writing circuit equations - information for educators...

54
2 C H A P T E R Writing Circuit Equations Objectives By the end of this chapter, you should be able to do the following: 1. Find the complete solution of a circuit using the exhaustive, node, and mesh methods. 2. Put a set of linear equations, such as the ones that define the constraints of a resistive circuit, into matrix-vector form. 3. Find the element variables in a circuit, using the method of superposition of sources. In the last chapter, we saw that, for some planar circuits containing only resistors and sources, we could find the complete solution by writing vi relations for all of the elements, KCL equations at all but one of the nodes, and KVL equations on closed paths around all of the meshes. This brute-force approach is a useful starting point, but it leaves many questions unanswered, not the least of which is, will it always work? It can also produce a very large number of equations to be solved to find the complete solution. A more serious limitation of the brute-force approach, however, is that it provides very little insight into the behavior of a circuit. This chapter seeks answers to three questions. Will our procedures for finding the solution of a circuit always work? How can we reduce the number of equations that we need to write? And how should the variables in those equations be chosen? We demonstrate that the first of these questions has an affirmative answer. Later we show that we can reduce the number of equations that we have to write if we do not need to find all of the variables in the circuit. By choosing the independent variables of the problem carefully, however, it is often possible to reduce the number of equations still further, so that, in 52

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2C H A P T E R

Writing CircuitEquations

Objectives

By the end of this chapter, you should be able to do thefollowing:

1. Find the complete solution of a circuit using theexhaustive, node, and mesh methods.

2. Put a set of linear equations, such as the onesthat define the constraints of a resistive circuit, intomatrix-vector form.

3. Find the element variables in a circuit, using themethod of superposition of sources.

In the last chapter, we saw that, for some planar circuitscontaining only resistors and sources, we could find thecomplete solution by writing v–i relations for all of theelements, KCL equations at all but one of the nodes, andKVL equations on closed paths around all of the meshes.

This brute-force approach is a useful starting point, but itleaves many questions unanswered, not the least of whichis, will it always work? It can also produce a very largenumber of equations to be solved to find the completesolution. A more serious limitation of the brute-forceapproach, however, is that it provides very little insightinto the behavior of a circuit.

This chapter seeks answers to three questions. Will ourprocedures for finding the solution of a circuit alwayswork? How can we reduce the number of equations thatwe need to write? And how should the variables in thoseequations be chosen? We demonstrate that the first ofthese questions has an affirmative answer. Later we showthat we can reduce the number of equations that we haveto write if we do not need to find all of the variablesin the circuit. By choosing the independent variables ofthe problem carefully, however, it is often possible toreduce the number of equations still further, so that, in

52

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2-1 THE EXHAUSTIVE METHOD FOR WRITING A SUFFICIENT SET OF CIRCUIT EQUATIONS 53

effect, we only have to write either the KCL equationsor the KVL equations. This is because the choice of thevariables guarantees that the other equations are satisfiedby construction. This choice of variables leads to twopopular formal techniques for analyzing circuits—nodeanalysis and mesh analysis. The chapter concludes with adevelopment and demonstration of these two techniques.Along the way, we also discover a few more simplifyingtechniques and useful properties of circuits.

All three of the formal methods that we develop,however, are still brute-force methods. They still providevery little insight that can be used in design, and they arestill tedious. The circuit simplifications that we developlater in Chapter 3 go a long way toward remedying thesedifficulties. The formal methods are still very important,however, for the following reasons: (1) We can prove thatthey will work. Thus, they serve as a starting point forother important simplification and analysis techniques.(2) They serve as our back-up for those situations whenad hoc methods fail or when we lack the insight to selectthe appropriate simplifications.

2-1 The Exhaustive Method for Writing aSufficient Set of Circuit Equations

The circuit analysis method that we used in Chapter 1is one that we call the exhaustive method, although italso has other names. It could, with some justification,also be called the exhausting method, because, of allthe methods we will consider, it results in the largestnumber of equations in the largest number of variables.We shall usually choose to use other methods, ones thatinvolve fewer variables and provide more insight, but theexhaustive method nevertheless does have several thingsgoing for it, not the least of which is that its equationsare the easiest to write. In addition, we can prove that themethod is guaranteed to produce a solution; as a result,we shall establish the adequacy of later methods by tying

them to the guarantees of the exhaustive method. Forsimple circuits involving only a few elements, it is oftenthe solution method of choice. As a starting point, let usformally restate the method.

EXHAUSTIVE METHOD:

1. Define a voltage variable and a currentvariable for each element in the circuit.Define a voltage variable for each(independent and dependent) currentsource and a current variable for eachvoltage source.

2. Write a v–i relation for each element.

3. Write a KVL equation for each mesh.

4. Write a KCL equation for all but oneof the nodes. (It does not matter whichone is excluded.)

5. Solve the resulting set of equations.

If the circuit has e elements and s sources, Step 1 ofthe procedure defines 2e + s variables. Steps 2, 3, and 4allow us to write 2e + s independent linear equations inthose variables, a fact that we shall demonstrate shortly.For a resistive circuit, the e element relations in Step 2are quite simple. We can use these e statements of Ohm’slaw to eliminate immediately either the resistor voltagesor the resistor currents from the set of variables. If we dothis, we are left with a set of e + s equations in e + sunknowns that need to be solved. Is this set of equationssufficient?

2-1-1 Proof of Sufficiency of the Method*

If we write a KCL equation at all but one of the nodesof a planar circuit, write a KVL equation at all of itsmeshes, and write a v–i relation for each element, can we

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54 CHAPTER 2 WRITING CIRCUIT EQUATIONS

be sure that we will have enough equations to specify thecomplete solution of the circuit? This is the question towhich this section finds an affirmative answer. It exploitsa fundamental property of mathematical graphs (of whichcircuits are a special case) that constrains the numbersof nodes, meshes, and branches (connections betweennodes) that a planar circuit can have. For those willing toaccept the method without justification, this section canbe omitted without disturbing the flow of the material.

We begin with a few definitions. As we have alreadyseen, a planar network is a circuit that we can draw ona flat sheet of paper without any elements crossing eachother. All the networks that we have drawn so far areplanar, and most of those that we shall see in what is tofollow are planar also.1 The exhaustive method, as wehave stated it, is applicable only to planar graphs. Let ebe the number of elements in the circuit, s be the numberof sources (independent or dependent), be the numberof meshes, and n be the number of nodes. A branch is aconnection between two nodes. In a circuit, a branch canbe either an element or a source; we let b = e+ s denotethe total number of branches.

The numbers of nodes, meshes, and branches in anyplanar graph are not independent; knowing any two ofthese specifies the other. To show this, we define atree as a connected network (or graph) that contains noclosed paths. A tree can be constructed from any circuitby selectively removing elements. Figure 2-1 shows anexample of a circuit and one possible tree that can beobtained from it. Because there are no closed paths, nocurrents can flow in a tree. Notice that in a tree thereis exactly one path connecting any two nodes. We canconstruct a tree by building the network in the followingfashion. We begin with the isolated nodes from thenetwork, and we construct the tree by inserting some of

1Planar networks are the only ones for which our definition of amesh is applicable. For dealing with the nonplanar case, see ProblemsP2-52 and P2-53.

(a)

(b)

Fig. 2-1. (a) A network. (b) A tree derived from thenetwork in (a).

the branches of the complete network one at a time. Forthe first branch, we select any branch from the circuit,which must connect two nodes. Then we select one ofthe isolated nodes and add a branch from the originalcircuit that will attach that node to the growing tree.We stop when there are no more isolated nodes. At thispoint we have a tree, plus several branches remaining tobe inserted. We have not yet created any closed paths,because every time that we added a branch, one terminalof that branch was attached to a node that was previouslyisolated and that now is attached only to the singleelement. Notice that we have created only one out of

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2-1 THE EXHAUSTIVE METHOD FOR WRITING A SUFFICIENT SET OF CIRCUIT EQUATIONS 55

many possible trees; but all of the tree possibilities willcontain exactly n − 1 branches, where n is the numberof nodes. This is because the first branch attached to thegrowing tree decreased the number of isolated nodes bytwo and each additional insertion reduced the numberof isolated nodes remaining by one until we ran out ofisolated nodes.

DEMO: Trees

Once we have the tree, we add the remaining branchesto the growing network one at a time. When the firstof these additional branches is added, a network withone mesh is created, because there are now two pathsthat connect the two nodes to which it is attached—onethrough the new element and one through the originaltree. The addition of another element either creates asecond mesh in the same fashion or divides a previouslyconstructed mesh into two smaller meshes. As this proce-dure is continued, the addition of each new branch addsone to the number of meshes, either by creating a newmesh or by dividing an existing mesh into two. The firstn−1 branches, which make up the tree, did not create anymeshes, but each branch added after the tree was createdadds exactly one additional mesh. Thus, the number ofmeshes created is b − (n − 1). This must be the totalnumber of meshes in the complete network. Therefore,

= b − n+ 1

or

b = n+ − 1. (2.1)

As an example, notice that, for the network inFigure 2-1(a), there are 12 branches, nine nodes, and fourmeshes, which agrees with the formula. You are also en-couraged to look at any of the other planar networks in thistext and verify that this formula is true for all of them also.

Now let us use this result to verify that the exhaustivemethod will produce enough independent equations to be

able to specify the complete solution. After we completeStep 1 of the procedure, each of the e elements in thecircuit has two unknown quantities associated with it, avoltage variable and a current variable, while each of thes sources has one unknown, the voltage in the case ofa current source or the current in the case of a voltagesource. By adding these up, we can see that the totalnumber of unknowns is

# unknowns = 2e + s.In order to find these unknowns, we write a number oflinear equations, each of which imposes a constraint.There are n− 1 KCL equations, KVL equations, and eelement relations. Thus, the total number of equations is

# equations = (n− 1)+ + e.Equation (2.1), however, tells us how the number of nodesand meshes is related to the number of elements andsources, since b = e + s = (n − 1) + . Substitutingthis relation, we see that the total number of equations is

# equations = 2e + s.Since the number of independent equations is equal tothe total number of unknowns, we are guaranteed thatthe exhaustive method will find the complete solution.

MULTIMEDIA TUTORIAL: Exhaustive Method

2-1-2 Examples of the Method

Example 2-1 The Exhaustive Method

As a simple first example, consider the circuit inFigure 2-2, which contains three resistors and a currentsource. The first step is to define seven variables: a voltageand a current for each of the three resistors and a voltage(potential difference) across the terminals of the current

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56 CHAPTER 2 WRITING CIRCUIT EQUATIONS

is(t)–0(t)+

+3(t)–

+1(t)

+2(t)

4

3

1

Fig. 2-2. A simple circuit to be analyzed via the exhaustivemethod. The circuit contains three nodes and two meshes(Example 2-1). All resistances are measured in ohms.

source. The four voltages are indicated on the figure. Leti1(t) be the current flowing into the + terminal of the 4resistor, i2(t) be the current flowing into the + terminalof the 3 resistor, and i3(t) be the current flowing intothe + terminal of the 1 resistor. The total number ofvariables is seven.

Step 2 of the procedure says that we should writeelement relations for all of the elements. Since in thiscase there are three elements (the resistors), we writethree statements of Ohm’s law:

v1(t)− 4i1(t)= 0

v2(t)− 3i2(t)= 0

v3(t)− i3(t)= 0.

The circuit contains three nodes and two meshes.Therefore, we should write two KCL equations and twoKVL equations. For the KCL equations written at the twoupper nodes in the circuit, we have

i1(t)+ i2(t)= is(t)

−i2(t)+ i3(t)= 0;

for the KVL equations written on the paths defined bythe two meshes, we have

v0(t)+ v1(t)= 0

−v1(t)+ v2(t)+ v3(t)= 0.

To solve these equations, we notice first that thetop KVL equation is the only equation involving v0(t).If we ignore it for the moment, we are left with sixequations in six unknowns. Using the three v–i relationsto eliminate the voltage variables gives for the remainingthree equations

i1(t)+ i2(t)= is(t)

−i2(t)+ i3(t)= 0

−4i1(t)+ 3i2(t)+ i3(t)= 0.

From the second of these equations, i2(t) = i3(t).Applying this fact to the other two equations gives

i1(t)+ i2(t)= is(t)

−i1(t)+ i2(t)= 0.

Solving these yields the result

i1(t) = i2(t) = i3(t) = 0.5 is(t).

Knowing the currents then gives

v1(t)= 4i1(t) = 2 is(t),

v2(t)= 3i2(t) = 1.5 is(t),

v3(t)= 1i3(t) = 0.5 is(t),

v0(t)=−v1(t) = −2.0 is(t).

To check your understanding, try DrillProblem P2-1.

WORKED SOLUTION: Exhaustive Method

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2-1 THE EXHAUSTIVE METHOD FOR WRITING A SUFFICIENT SET OF CIRCUIT EQUATIONS 57

Example 2-2 A Second Example of the ExhaustiveMethod

As a second and somewhat more complicated example,consider the circuit in Figure 2-3, which contains fourresistors, a voltage source, and a current source. Tosolve this circuit by using the exhaustive method willrequire setting up and solving 10 linear equations in 10unknowns. The unknowns are the voltages and currentsassociated with the four resistors, the current throughthe voltage source, i0(t), and the potential differenceacross the terminals of the current source, v0(t). Theresistor currents are shown. Assume that the four voltagesv1(t), v2(t), v3(t), and v4(t) are defined under the defaultsign convention on the appropriate resistors. The circuitcontains four nodes and three meshes. Therefore, the tenequations will consist of four element relations, threeKCL equations, and three KVL equations.

