Voltage and Current Laws

Transcription

1 CHAPTER 3 Voltage and Current Laws KEY CONCEPTS INTRODUCTION In Chap. 2 we were ntroduced to ndependent voltage and current sources, dependent sources, and resstors. We dscovered that dependent sources come n four varetes, and are controlled by a voltage or current whch exsts elsewhere. Once we know the voltage across a resstor, we know ts current (and vce versa); ths s not the case for sources, however. In general, crcuts must be analyzed to determne a complete set of voltages and currents. Ths turns out to be reasonably straghtforward, and only two smple laws are needed n addton to Ohm s law. These new laws are Krchhoff s current law (KCL) and Krchhoff s voltage law (KVL), and they are smply restatements of charge and energy conservaton, respectvely. They apply to any crcut we wll ever encounter, although n later chapters we wll learn more effcent technques for specfc types of stuatons. 3.1 NODES, PATHS, LOOPS, AND BRANCHES We now focus our attenton on the current-voltage relatonshps n smple networks of two or more crcut elements. The elements wll be connected by wres (sometmes referred to as leads ), whch have zero resstance. Snce the network then appears as a number of smple elements and a set of connectng leads, t s called a lumpedparameter network. A more dffcult analyss problem arses when we are faced wth a dstrbuted-parameter network, whch contans an essentally nfnte number of vanshngly small elements. We wll concentrate on lumped-parameter networks n ths text. New Crcut Terms: Node, Path, Loop, and Branch Krchhoff s Current Law (KCL) Krchhoff s Voltage Law (KVL) Analyss of Basc Seres and Parallel Crcuts Combnaton of Seres and Parallel Sources Reducton of Seres and Parallel Resstor Combnatons Voltage and Current Dvson Ground Connectons 39

2 40 CHAPTER 3 VOLTAGE AND CURRENT LAWS In crcuts assembled n the real world, the wres wll always have fnte resstance. However, ths resstance s typcally so small compared to other resstances n the crcut that we can neglect t wthout ntroducng sgnfcant error. In our dealzed crcuts, we wll therefore refer to zero resstance wres from now on A pont at whch two or more elements have a common connecton s called a node. For example, Fg. 3.1a shows a crcut contanng three nodes. Sometmes networks are drawn so as to trap an unwary student nto belevng that there are more nodes present than s actually the case. Ths occurs when a node, such as node 1 n Fg. 3.1a, s shown as two separate junctons connected by a (zero-resstance) conductor, as n Fg. 3.1b. However, all that has been done s to spread the common pont out nto a common zero-resstance lne. Thus, we must necessarly consder all of the perfectly conductng leads or portons of leads attached to the node as part of the node. Note also that every element has a node at each of ts ends. Suppose that we start at one node n a network and move through a smple element to the node at the other end. We then contnue from that node through a dfferent element to the next node, and contnue ths movement untl we have gone through as many elements as we wsh. If no node was encountered more than once, then the set of nodes and elements that we have passed through s defned as a path. If the node at whch we started s the same as the node on whch we ended, then the path s, by defnton, a closed path or a loop. For example, n Fg. 3.1a, f we move from node 2 through the current source to node 1, and then through the upper rght resstor to node 3, we have establshed a path; snce we have not contnued on to node 2 agan, we have not made a loop. If we proceeded from node 2 through the current source to node 1, down through the left resstor to node 2, and then up through the central resstor to node 1 agan, we do not have a path, snce a node (actually two nodes) was encountered more than once; we also do not have a loop, because a loop must be a path. Another term whose use wll prove convenent s branch. We defne a branch as a sngle path n a network, composed of one smple element and the node at each end of that element. Thus, a path s a partcular collecton of branches. The crcut shown n Fg. 3.1a and b contans fve branches. FIGURE 3.1 A crcut contanng three nodes and fve branches. Node 1 s redrawn to look lke two nodes; t s stll one node. 3.2 KIRCHHOFF S CURRENT LAW We are now ready to consder the frst of the two laws named for Gustav Robert Krchhoff (two h s and two f s), a German unversty professor who was born about the tme Ohm was dong hs expermental work. Ths axomatc law s called Krchhoff s current law (abbrevated KCL), and t smply states that The algebrac sum of the currents enterng any node s zero. Ths law represents a mathematcal statement of the fact that charge cannot accumulate at a node. A node s not a crcut element, and t certanly cannot store, destroy, or generate charge. Hence, the currents must sum to zero. A hydraulc analogy s sometmes useful here: for example, consder three water ppes joned n the shape of a Y. We defne three currents as flowng nto each of the three ppes. If we nsst that water s always flowng, then obvously we cannot have three postve water currents, or the ppes would burst. Ths s a result of our defnng currents ndependent of

3 SECTION 3.2 KIRCHHOFF S CURRENT LAW 41 the drecton that water s actually flowng. Therefore, the value of ether one or two of the currents as defned must be negatve. Consder the node shown n Fg The algebrac sum of the four currents enterng the node must be zero: A B ( C ) ( D ) = 0 However, the law could be equally well appled to the algebrac sum of the currents leavng the node: ( A ) ( B ) C D = 0 We mght also wsh to equate the sum of the currents havng reference arrows drected nto the node to the sum of those drected out of the node: D A FIGURE 3.2 Example node to llustrate the applcaton of Krchhoff s current law. B C A B = C D whch smply states that the sum of the currents gong n must equal the sum of the currents gong out. EXAMPLE 3.1 For the crcut n Fg. 3.3a, compute the current through resstor R 3 f t s known that the voltage source supples a current of 3 A. Identfy the goal of the problem. The current through resstor R 3, labeled as on the crcut dagram. Collect the known nformaton. The node at the top of R 3 s connected to four branches. Two of these currents are clearly labeled: 2 A flows out of the node nto R 2, and 5 A flows nto the node from the current source. We are told the current out of the 10 V source s 3 A. Devse a plan. If we label the current through (Fg. 3.3b), we may wrte a KCL equaton at the top node of resstors R 2 and R 3. Construct an approprate set of equatons. Summng the currents flowng nto the node: R1 2 5 = 0 The currents flowng nto ths node are shown n the expanded dagram of Fg. 3.3c for clarty. Determne f addtonal nformaton s requred. We have one equaton but two unknowns, whch means we need to obtan an addtonal equaton. At ths pont, the fact that we know the 10 V source s supplyng 3 A comes n handy: KCL shows us that ths s also the current R1. Attempt a soluton. Substtutng, we fnd that = = 6 A. Verfy the soluton. Is t reasonable or expected? It s always worth the effort to recheck our work. Also, we can attempt to evaluate whether at least the magntude of the soluton s 10 V 10 V R1 2 A R 2 R 3 2 A 2 A R 2 R 3 R1 ( R1 2 A) R 2 R 3 (c) 5 A FIGURE 3.3 Smple crcut for whch the current through resstor R 3 s desred. The current through resstor s labeled so that a KCL equaton can be wrtten. (c) The currents nto the top node of R 3 are redrawn for clarty. 5 A 5 A 5 A (Contnued on next page)

