Resistive Circuits. Preview The resistor, with resistance R, is an element commonly used in most electric CHAPTER

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1 esste Crcuts 3 CHAPTE Preew The resstor, wth resstance, s an element commonly used n most electrc crcuts. In ths chapter we consder the analyss of crcuts consstng of resstors and sources. In addton to Ohm s law, we need two laws for relatng (1) current flow at connected termnals and () the sum of oltages around a closed crcut path. These two laws were deeloped by Gusta Krchhoff n Usng Krchhoff s laws and Ohm s law, we are able to complete the analyss of resste crcuts and determne the currents and oltages at desred ponts n the crcut. Ths analyss may be accomplshed for crcuts wth both ndependent and dependent sources. 3.0 Preew 3.1 DESIGNCHALLENGE Adjustable Voltage Source 3. Electrc Crcut Applcatons 3.3 Krchhoff s Laws 3.4 A Sngle-Loop Crcut The Voltage Dder 3.5 Parallel esstors and Current Dson 3.6 Seres Voltage Sources and Parallel Current Sources 3.7 Crcut Analyss 3.8 Analyzng esste Crcuts Usng MATLAB 3.9 Verfcaton Example 3.10 DESIGN CHALLENGE SOLUTION Adjustable Voltage Source 3.11 Summary Problems Verfcaton Problems Desgn Problems 57

2 58 esste Crcuts 3.1 DESIGN CHALLENGE ADJUSTABLE VOLTAGE SOUCE A crcut s requred to prode an adjustable oltage. The specfcatons for ths crcut are: 1. It should be possble to adjust the oltage to any alue between 5 V and 5 V. It should not be possble to accdentally obtan a oltage outsde ths range.. The load current wll be neglgble. 3. The crcut should use as lttle power as possble. The aalable components are: 1. Potentometers: resstance alues of 10 k, 0 k, and 50 k are n stock.. A large assortment of standard percent resstors hang alues between 10 and 1M (see Appendx E). 3. Two power supples (oltage sources): one 1-V supply and one 1-V supply; each rated for a maxmum current of 100 ma (mllamps). DESCIBE THE SITUATION AND THE ASSUMPTIONS Fgure shows the stuaton. The oltage s the adjustable oltage. The crcut that uses the output of the crcut beng desgned s frequently called the load. In ths case, the load current s neglgble, so 0. STATE THE GOAL A crcut prodng the adjustable oltage 5V 5V must be desgned usng the aalable components. GENEATE A PLAN Make the followng obseratons. 1. The adjustablty of a potentometer can be used to obtan an adjustable oltage.. Both power supples must be used so that the adjustable oltage can hae both poste and negate alues. 3. The termnals of the potentometer cannot be connected drectly to the power supples because the oltage s not allowed to be as large as 1 V or 1 V. These obseratons suggest the crcut shown n Fgure 3.1-a. The crcut n Fgure 3.1-b s obtaned by usng the smplest model for each component n Fgure 3.1-a. Load current 0 FIGUE The crcut beng desgned prodes an adjustable oltage,, to the load crcut. Crcut beng desgned Load crcut

3 Electrc Crcut Applcatons 59 0 a p (1 a) p 0 p 1 1 V 1 V Load crcut 0 a V 1 V FIGUE 3.1- A proposed crcut for producng the arable oltage, and the equalent crcut after the potentometer s modeled. To complete the desgn, alues need to be specfed for 1,, and p. Then, seeral results need to be checked and adjustments made, f necessary. 1. Can the oltage be adjusted to any alue n the range 5 V to 5 V?. Are the oltage source currents less than 100 ma? Ths condton must be satsfed f the power supples are to be modeled as deal oltage sources. 3. Is t possble to reduce the power absorbed by 1,, and p? ACT ON THE PLAN Takng acton on ths plan requres analyzng the crcut n Fgure 3.1-b. Analyss of ths type of crcut s dscussed n ths chapter. We wll return to ths problem at the end of ths chapter. 3. Electrc Crcut Applcatons Oerseas communcatons hae always been of great mportance to natons. One of the most brllant chapters of the hstory of electrcal technology was the deelopment of underwater electrc cable crcuts. Underwater electrc cables were used to carry electrc telegraph communcatons. In late 185 England and Ireland were connected by cable, and a year later there was a cable between Scotland and Ireland. In June 1853 a cable was strung between England and Holland, a dstance of 115 mles. It was Cyrus Feld and Samuel Morse who saw the potental for a submarne cable across the Atlantc. By 1857 Feld had organzed a frm to complete the transatlantc telegraph cable and ssued a contract for the producton of 500 mles of cable. The cable layng began n June After seeral false starts, a cable was lad across the Atlantc by August 5, Howeer, ths cable faled after only a month of operaton. Another seres of cable-layng projects commenced, and by September 1865 a successful Atlantc cable was n place. Ths cable stretched oer 3000 mles, from England to eastern Canada. There followed a flurry of cable layng. Approxmately 150,000 km (90,000 m) were n use by 1870, lnkng all contnents and all major slands. An example of modern undersea cable s shown n Fgure One of the greatest uses of electrcty n the late 1800s was for electrc ralways. In 1884 the Sprague Electrc alway was ncorporated. Sprague bult an electrc ralway for chmond, Vrgna, n By 190 the horse-drawn street trolley was obsolete, and there were

4 60 esste Crcuts FIGUE 3.-1 Examples of undersea cable. Courtesy of Bell Laboratores. FIGUE 3.- Vew of the Sprague Electrc alway car on the Brooklne branch of the Boston system about Ths electrc ralway branch operates as an electrc trolley ralroad today wth modern electrc cars. Courtesy of General Electrc Company.,576 mles of electrc ralway track n the Unted States. A 1900 electrc ralway s shown n Fgure Krchhoff s Laws An electrc crcut conssts of crcut elements that are connected together. The places where the elements are connected to each other are called nodes. Fgure 3.3-1a shows an electrc crcut that conssts of sx elements connected together at four nodes. It s common practce to draw electrc crcuts usng straght lnes and to poston the elements horzontally or ertcally as shown n Fgure 3.3-1b. The crcut s shown agan n Fgure 3.3-1c, ths tme emphaszng the nodes. Notce that redrawng the crcut usng straght lnes and horzontal and ertcal elements has changed the way that the nodes are represented. In Fgure 3.3-1a, nodes are represented as ponts. In Fgures 3.3-1b,c, nodes are represented usng both ponts and straght-lne segments. The same crcut can be drawn n seeral dfferent ways. One drawng of a crcut mght look much dfferent from another drawng of the same crcut. How can we tell when two crcut drawngs represent the same crcut? Informally, we say that two crcut drawngs represent the same crcut f correspondng elements are connected to correspondng nodes. More formally, we say that crcut drawngs A and B represent the same crcut when the followng three condtons are met. 1. There s a one-to-one correspondence between the nodes of drawng A and the nodes of drawng B. (A one-to-one correspondence s a matchng. In ths one-to-one correspondence, each node n drawng A s matched to exactly one node of drawng B, and ce ersa. The poston of the nodes s not mportant.). There s a one-to-one correspondence between the elements of drawng A and the elements of drawng B. 3. Correspondng elements are connected to correspondng nodes.