Writing the equations is straightforward. The fourelement relations are

v1(t)− 4i1(t)= 0

v2(t)− 3i2(t)= 0

is(t)

i1(t)

i2(t) i4(t)

i3(t)

i0(t)

s(t)+

0(t)–

Fig. 2-3. Another circuit to illustrate the ExhaustiveMethod (Example 2-2).

v3(t)− 2i3(t)= 0

v4(t)− 2i4(t)= 0.

If we write KCL equations at the three nodes that line uphorizontally in the middle of the circuit, we have

i0(t)+ i1(t)+ i2(t)= 0

−i2(t)+ i3(t)+ i4(t)= 0

i1(t)+ i4(t)= is(t).

Finally, writing KVL equations over the three meshesgives

v1(t)− v2(t)− v4(t)= 0

v2(t)+ v3(t)= vs(t)

v0(t)− v3(t)+ v4(t)= 0.

Notice that i0(t) and v0(t) each occurs in only oneequation. These equations can be ignored for the timebeing, since each merely provides values for these twovariables in terms of the others. Using the elementvariables to eliminate the voltages as variables in theremaining equations reduces the number of equations andunknowns to four:

−i2(t)+ i3(t)+ i4(t)= 0

i1(t)+ i4(t)= is(t)

4i1(t)− 3i2(t)− 2i4(t)= 0

3i2(t)+ 2i3(t)= vs(t). (2.2)

Unfortunately, finding the solution of the final fourequations in four unknowns is algebraically tedious. Itcan be approached with a pencil and paper, using familiarmethods from algebra, or it can be done on a scientificcalculator or computer. (Section 2-3 will discuss how

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58 CHAPTER 2 WRITING CIRCUIT EQUATIONS

MATLAB can be used to solve equations like these.) Bywhatever means, the currents are seen to be

i1(t) = 1

12vs(t)+ 4

9is(t)

i2(t) = 1

6vs(t)+ 2

9is(t)

i3(t) = − 1

12vs(t)+ 5

9is(t)

i4(t) = 1

4vs(t)− 1

3is(t),

which means that the element voltages are

v1(t) = 4i1(t) = 1

3vs(t)+ 16

9is(t)

v2(t) = 3i2(t) = 1

2vs(t)+ 2

3is(t)

v3(t) = 2i3(t) = −1

6vs(t)+ 10

9is(t)

v4(t) = 2i4(t) = 1

2vs(t)+ 2

3is(t).

Finally, from the two equations that we omitted earlier,

v0(t) = v3(t)− v4(t) = −2

3vs(t)+ 4

9is(t)

and

i0(t) = −i1(t)− i2(t) = −1

4vs(t)− 2

3is(t).

Exhausting!

To check your understanding, try DrillProblem P2-2.

2-2 Supernodes and Supermeshes

In each of the two examples of the last section, wewere able to simplify the solution slightly by observingthat some of the equations (one in the first exampleand two in the second) could be decoupled from theothers. This reduced the size of the set of equationsthat had to be solved simultaneously. In both cases, theequations that could be set aside were those that involvedsource variables (voltages across the terminals of currentsources and currents flowing through voltage sources).There was nothing special about these two examples; wecan always decouple the equations involving the sourcevariables from the others, but to do this we could haveto write some KCL equations that are associated withisolated subnetworks rather than with simple nodes, andwe might have to write some KVL equations on closedpaths that are not meshes. To simplify the developmentin Sections 2-2-1 and 2-2-2, we assume initially that ourcircuits do not contain dependent sources. This is onlytemporary, and we will look at circuits that contain themin Section 2-2-3.

2-2-1 Supernodes

Let us first consider those equations that contain asvariables the currents that flow through voltage sources.We observe that all of these equations must be KCLequations, since these variables cannot be part of theelement relations for the resistors and the KVL equationscontain only voltage variables. Each voltage source mustnecessarily connect two nodes. Figure 2-4 shows anexample of a voltage source in a circuit that connectstwo nodes labelled a and b. The current flowing throughthe source is i. If KCL equations are written at both ofthese nodes, those equations are

node a: i1 + i2 − i = 0 (2.3)

node b: i3 + i4 + i = 0. (2.4)

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2-2 SUPERNODES AND SUPERMESHES 59

s(t)i3

i4i2

i1

a b

i

Fig. 2-4. A section of a network containing a voltagesource that connects two nodes.

Adding the two equations together gives

i1 + i2 + i3 + i4 = 0. (2.5)

This equation does not involve the source variable i.It is, however, the KCL equation for the surface thatencloses the voltage source and the two nodes to which itis attached! By replacing the pair of equations (2.3) and(2.4) by the single equation (2.5), we remove one variable(i) and reduce the number of equations by one. A surfacethat encircles a voltage source and its two attached nodes,such as this one, is called a supernode.

In the preceding argument, it was assumed that theexhaustive method wrote KCL equations at both a and b.Suppose instead that node b is the single node at whicha KCL equation is not written. In this case the KCLequation at node a,

i1 + i2 − i = 0,

is the only equation that involves the variable i and it onlyserves to tell us that

i = i1 + i2.Thus, unless we need to know the value of i, we can ignorethis equation, thereby again reducing both the number ofvariables and the number of equations by one. If we doneed the value of i, we can use the above equation afterwe have solved for i1 and i2.

is(t)–(t)+

+2(t)–

+1(t)

Fig. 2-5. A current source separating two meshes.

If we turn all of the voltage sources into supernodes andwrite KCL equations at all but one of the non-encirclednodes or supernodes, we will have eliminated all ofthe voltage source currents as variables and reduced thenumber of equations by the number of voltage sources.

2-2-2 Supermeshes

Next, we need to decouple the equations that involvethe voltages associated with current sources. The onlyequations that contain these voltages are KVL equations.

There are two cases that we need to consider. First,consider the case where the current source separates twomeshes, as in Figure 2-5.

The potential difference between the terminals of thecurrent source is v(t). The KVL equations for the twomeshes are

left mesh: −v1(t)− v(t) = 0 (2.6)

right mesh: v(t)+ v2(t) = 0. (2.7)

Adding the two equations together gives

−v1(t)+ v2(t) = 0. (2.8)

Eq. (2.8) is the KVL equation of the closed path formedfrom the union of these two meshes that are separated bythe current source. It is, of course, a valid KVL equationand it is consistent with (2.6) and (2.7). A path such as thisone that is formed from two adjacent meshes separated bya current source is called a supermesh. Replacing (2.6)

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60 CHAPTER 2 WRITING CIRCUIT EQUATIONS

is(t) s(t)+2(t)–

+(t)–

+1(t)–

Fig. 2-6. A circuit containing a current source that doesnot separate two meshes.

and (2.7) by (2.8) reduces the number of equations andthe number of variables by one (v(t)).

The second case that we need to consider occurs whena circuit contains a current source that does not separatetwo meshes, as in Figure 2-6. This will occur when thecurrent source lies on the outer “boundary” of the circuit.To give these sources a name, we shall call them exteriorcurrent sources. In this case, the KVL equation writtenover the left mesh,

−v(t)+ v2(t) = 0,

is the only one that contains the variable v(t). Ignoringthat equation (unless we need the value of v(t)) againreduces the number of equations and the number ofvariables by one.

If we define a supermesh for every pair of adjacentmeshes (or supermeshes) that are separated by a currentsource and then write a KVL equation for each meshor supermesh whose path does not contain an exteriorcurrent source, we will have decoupled all of the currentsource voltages as variables and reduced the number ofequations by the number of current sources.

By using both supernodes and supermeshes, weeliminate all of the source variables as unknowns andreduce the number of equations by the number of sources.If the circuit contains e elements, the resulting system ofequations contains 2e equations in 2e unknowns. e of the

unknowns are element voltages and e of them are elementcurrents. e of the equations are element relations and theremaining e are KCL and KVL equations.

The modifications to the exhaustive method thatincorporate supernodes and supermeshes to eliminatesource variables define the simplified exhaustivemethod. Even though we have yet to incorporatedependent sources, we shall formally state the methodnow. This statement is valid regardless of whether thereare dependent sources.

SIMPLIFIED EXHAUSTIVE METHOD:

1. Create a supernode around eachvoltage source and the nodes to whichit is attached.

2. Create a supermesh from each pair ofadjacent meshes that are separated bya current source.

3. Define a voltage variable and a currentvariable for each element (not thesources) in the circuit.

4. Write a v–i relation for each element.

5. Write a KVL equation for each mesh orsupermesh whose path does not containan exterior current source.

6. Write a KCL equation for all but one ofthe nodes or supernodes.

7. Solve the resulting set of equations forthe element variables.

8. If any of the source variables arerequired, solve for each of themafter the element variables have beendetermined.

Either of these methods—the exhaustive method or thesimplified exhaustive method—is guaranteed to find the

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2-2 SUPERNODES AND SUPERMESHES 61

solution of any planar circuit. The complete solution canbe found by solving even fewer linear equations by usingthe node method and the mesh method. These proceduresare developed later in the chapter.

Example 2-3 Using the Simplified Exhaustive Methodto Write KCL and KVL Equations

Let us illustrate the procedure we have just developedusing the simple circuit sketched in Figure 2-7, which wesolved via the exhaustive method in Example 2-1.

Since this circuit contains three resistors, there are sixelement variables: the currents through the three resistors,and the three associated voltages. The circuit does notcontain any voltage sources, so we do not need to createany supernodes, but it does contain a current source.Since that source is located on the exterior edge of thecircuit, we will not write a KVL equation on the pathcorresponding to the left mesh. This means that we needto write two KCL equations and one KVL equation. Tolimit the number of additional equations, we use only thevoltage variables and then use Ohm’s law and the defaultsign convention to express the element currents in termsof the voltages (i = v/R).

The KVL equation written around the right mesh is

−v1(t)+ v2(t)+ v3(t) = 0.

is(t)+3(t)–

+1(t)

+2(t)

4

3

1

Fig. 2-7. A simple circuit to illustrate how the simplifiedexhaustive method can be used to construct a minimal setof KVL and KCL equations. All resistances are measuredin ohms.

From KCL written at the upper left node, we have

1

4v1(t)+ 1

3v2(t) = is(t),

and from KCL at the upper right node, we have

1

3v2(t)− v3(t) = 0.

Solving these three linear equations in three unknownsgives the solution

v1(t) = 2 is(t); v2(t) = 3

2is(t); v3(t) = 1

2is(t).

The three element currents can be computed from thethree voltages:

i1(t) = 1

2is(t); i2(t) = 1

2is(t); i3(t) = 1

2is(t).

We observe that every voltage in the circuit and everycurrent is proportional to the single source signal, is(t).This is the same solution that we got in Example 2-1,except that, in that example, we set up and solved sevenequations in seven unknowns. (We very quickly reducedthose to three equations in three unknowns that werecomparable to the equations above.)

To check your understanding, try DrillProblem P2-3.

Example 2-4 Example 2-2 Revisited

The circuit in Figure 2-8 reexamines the slightly morecomplex example that we saw in Example 2-2. Since thecircuit contains a voltage source, we create a supernodethat encircles it. The circuit also contains a current source,but that source is located on the exterior boundary of thecircuit, so its effect will be to render unnecessary theKVL equation on its mesh. There are three nonencirclednodes or supernodes, as shown on the drawing, so wewill need to write two KCL equations in addition to

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62 CHAPTER 2 WRITING CIRCUIT EQUATIONS

is(t)s(t)

i1(t)

i2(t) i4(t)

i3(t)

2Ω3Ω

Fig. 2-8. Another circuit to illustrate writing a minimalset of KCL and KVL equations.

the two KVL equations on the two remaining meshes.Together with the four v–i relations for the resistors, thesewill be enough to specify completely the eight elementvariables associated with the four elements. (With theexhaustive method, we had to write three KCL equationsand three KVL equations.) As in the previous example,we can incorporate Ohm’s law directly into the writingof some of the equations to help reduce their number.We use the resistor currents as our variables, as we didin Example 2-2, being careful to adhere to the defaultsign convention. We arbitrarily select the supernode asthe node to be ignored when writing KCL equations. Thefour equations are as follows:

left node: −i2(t)+ i3(t)+ i4(t) = 0

right node: i1(t)+ i4(t) = is(t)

top mesh: 4i1(t)− 3i2(t)− 2i4(t) = 0

bottom mesh: 3i2(t)+ 2i4(t) = vs(t).

Since these are the same four equations that we had inequation (2.2) in Example 2-2, the solution is obviouslythe same as well.

To check your understanding, try DrillProblem P2-4.

2-2-3 Dependent Sources

To simplify our discussion to this point, we have assumedthat our networks were free of any dependent sources.Now it is time to remove that restriction. Fortunately,this is not difficult.

Dependent sources behave like independent ones; theyhave one variable that is controlled (the voltage in adependent voltage source or the current in a dependentcurrent source) and one variable that is uncontrolled (thecurrent in a dependent voltage source or the voltage in adependent current source). The value of the uncontrolledvariable is determined by the solution of the completecircuit. It can be found once the element variables areknown by writing a single KCL or KVL equation. Thevalue of the controlled variable is linked to anothervoltage or current in the circuit. As a result, the addition ofa dependent source does not introduce any new variableinto the network whose value needs to be computedinitially. Thus, it does not create the need for anyadditional equations. We cannot, of course, simply ignorethe dependent sources; we need to account for them inour equations, but it is important to note that we do notneed to add any more equations or variables to do thiswith the simplified exhaustive method.