4 42 CHAPTER 3 VOLTAGE AND CURRENT LAWS reasonable. In ths case, we have two sources one supples 5 A, and the other supples 3 A. There are no other sources, ndependent or dependent. Thus, we would not expect to fnd any current n the crcut n excess of 8 A. PRACTICE 3.1 Count the number of branches and nodes n the crcut n Fg If x = 3 A and the 18 V source delvers 8 A of current, what s the value of R A? (Hnt: You need Ohm s law as well as KCL.) 5 13 A 18 V R A 6 v x x FIGURE 3.4 Ans: 5 branches, 3 nodes, 1. A compact expresson for Krchhoff s current law s N n = 0 [1] n=1 whch s just a shorthand statement for N = 0 [2] When Eq. [1] or Eq. [2] s used, t s understood that the N current arrows are ether all drected toward the node n queston, or are all drected away from t. v 1 A 1 v 2 FIGURE 3.5 The potental dfference between ponts A and B s ndependent of the path selected. 2 B 3 C v KIRCHHOFF S VOLTAGE LAW Current s related to the charge flowng through a crcut element, whereas voltage s a measure of potental energy dfference across the element. There s a sngle unque value for any voltage n crcut theory. Thus, the energy requred to move a unt charge from pont A to pont B n a crcut must have a value ndependent of the path chosen to get from A to B (there s often more than one such path). We may assert ths fact through Krchhoff s voltage law (abbrevated KVL): The algebrac sum of the voltages around any closed path s zero. In Fg. 3.5, f we carry a charge of 1 C from A to B through element 1, the reference polarty sgns for v 1 show that we do v 1 joules of work. 1 Now (1) Note that we chose a 1 C charge for the sake of numercal convenence: therefore, we dd (1 C)(v 1 J/C) = v 1 joules of work.

5 SECTION 3.3 KIRCHHOFF S VOLTAGE LAW 43 f, nstead, we choose to proceed from A to B va node C, then we expend (v 2 v 3 ) joules of energy. The work done, however, s ndependent of the path n a crcut, and so any route must lead to the same value for the voltage. In other words, v 1 = v 2 v 3 [3] It follows that f we trace out a closed path, the algebrac sum of the voltages across the ndvdual elements around t must be zero. Thus, we may wrte or, more compactly, v 1 v 2 v 3 v N = 0 N v n = 0 [4] n=1 We can apply KVL to a crcut n several dfferent ways. One method that leads to fewer equaton-wrtng errors than others conssts of movng mentally around the closed path n a clockwse drecton and wrtng down drectly the voltage of each element whose () termnal s entered, and wrtng down the negatve of every voltage frst met at the ( ) sgn. Applyng ths to the sngle loop of Fg. 3.5, we have v 1 v 2 v 3 = 0 whch agrees wth our prevous result, Eq. [3]. In the crcut of Fg. 3.6, fnd v x and x. EXAMPLE 3.2 We know the voltage across two of the three elements n the crcut. Thus, KVL can be appled mmedately to obtan v x. Begnnng wth the bottom node of the 5 V source, we apply KVL clockwse around the loop: 5 7 v x = 0 so v x = 12 V. KCL apples to ths crcut, but only tells us that the same current ( x ) flows through all three elements. We now know the voltage across the 100 resstor, however. Invokng Ohm s law, x = v x 100 = 12 A = 120 ma 100 PRACTICE 3.2 Determne x and v x n the crcut of Fg Ans: v x = 4 V; x = 400 ma. 5 V 7 V 100 v x FIGURE 3.6 A smple crcut wth two voltage sources and a sngle resstor. 3 V x FIGURE V x 10 v x

6 44 CHAPTER 3 VOLTAGE AND CURRENT LAWS EXAMPLE 3.3 In the crcut of Fg. 3.8 there are eght crcut elements. Fnd v R2 (the voltage across R 2 ) and the voltage labeled v x. Ponts b and c, as well as the wre between them, are all part of the same node. The best approach for fndng v R2 s to look for a loop to whch we can apply KVL. There are several optons, but the leftmost loop offers a straghtforward route, as two of the voltages are clearly specfed. Thus, we fnd v R2 by wrtng a KVL equaton around the loop on the left, startng at pont c: 4 36 v R2 = 0 whch leads to v R2 = 32 V. 36 V 12 V a 14 V v 2 v s1 4 V R 2 v R2 v x v R1 c b FIGURE 3.8 A crcut wth eght elements for whch we desre v R2 and v x. To fnd v x, we mght thnk of ths as the (algebrac) sum of the voltages across the three elements on the rght. However, snce we do not have values for these quanttes, such an approach would not lead to a numercal answer. Instead, we apply KVL begnnng at pont c, movng up and across the top to a, through v x to b, and through the conductng lead to the startng pont: v x = 0 so that v x = 6V An alternatve approach: Knowng v R2, we mght have taken the shortcut through R 2 : v x = 0 yeldng v x = 6 V once agan. PRACTICE 3.3 For the crcut of Fg. 3.9, determne v R2 and v 2,fv R1 = 1V. 8 V c 12 V FIGURE 3.9 R 2 v R2 Ans: 4 V; 8 V. 7 V 9 V a v x b v 2 3 V v R1