5 Krchhoff s Laws a c b (c) 6 6 d FIGUE An electrc crcut. The same crcut, redrawn usng straght lnes and horzontal and ertcal elements. (c) The crcut after labelng the nodes and elements. Example Fgure 3.3- shows four crcut drawngs. Whch of these drawngs, f any, represent the same crcut as the crcut drawng n Fgure 3.3-1c? Soluton The crcut drawng shown n Fgure 3.3-a has fe nodes, labeled r, s, t, u and. The crcut drawng n Fgure 3.3-1c has four nodes. Snce the two drawngs hae dfferent numbers of nodes, there cannot be a one-to-one correspondence between the nodes of the two drawngs. Hence theses drawngs represent dfferent crcuts. The crcut drawng shown n Fgure 3.3-b has four nodes and sx elements, the same numbers of nodes and elements as the crcut drawng n Fgure 3.3-1c. The nodes n Fgure 3.3-b hae been labeled n the same way as the correspondng nodes n Fgure 3.3-1c. For example, node c n Fgure 3.3-b corresponds to node c n Fgure 3.3-1c. The elements n Fgure 3.3-b hae been labeled n the same way as the correspondng elements n Fgure 3.3-1c. For example, element 5 n Fgure 3.3-b corresponds to element 5 n Fgure 3.3-1c. Correspondng elements are ndeed connected to correspondng nodes. For example, element s connected to nodes a and b, both n Fgure 3.3-b and n Fgure 3.3-1c. Consequently, Fgure 3.3-b and Fgure 3.3-1c represent the same crcut. The crcut drawng shown n Fgure 3.3-c has four nodes and sx elements, the same numbers of nodes and elements as the crcut drawng n Fgure 3.3-1c. The nodes and elements n Fgure 3.3-c hae been labeled n the same

6 6 esste Crcuts r s t c a u d b d 5 c b 3 a (c) (d) FIGUE 3.3- Four crcut drawngs. way as the correspondng nodes and elements n Fgure 3.3-1c. Correspondng elements are ndeed connected to correspondng nodes. Therefore Fgure 3.3-c and Fgure 3.3-1c represent the same crcut. The crcut drawng shown n Fgure 3.3-d has four nodes and sx elements, the same numbers of nodes and elements as the crcut drawng n Fgure 3.3-1c. Howeer, the nodes and elements of Fgure 3.3-d cannot be labeled so that correspondng elements of Fgure 3.3-1c are connected to correspondng nodes. (For example, n Fgure 3.3-1c three elements are connected between the same par of nodes, a and b. That does not happen n Fgure 3.3-d.) Consequently, Fgure 3.3-d and Fgure 3.3-1c represent dfferent crcuts. In 1847, Gusta obert Krchhoff, a professor at the Unersty of Berln, formulated two mportant laws that prode the foundaton for analyss of electrc crcuts. These laws are referred to as Krchhoff s current law (KCL) and Krchhoff s oltage law (KVL) n hs honor. Krchhoff s laws are a consequence of conseraton of charge and conseraton of energy. Gusta obert Krchhoff s pctured n Fgure Krchhoff s current law states that the algebrac sum of the currents enterng any node s dentcally zero for all nstants of tme. Krchhoff s current law (KCL): The algebrac sum of the currents nto a node at any nstant s zero. The phrase algebrac sum ndcates that we must take reference drectons nto account as we add up the currents of elements connected to a partcular node. One way to take reference

7 Krchhoff s Laws 63 drectons nto account s to use a plus sgn when the current s drected away from the node and a mnus sgn when the current s drected toward the node. For example, consder the crcut shown n Fgure 3.3-1c. Four elements of ths crcut elements 1,, 3, and 4 are connected to node a. By Krchhoff s current law, the algebrac sum of the element currents 1,, 3, and 4 must be zero. Currents and 3 are drected away from node a, so we wll use a plus sgn for and 3. In contrast, currents 1 and 4 are drected toward node a, so we wll use amnus sgn for 1 and 4. The KCL equaton for node a of Fgure 3.3-1c s (3.3-1) An alternate way of obtanng the algebrac sum of the currents nto a node s to set the sum of all the currents drected away from node equal to the sum of all the currents drected toward that node. Usng ths technque, the KCL equaton for node a of Fgure 3.3-1c s (3.3-) Clearly, Eqs and 3.3- are equalent. Smlarly, the Krchhoff s current law equaton for node b of Fgure 3.3-1c s FIGUE Gusta obert Krchhoff ( ). Krchhoff stated two laws n 1847 regardng the current and oltage n an electrcal crcut. Courtesy of the Smthsonan Insttuton Before we can state Krchhoff s oltage law, we need the defnton of a loop. A loop s aclosed path through a crcut that does not encounter any ntermedate node more than once. For example, startng at node a n Fgure 3.3-1c, we can moe through element 4 to node c, then through element 5 to node d, through element 6 to node b, and fnally through element 3 back to node a. We hae a closed path, and we dd not encounter any of the ntermedate nodes b, c, or d more than once. Consequently, elements 3, 4, 5, and 6 comprse a loop. Smlarly, elements 1, 4, 5, and 6 comprse a loop of the crcut shown n Fgure 3.3-1c. Elements 1 and 3 comprse yet another loop of ths crcut. The crcut has three other loops: elements 1 and, elements and 3, and elements, 4, 5, and 6. We are now ready to state Krchhoff s oltage law. Krchhoff s oltage law (KVL): The algebrac sum of the oltages around any loop n a crcut s dentcally zero for all tme. The phrase algebrac sum ndcates that we must take polarty nto account as we add up the oltages of elements that comprse a loop. One way to take polarty nto account s to moe around the loop n the clockwse drecton whle obserng the polartes of the element oltages. We wrte the oltage wth a plus sgn when we encounter the of the oltage polarty before the. In contrast, we wrte the oltage wth a mnus sgn when we encounter the of the oltage polarty before the. For example, consder the crcut shown n Fgure 3.3-1c. Elements 3, 4, 5, and 6 comprse a loop of the crcut. By Krchhoff s oltage law, the algebrac sum of the element oltages 3, 4, 5, and 6 must be zero. As we moe around the loop n the clockwse drecton, we encounter the of 4 before the, the of 5 before the, the of 6 before the, and the of 3 before the. Consequently, we use a mnus sgn for 3, 5, and 6 and a plus sgn for 4. The KCL equaton for ths loop of Fgure 3.3-1c s Smlarly, the Krchhoff s oltage law equaton for the loop consstng of elements 1, 4, 5, and 6 s The Krchhoff s oltage law equaton for the loop consstng of elements 1 and s 1 0