The simplest way to deal with a dependent source is totreat it initially as if it were an independent source. Wecreate a supernode around each of the dependent voltagesources and we create a supermesh encircling any of theinterior dependent current sources. We then write a KVLequation around the meshes or supermeshes in the circuitthat do not contain (independent or dependent) currentsources and a KCL equation at all but one of the nodes orsupernodes. For the dependent sources, we next substitutethe controlling relation, which must be expressed in terms

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2-2 SUPERNODES AND SUPERMESHES 63

of element voltages and/or element currents. We illustratethis procedure in the following example.

Example 2-5 Simplified Exhaustive Method with aDependent Source

Consider the circuit drawn in Figure 2-9. If we create asupernode that encircles the voltage source and the twonodes to which it is attached, there will be two remainingisolated nodes, each of which is connected to one ofthe terminals of the 2 resistor. Thus, we need to writetwo KCL equations. Since the center and right mesh areseparated by a (dependent) current source, we create asupermesh from them and write two KVL equations—one around the left mesh and one around the supermesh.We also write four element relations for the four resistors.This results in a total of eight equations in the fourresistor voltages and the four resistor currents. In thiscircuit, since all of the elements are resistors, the elementrelations are simple:

v1(t)= 6i1(t)

v2(t)= 3i2(t)

v3(t)= 2i3(t)

v4(t)= 4i4(t).

We can substitute immediately for the element voltagesand write both the KVL and KCL equations in terms

s(t)

i1(t) i3(t)

i2(t) i4(t)

– i2(t)

6Ω 2Ω

3Ω 4Ω12

Fig. 2-9. An example of a network with one independentand one dependent source.

of the current variables alone. Doing this reduces thenumber of equations that we have to solve formally.

The first KVL equation is written around the left meshand uses the current variables. Remember to be carefulwith the sign conventions.

6i1(t)+ 3i2(t) = vs(t). (2.9)

The second KVL equation is written over the supermeshthat encircles the dependent current source:

−3i2(t)+ 2i3(t)+ 4i4(t) = 0. (2.10)

Since the network has three node/supernodes, we needto write KCL equations at two of them. We choose towrite them at the two isolated nodes. For the left node,we get

i1(t)− i2(t)− i3(t) = 0, (2.11)

and for the right one,

i3(t)− i4(t) = − 12 i2(t)

⇒ 12 i2(t)+ i3(t)− i4(t) = 0. (2.12)

These are our four equations in four unknowns. Byputting all of the variables on one side of the equationsand the source terms on the other, we have the equationsin a form that software packages, such as MATLAB, canaccept to provide a ready solution. It is also a form thatmakes a pencil-and-paper solution easier. By whatevermethod, for this problem the solution is

i1(t)= 7/60 vs(t) = 0.1167 vs(t)

i2(t)= 1/10 vs(t) = 0.1000 vs(t)

i3(t)= 1/60 vs(t) = 0.0167 vs(t)

i4(t)= 1/15 vs(t) = 0.0667 vs(t).

Section 2-3 reviews methods for writing linear equations,such as these, in matrix–vector form and shows howMATLAB or similar packages can be used to solve theseequations.

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64 CHAPTER 2 WRITING CIRCUIT EQUATIONS

To check your understanding, try DrillProblem P2-5.

2-3 Solving Circuit Equations

If a planar, resistive circuit contains e elements (resistors),then finding the complete solution of the network viathe simplified exhaustive method involves setting up andsolving a system of e linear equations in e unknownswhen Ohm’s law is used to eliminate either the voltagevariables or the current variables before the equations aresolved. We shall shortly explore some other techniquesthat reduce the number of equations further, but setting upand solving a set of linear equations is a key componentof all of these methods. This section looks at methods forwriting and manipulating the linear equations in matrix–vector form. This frames the problem in a form where aprogrammable calculator or a software package, such asMATLAB, can deal with it. In doing so, we will alsomake a key observation about circuits that will proveto be particularly useful. We begin by reviewing someproperties of matrices.

2-3-1 Matrices

An m × n matrix is a rectangular array of numbers orfunctions constructed to havem rows and n columns. Thenumbers that make up the array are called the elementsof the matrix, and the numbers m and n are called itsdimensions. As an example, the matrix

A =α11 α12 α13 α14

α21 α22 α23 α24

α31 α32 α33 α34

is a 3× 4 matrix consisting of the elements αik, where ispecifies the row in which the element is located and k

specifies its column. The special case of anm× 1 matrixis called a column vector. Column vectors take the form

b =

b1

b2...

bm

.

A 1× n matrix is called a row vector. The vector

c = [ c1 c2 · · · cn ]is an example of a row vector. Notice that, for row andcolumn vectors, a single-subscript notation is sufficientto represent the elements. Our convention will be to uselowercase boldface letters to denote vectors and upper-case boldface letters to denote more general matrices.

EXTRA INFORMATION: Matrix Equations

2-3-2 Matrix Operations

Matrix Addition: Two matrices can be added only ifthey have the same dimensions (the same numbers ofrows and columns.) If two matrices A and B are of thesame size, we define their sum, C = A+B, as the matrixformed from the sums of their corresponding elements.That is,

cik = aik + bik.Since row and column vectors are special cases ofmatrices, the same procedure can be used for summingvectors, provided that they are the same size.

Scalar Multiplication: Multiplication of a matrixAbya scalar α, denoted by D = αA, is done by multiplyingeach element in the matrix by the scalar. Thus, if aik isthe (i, k)th element of A and dik is the (i, k)th element ofD, then

dik = αaik.

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2-3 SOLVING CIRCUIT EQUATIONS 65

Matrix Multiplication: The product of two matricesE = AB is defined only if the two matrices havecompatible dimensions; the number of columns inAmustequal the number of rows in B. If A is anm×nmatrix andB is an n× p matrix, then the product E will be m× p.The element eik of E is computed from the formula

eik =n∑=1

aibk,

which is the inner product of the ith row of A with thekth column of B. Usually, matrix multiplication is notcommutative (i.e. AB = BA). In fact, if m = p, theproduct BA is not even defined. Even when m = p,however, these two products are generally different.

Matrix multiplication, like scalar multiplication, obeysthe distributive law with respect to addition. Thus,

A(B + C) = AB +AC

and

(D +E)F = DF +EF .

Matrix Inverse: The rows of a matrix are linearlyindependent if none of them can be expressed as a linearcombination of the others. This means that if ai is the ith

row of a matrix A, which contains m rows, there are noconstants ck such that

ai =m∑

j = 1j = i

cjaj .

If the rows of the matrix A are linearly independent and ifthe number of rows,m, of A is equal to the number of itscolumns, then A is invertible. The inverse of A, denotedas A−1, is a matrix that satisfies the matrix equation

A−1A = AA−1 = I ,

where I is them×m identity matrix. An identity matrixis a square matrix whose elements ik satisfy

ik =

1, k = 0, k = .

Example 2-6 Matrix Inverse

The matrix

A =1 0−1

1 1 01 0 1

has the inverse

A−1 = 0.5 0 0.5−0.5 1−0.5−0.5 0 0.5

.

It is straightforward to verify that

AA−1 =1 0−1

1 1 01 0 1

0.5 0 0.5−0.5 1−0.5−0.5 0 0.5

=

1 0 0

0 1 00 0 1

.

We also notice that

A−1A = 0.5 0 0.5−0.5 1−0.5−0.5 0 0.5

1 0−1

1 1 01 0 1

=

1 0 0

0 1 00 0 1

.

Any text in linear algebra will discuss a number ofefficient techniques for determining the inverse of aninvertible matrix. We shall not discuss any of these herebecause we do not need them. Computational packagessuch as MATLAB contain functions for invertingmatrices and we shall simply use them whenever we needto invert a matrix.

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66 CHAPTER 2 WRITING CIRCUIT EQUATIONS

2-3-3 Representing Linear Equations inMATLAB*

The system of m linear equations in n unknowns

a11x1 + b12x2 + · · · + a1nxn = b1

a21x1 + b22x2 + · · · + a2nxn = b2

...

am1x1 + bm2x2 + · · · + amnxn = bm

can be written in matrix–vector form as

Ax = b

if we define

A =

a11 a12 . . . a1n

a21 a22 . . . a2n...

.... . .

...

am1 am2 . . . amn

as the m× n matrix of coefficients,

x =

x1

x2...

xn

as an n× 1 column vector of variables, and

b =

b1

b2...

bm

as an m × 1 column vector of constants. The setof equations has a unique solution if the matrix ofcoefficients is invertible.

The default operations in MATLAB are matrixoperations. To enter a series of linear equations, such

as the ones above, one simply enters the matrix ofcoefficients A and the column vector of constants fromthe right-hand side, b. The solution can then be found byusing the MATLAB construction

x=A\b

Example 2-7 Solving Equations by Using MATLAB

As a simple example, we can use MATLAB to solve thesimple set of three linear equations in three unknowns

1.4x1 + 2.3x2 + 3.7x3 = 6.5

3.3x1 + 1.6x2 + 4.3x3 = 10.3

2.5x1 + 1.9x2 + 4.1x3 = 8.8.

We solve these by entering the 3×3 matrix of coefficients,entering the 3× 1 column vector of coefficients from theright-hand sides, and then solving the equations, usingthe “backslash” command. This is shown in the followingtranscription.

>> A = [1.4 2.3 3.7; 3.3 1.6 4.3;...2.5 1.9 4.1]

A =1.4000 2.3000 3.70003.3000 1.6000 4.30002.5000 1.9000 4.1000

>> b = [6.5; 10.3; 8.8]

b =6.500010.30008.8000

>> x=A\b

x =

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2-3 SOLVING CIRCUIT EQUATIONS 67

1.0000-1.00002.0000

>>

The commands are the expressions entered following theprompt >>. The values of the expressions are echoed byMATLAB. Those echoes are suppressed if a commandis followed by a semicolon (;). To enter a matrix (orvector), a space is used to delineate entries on a row anda semicolon is used to mark the end of a row.

DEMO: Solving Equations with MATLAB

2-3-4 Matrix Descriptions of Resistive Circuits

All of the circuit examples to this point have hadgeneric waveforms, such as vs(t) or is(t), as theirindependent sources. This allowed us to keep our resultscompletely general, and it also allowed us to make a keyobservation about resistive circuits: all of the element-variable waveforms are linear combinations of the sourcewaveforms. When the inputs are generic, we can stillsolve the resulting equations by using MATLAB, but weneed to manipulate them a bit more first.

To illustrate the approach, we use the network inFigure 2-10 as a working example. This network containstwo independent sources and five resistors. Thus, thereare 10 element variables, whose values can be foundfrom two KVL equations (one on the right mesh and oneon the supermesh that goes around the current source),three KCL equations (at, for example, the supernodethat encircles the voltage source and two of the threeremaining isolated nodes), and the five v–i relations.The element relations allows us to eliminate the voltagevariables immediately and to write both the KCL andKVL equations in terms of the five current variables

is(t)

s(t)

i4(t) i5(t)i1(t)i2(t) i3(t)

3Ω4Ω

1Ω2Ω

Fig. 2-10. A resistive network with two independentsources.

that are indicated. Anticipating putting these equationsinto matrix–vector form, we write them with the currentvariables on one side (the left) and the sources on theother (the right).

We begin with the KVL equations written on thesupermesh and the right mesh:

−2i1(t)+ 2i2(t)+ i3(t)+ 4i4(t)= 0

−4i4(t)+ 3i5(t)= vs(t).Then we add the KCL equations at the two nodes attachedto the horizontal 2 resistor and the supernode:

−i1(t)− i2(t)= 0

i2(t)− i3(t)=−is(t)i3(t)− i4(t)− i5(t)= 0.

Next, we put these equations into matrix–vector form:

Ci(t) = s(t). (2.13)

C is the 5 × 5 matrix of coefficients, i(t) is a 5 × 1column vector containing the current variables, and s(t)

is a 5 × 1 column vector of independent source termsfrom the right-hand sides. Notice that the coefficients ofthe matrix C are all constants; they do not vary with time.

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68 CHAPTER 2 WRITING CIRCUIT EQUATIONS

For our example, (2.13) takes the form−2 2 1 4 0

0 0 0−4 3−1−1 0 0 0

0 1−1 0 00 0 1−1−1

︸ ︷︷ ︸C

i1(t)

i2(t)

i3(t)

i4(t)

i5(t)

︸ ︷︷ ︸i(t)

=

0vs(t)

0−is(t)

0

.

︸ ︷︷ ︸s(t)

(2.14)

The rows of the C and s(t)matrices correspond to the fiveequations, and the columns of the C matrix correspondto the coefficients of the individual variables. That is,the first column of C contains the coefficients of i1(t)in the five equations, the second column contains thecoefficients of i2(t), etc. Notice that we must explicitlyinclude a coefficient value of zero for each current thatdoes not appear in a particular equation.

The right-hand side of (2.14) is still not in a formthat we can use. We can manipulate it into such aform, however, by rewriting it as the sum of columnvectors, each of which is proportional to one of the sourcewaveforms:−2 2 1 4 0

0 0 0−4 3−1−1 0 0 0

0 1−1 0 00 0 1−1−1

i1(t)

i2(t)

i3(t)

i4(t)

i5(t)

=

01000

vs +

000−1

0

is .