7 SECTION 3.3 KIRCHHOFF S VOLTAGE LAW 45 As we have just seen, the key to correctly analyzng a crcut s to frst methodcally label all voltages and currents on the dagram. Ths way, carefully wrtten KCL or KVL equatons wll yeld correct relatonshps, and Ohm s law can be appled as necessary f more unknowns than equatons are obtaned ntally. We llustrate these prncples wth a more detaled example. Determne v x n the crcut of Fg. 3.10a. EXAMPLE A V 10 2 v x x 5 A v 8 v 4 60 V v v x x 10 2 FIGURE 3.10 A crcut for whch v x s to be determned usng KVL. Crcut wth voltages and currents labeled. We begn by labelng voltages and currents on the rest of the elements n the crcut (Fg. 3.10b). Note that v x appears across the 2 resstor and the source x as well. If we can obtan the current through the 2 resstor, Ohm s law wll yeld v x. Wrtng the approprate KCL equaton, we see that 2 = 4 x Unfortunately, we do not have values for any of these three quanttes. Our soluton has (temporarly) stalled. Snce we were gven the current flowng from the 60 V source, perhaps we should consder startng from that sde of the crcut. Instead of fndng v x usng 2, t mght be possble to fnd v x drectly usng KVL. We can wrte the followng KVL equatons: 60 v 8 v 10 = 0 and v 10 v 4 v x = 0 [5] Ths s progress: we now have two equatons n four unknowns, an mprovement over one equaton n whch all terms were unknown. In fact, we know that v 8 = 40 V through Ohm s law, as we were told that 5 A flows through the 8 resstor. Thus, v 10 = = 20 V, (Contnued on next page)

8 46 CHAPTER 3 VOLTAGE AND CURRENT LAWS so Eq. [5] reduces to v x = 20 v 4 If we can determne v 4, the problem s solved. The best route to fndng a numercal value for the voltage v 4 n ths case s to employ Ohm s law, whch requres a value for 4. From KCL, we see that 4 = 5 10 = 5 v = = 3 so that v 4 = (4)(3) = 12 V and hence v x = = 8 V. PRACTICE 3.4 Determne v x n the crcut of Fg A V 10 2 v x x FIGURE 3.11 Ans: v x = 12.8 V. v s1 v s1 v R1 v s2 R 2 FIGURE 3.12 A sngle-loop crcut wth four elements. The crcut model wth source voltages and resstance values gven. (c) Current and voltage reference sgns have been added to the crcut. (c) v s2 v R2 R THE SINGLE-LOOP CIRCUIT We have seen that repeated use of KCL and KVL n conjuncton wth Ohm s law can be appled to nontrval crcuts contanng several loops and a number of dfferent elements. Before proceedng further, ths s a good tme to focus on the concept of seres (and, n the next secton, parallel) crcuts, as they form the bass of any network we wll encounter n the future. All of the elements n a crcut that carry the same current are sad to be connected n seres. As an example, consder the crcut of Fg The 60 V source s n seres wth the 8 resstor; they carry the same 5 A current. However, the 8 resstor s not n seres wth the 4 resstor; they carry dfferent currents. Note that elements may carry equal currents and not be n seres; two 100 W lght bulbs n neghborng houses may very well carry equal currents, but they certanly do not carry the same current and are not connected n seres. Fgure 3.12a shows a smple crcut consstng of two batteres and two resstors. Each termnal, connectng lead, and solder glob s assumed to have zero resstance; together they consttute an ndvdual node of the crcut dagram n Fg. 3.12b. Both batteres are modeled by deal voltage sources; any nternal resstances they may have are assumed to be small enough to neglect. The two resstors are assumed to be deal (lnear) resstors. We seek the current through each element, the voltage across each element, and the power absorbed by each element. Our frst step n the analyss s the assumpton of reference drectons for the unknown currents. Arbtrarly, let us select a clockwse current whch flows out of the upper termnal of the voltage source on the left. Ths choce s ndcated by an arrow labeled at that pont n the crcut, as shown n Fg. 3.12c. A trval

9 SECTION 3.4 THE SINGLE-LOOP CIRCUIT 47 applcaton of Krchhoff s current law assures us that ths same current must also flow through every other element n the crcut; we emphasze ths fact ths one tme by placng several other current symbols about the crcut. Our second step n the analyss s a choce of the voltage reference for each of the two resstors. The passve sgn conventon requres that the resstor current and voltage varables be defned so that the current enters the termnal at whch the postve voltage reference s located. Snce we already (arbtrarly) selected the current drecton,v R1 andv R2 are defned as n Fg. 3.12c. The thrd step s the applcaton of Krchhoff s voltage law to the only closed path. Let us decde to move around the crcut n the clockwse drecton, begnnng at the lower left corner, and to wrte down drectly every voltage frst met at ts postve reference, and to wrte down the negatve of every voltage encountered at the negatve termnal. Thus, v s1 v R1 v s2 v R2 = 0 [6] We then apply Ohm s law to the resstve elements: Substtutng nto Eq. [6] yelds v R1 = and v R2 = R 2 v s1 v s2 R 2 = 0 Snce s the only unknown, we fnd that = v s1 v s2 R 2 The voltage or power assocated wth any element may now be obtaned by applyng v = R, p = v, or p = 2 R. PRACTICE 3.5 In the crcut of Fg. 3.12b, v s1 = 120 V,v s2 = 30 V, = 30, and R 2 = 15. Compute the power absorbed by each element. Ans: p 120V = 240 W; p 30V =60 W; p 30 = 120 W; p 15 = 60 W. Compute the power absorbed n each element for the crcut shown n Fg. 3.13a. EXAMPLE V 30 2v A 15 v A 30 v V 2v A 15 v A FIGURE 3.13 A sngle-loop crcut contanng a dependent source. The current and voltage v 30 are assgned. (Contnued on next page)

10 48 CHAPTER 3 VOLTAGE AND CURRENT LAWS 30 v x 12 V 8 7 FIGURE 3.14 A smple loop crcut. 4v x We frst assgn a reference drecton for the current and a reference polarty for the voltage v 30 as shown n Fg. 3.13b. There s no need to assgn a voltage to the 15 resstor, snce the controllng voltage v A for the dependent source s already avalable. (It s worth notng, however, that the reference sgns for v A are reversed from those we would have assgned based on the passve sgn conventon.) Ths crcut contans a dependent voltage source, the value of whch remans unknown untl we determne v A. However, ts algebrac value 2v A can be used n the same fashon as f a numercal value were avalable. Thus, applyng KVL around the loop: 120 v 30 2v A v A = 0 [7] Usng Ohm s law to ntroduce the known resstor values: v 30 = 30 and v A = 15 Note that the negatve sgn s requred snce flows nto the negatve termnal of v A. Substtutng nto Eq. [7] yelds and so we fnd that = 0 = 8A Computng the power absorbed by each element: p 120v = (120)( 8) = 960 W p 30 = (8) 2 (30) = 1920 W p dep = (2v A )(8) = 2[( 15)(8)](8) = 1920 W p 15 = (8) 2 (15) = 960 W PRACTICE 3.6 In the crcut of Fg. 3.14, fnd the power absorbed by each of the fve elements n the crcut. Ans: (CW from left) W, 1.92 W, W, W, W. In the precedng example and practce problem, we were asked to compute the power absorbed by each element of a crcut. It s dffcult to thnk of a stuaton, however, n whch all of the absorbed power quanttes of a crcut would be postve, for the smple reason that the energy must come from somewhere. Thus, from smple conservaton of energy, we expect that the sum of the absorbed power for each element of a crcut should be zero. In