8 64 esste Crcuts Example 3.3- Consder the crcut shown n Fgure 3.3-4a. Determne the power suppled by element C and the power receed by element D. Soluton Fgure 3.3-4a prodes a alue for the current n element C but not for the oltage,, across element C. The oltage and current of element C gen n Fgure 3.3-4a adhere to the passe conenton, so the product of ths oltage and current s the power receed by element C. Smlarly, Fgure 3.3-4a prodes a alue for the oltage across element D but not for the current,, n element D. The oltage and current of element D gen n Fgure 3.3-4a do not adhere to the passe conenton, so the product of ths oltage and current s the power suppled by element D. We need to determne the oltage,, across element C and the current,, n element D. We wll use Krchhoff s laws to determne alues of and. Frst, we dentfy and label the nodes of the crcut as shown n Fgure 3.3-4b. Apply Krchhoff s oltage law (KVL) to the loop consstng of elements C, D, and B to get ( 4) 6 0 V The alue of the current n element C n Fgure 3.3-4b s 7 A. The oltage and current of element C gen n Fgure 3.3-4b adhere to the passe conenton, so p C (7) ( )(7) 14 W s the power receed by element C. Therefore element C supples 14 W. Next, apply Krchhoff s current law (KCL) at node b to get 7 ( 10) 0 3A The alue of the oltage across element D n Fgure 3.3-4b s 4 V. The oltage and current of element D gen n Fgure 3.3-4b do not adhere to the passe conenton, so the power suppled by element F s gen by p D ( 4) ( 4)(3) 1 W Therefore, element D recees 1 W. 4 V C E 6 V A 3 A 6 V B 4 A 7 A 10 A 4 V D 0 V F 10 A a b 4 V c C E 6 V A 3 A 6 V B 4 A 7 A 10 A 4 V D 0 V F 10 A FIGUE The crcut consdered n Example 3.3- and the crcut redrawn to emphasze the nodes. d Try It Yourself! More Problems and Worked Examples Are n the Electrc Crcut Study Applets

9 Krchhoff s Laws 65 Example Consder the crcut shown n Fgure Notce that the passe conenton was used to assgn reference drectons to the resstor oltages and currents. Ths antcpates usng Ohm s law. Fnd each current and each oltage when 1 8, 10 V, 3 A,and 3 1. Also, determne the resstance. Soluton The sum of the currents enterng node a s Usng Ohm s law for 3, we fnd that () V Krchhoff s oltage law for the bottom loop ncorporatng 1, 3, and the 10-V source s Therefore, Ohm s law for the resstor 1 s or V / 1 8/8 1A Snce we hae now found 1 1A and 3 A as orgnally stated, then We can now fnd the resstance from A a V 10 V FIGUE Crcut wth two constantoltage sources. or / 10/ 1 10 Example Determne the alue of the current, n amps, measured by the ammeter n Fgure 3.3-6a. Soluton An deal ammeter s equalent to a short crcut. The current measured by the ammeter s the current n the short crcut. Fgure 3.3-6b shows the crcut after replacng the ammeter by the equalent short crcut. The crcut has been redrawn n Fgure to label the nodes of the crcut. Ths crcut conssts of a oltage source, a dependent current source, two resstors, and two short crcuts. One of the short crcuts s the controllng element of the CCCS and the other short crcut s a model of the ammeter. Applyng KCL twce, once at node d and agan at node a shows that the current n the oltage source and the current n the 4- resstor are both equal to a. These currents are labeled n Fgure Applyng KCL agan, at node c, shows that the current n the resstor s equal to m. Ths current s labeled n Fgure Next, Ohm s law tells us that the oltage across the 4- resstor s equal to 4 a and that the oltage across the - resstor s equal to m. Both of these oltages are labeled n Fgure Applyng KCL at node b ges a 3 a m 0 Applyng KVL to closed path a-b-c-e-d-ages a m m m m

10 66 esste Crcuts 4 Ω Ω Ammeter 1 V 3 a a m 1 V 4 Ω Ω 3 a a FIGUE A crcut wth dependent source and an ammeter. The equalent crcut after replacng the ammeter by a short crcut. m a 4 Ω b Ω m a 4 a m 1 V 3 a a a d e FIGUE The crcut of Fgure after labelng the nodes and some element currents and oltages. c m Fnally, solng ths equaton ges m 4A Try It Yourself! More Problems and Worked Examples Are n the Electrc Crcut Study Applets Example Determne the alue of the oltage, n olts, measured by the oltmeter n Fgure 3.3-8a. Soluton An deal oltmeter s equalent to an open crcut. The oltage measured by the oltmeter s the oltage across the open crcut. Fgure 3.3-8b shows the crcut after replacng the oltmeter by the equalent open crcut. The crcut has been redrawn n Fgure to label the nodes of the crcut. Ths crcut conssts of a oltage source, a dependent oltage source, two resstors, a short crcut, and an open crcut. The short crcut s the controllng element of the CCVS and the open crcut s a model of the oltmeter. 1 V 4 Ω 5 Ω a 3 a m Voltmeter 1 V 4 Ω 5 Ω 3 a a FIGUE A crcut wth dependent source and a oltmeter. The equalent crcut after replacng the oltmeter by a open crcut. m a 4 Ω b 5 Ω 0 A a c 4 a 0 V 1 V 3 a m a a d f e FIGUE The crcut of Fgure 3.3-8b after labelng the nodes and some element currents and oltages.

11 Krchhoff s Laws 67 Applyng KCL twce, once at node d and agan at node a, shows that the current n the oltage source and the current n the 4- resstor are both equal to a. These currents are labeled n Fgure Applyng KCL agan, at node c, shows that the current n the 5- resstor s equal to the current n the open crcut, that s, zero. Ths current s labeled n Fgure Ohm s law tells us that the oltage across the 5- resstor s also equal to zero. Next, applyng KVL to the closed path b-c-f-e-bges m 3 a. Applyng KVL to the closed path a-b-e-d-ages 4 a 3 a 1 0 so a 1 A Fnally m 3 a 3( 1) 36 V Try It Yourself! More Problems and Worked Examples Are n the Electrc Crcut Study Applets Exercse Determne the alues of 3, 4, 6,, 4, and 6 n Fgure E Answer: 3 3A, 4 3A, 6 4A, 3V, 4 6V, 6 6V 3 V C 3 V A A 3 B 1 A 6 V E 1 A 4 6 F 6 D 4 FIGUE E Exercse 3.3- Determne the current n Fgure E Answer: 4A 4 Ω 4 V A 4 Ω FIGUE E 3.3- Exercse Determne the alue of the current m n Fgure E 3.3-3a. Hnt: Apply KVL to the closed path a-b-d-c-a n Fgure E 3.3-3b to determne a. Then apply KCL at node b to fnd m Answer: m 9A. a 3 A 6 Ω 4 Ω 5 a m a a c 3 A 6 Ω 18 V 3 A 1 V 3 A 4 Ω b d 5 a m FIGUE E A crcut contanng avccs. The crcut after labelng the nodes and some element currents and oltages.