(2.15)

If we define sv and si as the two (constant) columnvectors on the right-hand side of (2.15), namely,

sv =

01000

and si =

000−1

0

,

then the system of equations can be written as

Ci(t) = sv vs(t)+ si is(t). (2.16)

Since this system of equations possesses a solution, C isinvertible, and we can write the solution in the form

i(t)= C−1sv vs(t)+ C−1si is(t)

= iv vs(t)+ ii is(t),

where we have defined

iv = C−1sv and ii = C−1si .

Notice that iv and ii are constant column vectors and thatthey can be found by solving the following two sets oflinear equations:

Civ = sv; Cii = si .

Because both sets of equations have the same coefficientmatrix C, we can solve both of them at the same time bycreating a matrix s with two columns, one containing svand the other containing si . The resulting solution vectori will also contain two columns, one containing iv and theother containing ii . (See the MATLAB documentation.)This is preferable to computing C−1 explicitly, since itrequires less computation if properly programmed. Thecomplete solution can be found by using the followingMATLAB dialog:

>> C=[-2 2 1 4 0; 0 0 0 -4 3; ...-1 -1 0 0 0; 0 1 -1 0 0; ...0 0 1 -1 -1];>> s=[0 0; 1 0; 0 0; 0 -1; 0 0];>> i=C\s

i =

-0.0851 0.40430.0851 -0.40430.0851 0.5951-0.1064 0.25530.1915 0.3404

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2-3 SOLVING CIRCUIT EQUATIONS 69

Therefore, the five currents are

i1(t) = −0.0851 vs(t)+ 0.4043 is(t)

i2(t) = 0.0851 vs(t)− 0.4043 is(t)

i3(t) = 0.0851 vs(t)+ 0.5951 is(t)

i4(t) = −0.1064 vs(t)+ 0.2553 is(t)

i5(t) = 0.1915 vs(t)+ 0.3404 is(t).

Example 2-8 Using MATLAB to Find Circuit Variables

To illustrate the procedure with a second example,consider the circuit in Figure 2-11. Since the two currentsources lie on the outer boundary of the circuit, only oneKVL equation is necessary (on the center mesh). We alsoneed to write KCL equations at two of the three nodes.Using the three resistor currents as variables and selectingthe two upper nodes makes the three equations

KVL: −2i1(t)+ i2(t)+ 3i3(t) = 0

KCL1: i1(t)+ i2(t) = is(t)

KCL2: i2(t)+ 2i2(t)− i3(t) = 0.

Again, all of the variables have been placed on the left-hand sides, and the independent source terms have beenplaced on the right. In matrix–vector form, these become

−2 1 31 1 00 3−1

i1(t)i2(t)

i3(t)

=

0

10

is(t).

is(t)

i3(t)i1(t)

i2(t)

2i2(t)3Ω2Ω

Fig. 2-11. Circuit for Example 2-6.

The solution obtained from MATLAB is

i1(t) = 0.8333 is(t)

i2(t) = 0.1667 is(t)

i3(t) = 0.5000 is(t).

WORKED SOLUTION: Circuit Variablesin MATLAB

To check your understanding, try DrillProblem P2-6.

2-3-5 Superposition of Independent Sources

In addition to providing a means to enable us to useMATLAB to find the element variables in a circuit, thematrix representation of the circuit equations allows usto make several useful observations about circuits. Oneof these is an analysis method known as superpositionof sources.

Suppose that a network is excited by more than oneindependent source, such as the network in Figure 2-10,which has two independent sources. Then the sourcevector s(t)—the vector from the right-hand sides ofthe equations that contains the source terms—can beseparated into components from each of the sources.In the example of Figure 2-10, when the equations thatdefined the solution were put into matrix–vector form,we saw that the right-hand side could be written in theform

s(t)=

0vs(t)

0−is(t)

0

=

01000

vs(t)+

000−1

0

is(t)

= svvs(t)+ si is(t).

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70 CHAPTER 2 WRITING CIRCUIT EQUATIONS

Thus,

Ci(t) = svvs(t)+ si is(t).

The vector of solutions also divides into two components:

i(t)= C−1s(t)

= C−1[sv vs(t)+ si is(t)]= C−1sv vs(t)+ C−1si is(t)

= iv vs(t)+ ii is(t). (2.17)

Here,

iv= C−1sv

ii= C−1si .

We used this relationship in the previous section todescribe how to obtain a solution by using MATLAB.Now, let us look at it more closely. It says that the valueof each current (and each voltage) in the network containstwo components, one of which is proportional to thevoltage source and one of which is proportional to thecurrent source. Notice that the value of si is independentof the value of sv, and vice versa. This means that wecan turn off si (replace the current source by an opencircuit) and solve for iv. We can then turn off the voltagesource (replace it by a short circuit) and solve for ii .The resulting vectors iv and ii define the solution to theoriginal circuit. This property is known as the sourcesuperposition principle. Source superposition is clearlypossible for any network that has multiple independentsources. For a network with N independent sources, theoutput can be expressed as the sum ofN components fromthe sources acting independently. Source superpositionimplies that we can solve for the network variables byapplying the sources one at a time and then summingthe results. Chapter 3 presents some techniques fornetwork simplification that will make this approachparticularly useful. Source superposition cannot be used

for dependent sources, however, since turning off adependent source would affect the coefficient matrix C.

Example 2-9 Source Superposition

As an example of how source superposition can beapplied, consider the network in Figure 2-12(a), whichcontains two independent sources. Because this networkis fairly simple, it is not difficult to solve it for the currenti(t). We need to write one KVL equation over the path

is(t)

ib(t)

is(t)

ia(t)

s(t)

i1(t)

s(t)

i(t)

R2

R2

R1

R1

i2(t)

R2

R1

(a)

(b)

(c)

Fig. 2-12. An illustration of source superposition. (a) Anetwork containing two independent sources. (b) Thenetwork in (a) with the current source turned off. (c) Thenetwork in (a) with the voltage source turned off.

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2-3 SOLVING CIRCUIT EQUATIONS 71

defined by the supermesh and one KCL equation on eitherthe supernode or the isolated node. Either KCL equationgives

i(t)− ia(t) = −is(t).

Applying KVL around the supermesh and using thecurrents as the variables results in the equation

R1i(t)+ R2ia(t) = vs(t).

From these, we can solve for i(t) by eliminating ia(t):

i(t) = 1

R1 + R2[vs(t)− R2is(t)] , (2.18)

which contains one component that is proportional toeach source.

As an alternative, we can consider the two networksshown in Figure 2-12(b) and (c). In Figure 2-12(b), thecurrent source from the original network is turned off;in Figure 2-12(c), the voltage source is turned off. If wesolve for i1(t) in the network in Figure 2-12(b) and fori2(t) in the network in Figure 2-12(c), then, by sourcesuperposition, we must have

i(t) = i1(t)+ i2(t).

For the circuit in Figure 2-12(b), the current i1(t) flowsthrough both resistorsR1 andR2, which are connected inseries. Applying KVL around the loop, we have

R1i1(t)+ R2i1(t) = vs(t)

or

i1(t) = 1

R1 + R2vs(t).

In part (c) of the figure, the voltage source is replacedby a short circuit. If we let the current through the resistorR2 be denoted by ib(t), then, from KCL at the upper node,

ib(t) = i2(t)+ is(t),

and from KVL applied around the outer path,

R1i2(t)+ R2ib(t) = 0.

We can solve these two equations for i2(t) by eliminatingib(t):

i2(t) = − R2

R1 + R2is(t).

Then

i1(t)+ i2(t)= 1

R1 + R2[vs(t)− R2is(t)]

= i(t),

which agrees with (2.18).

WORKED SOLUTION: Source Superposition

To check your understanding, try DrillProblem P2-7.

Two additional observations can be made from (2.17):(1) For a resistive circuit, all of the element variableswill be equal to a superposition of the source waveforms;(2) When all of the independent sources are turned off,all of the element variables will be zero.

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72 CHAPTER 2 WRITING CIRCUIT EQUATIONS

2-4 The Node Method

The exhaustive method and the simplified exhaustivemethod are systematic procedures for writing sets oflinear equations that can be solved to find the completesolution of a planar resistive network. The simplifiedexhaustive method results in a complete set of 2eequations in the 2e element variables, when e is thenumber of elements in the network. As we saw, however,the matrix of coefficients in those equations is fairlysparse. This suggests that we might be able to reducethe number of equations if we use some of the simplerequations to reduce the number of variables. Onepossibility is to exploit the simplicity of the elementrelations when the elements are resistors, since, for aresistor,

vk(t) = Rkik(t).Clearly, if we know all of the resistor currents, we cantrivially determine all of the resistor voltages or viceversa. This suggests that we might use Ohm’s law toexpress all of the element voltages in terms of the elementcurrents, solve a reduced set of e equations to calculatethe e currents, and then use the calculated current valuesto find the voltages. We did this in many of the examples.

Actually, we can go farther than this. If we choose ourvariables carefully, we can incorporate both the elementrelations and the KVL equations into the KCL equations,to reduce the number of variables and equations to n−1,where n is the number of isolated nodes or supernodesin the circuit after supernodes have been created at thevoltage sources. This approach is called the node method.It is the subject of this section. Alternatively, we canincorporate the element relations and the KCL equationsinto the KVL equations. This reduces the number ofvariables and equations to , where is the numberof meshes or supermeshes that do not contain exteriorcurrent sources in the network. This latter approach iscalled the mesh method; it is the subject of the next

eb(t)

ec(t)

ea(t)

ed(t)

+2(t)–

+4(t)

+1(t)–

+3(t)

Fig. 2-13. A mesh extracted from a larger network withthe node potentials indicated.

section. Which approach is to be preferred depends upona number of factors, one of which is the relative values ofn and . All other factors being equal, we would normallyprefer the approach that leads to the smaller number ofequations to be solved, particularly if we are solving thoseequations by hand.

The key to the node method is identifying an alternativeset of fewer than e variables from which all of the elementvoltages and currents can be calculated. One way to dothis is to find a set of variables that implicitly incorporatesthe KVL constraints. One such set is the set of nodepotentials, the electrical potentials associated with eachnode measured with respect to a reference node. To seehow these have the desired attributes, consider the meshextracted from a larger circuit in Figure 2-13, where thenode potentials are denoted by ea(t), eb(t), ec(t), anded(t). The voltage across each element is the differencein potential between its two terminals. Thus, for example,

v1(t) = ea(t)− eb(t).

Recall that, in Chapter 1, we argued that the voltageassociated with an element is the potential at the +

terminal minus the potential at the − terminal. If we sum

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2-4 THE NODE METHOD 73

the voltages around the mesh, we obtain

v1(t)+ v2(t)− v3(t)− v4(t)

= [ea(t)− eb(t)] + [eb(t)− ec(t)]− [ed(t)− ec(t)] − [ea(t)− ed(t)] = 0.

As we traverse any path in the network, the potential atthe current node is equal to the potential at the startingnode plus the accumulated potential differences of theelements that have been encountered. When we returnto the starting node, the potential must be the same asthe initial potential, which means that the sum of theaccumulated voltages must be zero.

When we use the node potentials asvariables, KVL will always be satisfiedautomatically for all closed paths!

If we know all of the node potentials, we can computeall of the element voltages as their differences; once weknow all of the element voltages, we can calculate theelement currents by using Ohm’s law. Thus, knowingthe node potentials is enough to specify the completesolution. The next example illustrates the process ofcomputing element voltages and currents from the nodepotentials.

Example 2-10 Finding Element Variables from NodePotentials

The circuit in Figure 2-14 contains three nodes, with nodepotentials ea(t), eb(t), and ec(t), measured with respectto some reference, as indicated. The voltage across theterminals of the 1 resistor, v1(t), is simply the differencebetween two node potentials. Thus,

v1(t) = ea(t)− ec(t).

is(t)

eb(t)ea(t)

ec(t)

i2(t)

i3(t)+1(t)–

1Ω 3Ω

Fig. 2-14. A simple circuit with three nodes (Exam-ple 2-7).

The current i2(t), which flows through the 2 resistor,has its reference direction defined as pointing from nodeb to node a. Therefore,

v2(t) = eb(t)− ea(t)and

i2(t) = 1

2[eb(t)− ea(t)].

Similarly,

i3(t) = 1

3[eb(t)− ec(t)].

Every voltage in the circuit and every current can beexpressed in terms of the three node potentials ea(t),eb(t), and ec(t) and in terms of is(t).

When we defined the node potentials, we were verycareful to state that they were a measure of the electricalpotential measured with respect to a reference node.This is because we cannot compute the absolute valuesof these potentials, only their relative values. We cansee this by observing that, if we were to add a constantvalue to each node potential, this would not affect thevoltages across the terminals of any of the elements ortheir currents. Therefore, it is customary to arbitrarilyset one of the node potentials to zero and measure allof the remaining potentials relative to it. This reference

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74 CHAPTER 2 WRITING CIRCUIT EQUATIONS

node is designated as the ground; on a circuit diagram,it is indicated by the symbol shown in Figure 2-15. Theground node is frequently chosen to be the one that isomitted when writing KCL equations, but this is notnecessary; any of the n nodes (or supernodes) of thecircuit can be designated as the ground and any nodecan be the one omitted when writing KCL equations.

When the network contains voltage sources, there willbe supernodes created. We shall examine what to do inthese cases in one of the examples that follows. First,however, we state the node method as a formal procedure.

