 and 3: and = 0 [2-8] = 0. [2-9] Note 1 = 3 is obvious. Substituting 1 = 3 esults in two equations in two unknowns. The esults ae: 1= 3, 2 = 36 17, 3 = CHLLENGE: How would you solve this poblem any othe way? Method of Supeposition The method of supeposition is also a vaiation of Kichoff s ules, o at the vey least, involves Kichoff s ules. oltages and cuents obey a pinciple of supeposition which can be stated as follows: The cuent in any banch of a cicuit equals the sum of the cuents poduced in that banch by each individual independent souce with all othes zeoed. The voltage at any node of a cicuit equals the sum of the voltages poduced at that node by each individual independent souce with all othes zeoed. Let us conside an example. Example Poblem 2-5 The Method of Supeposition Solve fo the cuent 1 in the cicuit in Figue 2-11a using the method of supeposition. This method equies that the souces be zeoed. Recall that zeoing a voltage souce amounts to eplacing the souce with a shot cicuit (Figue 2-3) and zeoing a cuent souce amounts to eplacing the souce with an open cicuit (Figue 2-6). The cicuit of Figue 2-11a with the souces zeoed esults in the cicuits in Figues 2-11b and c. Fom Figue 2-11b we get 1 ' = 3, N2-5

6 zeo cuent souce 1 Ω 1 Ω 1 Ω 1 Ω 1 Ω 2 1 (a) zeo voltage souce 1 Ω P 1 Ω Thevenin s Theoem Thévenin s theoem can be stated in these wods: 5 ny two teminal netwok is composed of an ideal voltage souce TH and a seies esisto R TH. The theoem states, in effect, that the cicuit in Figue 2-12a is equivalent to the one in Figue 2-12b. 1 ' 1 Ω 2 (b) (c) 1 Ω ' 1 Figue The pocess of zeoing the cuent and voltage souces in (a) esults in (b) and (c) espectively. netwok composed of unknown voltage and/o cuent souces, esistos etc. (a) by applying Ohm s law, and fom Figue 2-11c we get N2-6 1 '' = 2 3, by applying the method of node voltages to node P. The total cuent is the sum of the two cuents: ' '' 1 = 1 1 = 2. Clealy, the methods of loop cuents, node voltages and supeposition ae quite equivalent. You should ty to maste all of these methods as each has its own advantages in paticula situations. Equivalent Cicuits Thus fa we have consideed techniques fo analyzing cicuits whose cicuit diagam o intenal stuctue is known. Sometimes, we ae foced to deal with a cicuit whose intenal stuctue is not known. Such a cicuit is often called a black box. Fotunately, in these cases we can usually extact infomation about the cicuit using the Thévenin and Noton theoems. These theoems (which ae eally statements given without poof) enable us to expess a netwok in tems of equivalent souces and esistos. The theoems ae useful even when the netwok s cicuit diagam is known, fo they epesent one moe way of educing a cicuit s compexity. You must emembe, howeve, that they ae applicable only to netwoks consisting of linea elements (like voltage souces and esistos). R TH TH N R N (b) (c) Figue ccoding to the theoems the cicuit in (a) is electically the same as the Thévenin equivalent in (b) and the Noton equivalent in (c). TH and R TH may be found as follows: 1 Calculate (o measue with a suitable instument) the open cicuit voltage OC. Then TH = OC. The method of finding R TH depends on whethe the cicuit diagam is known o not. f unknown then 2a Shot the output and measue the shot-cicuit cuent SC with a suitable instument. Then R TH = TH / SC. f the cicuit diagam is known then 2b Zeo the voltage and/o cuent souces and fom the esult calculate the effective esistance R between the output teminals. Then R TH = R. 5 n some places in the documentation fo this couse you will find R TH and TH witten R eq and eq.