12 68 esste Crcuts Exercse Determne the alue of the oltage m n Fgure E 3.3-4a. Hnt: Apply KVL to the closed path a-b-d-c-an Fgure E 3.3-4b to determne a. Answer: m 4 V a A 5 Ω 4 Ω 4 a 10 V A 8 V FIGUE E A crcut contanng a VCVS. The crcut after labelng the nodes and some element currents and oltages. m a a c A 5 Ω A 4 Ω b d 4 a m 3.4 A Sngle-Loop Crcut The Voltage Dder a s s 3 d 3 3 b c FIGUE Sngle-loop crcut wth a oltage source s. Let us consder a sngle-loop crcut, as shown n Fgure In antcpaton of usng Ohm s law, the passe conenton has been used to assgn reference drectons to resstor oltages and currents. Usng KCL at each node, we obtan a: s 1 0 (3.4-1) b: 1 0 (3.4-) c: 3 0 (3.4-3) d: 3 s 0 (3.4-4) We hae four equatons, but any one of the four can be dered from the other three equatons. In any crcut wth n nodes, n 1 ndependent current equatons can be dered from Krchhoff s current law. Of course, we also note that s 1 3 so that the current 1 can be sad to be the loop current and flows contnuously around the loop from a to b to c to d and back to a. The resstors n Fgure are connected n seres. Notce, for example, that resstors 1 and are both connected to node b and that no other crcut elements are connected to node b. Consequently, 1, so both resstors hae the same current. A smlar argument shows that resstors and 3 are also connected n seres. Notcng that s connected n seres wth both 1 and 3, we say that all three resstors are connected n seres. The order of seres resstors s not mportant. For example, the oltages and currents of the three resstors n Fgure wll not change f we nterchange the postons and 3. The defnng characterstc of seres elements s that they hae the same current. To dentfy a par of seres elements, we look for two elements connected to sngle a node that has no other elements connected to t. In order to determne 1, we use KVL around the loop to obtan s (3.4-5) where 1 s the oltage across the resstor 1. Usng Ohm s law for each resstor, Eq can be wrtten as s

13 A Sngle-Loop Crcut The Voltage Dder 69 Solng for 1, we hae s Thus, the oltage across the nth resstor n s n and can be obtaned as s n (3.4-6) n 1 n 1 3 For example, the oltage across resstor s s 1 3 Thus, the oltage appearng across one of a seres connecton of resstors connected n seres wth a oltage source wll be the rato of ts resstance to the total resstance tmes the source oltage. Ths crcut demonstrates the prncple of oltage dson, and the crcut s called a oltage dder. In general, we may represent the oltage dder prncple by the equaton n (3.4-7) n s 1 N where the oltage s across the nth resstor of N resstors connected n seres. Example Let us consder the crcut shown n Fgure 3.4- and determne the resstance requred so that the oltage across wll be one-fourth of the source oltage when 1 9. Determne the current flowng when s 1 V. Soluton The oltage across resstor wll be Snce we desre / s 1/4, we hae s or 1 3 Snce 1 9, we requre that 3. Usng KVL around the loop, we hae s 1 0 or s 1 s Therefore, (3.4-8) 1 or 1 1 1A 1 1 s FIGUE 3.4- Voltage dder crcut wth 1 9.

14 70 esste Crcuts s s FIGUE Equalent crcut for a seres connecton of resstors. Let us consder the smple crcut of the oltage source connected to a resstance s as shown n Fgure For ths crcut s s Comparng Eqs and 3.4-9, we see that the currents are dentcal when s 1 (3.4-9) The resstance s s sad to be an equalent resstance of the seres connecton of resstors 1 and. In general, the equalent resstance of a seres of N resstors s s 1... N (3.4-10) In ths specfc case s Each resstor n Fgure 3.4- absorbs power, so the power absorbed by 1 s and the power absorbed by the second resstor s The total power absorbed by the two resstors s Howeer, accordng to the oltage dder prncple, p p p p1 p 1 1 (3.4-11) Then we may rewrte Eq as p 1 s ( ) ( ) 1 1 Snce 1 s, the equalent seres resstance, we hae n n s 1 s p 1 s s s s Thus, the total power absorbed by the two seres resstors s equal to the power absorbed by the equalent resstance s. The power absorbed by the two resstors s equal to that suppled by the source s. We show ths by notng that the current s and therefore the power suppled s s s p s s s

15 A Sngle-Loop Crcut The Voltage Dder 71 Example 3.4- For the crcut of Fgure 3.4-4a, fnd the current measured by the ammeter. Then show that the power absorbed by the two resstors s equal to that suppled by the source. Soluton Fgure 3.4-4b shows the crcut after the deal ammeter has been replaced by the equalent short crcut and a label has been added to ndcate the current measured by the ammeter, m. Applyng KVL ges 15 5 m 10 m 0 The current measured by the ammeter s m 15 1A 5 10 (Why s m negate? Why can t we just dde the source oltage by the equalent resstance? ecall that when we use Ohm s law, the oltage and current must adhere to the passe conenton. In ths case, the current calculated by ddng the source oltage by the equalent resstance does not hae the same reference drecton as m, and so we need a mnus sgn.) The total power absorbed by the two resstors s The power suppled by the source s p 5 m 10m 15(1 ) 15 W p s s m 15( 1) 15 W Thus, the power suppled by the source s equal to that absorbed by the seres connecton of resstors. 15 V Ammeter 15 V 5 Ω 10 Ω 5 Ω 10 Ω m FIGUE A crcut contanng seres resstors. The crcut after the deal ammeter has been replaced by the equalent short crcut and a label has been added to ndcate the current measured by the ammeter, m. Try It Yourself! More Problems and Worked Examples Are n the Electrc Crcut Study Applets 6 Ω Exercse For the crcut of Fgure E 3.4-1, fnd the oltage 3 and the current and show that the power delered to the three resstors s equal to that suppled by the source. Answer: 3 3V, 1A 1 V 1 3 Ω 3 Exercse 3.4- Consder the oltage dder shown n Fgure E 3.4- when 1 6. It s desred that the output power absorbed by 1 6 be 6 W. Fnd the oltage o and the requred source s. Answer: s 14 V, o 6V 3 Ω FIGUE E Crcut wth three seres resstors (for Exercse 3.4-1).