THE NODE METHOD:

1. Create a supernode encircling eachindependent or dependent voltagesource and the two nodes to which itis attached.

2. Select one of the nodes of the circuit asthe ground node.

3. Define n−1 node potential variables atthe remaining nodes or supernodes ofthe circuit.

4. Set up KCL equations at n − 1 ofthe nodes/supernodes in the network.The currents in these equations mustbe expressed in terms of the nodepotentials.

5. Solve those n−1 equations for the n−1node potentials.

6. Calculate the element voltages andcurrents of interest from the nodepotentials.

The whole procedure will be clearer after we look at afew examples.

Fig. 2-15. A ground node.

FLUENCY EXERCISE: Node Potentials

Example 2-11 The Node Method with Current Sources

This first example, the circuit drawn in Figure 2-16,looks at the node method for a circuit that does notcontain any voltage sources. This is the simplest case,because, for such networks, there are no supernodes.Thefour nodes are indicated by colored dashed lines. Three ofthe nodes are labelled a, b, and c; the fourth is identifiedas the ground. Let the node potentials on the nongroundnodes be denoted by ea(t), eb(t), and ec(t), respectively.(The potential at the ground node is zero.) Notice that,when we use the node potentials, it is not necessary toindicate the reference directions for the element voltages

is1(t)

is2(t)2Ω 4Ω

2Ωa c

b

Fig. 2-16. An example of a circuit with no voltage sourcesto illustrate the use of the node method.

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2-4 THE NODE METHOD 75

and currents, unless we desire the value of a particularone. The voltage across the 2 resistor measured fromnode a to node b, for example, is simply ea(t) − eb(t)and the current that flows through that resistor from nodea to node b is simply [ea(t) − eb(t)]/2. Similarly, thecurrent flowing in that resistor in the direction from nodeb to node a is [eb(t) − ea(t)]/2 = −[ea(t) − eb(t)]/2,which is the negative of the current flowing in the oppositedirection.

The next step, Step 4 of the formal procedure, involveswriting three KCL equations by summing the currentsthat leave (or enter) nodes a, b, and c. (Although it isnot required, it is recommended that all of the equationsbe written the same way, i.e., by summing the currentsentering each node or by summing the currents leavingeach node.) These need to be written by using the nodepotentials as variables. Summing the currents that leavenode a, we have

12 [ea(t)− 0]+ 1

2 [ea(t)− eb(t)] = is1(t).After we regroup the terms, this simplifies to

ea(t)− 12eb(t) = is1(t). (2.19)

Moving on to node b, we write

12 [eb(t)− ea(t)]+ 1

4 [eb(t)− 0]+ [eb(t)− ec(t)] = is2(t),which simplifies to

− 12ea(t)+ 7

4eb(t)− ec(t) = is2(t). (2.20)

Finally, at node c, we get the final KCL equation,

[ec(t)− eb(t)]+ 13 [ec(t)− 0] = −is1(t),

which reduces to

−eb(t)+ 43ec(t) = −is1(t). (2.21)

Notice that there is a regular structure to these equationsif they are written at the nonground nodes and if theyare written consistently. In the KCL equation for node

k, each of the current terms from the elements includesa contribution from node potential ek(t). Furthermore,all of these terms have the same algebraic sign, whereasall of the terms involving other node potentials in thatequation occur with the opposite sign.

We now have three equations in three unknowns. Thenext step is to solve them. This is facilitated by firstputting them into matrix–vector form: 1− 1

2 0

− 12

74 −1

0 −1 43

ea(t)eb(t)

ec(t)

=

1

0

−1

is1(t)+

0

1

0

is2(t).

Then we use MATLAB (or a calculator or handcalculation) to get the solution for the node potentials:

ea(t)= 56 is1(t)+ 2

3 is2(t)

eb(t)=− 13 is1(t)+ 4

3 is2(t) (2.22)

ec(t)=−is1(t)+ is2(t). (2.23)

Notice that, once again, each independent sourceproduces a component in each of the variables.

We can use the node potentials to find any of the othervariables in the circuit. For example, the current flowingfrom node a to node b through the 2 resistor is

iab(t) = 1

2[ea(t)− eb(t)] = 7

12is1(t)−

1

3is2(t).

Using the node method required that we solve three linearalgebraic equations in three unknowns. The simplifiedexhaustive method would have required solving fiveequations in five unknowns; the exhaustive methodwould have required solving seven equations in sevenunknowns.

To check your understanding, try DrillProblems P2-8 and P2-9.

The node method is valid for all circuits, but circuitsthat contain voltage sources can be trickier to deal with,

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76 CHAPTER 2 WRITING CIRCUIT EQUATIONS

s(t)

ea(t) + s(t)

ea(t)

Fig. 2-17. A supernode containing a voltage source hastwo node potentials, but these are not independent. Theirdifference is equal to the source voltage.

because they contain supernodes. Consider the supernodein Figure 2-17. It contains two nodes within it, buttheir two node potentials are not independent, since thedifference of these potentials must be equal to the sourcevoltage, vs(t). Thus, we treat one of the node potentialsas an unknown, ea(t) in Figure 2-17, and we write theother as ea(t)+ vs(t), where vs(t) is the source voltage.There is only one variable associated with the supernode,and we write only one KCL equation for it. In the writingof that equation, however, some of the currents will becalculated by using the node potential ea(t) and some willuse ea(t)+ vs(t), depending on which of the two nodesreceives the current. (If one of the nodes in the supernodeis the ground node, then ea(t) = 0.) The whole procedureis illustrated in the next example.

Example 2-12 The Node Method with Voltage Sources

To illustrate how to use the node method when thereare voltage sources, consider the network in Figure 2-18,which contains two of them. After Step 1 of theprocedure, we have two supernodes and one isolatednode, which are indicated by the colored dashed lines. Toaid in the discussion, the isolated node is node b, nodea is one of the two nodes within the left supernode (the

s2(t)

s1(t) 2Ω4Ω4Ω

6Ω2Ωa bc d

Fig. 2-18. An example to illustrate the use of the nodemethod when voltage sources are present (Example 2-12).

other is the ground), and nodes c and d are the two nodeswithin the other supernode. Although the complete circuitcontains four different nonzero node potentials, denotedas ea(t), eb(t), ec(t), and ed(t), only two of these areindependent. We can see this by looking at the networkdiagram. Notice that

ea(t) = vs1(t), (2.24)

because vs1(t) represents the difference in potentialbetween node a and the ground. Similarly,

ec(t) = ed(t)+ vs2(t). (2.25)

For this network, we can treat eb(t) and either ed(t) orec(t) as the independent variables and write our KCLequations in terms of these. We shall write equations atnode b and at the supernode enclosing nodes c, d , and theassociated voltage source. As before, we sum the currentsleaving the nodes (and surfaces).

At node b, we write

12

[eb(t)− vs1(t)

]+ 14eb(t)

+ 16

eb(t)− [ed(t)+ vs2(t)]

= 0.

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2-4 THE NODE METHOD 77

Here we used (2.24) and (2.25) to eliminate ea(t) andec(t) as variables. This equation can be simplified to

1112eb(t)− 1

6ed(t) = 12vs1(t)+ 1

6vs2(t). (2.26)

At the surface corresponding to the supernode containingnodes c and d of the original circuit, three currents crossthe surface. Thus,

16

[ed(t)+ vs2(t)] − eb(t)+ 1

4

[ed(t)+ vs2(t)

]+ 12ed(t) = 0.

Notice that, for the currents that originate from node c,the node potential is ec(t) = ed(t)+ vs1(t), and for thosethat originate at node d, ed(t) is the node potential. Thisequation can now be simplified to

− 16eb(t)+ 11

12ed(t) = − 512vs2(t). (2.27)

Solving (2.26) and (2.27) gives

eb(t)= 0.5641 vs1(t)+ 0.1026 vs2(t) (2.28)

ed(t)= 0.1026 vs1(t)− 0.4359 vs2(t). (2.29)

From eb(t) and ed(t), we can compute any of the othervoltages and currents in the network.

The important facts to remember when we use the nodemethod with voltage sources present are (1) that some ofthe surfaces at which we write KCL equations are notnodes, but supernodes, and (2) that the node potential forthe node at one terminal of a voltage source is tied to thepotential of the node at the other terminal by the sourcevoltage.

To check your understanding, try DrillProblem P2-10.

When the node method is used to solve a circuitcontaining a dependent source, there is an additional step

that is required. Consider a current-controlled voltagesource whose source voltage is governed by the relation

vs(t) = ri(t),where i(t), the controlling variable, is some other currentin the circuit. Since the node method requires that wewrite KCL equations by using only the node potentialsas variables, we need to first express i(t) in terms ofthe node potentials in the circuit before writing the KCLequations. This step is performed in the next example.

Example 2-13 Node Method with a Dependent Source

As a final example of the node method, consider thenetwork in Figure 2-19. This contains both a voltagesource and a dependent current source. When a dependentsource is present, it is necessary to restate the dependencerelation in terms of the variables of the problem, whichin this case are the node potentials.

The circuit contains one isolated node and onesupernode, so we will need only one unknown nodepotential and one KCL equation. In this network, thequantity of interest is the voltage across the load resistor,RL, which is denoted as vout(t). Since the value of thisvoltage is the same as the node potential, we denotethe node potential as vout(t). The potential at the nodethat connects the voltage source with resistor Rs is vs(t).

s(t)

ie(t)

αie(t)+out(t)–

Rs

RL

Fig. 2-19. An example of the node method with adependent source.

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78 CHAPTER 2 WRITING CIRCUIT EQUATIONS

Now that the node potentials have been labelled, we canexpress the current ie(t) in terms of them as

ie(t) = 1

Rs[ vs(t)− vout(t)].

The current through the dependent source is, therefore,equal to αie(t) = α[ vs(t)− vout(t)]/Rs . Writing a KCLequation at the circled node by summing the currents thatleave that node gives

1

Rs[vout(t)− vs(t)] + 1

RLvout(t)

= −α 1

Rs[ vs(t)− vout(t)]. (2.30)

Solving for vout(t) gives

vout(t)=1−αRs

1−αRs+ 1

RL

vs(t)

= 1

1+ RsRL(1−α)

vs(t).

To check your understanding, try DrillProblem P2-11.

2-5 The Mesh Method

The mesh method is very similar to the node method,except that the roles of voltages and currents are reversed.The latter uses the definitions of the node potentialsto guarantee that KVL is satisfied for all closed paths.Then, finding the complete solution becomes equivalentto finding a solution of the KCL equations, once thoseare expressed in terms of the node potentials. The meshmethod makes use of a different set of variables thatguarantees that KCL is satisfied for all nodes. It isthen sufficient merely to set up and solve the resulting

set of KVL equations on the appropriate meshes andsupermeshes. The mesh method produces the samevalues for all the element variables that the node methoddoes, but it results in fewer equations to be solved whenthe number of meshes/supermeshes is fewer than thenumber of nodes/supernodes ( ≤ n − 1); when thiscondition is not true, the node method requires fewerequations.

How can we find the appropriate variables? We arelooking to repeat the success of the node method, but withthe roles of KVL and KCL, the roles of nodes and meshes,and the roles of voltages and currents reversed. With thenode method, a node potential was a voltage that wasassociated with each node, such that each element voltagein the network could be expressed as the difference of twonode potentials. For the mesh method, we would like tofind a variable that is a current that we can associate witheach mesh, such that all of the element currents can beexpressed as differences of these new variables. We callthese variables mesh currents, or circulating currents.In the network shown in Figure 2-20, they are shownas the currents iα(t), iβ(t), and iγ (t) that appear in thecenters of the meshes.

i3(t)

i2(t)

i1(t)

Fig. 2-20. A network with three meshes and theirassociated mesh currents. (The time dependence of themesh currents has been suppressed.)

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2-5 THE MESH METHOD 79

We defined the mesh currents in Figure 2-20 to havea clockwise direction. This is not required, and for someproblems, it could be more convenient for one or moreof them to be defined to be counterclockwise. For thoseelements not on the exterior boundary of the network,the element current is the sum or difference of the meshcurrents of the meshes separated by the element. If thereference direction of the element current flows in thedirection of the mesh current, it is added (receives a +sign); if it is in the opposing direction, it is subtracted(receives a − sign). For example, for the currents i1(t),i2(t), and i3(t) in Figure 2-20,

i1(t)= iγ (t)− iα(t)i2(t)= iα(t)− iβ(t)i3(t)= iβ(t)− iγ (t).

For the elements on the perimeter, the element currentsare equal either to the appropriate mesh currents or totheir negatives.2

If the currents in the network are specified by usingthe mesh currents as variables, then KCL is satisfiedautomatically. For example, consider the network inFigure 2-20. If we sum the currents entering the dashednode in the center of the network, we see that

i1(t)+ i2(t)+ i3(t) = [iγ (t)− iα(t)]+ [iα(t)− iβ(t)] + [iβ(t)− iγ (t)] = 0. (2.31)

It is similar for the node on the right. The current enteringthe node through the upper element is iβ(t), the currententering through the lower element is −iγ (t), and thatentering through the horizontal element is iγ (t)− iβ(t).These three currents also sum to zero.

2For symmetry, we could define an additional mesh correspondingto the exterior of the network. We could then assign a(counterclockwise) mesh current with a value of zero to this mesh,which would serve as a “ground mesh.” Then all element currentscould be expressed as a sum or difference of two mesh currents andthe mesh method would be a closer parallel to the node method.

When we use the mesh currents as variables,KCL is satisfied automatically at all nodes!

Once we know all of the mesh currents, we cancompute all of the element currents, and once we knowall of the element currents, we can calculate all of theelement voltages, using Ohm’s law.