11 SECTION 3.5 THE SINGLE-NODE-PAIR CIRCUIT 49 other words, at least one of the quanttes should be negatve (neglectng the trval case where the crcut s not operatng). Stated another way, the sum of the suppled power for each element should be zero. More pragmatcally, the sum of the absorbed power equals the sum of the suppled power, whch seems reasonable enough at face value. Let s test ths wth the crcut of Fg from Example 3.5, whch conssts of two sources (one dependent and one ndependent) and two resstors. Addng the power absorbed by each element, we fnd p absorbed = = 0 all elements In realty (our ndcaton s the sgn assocated wth the absorbed power) the 120 V source supples 960 W, and the dependent source supples 1920 W. Thus, the sources supply a total of = 2880 W. The resstors are expected to absorb postve power, whch n ths case sums to a total of = 2880 W. Thus, f we take nto account each element of the crcut, pabsorbed = p suppled as we expect. Turnng our attenton to Practce Problem 3.6, the soluton to whch the reader mght want to verfy, we see that the absorbed powers sum to = 0. Interestngly enough, the 12 V ndependent voltage source s absorbng 1.92 W, whch means t s dsspatng power, not supplyng t. Instead, the dependent voltage source appears to be supplyng all the power n ths partcular crcut. Is such a thng possble? We usually expect a source to supply postve power, but snce we are employng dealzed sources n our crcuts, t s n fact possble to have a net power flow nto any source. If the crcut s changed n some way, the same source mght then be found to supply postve power. The result s not known untl a crcut analyss has been completed. 3.5 THE SINGLE-NODE-PAIR CIRCUIT The companon of the sngle-loop crcut dscussed n Sec. 3.4 s the snglenode-par crcut, n whch any number of smple elements are connected between the same par of nodes. An example of such a crcut s shown n Fg. 3.15a. KVL forces us to recognze that the voltage across each branch s the same as that across any other branch. Elements n a crcut havng a common voltage across them are sad to be connected n parallel. Fnd the voltage, current, and power assocated wth each element n the crcut of Fg. 3.15a. We frst defne a voltage v and arbtrarly select ts polarty as shown n Fg. 3.15b. Two currents, flowng n the resstors, are selected n conformance wth the passve sgn conventon, as shown n Fg. 3.15b. (Contnued on next page) EXAMPLE 3.6

12 50 CHAPTER 3 VOLTAGE AND CURRENT LAWS 120 A 1 30 A 1 30 R A v A 1 15 R FIGURE 3.15 A sngle-node-par crcut. A voltage and two currents are assgned. Determnng ether current 1 or 2 wll enable us to obtan a value for v. Thus, our next step s to apply KCL to ether of the two nodes n the crcut. Equatng the algebrac sum of the currents leavng the upper node to zero: = 0 Wrtng both currents n terms of the voltage v usng Ohm s law we obtan 1 = 30v and 2 = 15v v 30 15v = 0 Solvng ths equaton for v results n and nvokng Ohm s law then gves v = 2V 1 = 60 A and 2 = 30 A The absorbed power n each element can now be computed. In the two resstors, p R1 = 30(2) 2 = 120 W and p R2 = 15(2) 2 = 60 W and for the two sources, p 120A = 120( 2) = 240 W and p 30A = 30(2) = 60 W Snce the 120 A source absorbs negatve 240 W, t s actually supplyng power to the other elements n the crcut. In a smlar fashon, we fnd that the 30 A source s actually absorbng power rather than supplyng t. PRACTICE 3.7 Determne v n the crcut of Fg A 10 v 1 A 10 6 A FIGURE 3.16 Ans: 50 V.

13 SECTION 3.6 SERIES AND PARALLEL CONNECTED SOURCES 51 Determne the value of v and the power suppled by the ndependent current source n Fg EXAMPLE 3.7 x 6 6 k 2 x v 24 ma 2 k FIGURE 3.17 A voltage v and a current 6 are assgned n a sngle-node-par crcut contanng a dependent source. By KCL, the sum of the currents leavng the upper node must be zero, so that 6 2 x x = 0 Agan, note that the value of the dependent source (2 x ) s treated the same as any other current would be, even though ts exact value s not known untl the crcut has been analyzed. We next apply Ohm s law to each resstor: 6 = v 6000 and x = v 2000 Therefore, ( ) ( ) v v v = and so v = (600)(0.024) = 14.4 V. Any other nformaton we may want to fnd for ths crcut s now easly obtaned, usually n a sngle step. For example, the power suppled by the ndependent source s p 24 = 14.4(0.024) = W (345.6mW). PRACTICE 3.8 For the sngle-node-par crcut of Fg. 3.18, fnd A, B, and C. A B C 5.6 A v x v x 9 2 A FIGURE 3.18 Ans: 3 A; 5.4 A; 6 A. 3.6 SERIES AND PARALLEL CONNECTED SOURCES It turns out that some of the equaton wrtng that we have been dong for seres and parallel crcuts can be avoded by combnng sources. Note, however, that all the current, voltage, and power relatonshps n the remander of the crcut wll be unchanged. For example, several voltage

14 52 CHAPTER 3 VOLTAGE AND CURRENT LAWS v 1 v 2 = v 1 v 2 v = v 3 FIGURE 3.19 Seres-connected voltage sources can be replaced by a sngle source. Parallel current sources can be replaced by a sngle source. sources n seres may be replaced by an equvalent voltage source havng a voltage equal to the algebrac sum of the ndvdual sources (Fg. 3.19a). Parallel current sources may also be combned by algebracally addng the ndvdual currents, and the order of the parallel elements may be rearranged as desred (Fg. 3.19b). EXAMPLE 3.8 Determne the current n the crcut of Fg. 3.20a by frst combnng the sources nto a sngle equvalent voltage source. To be able to combne the voltage sources, they must be n seres. Snce the same current () flows through each, ths condton s satsfed. Startng from the bottom left-hand corner and proceedng clockwse, = 16 V so we may replace the four voltage sources wth a sngle 16 V source havng ts negatve reference as shown n Fg. 3.20b. KVL combned wth Ohm s law then yelds or = 0 = 16 = 50 ma 320 We should note that the crcut n Fg. 3.20c s also equvalent, a fact easly verfed by computng V 3 V 5 V 16 V 16 V V (c) FIGURE 3.20