16 7 esste Crcuts Ω s o 4 Ω 1 Exercse Determne the oltage measured by the oltmeter n the crcut shown n Fgure E 3.4-3a. Hnt: Fgure E 3.4-3b shows the crcut after the deal oltmeter has been replaced by the equalent open crcut and a label has been added to ndcate the oltage measured by the oltmeter, m. Answer: m V Ω FIGUE E 3.4- Voltage dder for Exercse Ω 8 V 5 Ω Voltmeter 75 Ω 8 V 5 Ω m FIGUE E A oltage dder. The oltage dder after the deal oltmeter has been replaced by the equalent open crcut and a label has been added to ndcate the oltage measured by the oltmeter, m. Exercse Determne the oltage measured by the oltmeter n the crcut shown n Fgure E 3.4-4a. Hnt: Fgure E 3.4-4b shows the crcut after the deal oltmeter has been replaced by the equalent open crcut and a label has been added to ndcate the oltage measured by the oltmeter, m. Answer: m V 75 Ω Voltmeter 75 Ω 8 V 5 Ω 8 V 5 Ω m FIGUE E A oltage dder. The oltage dder after the deal oltmeter has been replaced by the equalent open crcut and a label has been added to ndcate the oltage measured by the oltmeter, m. 3.5 Parallel esstors and Current Dson Crcut elements, such as resstors, are connected n parallel when the oltage across each element s dentcal. The resstors n Fgure are connected n parallel. Notce, for example, that resstors 1 and are each connected to both node a and node b. Consequently, 1, so both resstors hae the same oltage. A smlar argument shows that resstors and 3 are also connected n parallel. Notcng that s connected n parallel wth both 1 and 3, we say that all three resstors are connected n parallel. The order of parallel resstors s not mportant. For example, the oltages and currents of the three resstors n Fgure wll not change f we nterchange the postons and 3.

17 Parallel esstors and Current Dson 73 a s b FIGUE A crcut wth parallel resstors. The defnng characterstc of parallel elements s that they hae the same oltage. To dentfy a par of parallel elements, we look for two elements connected between the same par of nodes. Consder the crcut wth two resstors and a current source shown n Fgure Note that both resstors are connected to termnals a and b and that the oltage appears across each parallel element. In antcpaton of usng Ohm s law, the passe conenton s used to assgn reference drectons to the resstor oltages and currents. We may wrte KCL at node a (or at node b) to obtan s 1 0 or s 1 Howeer, from Ohm s law and Then s (3.5-1) 1 ecall that we defned conductance G as the nerse of resstance. We may therefore rewrte Eq as s G 1 G (G 1 G ) (3.5-) Thus, the equalent crcut for ths parallel crcut s a conductance G p, as shown n Fgure 3.5-3, where G p G 1 G The equalent resstance for the two-resstor crcut s found from Snce G p 1/ p, we hae 1 1 Gp s FIGUE 3.5- Parallel crcut wth a current source. s FIGUE Equalent crcut for a parallel crcut. a b 1 1 G p p 1 or p 1 (3.5-3) 1 Note that the total conductance, G p, ncreases as addtonal parallel elements are added and that the total resstance, p, declnes as each resstor s added. The crcut shown n Fgure 3.5- s called a current dder crcut snce t ddes the source current. Note that 1 G 1 (3.5-4)

18 74 esste Crcuts Also, snce s (G 1 G ), we sole for, obtanng G s G 1 Substtutng from Eq nto Eq , we obtan (3.5-5) 1 G 1 s G G 1 (3.5-6) Smlarly, Note that we may use G 1/ and G 1 1/ 1 to obtan the current n terms of two resstances as follows: G s G G 1 1 s 1 The current of the source ddes between conductances G 1 and G n proporton to ther conductance alues. Let us consder the more general case of current dson wth a set of N parallel conductors as shown n Fgure The KCL ges s 1 3 N (3.5-7) for whch n G n (3.5-8) for n 1,,N. We may wrte Eq as s (G 1 G G 3 G N ) (3.5-9) N G N Therefore G n (3.5-10) N s n 1 Snce n G n, we may obtan from Eq and substtute t n Eq , obtanng 3 G 3 n G n s N n 1 G n (3.5-11) G 1 G 1 s FIGUE Set of N parallel conductances wth a current source s. ecall that the equalent crcut, Fgure 3.5-3, has an equalent conductance G p such that G N p G n n 1 (3.5-1) G n s Therefore n (3.5-13) Gp whch s the basc equaton for the current dder wth N conductances. Of course, Eq can be rewrtten as N 1 1 p n 1 n (3.5-14)

19 Parallel esstors and Current Dson 75 Example For the crcut n Fgure fnd the current n each branch, the equalent crcut, and (c) the oltage. The resstors are ,, Soluton The current dder follows the equaton n G n G s p so t s wse to fnd the equalent crcut, as shown n Fgure 3.5-6, wth ts equalent conductance G p. We hae G G G1 G G S ecall that the unts for conductance are semens (S). Then p N n 1 n Smlarly, 1 G 1 s G ( 14 8 ) 4 A p G s 48 ( ) 8 A G 14 p and Snce n G n, we hae 3 G 3 s 16 A G p 1 4 V G 1 8 A 1 FIGUE Parallel crcut for Example A G p FIGUE Equalent crcut for the parallel crcut of Fgure Example 3.5- For the crcut of Fgure 3.5-7a, fnd the oltage measured by the oltmeter. Then show that the power absorbed by the two resstors s equal to that suppled by the source. Soluton Fgure 3.5-7b shows the crcut after the deal oltmeter has been replaced by the equalent open crcut and alabel has been added to ndcate the oltage measured by the oltmeter, m. The two resstors are connected

20 76 esste Crcuts n parallel and can be replaced wth a sngle equalent resstor. The resstance of ths equalent resstor s calculated as Fgure 3.5-7c shows the crcut after the parallel resstors hae been replaced by the equalent resstor. The current n the equalent resstor s 50 ma, drected upward. Ths current and the oltage m do not adhere to the passe conenton. The current n the equalent resstance can also be expressed as 50 ma, drected downward. Ths current and the oltage m do adhere to the passe conenton. Ohm s law ges m 8( 0.5) V The oltage m n Fgure 3.5-7b s equal to the oltage m n Fgure 3.5-7c. Ths s a consequence of the equalence of the 8- resstor to the parallel combnaton of the 40- and 10- resstors. Lookng at Fgure 3.5-7b, the power absorbed by the resstors s The power suppled by the current source s p m m W p s (0.5) 0.5 W Thus, the power absorbed by the two resstors s equal to that suppled by the source. Voltmeter 50 ma 40 Ω 10 Ω FIGUE A crcut contanng parallel resstors. The crcut after the deal oltmeter has been replaced by the equalent open crcut and a label has been added to ndcate the oltage measured by the oltmeter, m. (c) The crcut after the parallel resstors hae been replaced by an equalent resstance. 50 ma 40 Ω m 10 Ω 50 ma (c) m 8 Ω Try It Yourself! More Problems and Worked Examples Are n the Electrc Crcut Study Applets Exercse A resstor network consstng of parallel resstors s shown n a package used for prnted crcut board electroncs n Fgure E 3.5-1a. Ths package s only cm 0.7 cm, and each resstor s 1 k. The crcut s connected to use four resstors as shown n Fgure E 3.5-1b. Fnd the equalent crcut for ths network. Determne the current n each resstor when s 1 ma. Answer: p 50