Example 2-14 Finding Element Variables from MeshCurrents

The simple circuit in Figure 2-21 contains two meshesand two mesh currents, labelled iα(t) and iβ(t). It isstraightforward to express the indicated element variablesin terms of iα(t) and iβ(t). The voltage v1(t) associatedwith the 3 resistor is

v1(t) = 3iα(t),

since the reference direction for the current (by the defaultsign convention) is in the same direction as the meshcurrent. On the other hand,

v3(t) = −5iβ(t),

because the reference direction for this current is oppositeto the direction of the mesh current. Finally,

i2(t) = iα(t)− iβ(t).

s1(t) s2

(t)

i2(t)

iα iβ

+1(t)

––3(t)+

3

4

5

Fig. 2-21. A circuit with element variables to be expressedin terms of its two mesh currents. All resistances aremeasured in ohms.

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80 CHAPTER 2 WRITING CIRCUIT EQUATIONS

Since this current is associated with an element thatseparates two meshes, it is the difference of two meshcurrents.

The paths over which we write KVL equations willbe either meshes or supermeshes; as in the simplifiedexhaustive method, we also omit meshes or supermeshesthat contain exterior current sources. Just as the nodemethod requires special care when a circuit containsvoltage sources, so the mesh method requires special carewhen the circuit contains current sources. As before, weaddress the details of this method with examples afterfirst formally stating the mesh method as a procedure.

THE MESH METHOD:

1. Create a supermesh from any pair ofadjacent meshes that are separated byan independent or dependent currentsource.

2. Define a mesh current for each meshor supermesh that does not contain acurrent source.

3. Write KVL equations, each of whichis written over one of the paths overwhich the mesh currents mesh havebeen defined. The voltages in theseequations must be expressed in termsof the mesh currents.

4. Solve those equations for the meshcurrents.

5. Compute the element currents andvoltages of interest from the meshcurrents.

FLUENCY EXERCISE: Mesh Currents

Example 2-15 The Mesh Method with Voltage Sources

As a first example of the method, let’s consider thenetwork in Figure 2-18, which we solved earlier by thenode method. This circuit is redrawn in Figure 2-22 toshow the mesh currents explicitly. We have defined threemesh currents, because the circuit contains three meshesand no current sources.

The next step of the procedure is to write three KVLequations. As we prefer to do, we write these with theelement variables on the left-hand side and the sourceterms on the right. For the left mesh, the KVL equationcan be written as

2iα(t)+ 4[iα(t)− iβ(t)] = vs1(t).After regrouping terms, we find that this simplifies to

6iα(t)− 4iβ(t) = vs1(t).Similarly, for the center and right meshes, the KVLequations are

4[iβ(t)− iα(t)] + 6iβ(t)+ 4[iβ(t)− iγ (t)] = 0

4[iγ (t)− iβ(t)] + 2iγ (t) = −vs2(t).

s2(t)

iα iβ iγ

2

4

6

4 2

b d

s1(t)

Fig. 2-22. A redrawing of the network of Figure 2-18with the three mesh currents identified. All resistancesare measured in ohms.

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2-5 THE MESH METHOD 81

We can combine these three KVL equations into a singlematrix–vector equation 6−4 0

−4 14−4

0−4 6

iα(t)iβ(t)

iγ (t)

=

1

0

0

vs1(t)+

0

0

−1

vs2(t),

whose solution is

iα(t)= 0.2179 vs1(t)− 0.0513 vs2(t)

iβ(t)= 0.0769 vs1(t)− 0.0769 vs2(t)

iγ (t)= 0.0513 vs1(t)− 0.2179 vs2(t).

Since we previously solved this network with the nodemethod, it is instructive to compare these answers withthe ones in (2.28) and (2.29). The potential at node b

relative to the ground is the same as the voltage acrossthe leftmost 4 resistor. This can be computed, using themesh currents, as

eb(t) = 4[iα(t)− iβ(t)]= 0.5641vs1(t)+ 0.0126vs2(t)

and the potential at node d is the voltage across therightmost 2 resistor, which is

ed(t) = 2iγ (t) = 0.1026vs1(t)− 0.4359vs2(t).

Both of these agree with the earlier values, as indeed theymust.

The node method requires solving two equationsin two unknowns, whereas the mesh method requiressolving three equations in three unknowns, although theequations were easier to write with the mesh method thanwith the node method. Clearly, either method can be used,and both give the same result. Which method shouldbe selected is largely a matter of personal taste. If theequations will be solved by hand, the method that yieldsfewer equations is probably to be preferred; if a computeror calculator will be used to solve the equations, the one

that simplifies writing the equations might be the bettercandidate.

To check your understanding, try DrillProblem P2-12.

When a circuit contains current sources, supermeshesmight be generated and/or meshes containing exteriorcurrent sources will be ignored. In either case, there willbe fewer KVL equations and fewer independent meshcurrents than there are meshes in the circuit. The nextexample shows how to deal with this situation.

Example 2-16 The Mesh Method with Current Sources

The mesh method is complicated by the presence ofcurrent sources, just as the node method was complicatedby the presence of voltage sources. To explore this case,let us reexamine the network in Figure 2-16, using themesh method. The circuit is redrawn in Figure 2-23, withmesh currents shown in each of the four meshes of thecomplete network.

is1(t)

is2(t)

iβ iγ iδ 3Ω

1Ω2Ω

2Ω 4Ω

Fig. 2-23. A repetition of the network of Figure 2-16 toillustrate the mesh method. The time dependence of themesh currents has been suppressed for clarity.

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82 CHAPTER 2 WRITING CIRCUIT EQUATIONS

Even though we have indicated a mesh current foreach mesh of the complete network, there are only twoindependent mesh currents, and we will write only twoKVL equations. The independent mesh currents are iβ(t)and either iγ (t) or iδ(t). The other two mesh currentsshown in the complete circuit are constrained by theseand the current sources. If we look closely at the circuitdrawing, we notice that

iα(t) = −is1(t)iδ(t) = iγ (t)+ is2(t). (2.32)

In writing the KVL equations, it is often helpful to definea mesh current for each mesh in the circuit, as we havedone, and then to replace those that are not independentby their equivalent values in a second step. In this case,this means adding the mesh currents iα(t) and iδ(t). Then,once these equations have been written, we substitute(2.32) to reduce the number of unknowns. Writing KVLaround the left mesh gives

2iβ(t)+ 2[iβ(t)− iα(t)] + 4[iβ(t)− iγ (t)] = 0.

Setting iα(t) = −is1(t), this becomes

2iβ(t)+ 2[iβ(t)+ is1(t)] + 4[iβ(t)− iγ (t)] = 0.(2.33)

For the supermesh, we have

4[iγ (t)− iβ(t)] + [iδ(t)− iα] + 3iδ(t) = 0.

After we substitute iα(t) = −is1(t) and iδ(t) =iγ (t)+ is2(t), this becomes

4[iγ (t)− iβ(t)] + [iγ (t)+ is2(t)+ is1(t)]+ 3[iγ (t)+ is2(t)] = 0. (2.34)

We can collect terms to put these equations into our morestandard form:

8iβ(t)− 4iγ (t)=−2is1(t)

−4iβ(t)+ 8iγ (t)=−is1(t)− 4is2(t).

Or, in matrix–vector form,[8−4

−4 8

][iβ(t)

iγ (t)

]=

[−2

−1

]is1(t)+

[0

−4

]is2(t).

The solution is

iβ(t)=−0.4167 is1(t)− 0.3333 is2(t)

iγ (t)=−0.3333 is1(t)+ 0.6667 is2(t).

We leave it as an exercise for the reader to verify that thissolution is identical to the one we obtained for this circuitvia the node method in (2.23).

To check your understanding, try DrillProblem P2-13.

Circuits that contain dependent sources require anextra step when you are using the mesh method foranalysis, just as the node method required an extra step.Before we can write down the KVL equations, we need torewrite the constraint expression for the source in termsof the mesh currents. The next example illustrates theprocedure.

Example 2-17 Mesh Method with a Dependent Source

As a final example of the mesh method, let us reconsiderthe circuit with a dependent source that we analyzedvia the node method. This is redrawn in Figure 2-24.Although we have drawn the network with two meshcurrents, the circuit contains a single supermesh. Thus,there will be only one KVL equation and one of the meshcurrents depends upon the other. Looking at the currentthrough the dependent source, we see that

iα(t)− iβ(t) = αie(t).Furthermore, the controlling variable, ie(t) can easily bewritten in terms of iα(t), since

ie(t) = iα(t).

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2-6 CONSERVATION OF POWER 83

s(t) iα iβ

ie(t)

αie(t)

+out(t)–

Rs

RL

Fig. 2-24. An example of the mesh method with adependent source.

Putting these two relations together gives

iα(t)− iβ(t) = αiα(t)⇒ iα(t) = 1

1− α iβ(t). (2.35)

Now we can write the KVL equation around the outerloop, which corresponds to the supermesh:

Rsiα(t)+ RLiβ(t) = vs(t).Substituting for iα(t) yields

Rs

[1

1− α iβ(t)]+ RLiβ(t) = vs(t),

or

iβ(t) = 1

RL + Rs1−α

vs(t).

Now that we know the mesh current, we can use it tocalculate the output voltage:

vout(t)= RLiβ(t) = RL

RL + Rs1−α

vs(t)

= 1

1+ RsRL(1−α)

vs(t).

For this example, both the node method and the meshmethod required solving only a single equation in a singleunknown.

WORKED SOLUTION: Mesh Method

To check your understanding, try DrillProblem P2-14.

Both the node method and the mesh method aresystematic procedures for finding the complete solutionof a resistive network. We shall see later that theyalso serve as useful tools for more general networks.For simple networks, such as the final example above,however, ad hoc methods can be just as easy. In thenext chapter, we shall look at a number of techniques forsimplifying networks to make them easier to understandand easier to analyze.

2-6 Conservation of Power

The net power absorbed in any circuit is zero.

This law follows as a direct consequence of Kirchhoff’slaws, by using the definitions of the node potentialsor the mesh currents. (See Problem P2-57.) This resultis included in this chapter because it illustrates anapplication of node potentials that is different fromsimply analyzing a circuit.

For every watt of power absorbed in the resistors ina circuit, a watt must be supplied, either by the sourcesor by the other elements. If vk(t) is the voltage acrossthe terminals of the kth element or source and ik(t) isthe current flowing into its + terminal, then the formalstatement of the conservation of power for a circuit withN elements (or sources) is

N∑k=1

vk(t)ik(t) = 0. (2.36)

In this section, we demonstrate this result for thespecific circuit shown in Figure 2-25, although the resultis completely general.

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84 A CONFUSING ISSUE: PROPER TREATMENT OF SOURCES

A Confusing Issue: Proper Treatment ofSources

Most errors made by beginning students when theyuse the node and mesh methods occur becausethey treat the source variables improperly. Considerthe following circuit, in which all of the resistancesare measured in ohms.

s(t) iα iβ iγ

is(t)

4

6 12

If the mesh method is applied as described inSection 2-5, we would write two KVL equations—one around the mesh on the left and one over thesupermesh formed from the other two meshes witha path that encircles the current source. Becauseof the current source, mesh currents iβ(t) and iγ (t)are not independent: iγ (t) − iβ(t) = is(t); hence,iγ (t) = iβ(t)+ is(t). The two KVL equations are

− vs(t)+ 4iα(t)+ 6[iα(t)− iβ(t)] = 0

6[iβ(t)− iα(t)] + 12[iβ(t)+ is(t)] = 0.

Algebraic simplification turns these into

10iα(t)− 6iβ(t)= vs(t)

−6iα(t)+ 18iβ(t)=−12is(t),

and their solution is readily seen to be

iα(t)= 18 vs(t)− 1

2 is(t)

iβ(t)= 124 vs(t)− 5

6 is(t),

from which all of the other variables in the circuit canbe calculated.

Errors frequently occur when students try towrite KVL equations on all three meshes. Doing so

requires assigning a potential difference across theterminals of the current source. Remember, thisvoltage is almost never zero, nor is it equal tois(t). To get the correct solution in this situation, it isnecessary to assign a variable to this voltage andthen solve for it. For this example, let v(t) be thevoltage across the terminals of the current source(+ sign at the top). The three KVL equations are then

− vs(t)+ 4iα(t)+ 6[iα(t)− iβ(t)] = 0

6[iβ(t)− iα(t)] + v(t)= 0

−v(t)+ 12iγ (t)= 0.

To these we must add the constraint between iβand iγ :

iγ (t)− iβ(t) = is(t).

These four equations reduce to the ones that wederived earlier and, of course, they have the same(correct) solution. While this approach can be madeto work, it results in more equations to solve andprobably makes errors more likely. It is for thisreason that we prefer writing a minimal set ofequations by creating the appropriate supermeshesand supernodes, which eliminates the need todefine any source variables.

As a final note, observe that the potentialdifference across the current source is v(t) =12iγ (t) from the third KVL equation. Thus, v(t) =12[iβ(t) + is(t)] = 12[ 1

24 vs(t) + 16 is(t)] =

12 vs(t)+ 2 is(t), which is not zero!

Similar errors occur with the node method ifimproper assumptions are made about the currentflowing through the voltage sources. The best way todeal with this situation is to follow the procedure aswe have presented it in Section 2-4. Alternatively,the correct solution can be obtained if auxiliaryvariables are defined for these currents and thensolved for along with the node potentials.