15 SECTION 3.6 SERIES AND PARALLEL CONNECTED SOURCES 53 PRACTICE 3.9 Determne the current n the crcut of Fg after frst replacng the four sources wth a sngle equvalent source V 5 V 1 V 3 V FIGURE Ans: 54 A. EXAMPLE 3.9 Determne the voltage v n the crcut of Fg. 3.22a by frst combnng the sources nto a sngle equvalent current source. The sources may be combned f the same voltage appears across each one, whch we can easly verfy s the case. Thus, we create a new source, arrow pontng upward nto the top node, by addng the currents that flow nto that node: = 3 A One equvalent crcut s shown n Fg. 3.22b. KCL then allows us to wrte 3 v 5 v 5 = 0 Solvng, we fnd v = 7.5 V. Another equvalent crcut s shown n Fg. 3.22c. 2.5 A 5 v 2.5 A 5 3 A 3 A 5 v 5 5 v 3 A 5 (c) FIGURE 3.22 (Contnued on next page)

16 54 CHAPTER 3 VOLTAGE AND CURRENT LAWS PRACTICE 3.10 Determne the voltage v n the crcut of Fg after frst replacng the three sources wth a sngle equvalent source. 5 A 10 v 1 A 10 6 A FIGURE 3.23 Ans: 50 V. To conclude the dscusson of parallel and seres source combnatons, we should consder the parallel combnaton of two voltage sources and the seres combnaton of two current sources. For nstance, what s the equvalent of a 5 V source n parallel wth a 10 V source? By the defnton of a voltage source, the voltage across the source cannot change; by Krchhoff s voltage law, then, 5 equals 10 and we have hypotheszed a physcal mpossblty. Thus, deal voltage sources n parallel are permssble only when each has the same termnal voltage at every nstant. In a smlar way, two current sources may not be placed n seres unless each has the same current, ncludng sgn, for every nstant of tme. EXAMPLE 3.10 Determne whch of the crcuts of Fg are vald. The crcut of Fg. 3.24a conssts of two voltage sources n parallel. The value of each source s dfferent, so ths crcut volates KVL. For example, f a resstor s placed n parallel wth the 5 V source, t s also n parallel wth the 10 V source. The actual voltage across t s therefore ambguous, and clearly the crcut cannot be constructed as ndcated. If we attempt to buld such a crcut n real lfe, we wll fnd t mpossble to locate deal voltage sources all real-world sources have an nternal resstance. The presence of such resstance allows a voltage dfference between the two real sources. Along these lnes, the crcut of Fg. 3.24b s perfectly vald. R 1 A 5 V 10 V 2 V 14 V 1 A R FIGURE 3.24 to (c) Examples of crcuts wth multple sources, some of whch volate Krchhoff s laws. (c)

17 SECTION 3.7 RESISTORS IN SERIES AND PARALLEL 55 The crcut of Fg. 3.24c volates KCL: t s unclear what current actually flows through the resstor R. PRACTICE 3.11 Determne whether the crcut of Fg volates ether of Krchhoff s laws. 5 A 3 A R FIGURE 3.25 Ans: No. If the resstor were removed, however, the resultng crcut would. 3.7 RESISTORS IN SERIES AND PARALLEL It s often possble to replace relatvely complcated resstor combnatons wth a sngle equvalent resstor. Ths s useful when we are not specfcally nterested n the current, voltage, or power assocated wth any of the ndvdual resstors n the combnatons. All the current, voltage, and power relatonshps n the remander of the crcut wll be unchanged. Consder the seres combnaton of N resstors shown n Fg. 3.26a. We want to smplfy the crcut wth replacng the N resstors wth a sngle resstor R eq so that the remander of the crcut, n ths case only the voltage source, does not realze that any change has been made. The current, voltage, and power of the source must be the same before and after the replacement. Frst, apply KVL: and then Ohm s law: v s = v 1 v 2 v N v s = R 2 R N = ( R 2 R N ) Now compare ths result wth the smple equaton applyng to the equvalent crcut shown n Fg. 3.26b: v s = R eq Helpful Tp: Inspecton of the KVL equaton for any seres crcut wll show that the order n whch elements are placed n such a crcut makes no dfference. R 2 R N v s v 1 v 2 v N v s R eq FIGURE 3.26 Seres combnaton of N resstors. Electrcally equvalent crcut.

18 56 CHAPTER 3 VOLTAGE AND CURRENT LAWS Thus, the value of the equvalent resstance for N seres resstors s R eq = R 2 R N [8] We are therefore able to replace a two-termnal network consstng of N seres resstors wth a sngle two-termnal element R eq that has the same v- relatonshp. It should be emphaszed agan that we mght be nterested n the current, voltage, or power of one of the orgnal elements. For example, the voltage of a dependent voltage source may depend upon the voltage across R 3. Once R 3 s combned wth several seres resstors to form an equvalent resstance, then t s gone and the voltage across t cannot be determned untl R 3 s dentfed by removng t from the combnaton. In that case, t would have been better to look ahead and not make R 3 a part of the combnaton ntally. EXAMPLE 3.11 Use resstance and source combnatons to determne the current n Fg. 3.27a and the power delvered by the 80 V source. We frst nterchange the element postons n the crcut, beng careful to preserve the proper sense of the sources, as shown n Fg. 3.27b. The V 30 V 8 20 V 20 V V 30 V V 30 (c) FIGURE 3.27 A seres crcut wth several sources and resstors. The elements are rearranged for the sake of clarty. (c) A smpler equvalent.

19 SECTION 3.7 RESISTORS IN SERIES AND PARALLEL 57 next step s to then combne the three voltage sources nto an equvalent 90 V source, and the four resstors nto an equvalent 30 resstance, as n Fg. 3.27c. Thus, nstead of wrtng we have smply and so we fnd that = = 0 = 3A In order to calculate the power delvered to the crcut by the 80 V source appearng n the gven crcut, t s necessary to return to Fg. 3.27a wth the knowledge that the current s 3 A. The desred power s then 80 V 3A 240 W. It s nterestng to note that no element of the orgnal crcut remans n the equvalent crcut. PRACTICE 3.12 Determne n the crcut of Fg V V 5 V 5 FIGURE 3.28 Ans: 333 ma. Smlar smplfcatons can be appled to parallel crcuts. A crcut contanng N resstors n parallel, as n Fg. 3.29a, leads to the KCL equaton or Thus, s = 1 2 N s = v v R 2 v R N = v R eq 1 = R eq R 2 R N [9] s v s 1 v 2 R 2 R eq N R N FIGURE 3.29 A crcut wth N resstors n parallel. Equvalent crcut.