21 Seres Voltage Sources and Parallel Current Sources 77 s FIGUE E A parallel resstor network. Courtesy of Dale Electroncs. The connected crcut uses four resstors where 1k. Exercse 3.5- Determne the current measured by the ammeter n the crcut shown n Fgure E 3.5-a. Hnt: Fgure E 3.5-b shows the crcut after the deal ammeter has been replaced by the equalent short crcut and a label has been added to ndcate the current measured by the ammeter, m. Answer: m 1A 40 Ω Ammeter 5 A 10 Ω 5 A 40 Ω 10 Ω m FIGUE E 3.5- A current dder. The current dder after the deal ammeter has been replaced by the equalent short crcut and a label has been added to ndcate the current measured by the ammeter, m. 3.6 Seres Voltage Sources and Parallel Current Sources Voltage sources connected n seres are equalent to a sngle oltage source. The oltage of the equalent oltage source s equal to the sum of oltages of the seres oltage sources. Consder the crcut shown n Fgure 3.6-1a. Notce that the currents of both oltage sources are equal. Accordngly, defne the current, s, to be s a b (3.6-1) Next, defne the oltage, s, to be s a b (3.6-)

22 78 esste Crcuts a b c c a b 3 3 s FIGUE A crcut contanng oltage sources connected n seres and an equalent crcut. c 1 3 c 1 s Usng KCL, KVL, and Ohm s law, we can represent the crcut n Fgure 3.6-1a by the equatons c c b a a b FIGUE 3.6- A crcut contanng parallel current sources and an equalent crcut. c 1 s 1 (3.6-3) s 3 (3.6-4) c 1 (3.6-5) 1 s (3.6-6) 3 3 (3.6-7) where s a b and s a b. These same equatons result from applyng KCL, KVL, and Ohm s law to the crcut n Fgure 3.6-1b. If s a b and s a b, then the crcuts shown n Fgures 3.6-1a and 3.6-1b are equalent because they are both represented by the same equatons. For example, suppose that c 4A, 1, 6, 3 3, a 1V,and b 3V. The equatons descrbng the crcut n Fgure 3.6-1a become 1 4 s (3.6-8) s 6 3 (3.6-9) c 1 (3.6-10) 1 4 (3.6-11) 3 3 (3.6-1) The soluton to ths set of equatons s 1 6V, s 1A, A, V,and c 6V. Eqs to also descrbe the crcut n Fgure 3.6-1b. Thus, 1 6V, s 1A, A, V, and c 6V n both crcuts. eplacng seres oltage sources by a sngle, equalent oltage source does not change the oltage or current of other elements of the crcut. Fgure 3.6-a shows a crcut contanng parallel current sources. The crcut n Fgure 3.6-b s obtaned by replacng these parallel current sources by a sngle, equalent current source. The current of the equalent current source s equal to the sum of the currents of the parallel current sources.

23 Crcut Analyss 79 Table 3.61 Parallel and Seres Voltage and Current Sources CICUIT EQUIVALENT CICUIT CICUIT EQUIVALENT CICUIT a b a b a b a b a b a b a b Not allowed a b a b a b Not allowed We are not allowed to connect ndependent current sources n seres. Seres elements hae the same current. Ths restrcton preents seres current sources from beng ndependent. Smlarly, we are not allowed to connect ndependent oltage sources n parallel. Table summarzes the parallel and seres connectons of current and oltage sources. 3.7 Crcut Analyss In ths secton we consder the analyss of a crcut by replacng a set of resstors wth an equalent resstance, thus reducng the network to a form easly analyzed. Consder the crcut shown n Fgure Note that t ncludes a set of resstors that s n seres and another set of resstors that s n parallel. It s desred to fnd the output oltage o, so we wsh to reduce the crcut to the equalent crcut shown n Fgure Two crcuts are equalent f they exhbt dentcal characterstcs at the same two termnals. s o 6 FIGUE Crcut wth a set of seres resstors and a set of parallel resstors. s s o FIGUE 3.7- Equalent crcut for the crcut of Fgure P

24 80 esste Crcuts We note that the equalent seres resstance s s 1 3 and the equalent parallel resstance s where G p G 4 G 5 G 6 Then, usng the oltage dder prncple, wth Fgure 3.7-, we hae o p 1 G s p If a crcut has seeral combnatons of sets of parallel resstors and sets of seres resstors, one works through seeral steps, reducng the network to ts smplest form. p p s Example Consder the crcut shown n Fgure Fnd the current 1 when 4 and 3 8 Soluton Snce the objecte s to fnd 1, we wll attempt to reduce the crcut so that the 3- resstor s n parallel wth one resstor and the current source s. Then we can use the current dder prncple to obtan 1. Snce and 3 are n parallel, we fnd an equalent resstance as p1 3 Then addng p1 to 4, we hae a seres equalent resstor 3 4 s 4 p1 4 6 Now the s resstor s n parallel wth three resstors as shown n Fgure 3.7-3b. Howeer, we wsh to obtan the equalent crcut as shown n Fgure 3.7-4, so that we can fnd 1. Therefore, we combne the 9- resstor, the 18- resstor, and s shown to the rght of termnals ab n Fgure 3.7-3b nto one parallel equalent conductance G p. Thus, we fnd G p s 1 S a s 3 Ω 9 Ω 18 Ω 1 3 s 3 Ω 9 Ω s 18 Ω FIGUE Crcut for Example Partally reduced crcut for Example b

25 Crcut Analyss 81 1 s 3 Ω G P FIGUE Equalent crcut for Fgure Then, usng the current dder prncple, 1 G 1 G s p where 1 1 Gp G1 Gp / 1 Therefore, 1 s 3 / s Example 3.7- The crcut n Fgure 3.7-5a contans an ohmmeter. An ohmmeter s an nstrument that measures resstance n ohms. The ohmmeter wll measure the equalent resstance of the resstor crcut connected to ts termnals. Determne the resstance measured by the ohmmeter n Fgure 3.7-5a. Soluton Workng from left to rght, the 30- resstor s parallel to the 60- resstor. The equalent resstance s In Fgure 3.7-5b the parallel combnaton of the 30- and 60- resstors has been replaced wth the equalent 0- resstor. Now the two 0- resstors are n seres. The equalent resstance s Ω Ohmmeter 0 Ω Ohmmeter 60 Ω 30 Ω 10 Ω 0 Ω 10 Ω Ohmmeter Ohmmeter 40 Ω 10 Ω 8 Ω FIGUE (c) (d)