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2-6 CONSERVATION OF POWER 85

eceaeb

ed

+1(t)–

+4(t)–

+6(t)–

+3(t)–

+2(t)– +

5(t)–

Fig. 2-25. Circuit to demonstrate conservation of power.

Let P be the total net power:

P =6∑k=1

vkik (2.37)

= v1i1 + v2i2 + v3i3 + v4i4 + v5i5 + v6i6.

We need to show that P = 0. In writing (2.37), wehave suppressed the dependence on time to simplify theresulting expressions. Using Figure 2-25, we next expresseach of the voltages in terms of the four indicated nodepotentials:

P = (ea − ed)i1 + (ea − eb)i2 + (ea − ec)i3+(eb − ed)i4 + (eb − ec)i5 + (ec − ed)i6.

Then we regroup the terms by node potential, instead ofby current. This gives the alternative expression

P = ea(i1 + i2 + i3)+ eb(−i2 + i4 + i5)+ ec(−i3 − i5 + i6)

+ ed(−i1 − i4 − i6). (2.38)

Now we write KCL equations for the four nodes,summing the currents that leave the node in each caseand paying close attention to the default sign convention:

node a: i1 + i2 + i3 = 0

node b: −i2 + i4 + i5 = 0

node c: −i3 − i5 + i6 = 0

node d: −i1 − i4 − i6 = 0.

Notice that the expression that multiplies ea in (2.38) isthe sum of the currents that leave node a, which must bezero because of KCL. Similarly, each of the other nodepotentials is multiplied by the sum of the currents leavingits particular node. Therefore,

P = ea · 0+ eb · 0+ ec · 0+ ed · 0 = 0,

which establishes the result. The net power absorbed inany circuit is zero.

Notice that this result depends only upon KVL andKCL. It places no restrictions on the elements, whichmay be resistors, inductors, capacitors, sources, or someother elements not yet defined.

DEMO: Conservation of Power

Example 2-18 Conservation of Power

To demonstrate the conservation of power, consider thecircuit in Figure 2-26. This is a simple circuit, but weallow it to have arbitrary source waveforms. The firststep is to solve the circuit. The circuit has one supernode,one isolated node, and one supermesh, so either the nodemethod or the mesh method will require only a singleequation. Let us use the node method. Let the bottomnode be the ground and the one at the other end of the 2resistor have node potential v1(t). Then, writing a KCLequation, we get

1

2v1(t)+ 1

4[v1(t)− vs(t)] = is(t).

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86 CHAPTER 2 WRITING CIRCUIT EQUATIONS

s(t)is(t)+1(t)–

−2(t)+

Fig. 2-26. Circuit to demonstrate conservation of power.

Thus,

3

4v1(t)= 1

4vs(t)+ is(t)

⇒v1(t)= 1

3vs(t)+ 4

3is(t).

We can use v1(t) to find the other three element variables:

v2(t) = vs(t)− v1(t) = 2

3vs(t)− 4

3is(t)

i1(t) = 1

2v1(t) = 1

6vs(t)+ 2

3is(t)

i2(t) = 1

4v2(t) = 1

6vs(t)− 1

3is(t).

Next, we need to compute the power absorbed in thefour elements. The power absorbed in the vertical resistoris

P1 = v1(t)i1(t)

= 1

18v2s (t)+

8

9i2s (t)+

4

9vs(t) is(t),

and the power absorbed in the horizontal resistor is

P2 = v2(t)i2(t)

= 1

9v2s (t)+

4

9i2s (t)−

4

9vs(t) is(t).

Thus, the total power absorbed in the two resistors is

P1 + P2 = 1

6v2s (t)+

4

3i2s (t).

The voltage across the terminals of the current source is−v1(t) if we adhere to the default sign convention, andits current is 4is(t). Therefore, the power absorbed by thecurrent source is

P3 =−v1(t) is(t)

=−1

3vs(t) is(t)− 4

3i2s (t).

The current entering the + terminal of the voltagesource is−i2(t) and its voltage is vs(t). Thus, the powerabsorbed by the voltage source is

P4 =−i2(t) vs(t)=−1

6v2s (t)+

1

3vs(t) is(t).

Therefore, the total power absorbed by the two sourcesis

P3 + P4 = −1

6v2s (t)−

4

3i2s (t) = −P1 − P2,

so that

P1 + P2 + P3 + P4 = 0,

and the net power absorbed is zero. Notice that the twosources have a negative absorbed power, which meansthat they are supplying the power that is absorbed by theresistors.

To check your understanding, try DrillProblem P2-15.

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2-7 CHAPTER SUMMARY 87

2-7 Chapter Summary

2-7-1 Important Points Introduced

• A planar network is a circuit that can be drawnon a sheet of paper without any elements or wirescrossing each other.

• The exhaustive method for finding the completesolution finds every voltage and current, includingthe unconstrained source variables, by solving 2e+slinear equations in 2e+ s unknowns, where e is thenumber of elements and s is the number of sources.

• For a planar network, there is a constraint betweenthe number of branches, b, the number of nodes, n,and the number of meshes, : b = n+ − 1.

• To calculate the values of the element variables in acircuit without solving for the unconstrained sourcevariables, it is sufficient to create a supernode aroundeach voltage source and a supermesh surroundingeach interior current source, then to write a v–irelation for each element, a KCL equation at everynode or supernode but one, and a KVL equation forevery mesh or supermesh in the circuit that does notcontain an exterior current source.

• The simplified exhaustive method finds the valuesof all of the element voltages and currents by solving2e linear equations in 2e unknowns.

• When the node potentials are used as theindependent variables in a circuit, there is no needto write KVL equations; KVL is automaticallysatisfied on any closed path.

• Using the node method, we need solve only n − 1equations in n − 1 node potentials, where n is thenumber of nodes or supernodes remaining aftersupernodes have been created around each of thevoltage sources.

• When the mesh currents are used as the independentvariables in a circuit, there is no need to write

KCL equations; KCL is automatically satisfied atall nodes.

• Using the mesh method, we need solve only

equations in mesh currents, where is the numberof meshes or supermeshes that do not containexterior current sources.

• The net power absorbed in a circuit is always zero.

2-7-2 New Abilities Acquired

You should now be able to do the following:

(1) Identify the nodes or supernodes in the circuit wherea minimal set of KCL equations can be written.

(2) Use supermeshes to identify the closed paths in thecircuit where a minimal set of KVL equations canbe written.

(3) Set up and solve the set of linear equations thatdefines the complete solution of a circuit using eitherthe exhaustive method or the simplified exhaustivemethod.

(4) Put a set of linear equations, such as the ones thatdefine the constraint equations of a resistive circuit,into matrix–vector form.

(5) Solve a set of linear equations, using MATLAB.

(6) Use superposition of sources to find the completesolution of a circuit with multiple independentsources.

(7) Express all element voltages and currents in termsof the node potentials.

(8) Find the complete solution of a circuit, using thenode method.

(9) Express all element voltages and currents in termsof a set of mesh currents.

(10) Find the complete solution of a circuit using themesh method.

(11) Compute the total amount of power absorbed andsupplied in a circuit.

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88 CHAPTER 2 WRITING CIRCUIT EQUATIONS

2-8 Problems

2-8-1 Drill Problems

All drill problems have solutions on the web site.

P2-1 Solve for the element voltages in the circuit inFigure P2-1, using the exhaustive method. This problemis similar to Example 2-1.

is3(t)

is2(t)

is1(t)

+3(t)–

+1(t)–

+2(t)–

5Ω 2Ω

10Ω

Fig. P2-1. Circuit for Problem P2-1.

P2-2 This problem, which is similar to Example 2-2, isconcerned with the circuit in Figure P2-2. It should besolved by using the exhaustive method.

(a) Define a complete set of variables for this circuit.

(b) How many nodes are present in this circuit?

(c) How many meshes are present in this circuit?

(d) Write a sufficient set of KCL equations for thiscircuit. Your equations should be written in termsof the indicated variables.

(e) Write a sufficient set of KVL equations. Theseshould also be written in terms of the indicatedvariables.

(f) Solve these equations for i1(t).

s(t)

i1(t)

is(t)+2(t)–

+3(t)–

2Ω 2Ω

Fig. P2-2. Circuit for Problem P2-2.

P2-3 Find the voltages v1(t), v2(t), and v3(t) for thecircuit in Problem P2-1, using the simplified exhaustivemethod (Figure P2-1). This is similar to Example 2-3.

P2-4 Define an appropriate set of variables and solvefor i1(t) and v2(t) in the circuit of Figure P2-2 using thesimplified exhaustive method.

P2-5 This problem is similar to Example 2-5. For thecircuit in Figure P2-5,

(a) Write a minimal set of KCL equations to specifythe element variables in the circuit. These should beexpressed in terms of the variables indicated on thefigure. Define supernodes as appropriate.

(b) Write a minimal set of KVL equations to specifythe element variables in the circuit. These should beexpressed in terms of the indicated variables.

(c) Put these equations into matrix–vector form andsolve for v1(t) and i2(t).

i2(t)

is(t)

21(t)

+1(t)

–20kΩ10kΩ

Fig. P2-5. Circuit for Problem P2-5.

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PROBLEMS 89

P2-6 This problem is concerned with the three networks,shown in Figure P2-6. For each of the networks, the goalis to solve for the element voltages and currents. Theuncontrolled source variables are not needed.

(a) Identify the closed paths in the circuit that can beused to write a minimal set of KVL equations.

s(t)

s(t)

i2(t)i1(t)

i3(t)i1(t)

i2(t)

i2(t) i3(t)

2i1(t)

81(t)

is(t)

is(t)

s(t)

+1(t)–

+3(t)–

(a)

(b)

(c)

Fig. P2-6. Networks for Problem P2-6.

(b) Identify the nodes and supernodes in the circuit thatcan be used to write a minimal set of KCL equations.

(c) Write the set of linear equations associated withthe paths and closed surfaces that you found in (a)and (b).

(d) Put these equations into matrix–vector form andsolve them for the indicated variables, usingMATLAB.

P2-7 The purpose of this problem is to demonstratethe superposition principle. We shall use the circuit inFigure P2-7.

(a) Defining an appropriate set of variables wherenecessary, use one of the familiar methods to expressv(t) in terms of vs(t) and is(t).

(b) Now turn off the voltage source by replacing it bya short circuit. Solve the resulting circuit for thevoltage v(t).

(c) Next, turn the voltage source back on, and turn offthe current source by replacing it by an open circuit.Solve the resulting circuit for the voltage v(t).

(d) Show that your result in part (a) is equal to the sumof your results in parts (b) and (c).

s(t)

is(t)+(t)–

3Ω3Ω

Fig. P2-7. Circuit for Problem P2-7.

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90 CHAPTER 2 WRITING CIRCUIT EQUATIONS

P2-8

(a) For the circuit in Figure P2-8, what is the minimumnumber of KCL equations that need to be written tospecify the complete solution?

(b) Select one of the nodes of the circuit as the groundnode and label it. Define a supernode around eachof the voltage sources. Define an appropriate set ofindependent node potentials, and indicate these onthe circuit drawing. Label all of the non-independentnode potentials as well. Next, write a KCL equationat each of the nonground nodes or supernodesin terms of the node potentials (and the sourcewaveforms).

(c) Put your equations into matrix–vector form with thenode potentials as unknowns. You do not need tosolve the equations.

s(t) is(t)

R2

R1

R3

R4

Fig. P2-8. Circuit for Problem P2-8.

P2-9

(a) Write KCL equations at nodes a and b in the circuitin Figure P2-9, using their node potentials ea(t) andeb(t) as variables.

(b) Solve your equations for ea(t) and eb(t).

(c) What is i(t)?

is2(t)

i(t)

is1(t)

ea(t) eb(t)

Fig. P2-9. Circuit for Problem P2-9.

P2-10

(a) Denote the potentials at nodes a, b, and c in thecircuit in Figure P2-10 as ea(t), eb(t), and ec(t).Express ea(t) and ec(t) in terms of eb(t) and thesource voltages.

(b) Define a supernode surrounding each voltagesource. How many KCL equations will have to bewritten?

(c) Write a KCL equation at the supernode that encirclesnodes a, b, and the voltage source on the left.

(d) Solve for eb(t).

s2(t)

s1(t)

2Ω2Ω

a b c

Fig. P2-10. Circuit for Problem P2-10.

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PROBLEMS 91

P2-11

(a) Using the node potentials ea(t) and v(t) inFigure P2-11 as variables, write KCL equations atnodes a and b.

(b) Solve your equations for v(t).

s(t)

ea(t)

3(t)+(t)–

2Ω 2Ω

b

Fig. P2-11. Circuit for Problem P2-11.

P2-12

(a) Write KVL equations at the two meshes in the circuitin Figure P2-12, using the mesh currents iα(t) andiβ(t) as variables.

(b) Solve your equations for iα(t) and iβ(t).

(c) What is v(t)?

s2(t)

s1(t)

iα iβ+

(t)–

4Ω4Ω

Fig. P2-12. Circuit for Problem P2-12.

P2-13

(a) Denote the mesh current in the lower left mesh inthe circuit in Figure P2-13 as iα(t). Express each ofthe other mesh currents in that circuit in terms ofiα(t) and the source currents.

(b) Write a KVL equation on the closed path thatencircles the current source is1(t) in terms of iα(t)and the source currents.

(c) Solve for iα(t).

(d) Use your result to compute i(t).

is2(t)

is1(t)

i(t)

1Ω1Ω

Fig. P2-13. Circuit for Problem P2-13.