20 58 CHAPTER 3 VOLTAGE AND CURRENT LAWS whch can be wrtten as R 1 eq or, n terms of conductances, as = R 1 1 R 1 2 R 1 N G eq = G 1 G 2 G N The smplfed (equvalent) crcut s shown n Fg. 3.29b. A parallel combnaton s routnely ndcated by the followng shorthand notaton: R eq = R 2 R 3 The specal case of only two parallel resstors s encountered farly often, and s gven by Or, more smply, R eq = R 2 1 = 1 1 R 2 R eq = R 2 R 2 [10] The last form s worth memorzng, although t s a common error to attempt to generalze Eq. [10] to more than two resstors, e.g., R eq = R 2 R 3 R 2 R 3 A quck look at the unts of ths equaton wll mmedately show that the expresson cannot possbly be correct. PRACTICE 3.13 Determne v n the crcut of Fg by frst combnng the three current sources, and then the two 10 resstors. 5 A 10 v 1 A 10 6 A FIGURE 3.30 Ans: 50 V.

21 SECTION 3.7 RESISTORS IN SERIES AND PARALLEL 59 Calculate the power and voltage of the dependent source n Fg. 3.31a. EXAMPLE A v x A A v v 2 A 3 6 (c) FIGURE 3.31 A multnode crcut. The two ndependent current sources are combned nto a 2 A source, and the 15 resstor n seres wth the two parallel 6 resstors are replaced wth a sngle 18 resstor. (c) A smplfed equvalent crcut. We wll seek to smplfy the crcut before analyzng t, but take care not to nclude the dependent source snce ts voltage and power characterstcs are of nterest. Despte not beng drawn adjacent to one another, the two ndependent current sources are n fact n parallel, so we replace them wth a 2 A source. The two 6 resstors are n parallel and can be replaced wth a sngle 3 resstor n seres wth the 15 resstor. Thus, the two 6 resstors and the 15 resstor are replaced by an 18 resstor (Fg. 3.31b). No matter how temptng, we should not combne the remanng three resstors; the controllng varable 3 depends on the 3 resstor and so that resstor must reman untouched. The only further smplfcaton, then, s 9 18 = 6, as shown n Fg. 3.31c. (Contnued on next page)

22 60 CHAPTER 3 VOLTAGE AND CURRENT LAWS Applyng KCL at the top node of Fg. 3.31c, we have v 6 = 0 Employng Ohm s law, whch allows us to compute v = = 10 3 A Thus, the voltage across the dependent source (whch s the same as the voltage across the 3 resstor) s v = 3 3 = 10 V The dependent source therefore furnshes v = 10(0.9)(10/3) = 30 W to the remander of the crcut. Now f we are later asked for the power dsspated n the 15 resstor, we must return to the orgnal crcut. Ths resstor s n seres wth an equvalent 3 resstor; a voltage of 10 V s across the 18 total; therefore, a current of 5/9 A flows through the 15 resstor and the power absorbed by ths element s (5/9) 2 (15) or 4.63 W. PRACTICE 3.14 For the crcut of Fg. 3.32, calculate the voltage v x A v x FIGURE 3.32 Ans: V.

23 SECTION 3.8 VOLTAGE AND CURRENT DIVISION 61 R 2 R 3 v s R R 7 R5 R4 R 8 v s R 6 A B R A s R B R C v s R D R E FIGURE 3.33 These two crcut elements are both n seres and n parallel. R 2 and R 3 are n parallel, and and R 8 are n seres. (c) There are no crcut elements ether n seres or n parallel wth one another. (c) Three fnal comments on seres and parallel combnatons mght be helpful. The frst s llustrated by referrng to Fg. 3.33a and askng, Are v s and R n seres or n parallel? The answer s Both. The two elements carry the same current and are therefore n seres; they also enjoy the same voltage and consequently are n parallel. The second comment s a word of cauton. Crcuts can be drawn n such a way as to make seres or parallel combnatons dffcult to spot. In Fg. 3.33b, for example, the only two resstors n parallel are R 2 and R 3, whle the only two n seres are and R 8. The fnal comment s smply that a smple crcut element need not be n seres or parallel wth any other smple crcut element n a crcut. For example, R 4 and R 5 n Fg. 3.33b are not n seres or parallel wth any other smple crcut element, and there are no smple crcut elements n Fg. 3.33c that are n seres or parallel wth any other smple crcut element. In other words, we cannot smplfy that crcut further usng any of the technques dscussed n ths chapter. 3.8 VOLTAGE AND CURRENT DIVISION By combnng resstances and sources, we have found one method of shortenng the work of analyzng a crcut. Another useful shortcut s the applcaton of the deas of voltage and current dvson. Voltage dvson s used to express the voltage across one of several seres resstors n terms of the

24 62 CHAPTER 3 VOLTAGE AND CURRENT LAWS v v 1 R 2 v 2 FIGURE 3.34 An llustraton of voltage dvson. voltage across the combnaton. In Fg. 3.34, the voltage across R 2 s found va KVL and Ohm s law: so or Thus, v = v 1 v 2 = R 2 = ( R 2 ) = v R 2 ( ) v v 2 = R 2 = R 2 R 2 v 2 = and the voltage across s, smlarly, v 1 = R 2 R 2 v R 2 v If the network of Fg s generalzed by removng R 2 and replacng t wth the seres combnaton of R 2, R 3,...,R N, then we have the general result for voltage dvson across a strng of N seres resstors v k = R k R 2 R N v [11] whch allows us to compute the voltage v k that appears across an arbtrary resstor R k of the seres. EXAMPLE 3.13 Determne v x n the crcut of Fg. 3.35a Ω 12 sn t V 6 3 v x 12 sn t V 2 v x FIGURE 3.35 A numercal example llustratng resstance combnaton and voltage dvson. Orgnal crcut. Smplfed crcut. We frst combne the 6 and 3 resstors, replacng them wth (6)(3)/(6 3) = 2. Snce v x appears across the parallel combnaton, our smplfcaton has not lost ths quantty. However, further smplfcaton of the crcut by replacng the seres combnaton of the 4 resstor wth our new 2 resstor would.