7 Noton s Theoem Noton s theoem can be stated in these wods: ny two teminal netwok is equivalent to an ideal cuent souce N and a paallel esisto R N. The meaning of the theoem is sketched in Figues 2-12a and c. N and R N may be found as follows: 1 Calculate (o measue with a suitable instument) the shot cicuit cuent SC. Then N = SC. s in the case of Thévenin s theoem, the method of finding R N depends on whethe the cicuit diagam is known o not. f known o unknown 2a Calculate (o measue with a suitable instument) the open cicuit voltage, i.e., TH. Then R N = TH / SC. f the cicuit diagam is known, the option is to 2b Zeo the voltage and/o cuent souces and fom the esult calculate the effective esistance R between the output teminals. Then R N = R. t follows fom this and the pevious section that R N = R TH. t should be emphasized hee that the theoems ae most useful when the cicuit is a complicated one o when the cicuit diagam is unknown. These cases we exploe in Example Poblems 2-7 and 2-8. ut fist, a simple equivalent. Example Poblem 2-6 Simple Equivalent Daw the Thevenin and Noton equivalents of the eal enegy souce of Figue 2-2a. ε ε Note 02 Figue Thevenin and Noton equivalents of a eal enegy souce (Figue 2-2a). Let us now conside a moe complex example. Example Poblem 2-7 Thévenin and Noton Theoems: n Example with oltage and Cuent Souces. Find the Thévenin and Noton equivalents of the two teminal netwok shown within the dashed outline of Figue 2-1a. zeo souces 1 Ω 1 Ω 1 Ω P 1 Ω 1 R TH = o (a) ➁ 2 shot output 1 Ω P 1 Ω N 2 (b) (d) 2 TH N The open-cicuit voltage of Figue 2-2a is automatically OC = TH = ε. Zeoing the voltage souce and calculating the effective esistance between the output teminals we get 6 3 R TH = = R N. (c) (e) Now shoting the output and calculating the shot cicuit cuent we have Figue 2-1. The Thévenin and Noton equivalent cicuits of (a) ae shown in (c) and (e), espectively. SC = N = ε. Thus the equivalents ae as dawn in Figue The steps involved in finding the Thévenin equivalent ae sketched in Figues 2-1a, b and c, fo the Noton equivalent in Figues 2-1a, d and e. N2-7

8 Thévenin Equivalent pplying KCR to point P in Figue 2-1a when is open we get 2 = 0, so that = 2. pplying KR aound loops 1 and 2 we get (2)1 0 = 0 and 0 0 TH = 0. Solving these equations we get TH = 6. Now zeoing the souces of Figue 2-1a we get the cicuit in Figue 2-1b. Thus R TH = 2Ω. Thus the Thévenin equivalent is as given in Figue 2-1c. Noton Equivalent y shot-cicuiting the output of Figue 2-1a we get the cicuit in Figue 2-1d. The shot cicuit cuent is the Noton cuent N. Theefoe applying KCR to node P we see that the cuent deliveed by the battey is N 2. pplying KR aound the loop (the peiphey of Figue 2-1d) we get ( N 2)1 ( N )1 = 0. Fom which N = 3. Fom the Thévenin equivalent R N = R TH = 2Ω. The Noton equivalent is as shown in Figue 2-1e. Example Poblem 2-8 Finding a Thévenin Equivalent Expeimentally Explain how the Thévenin equivalent of an active DC black box might be found expeimentally. This is done using a voltmete with a lage intenal esistance (DMM) and a vaiable esisto (Figue 2-15). With R L disconnected (o of vey lage value) use the voltmete to measue the open-cicuit voltage, TH. Connect the load esisto and vay it until the output voltage o ead on the mete dops to 1/2 TH. The coesponding value of R L then equals R TH. You will follow this pocedue in Expeiment 01. DC lack ox DMM Figue setup fo finding the Thévenin equivalent of a DC black box expeimentally. The pocedue just descibed applies equally well to any kind of active black box fom a battey to a powe supply. t does not wok fo a passive element like a d sonval mete movement. 6 R L Pactice Poblems Note that you can deive the Thévenin equivalent fom the Noton equivalent and vice vesa. Fo example, in Figue 2-1c you can find the Noton cuent by calculating the cuent when is shoted. Convesely, the Thévenin voltage is the voltage between and acoss the esisto in Figue 2-1e. You should keep both appoaches in mind when solving poblems of this kind; you can often find a Noton equivalent easie by fist solving fo the Thévenin equivalent and vice vesa. N2-8 6 method of doing this is descibed in Expeiment 01, DC Cicuits and Measuements.

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