26 8 esste Crcuts In Fgure 3.7-5c the seres combnaton of the two 0- resstors has been replaced wth the equalent 40- resstor. Now the 40- resstor s parallel to the 10- resstor. The equalent resstance s In Fgure 3.7-5d the parallel combnaton of the 40- and 10- resstors has been replaced wth the equalent 8- resstor. Thus, the ohmmeter measures a resstance equal to 8. Try It Yourself! More Problems and Worked Examples Are n the Electrc Crcut Study Applets In general, we may fnd the equalent resstance (or conductance) for a porton of a crcut consstng only of resstors and then replace that porton of the crcut wth the equalent resstance. For example, consder the crcut shown n Fgure We use eq x y to denote the equalent resstance seen lookng nto termnals xy. We note that the equalent resstance to the rght of termnals cd s 15( 6 4) eq cd 15 ( 6 4) Then the equalent resstance of the crcut to the rght of termnals ab s eq a b 4 eq c d a 4 Ω c 6 Ω 15 Ω 4 Ω eq ab b eq cd d FIGUE The equalent resstance lookng nto termnals cd s denoted as eq c d. Exercse Determne the resstance measured by the ohmmeter n Fgure E Answer: ( 30 30) ( 30 30) Ω 30 Ω Ohmmeter 30 Ω 30 Ω FIGUE E 3.7-1

27 Analyzng esste Crcuts Usng MATLAB 83 Exercse 3.7- Determne the resstance measured by the ohmmeter n Fgure E Answer: Ω Ohmmeter 40 Ω 10 Ω 4 Ω FIGUE E 3.7- Exercse Determne the resstance measured by the ohmmeter n Fgure E Answer: ( ) 60 ( ) Ω 60 Ω Ohmmeter 60 Ω 60 Ω FIGUE E Analyzng esste Crcuts Usng MATLAB The computer program MATLAB s a tool for makng mathematcal calculatons. In ths secton MATLAB s used to sole the equatons encountered when analyzng a resste crcut. Consder the resste crcut shown n Fgure 3.8-1a. The goal s to determne the alue of the nput oltage, V s,requred to cause the current I to be 1 A. (Subscrpts can t be used n the MATLAB nput fle. Thus V s and p n Fgure become Vs and p n the MATLAB nput fle. We hae been usng lowercase letters to represent element oltages and currents, but n MATLAB examples we wll use captal 1 1 Ω Ω 3 3 Ω I s V s 4 6 Ω 5 3 Ω 6 Ω V o V s P V o FIGUE A resste crcut and an equalent crcut.

28 84 esste Crcuts letters to represent currents and oltages to mproe the readablty of the MATLAB nput fle. For example, the nput and output oltage are denoted as V s and V o rather than s and o.) esstors 1,, and 3 are connected n seres and can be replaced by an equalent resstor, s,gen by s 1 3 (3.8-1) esstors 4, 5, and 6 are connected n parallel and can be replaced by an equalent resstor, p,gen by p (3.8-) Fgure 3.8-1b shows the crcut after 1,, and 3 are replaced by s and 4, 5, and 6 are replaced by P. Applyng oltage dson to the crcut n Fgure 3.8-1b ges V o s p V s p (3.8-3) where V o s the oltage across p n Fgure 3.8-1b and s also the oltage across the parallel resstors n Fgure 3.8-1a. Ohm s law ndcates that the current n 6 s gen by I V o 6 (3.8-4) Fgure 3.8- shows a plot of the output current I ersus the nput oltage V s. Ths plot shows that I wll be 1 A when V s 14 V. Fgure shows the MATLAB nput fle that was used to obtan Fgure The MATLAB program frst causes V s to ary oer a range of oltages. Next, MATLAB calculates the alue of I correspondng to each alue of V s usng Eqs through Fnally, MATLAB plots the current I ersus the oltage V s. 1. Current n I, A FIGUE 3.8- Plot of I ersus V s for the crcut shown n Fgure V s, V

29 Verfcaton Example 85 FIGUE MATLAB nput fle used to obtan the plot of I ersus V s shown n Fgure Verfcaton Example The crcut shown n Fgure 3.9-1a was analyzed by wrtng and solng a set of smultaneous equatons: 1 43, , 5 A computer and the program Mathcad (Mathcad User s Gude, 1991) was used to sole the equatons as shown n Fgure 3.9-1b. It was determned that 60 V, 3 18 A, 4 6A, 5 7 V Are these currents and oltages correct? The current can be calculated from, 3, 4, and 5 n a couple of dfferent ways. Frst, Ohm s law ges Next, applyng KCL at node b ges A 60 1 A

30 86 esste Crcuts C 4 c Ω 5 B b 5 Ω a 4 Ω 3 A V d FIGUE An example crcut and computer analyss usng Mathcad. Clearly, cannot be both 1 and 4 A, so the alues calculated for, 3, 4, and 5 cannot be correct. Checkng the equatons used to calculate, 3, 4, and 5, we fnd a sgn error n the KCL equaton correspondng to node b. Ths equaton should be After makng ths correcton,, 3, 4, and 5 are calculated to be Now 7.5 V, A, A, V and Ths checks as we expected. As an addtonal check, consder 3. Frst, Ohm s law ges (1.15) 4.5 Next, applyng KVL to the loop consstng of the oltage source and the 4- and 5- resstors ges Fnally, applyng KVL to the loop consstng of the - and 4- resstors ges The results of these calculatons agree wth each other, ndcatng that are the correct alues. 7.5 V, A, A, V

31 Desgn Challenge Soluton DESIGN CHALLENGE SOLUTION ADJUSTABLE VOLTAGE SOUCE A crcut s requred to prode an adjustable oltage. The specfcatons for ths crcut are: 1. It should be possble to adjust the oltage to any alue between 5 V and 5 V. It should not be possble to accdentally obtan a oltage outsde ths range.. The load current wll be neglgble. 3. The crcut should use as lttle power as possble. The aalable components are: 1. Potentometers: resstance alues of 10 k, 0k, and 50 k are n stock. A large assortment of standard percent resstors hang alues between 10 and 1M (see Appendx E) 3. Two power supples (oltage sources): one 1-V supply and one 1-V supply, both rated at 100 ma (maxmum) DESCIBE THE SITUATION AND THE ASSUMPTIONS Fgure shows the stuaton. The oltage s the adjustable oltage. The crcut that uses the output of the crcut beng desgned s frequently called the load. In ths case, the load current s neglgble, so 0. STATE THE GOAL A crcut prodng the adjustable oltage 5 V 5V must be desgned usng the aalable components. GENEATE A PLAN Make the followng obseratons. 1. The adjustablty of a potentometer can be used to obtan an adjustable oltage.. Both power supples must be used so that the adjustable oltage can hae both poste and negate alues. 3. The termnals of the potentometer cannot be connected drectly to the power supples because the oltage s not allowed to be as large as 1 V or 1 V. Load current 0 Crcut beng desgned Load crcut FIGUE The crcut beng desgned prodes an adjustable oltage,, to the load crcut.