25 SECTION 3.8 VOLTAGE AND CURRENT DIVISION 63 Thus, we proceed by smply applyng voltage dvson to the crcut n Fg. 3.35b: 2 v x = (12 sn t) = 4 sn t volts 4 2 PRACTICE 3.15 Use voltage dvson to determne v x n the crcut of Fg v x 10 V FIGURE 3.36 Ans: 2 V. The dual 2 of voltage dvson s current dvson. We are now gven a total current suppled to several parallel resstors, as shown n the crcut of Fg The current flowng through R 2 s v 1 2 R 2 or 2 = v R 2 = ( R 2 ) R 2 = R 2 R 2 R 2 FIGURE 3.37 An llustraton of current dvson. 2 = R 2 [12] and, smlarly, R 2 1 = R 2 [13] Nature has not smled on us here, for these last two equatons have a factor whch dffers subtly from the factor used wth voltage dvson, and some effort s gong to be needed to avod errors. Many students look on the expresson for voltage dvson as obvous and that for current dvson as beng dfferent. It helps to realze that the larger of two parallel resstors always carres the smaller current. For a parallel combnaton of N resstors, the current through resstor R k s 1 k = R k 1 1 R 2 1 R N [14] (2) The prncple of dualty s encountered often n engneerng. We wll consder the topc brefly n Chap. 7 when we compare nductors and capactors.

26 64 CHAPTER 3 VOLTAGE AND CURRENT LAWS Wrtten n terms of conductances, G k k = G 1 G 2 G N whch strongly resembles Eq. [11] for voltage dvson. EXAMPLE sn t V 6 3 FIGURE 3.38 A crcut used as an example of current dvson. The wavy lne n the voltage source symbol ndcates a snusodal varaton wth tme. 3 v x Wrte an expresson for the current through the 3 resstor n the crcut of Fg The total current flowng nto the 3 6 combnaton s (t) = 12 sn t 12 sn t = = 2 sn t A and thus the desred current s gven by current dvson: ( ) 6 3 (t) = (2 sn t) = sn t A Unfortunately, current dvson s sometmes appled when t s not applcable. As one example, let us consder agan the crcut shown n Fg. 3.33c, a crcut that we have already agreed contans no crcut elements that are n seres or n parallel. Wthout parallel resstors, there s no way that current dvson can be appled. Even so, there are too many students who take a quck look at resstors R A and R B and try to apply current dvson, wrtng an ncorrect equaton such as R B A = S R A R B Remember, parallel resstors must be branches between the same par of nodes. PRACTICE 3.16 In the crcut of Fg. 3.39, use resstance combnaton methods and current dvson to fnd 1, 2, and v ma v 3 FIGURE 3.39 Ans: 100 ma; 50 ma; 0.8 V.

27 Up to now, we have been drawng crcut schematcs n a fashon smlar to that of the one shown n Fg. 3.40, where voltages are defned across two clearly marked termnals. Specal care was taken to emphasze the fact that voltage cannot be defned at a sngle pont t s by defnton the dfference n potental between two ponts. However, many schematcs make use of the conventon of takng the earth as defnng zero volts, so that all other voltages are mplctly referenced to ths potental. The concept s often referred to as earth ground, and s fundamentally ted to safety regulatons desgned to prevent fres, fatal electrcal shocks, and related mayhem. The symbol for earth ground s shown n Fg. 3.41a. Snce earth ground s defned as zero volts, t s often convenent to use ths as a common termnal n schematcs. The crcut of Fg s shown redrawn n ths fashon n Fg. 3.42, where the earth ground symbol represents a common node. It s mportant to note that the two crcuts are equvalent n terms of our value for v a (4.5 V n ether case), but are no longer exactly the same. The crcut n Fg s sad to be floatng n that t could for all practcal purposes be nstalled on a crcut board of a satellte n geosynchronous orbt (or on ts way to Pluto). The crcut n Fg. 3.42, however, s somehow physcally connected to the ground through a conductng path. For ths reason, there are two other symbols that are occasonally used to denote a common termnal. Fgure 3.41b shows what s commonly referred to as sgnal ground; there can be (and often s) a large voltage between earth ground and any termnal ted to sgnal ground. The fact that the common termnal of a crcut may or may not be connected by some low-resstance pathway to earth ground can lead to potentally dangerous stuatons. Consder the dagram of Fg. 3.43a, whch depcts an nnocent bystander about to touch a pece of equpment powered by an ac outlet. Only two termnals have been used from the wall socket; the round ground pn 4.7 k PRACTICAL APPLICATION Not the Earth Ground from Geology of the receptacle was left unconnected. The common termnal of every crcut n the equpment has been ted together and electrcally connected to the conductng equpment chasss; ths termnal s often denoted usng the chasss ground symbol of Fg. 3.41c. Unfortunately, a wrng fault exsts, due to ether poor manufacturng or perhaps just wear and tear. At any rate, the chasss s not grounded, so there s a very large resstance between chasss ground and earth ground. A pseudo-schematc (some lberty was taken wth the person s equvalent resstance symbol) of the stuaton s shown n Fg. 3.43b. The electrcal path between the conductng chasss and ground may n fact be the table, whch could represent a resstance of hundreds of megaohms or more. The resstance of the person, however, s many orders of magntude lower. Once the person taps on the equpment to see why t sn t workng properly... well, let s just say not all stores have happy endngs. The fact that ground s not always earth ground can cause a wde range of safety and electrcal nose problems. One example s occasonally encountered n older buldngs, where plumbng orgnally conssted of electrcally conductng copper ppes. In such buldngs, any water ppe was often treated as a low-resstance path to earth ground, and therefore used n many electrcal connectons. However, when corroded ppes are replaced wth more modern and cost-effectve 9 V (c) FIGURE 3.41 Three dfferent symbols used to represent a ground or common termnal: earth ground; sgnal ground; (c) chasss ground. 4.7 k 4.7 k v a 9 V 4.7 k v a FIGURE 3.40 A smple crcut wth a voltage v a defned between two termnals. FIGURE 3.42 The crcut of Fg. 3.40, redrawn usng the earth ground symbol. The rghtmost ground symbol s redundant; t s only necessary to label the postve termnal of v a ; the negatve reference s then mplctly ground, or zero volts. (Contnued on next page)