32 88 esste Crcuts 0 a p (1 a) p 0 1 V 1 p 1 V Load crcut 0 a V 1 V FIGUE A proposed crcut for producng the arable oltage,, and the equalent crcut after the potentometer s modeled. These obseratons suggest the crcut shown n Fgure 3.10-a. The crcut n Fgure 3.10-b s obtaned by usng the smplest model for each component n Fgure 3.10-a. To complete the desgn, alues need to be specfed for 1,, and p. Then, seeral results need to be checked and adjustments made, f necessary. 1. Can the oltage be adjusted to any alue n the range 5 V to 5 V?. Are the oltage source currents less than 100 ma? Ths condton must be satsfed f the power supples are to be modeled as deal oltage sources. 3. Is t possble to reduce the power absorbed by 1,, and p? ACT ON THE PLAN It seems lkely the 1 and wll hae the same alue, so let 1. Then t s conenent to redraw Fgure 3.10-b as shown n Fgure Applyng KVL to the outsde loop yelds 1 a a p a (1 a) p a a 1 0 so a 4 p Next, applyng KVL to the left loop ges Substtutng for a ges 1 ( a p ) a 1 4( a ) p p a p (1 a) p FIGUE The crcut after settng 1. 1 V 1 V a

33 Desgn Challenge Soluton 89 When a 0, must be 5 V, so p Solng for ges 0.7 p Suppose the potentometer resstance s selected to be p 0 k, the mddle of the three aalable alues. Then VEIFY THE POPOSED SOLUTION As a check, notce that when a 1 as requred. The specfcaton that 14 k 5 V 5V has been satsfed. The power absorbed by the three resstances s so p 1 mw Notce that ths power can be reduced by choosng p to be as large as possble, 50 k n ths case. Changng p to 50 k requres a new alue of : Snce the specfcaton that 14 k 0 k k 0 k p ( ) 0.7 p 35 k 35 k 50 k 35 k 5V V 70 k 50 k 70 k 50 k 5 V 5V 4 has been satsfed. The power absorbed by the three resstances s now a p p p 4 50 k 70 k 5 mw Fnally, the power supply current s 4 a 50 k 70 k 0. ma whch s well below that 100 ma that the oltage sources are able to supply. The desgn s complete.

34 90 esste Crcuts 3.11 SUMMAY Gusta obert Krchhoff formulated the laws that enable us to study a crcut. Krchhoff s current law (KCL) states that the algebrac sum of all the currents enterng a node s zero. Krchhoff s oltage law (KVL) states that the algebrac sum of all the oltage around a closed path (loop) s zero. Two specal crcuts of nterest are the seres crcut and the parallel crcut. Table summarzes the results from ths chapter regardng seres and parallel elements. The frst row of the table shows seres resstors connected to a crcut. These seres resstors can be replaced by an equalent resstor. Dong so does not dsturb the crcut connected to the seres resstors. The element oltages and currents n ths crcut don t change. In partcular, the termnal oltage and current (labeled and, respectely, n Table ) do not change when the seres resstors are replaced by the equalent resstor. Indeed, ths s what s meant by equalent. eplacng parallel resstors by an equalent resstor does not change any oltage or current n the crcut connected to the parallel resstors. eplacng seres oltage sources by an equalent oltage source does not dsturb the crcut connected to the seres oltage sources. eplacng parallel current sources by an equalent current source does not dsturb the crcut connected to the parallel current sources. A crcut consstng of seres resstors s sometmes called a oltage dder because the oltage across the seres resstors ddes between the nddual resstors. Smlarly, a crcut consstng of parallel resstors s called a current dder because the current through the parallel resstors ddes between the nddual resstors. Table prodes the equatons descrbng oltage dson for seres resstors and current dson for parallel resstors. Table Equalent Crcuts for Seres and Parallel Elements 1 1 Seres resstors Crcut 1 Crcut s 1, 1 1 1, and 1 s 1 and s Parallel resstors Crcut 1 1 Crcut 1 p 1, 1 1, and 1 1 p 1 1 and p Seres oltage sources Crcut 1 1 Crcut s 1 and 1 s 1 Parallel current sources Crcut 1 1 Crcut p 1 and 1 p 1

35 Problems 91 POBLEMS Secton 3.3 Krchhoff s Laws P Consder the crcut shown n Fgure P Determne the alues of the power suppled by branch B and the power suppled by branch F. 3 A 1 V 6 Ω 4 Ω 0 V 4 V D 1 A 1 V A A 1 V B 1 V C 4 A 5 V E F 1 A FIGUE P Ω P Determne the power absorbed by each of the resstors n the crcut shown n Fgure P Answer: The 4- resstor absorbs 16 W, the 6- resstor absorbs 4 W, and the 8- resstor absorbs 8 W. FIGUE P A 4 Ω 8 V P 3.3- Determne the alues of, 4,, 3, and 6 n Fgure P V A A FIGUE P C 6 A B V D 4 4 V E 3 A 6 F 3 A 6 V 6 Ω 8 Ω 8 V FIGUE P A P Determne the power suppled by each current source n the crcut of Fgure P Answer: The -ma current source supples 6 mw and the 1-mA current source supples 7 mw. 5 V ma 1 V P Consder the crcut shown n Fgure P Suppose that 1 6 and 3. Fnd the current and the oltage. Suppose, nstead, that 1.5 A and V. Determne the resstances 1 and. (c) Suppose, nstead, that the oltage source supples 4 W of power and that the current source supples 9 W of power. Determne the current, the oltage, and the resstances 1 and. FIGUE P V 1 ma 8 V P Determne the power suppled by each oltage source n the crcut of Fgure P Answer: The -V oltage source supples mw and the 3-V oltage source supples 6 mw. 1 V 1 3 A 3 V V FIGUE P ma ma P Determne the power absorbed by each of the resstors n the crcut shown n Fgure P Answer: The 4- resstor absorbs 100 W, the 6- resstor absorbs 4 W, and the 8- resstor absorbs 7 W. FIGUE P ma