28 nonconductng PVC ppng, the low-resstance path to earth ground no longer exsts. A related problem occurs when the composton of the earth vares greatly over a partcular regon. In such stuatons, t s possble to actually have two separated buldngs n whch the two earth grounds are not equal, and current can flow as a result. Wthn ths text, the earth ground symbol wll be used exclusvely. It s worth rememberng, however, that not all grounds are created equal n practce. Wall outlet R equpment 115 V R to ground FIGURE 3.43 A sketch of an nnocent person about to touch an mproperly grounded pece of equpment. It s not gong to be pretty. A schematc of an equvalent crcut for the stuaton as t s about to unfold; the person has been represented by an equvalent resstance, as has the equpment. A resstor has been used to represent the nonhuman path to ground. SUMMARY AND REVIEW We began ths chapter by dscussng connectons of crcut elements, and ntroducng the terms node, path, loop, and branch. The next two topcs could be consdered the two most mportant n the entre textbook, namely, Krchhoff s current law (KCL) and Krchhoff s voltage law. The frst s derved from conservaton of charge, and can be thought of n terms of what goes n (current) must come out. The second s based on conservaton of energy, and can be vewed as what goes up (potental) must come down. These two laws allow us to analyze any crcut, lnear or otherwse, provded we have a way of relatng the voltage and current assocated wth passve elements (e.g., Ohm s law for the resstor). In the case of a sngle-loop crcut, the elements are connected n seres and hence each carres the same current. The sngle-node-par crcut, n whch elements are connected n parallel wth one another, s characterzed by a sngle voltage common to each element. Extendng these concepts allowed us to develop a means of smplfyng voltage sources connected n seres, or current sources n parallel; subsequently we obtaned classc expressons for seres and parallel connected resstors. The fnal topc, that of voltage and current dvson, fnds consderable use n the desgn of crcuts where a specfc voltage or current s requred but our choce of source s lmted. Let s conclude wth key ponts of ths chapter to revew, hghlghtng approprate examples.

29 EXERCISES 67 Krchhoff s current law (KCL) states that the algebrac sum of the currents enterng any node s zero. (Examples 3.1, 3.4) Krchhoff s voltage law (KVL) states that the algebrac sum of the voltages around any closed path n a crcut s zero. (Examples 3.2, 3.3) All elements n a crcut that carry the same current are sad to be connected n seres. (Example 3.5) Elements n a crcut havng a common voltage across them are sad to be connected n parallel. (Examples 3.6, 3.7) Voltage sources n seres can be replaced by a sngle source, provded care s taken to note the ndvdual polarty of each source. (Examples 3.8, 3.10) Current sources n parallel can be replaced by a sngle source, provded care s taken to note the drecton of each current arrow. (Examples 3.9, 3.10) A seres combnaton of N resstors can be replaced by a sngle resstor havng the value R eq = R 2 R N. (Example 3.11) A parallel combnaton of N resstors can be replaced by a sngle resstor havng the value 1 R eq = 1 1 R 2 1 R N (Example 3.12) Voltage dvson allows us to calculate what fracton of the total voltage across a seres strng of resstors s dropped across any one resstor (or group of resstors). (Example 3.13) Current dvson allows us to calculate what fracton of the total current nto a parallel strng of resstors flows through any one of the resstors. (Example 3.14) READING FURTHER A dscusson of the prncples of conservaton of energy and conservaton of charge, as well as Krchhoff s laws, can be found n R. Feynman, R. B. Leghton, and M. L. Sands, The Feynman Lectures on Physcs. Readng, Mass.: Addson-Wesley, 1989, pp. 4-1, 4-7, and Detaled dscussons of numerous aspects of groundng practces consstent wth the 2008 Natonal Electrcal Code can be found throughout J. E. McPartland, B. J. McPartland, and F. P. Hartwell, McGraw-Hll s Natonal Electrcal Code 2008 Handbook, 26th ed. New York, McGraw-Hll, EXERCISES 3.1 Nodes, Paths, Loops, and Branches 1. Referrng to the crcut depcted n Fg. 3.44, count the number of nodes; elements; (c) branches. 5 A 14 FIGURE A

30 68 CHAPTER 3 VOLTAGE AND CURRENT LAWS 2. Referrng to the crcut depcted n Fg. 3.45, count the number of nodes; elements; (c) branches A A A B C F FIGURE 3.46 A B E D FIGURE 3.47 C D E G FIGURE For the crcut of Fg. 3.46: Count the number of nodes. In movng from A to B, have we formed a path? Have we formed a loop? (c) In movng from C to F to G, have we formed a path? Have we formed a loop? 4. For the crcut of Fg. 3.46: Count the number of crcut elements. If we move from B to C to D, have we formed a path? Have we formed a loop? (c) If we move from E to D to C to B, have we formed a path? Have we formed a loop? 5. Refer to the crcut of Fg. 3.47, and answer the followng: How many dstnct nodes are contaned n the crcut? How many elements are contaned n the crcut? (c) How many branches does the crcut have? (d) Determne f each of the followng represents a path, a loop, both, or nether: () A to B () B to D to C to E () C to E to D to B to A to C (v) C to D to B to A to C to E 3.2 Krchhoff s Current Law 6. A local restaurant has a neon sgn constructed from 12 separate bulbs; when a bulb fals, t appears as an nfnte resstance and cannot conduct current. In wrng the sgn, the manufacturer offers two optons (Fg. 3.48). From what you ve learned about KCL, whch one should the restaurant owner select? Explan. EAT AT RALPH S FIGURE 3.48 EAT AT RALPH S

31 EXERCISES Referrng to the sngle node dagram of Fg. 3.49, compute: B, f A = 1 A, D = 2 A, C = 3 A, and E = 0; E, f A = 1 A, B = 1 A, C = 1 A, and D = 1 A. A B D C E FIGURE Determne the current labeled I n each of the crcuts of Fg A I V 6 A 3 A I 2 A 9 A 1 5 I 3 A FIGURE 3.50 (c) 9. In the crcut shown n Fg. 3.51, the resstor values are unknown, but the 2 V source s known to be supplyng a current of 7 A to the rest of the crcut. Calculate the current labeled 2. 2 V 2 R 2 R 3 1 A 3 A FIGURE The voltage source n the crcut of Fg has a current of 1 A flowng out of ts postve termnal nto resstor. Calculate the current labeled In the crcut depcted n Fg. 3.53, x s determned to be 1.5 A, and the 9 V source supples a current of 7.6 A (that s, a current of 7.6 A leaves the postve reference termnal of the 9 V source). Determne the value of resstor R A. 12. For the crcut of Fg (whch s a model for the dc operaton of a bpolar juncton transstor based n forward actve regon), I B s measured to be 100 μa. Determne I C and I E. 2 V FIGURE R 2 R 3 I C 3 A 7 A 1.6 A 9 V 5 R A 6 x v x I B 1 k 150I B V 1 I E V 2 R 2 1 k FIGURE 3.53 FIGURE 3.54