P2-14

(a) For the circuit in Figure P2-14, what is the minimumnumber of KVL equations that need to be written tospecify the solution of the circuit?

2i(t)

i(t)

is(t)

1Ω 2Ω

Fig. P2-14. Circuit for Problem P2-14.

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92 CHAPTER 2 WRITING CIRCUIT EQUATIONS

(b) Define a number of mesh currents equal to yourminimum number of equations from part (a). Writea sufficient set of KVL equations in terms of themesh currents and the source waveforms.

(c) Solve your equations, and use the results to find i(t).

P2-15 In the circuit in Figure P2-15, the voltage sourceis a DC source with vs = 60V, and the current sourcesupplies a constant current of 3mA.

s is+

4(t)–

+2(t)

18kΩ10kΩ

15kΩ 12kΩ

Fig. P2-15. Circuit for Problem P2-15.

(a) Compute the voltages v2 and v4.(b) Compute the amount of power supplied by each

source.(c) Compute the amount of power absorbed by each of

the four resistors.(d) Verify that the total power supplied by the two

sources is equal to the total power absorbed by thefour resistors.

2-8-2 Basic Problems

P2-16 In the circuit in Figure P2-16, the only variablesof interest are i(t) and v(t).

(a) Write a minimal set of KCL equations in terms ofi(t) and v(t) to specify the solution of the circuit.

(b) Write a similar set of KVL equations.(c) Solve these for i(t) and v(t).

s(t) 9i(t)

i(t)

+(t)–

50Ω

Fig. P2-16. Circuit for Problem P2-16.

P2-17 Set up a set of minimal KCL and KVL equationsto find the element currents in the circuit of Figure P2-17,using the simplified exhaustive method.

i1(t)

i2(t) i4(t)

i3(t) i5(t)

s(t)is(t)

1Ω1Ω

Fig. P2-17. Circuit for Problems P2-17 and P2-23.

P2-18 For the circuit in Figure P2-18,

(a) Write a minimal set of KVL equations to specifythe complete solution. These should be expressedin terms of the variables indicated on the figure.

(b) Write a minimal set of KCL equations to specify thecomplete solution. These should also be expressedin terms of the indicated voltage variables.

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PROBLEMS 93

s(t)

is(t)+4(t)–

+1(t)–

+2(t)– +

3(t)–

3Ω3Ω

6Ω 6Ω

Fig. P2-18. Circuit for Problem P2-18.

(c) Put these equations into matrix–vector form andsolve for the four voltages v1(t), v2(t), v3(t), andv4(t).

P2-19 For the circuit in Figure P2-19,

(a) Write a minimal set of KCL equations to specifythe complete solution. These should be expressedin terms of the variables indicated on the figure.

(b) Write a minimal set of KVL equations to specifythe complete solution. These should be expressedin terms of the indicated current variables.

(c) Put these equations into matrix–vector form andsolve for the four currents i1(t), i2(t), i3(t), and i4(t).

i1(t)

i2(t) i4(t)

i3(t)

s(t) is(t)

2Ω2Ω

Fig. P2-19. Circuit for Problem P2-19.

P2-20

(a) In the circuit in Figure P2-20, what is the minimumnumber of KCL equations that you need to write tocomplete the solution?

(b) What is the minimum number of KVL equationsthat you will need to write to complete the solution?

(c) Write the KCL equations, using the current variablesdefined on the figure.

(d) Write the KVL equations in terms of the currentsand vs(t) by incorporating the element relations.

(e) Write the complete set of KCL and KVL equationsthat is sufficient to calculate i1(t), i2(t), i3(t), i4(t),and i5(t), and put them into matrix–vector form.

s(t)

i1(t)

i3(t) i5(t) i4(t)

i2(t)

is(t)

3Ω2Ω

Fig. P2-20. Circuit for Problem P2-20.

P2-21 This problem will solve the circuit inFigure P2-21 via the exhaustive method by usingboth voltage and current variables. In addition to thethree voltage variables that are indicated, define currentvariables i1(t), i2(t), and i3(t) flowing through the threeresistors, with reference directions defined according tothe default sign conventions. Let i1(t) be the currentthrough the resistor with voltage v1(t), etc.

(a) Write a sufficient set of KCL equations to constrainthe currents. Write these equations in terms of thecurrent variables i1(t), i2(t), and i3(t).

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94 CHAPTER 2 WRITING CIRCUIT EQUATIONS

s1(t) s2

(t)+2(t)–

+1(t)

– +3(t)

1Ω1Ω

Fig. P2-21. Circuit for Problem P2-21.

(b) Write a sufficient set of KVL equations to constrainthe voltages. Write these equations in terms of thevoltage variables v1(t), v2(t), and v3(t).

(c) Write the element relations for the three resistors.Write these with both the voltage and currentvariables on the left-sides of the equations, sincethese are both element variables.

(d) Put these equations into matrix–vector form. Definethe vector of variables as

x =

v1(t)

v2(t)

v3(t)

i1(t)

i2(t)

i3(t)

.

Solve these equations for the complete set ofelement variables.

P2-22 Repeat Problem P2-21 for the circuit inFigure P2-22.

P2-23 Find the currents i1(t), i2(t), i3(t), i4(t), and i5(t)for the circuit in Problem P2-17 (Figure P2-17).

i1(t) i3(t)

i2(t)

is1(t) is2

(t)3Ω

Fig. P2-22. Circuit for Problem P2-22.

P2-24 In the circuit in Figure P2-24, the voltage v(t) isthe variable of interest.

s(t)

i1(t) i3(t)

i4(t)i2(t)

10i2(t)+(t)–

Fig. P2-24. Circuit for Problem P2-24.

(a) Show that, once the variables i1(t), i2(t), i3(t), andi4(t) are known, then all of the other variables in thecircuit are known. You can do this by expressing allof the other voltages and currents in terms of these.

(b) Indicate on a drawing of the circuit a sufficient setof paths and nodes (or supernodes) on which KVLand KCL equations should be written to solve forthe four indicated variables.

(c) Write the KVL and KCL equations that youidentified in part (b).

(d) Solve your equations for v(t). Your answer shouldbe in terms of vs(t).

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PROBLEMS 95

P2-25 The circuit in Figure P2-25 is to be analyzed byusing the simplified exhaustive method.

is1(t) is2

(t)

s2(t)

s1(t)

+4(t)–

+1(t)

+2(t)–

+3(t)–

Fig. P2-25. Circuit for Problem P2-25.

(a) Indicate on a drawing of the original circuit theclosed paths on which the KVL equations shouldbe written.

(b) Also indicate on that drawing a sufficient set ofsurfaces (minimal number) on which the KCLequations should be written. These may correspondto nodes or to supernodes.

(c) Write the KVL and KCL equations correspondingto your selections in parts (a) and (b). These shouldbe written in terms of the indicated voltages only,by incorporating Ohm’s law when appropriate. Youdo not need to solve the equations, but you shouldverify that the number of equations is equal to thenumber of unknowns.

P2-26 Use the simplified exhaustive method to analyzethe circuit in Figure P2-26.

(a) Indicate on a drawing of the original circuit theclosed paths on which the KVL equations shouldbe written.

(b) Also indicate on that drawing a sufficient set ofsurfaces (minimal number) on which the KCLequations should be written. They may correspondto nodes or to supernodes.

(c) Write the KVL and KCL equations correspondingto your selections in parts (a) and (b). These shouldbe written in terms of the indicated currents only, byincorporating Ohm’s law when appropriate.

(d) Solve your equations, using MATLAB, to expressthe current i1(t) in terms of the two sourcewaveforms vs(t) and is(t).

5i1(t)

s(t)

i1(t)

i3(t)

i4(t)

i2(t)

is(t)

1Ω 4Ω

Fig. P2-26. Circuit for Problem P2-26.

P2-27 Use the simplified exhaustive method to solve forthe voltage v(t) in the circuit in Figure P2-27.

is1(t) is2

(t)+(t)–

1Ω1Ω

Fig. P2-27. Circuit for Problem P2-27.

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96 CHAPTER 2 WRITING CIRCUIT EQUATIONS

P2-28

(a) For the circuit in Figure P2-28, write a minimalsufficient set of KVL equations that will provideindependent constraints on the element voltages thatare labelled. Write your equations in terms of theelement voltages and the source signals only.

(b) For the same circuit, write a minimal sufficient setof KCL equations that will provide independentconstraints on the same element voltages. Again,write your equations in terms of the element voltagesand source signals only.

s(t)

is(t)

+4(t)–

+3(t)–

+2(t)–

+1(t)–

20Ω

20Ω

10Ω 10Ω

Fig. P2-28. Circuit for Problem P2-28.

P2-29 This problem analyzes the circuit in Figure P2-29.

(a) Write a minimal sufficient set of KCL equations thatwill provide independent constraints on the elementcurrents that are labelled. Write your equations interms of the element currents and the source signalsonly.

(b) For the same circuit, write a minimal sufficient setof KVL equations that will provide independentconstraints on the same element currents. Again,write your equations in terms of the element currentsand source signals only.

(c) Put your equations into matrix–vector form byfilling in the empty matrix and vector below.

i1(t)

i2(t)

i3(t)

i4(t)

=

vs(t).

s(t)

i1(t)

5i1(t)

i3(t)

i4(t)

i2(t)

Fig. P2-29. Circuit for Problem P2-29.

P2-30

(a) For the circuit in Figure P2-30, write a sufficient setof KCL equations to find the solution of the circuit.These should be in terms of the variables i1(t), i2(t),and i3(t).

i1(t)i3(t)

i2(t)

s1(t) s2

(t)

3Ω2Ω

Fig. P2-30. Circuit for Problem P2-30.

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PROBLEMS 97

(b) Write a sufficient set of KVL equations to findthe solution. These should also be in terms of thevariables i1(t), i2(t), and i3(t).

(c) Express your complete set of equations in matrix–vector form. You do not need to solve them.

P2-31 In the circuit in Figure P2-31, each of the currentsources generates a constant current. Compute the valuesof the four resistor currents, i1, i2, i3, and i4.

i3i1i4i2

2Ω2Ω

3Ω1Ω2A 3A

Fig. P2-31. Circuit for Problem P2-31.

P2-32

(a) Using the simplified exhaustive method and eitherthe resistor voltages or the resistor currents asvariables, write a set of three KCL and three KVLequations that will specify the solution of the circuitin Figure P2-32. v(t) is the voltage across theterminals of the rightmost current source and theindicated open circuit.

s(t)

is1(t) is2(t)

+

(t)

3Ω 10Ω10Ω

2Ω 1Ω

Fig. P2-32. Circuit for Problem P2-32.

(b) Express your equations in matrix–vector form.

(c) Use MATLAB to solve for your variables, and thenuse those to express v(t). Your answer should be inthe form v(t) = K1is1(t)+K2is2(t)+K3 vs(t).

P2-33 Use the node method to write a set of constraintequations for the network of Figure P2-33. Use asvariables the voltages ea(t), eb(t), and ec(t). Do not solvethe set of equations.

is1(t)

is2(t)

eb(t)ea(t) ec(t)

R6R4

R3 R5

R2R1

Fig. P2-33. Circuit for Problem P2-33.

P2-34 In this problem, we solve the circuit inFigure P2-34 by using the node method.

(a) Write the KCL equations at nodes a and b in termsof the node potentials at those nodes, ea(t) and eb(t).

(b) Put your equations into matrix–vector form bysupplying the missing constants in the frameworkbelow.

ea(t)eb(t)

=

is(t)+

vs(t).

(c) Solve them for ea(t) and eb(t).

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98 CHAPTER 2 WRITING CIRCUIT EQUATIONS

s(t)

is(t)

4Ω 2Ω

a b

Fig. P2-34. Circuit for Problem P2-34.

P2-35 Find the voltage at each connection point inthe circuit in Figure P2-35 with respect to the indicatedground.

is(t)

s(t)

eb(t)

ea(t)

ec(t)

ed(t)

1Ω1Ω

Fig. P2-35. Circuit for Problem P2-35.

P2-36 We wish to solve the circuit in Figure P2-36 byusing the node method. Let ea(t) be the node potential atthe indicated node.

3i(t)i(t)

s(t)

ea(t)

Ω–12

Fig. P2-36. Circuit for Problem P2-36.

(a) Express i(t) in terms of ea(t) and vs(t).(b) Write a KCL equation at the supernode that encircles

the dependent voltage source. This equation shouldinvolve only the variables ea(t) and vs(t).

(c) Find ea(t).

P2-37

(a) Define and label an appropriate set of nodepotentials for the circuit in Figure P2-37.

(b) Write a set of KCL equations that can be solved forthe node potentials that you defined in part (a), usingthe node potentials as variables.

(c) Solve your equations to express v(t) in termsof vs(t).

s(t)+(t)–

3Ω 3Ω

1Ω 1Ω

Fig. P2-37. Circuit for Problem P2-37.

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PROBLEMS 99

P2-38 Consider the circuit with a CCVS shown inFigure P2-38.

s(t)

i(t)

10i(t)2Ω

10Ω 2Ωa

Fig. P2-38. Circuit for Problem P2-38.

(a) Redraw the circuit with the CCVS replaced byan equivalent VCVS that depends upon the nodepotential ea(t).

(b) Write a KCL equation at node a in terms of thevariable ea(t).

(c) Solve for ea(t).(d) What is the instantaneous power absorbed by the

independent voltage source?

P2-39 Solve for v(t) in terms of is(t) in the circuit inFigure P2-39.

i(t)

2i(t)is(t)+(t)–

3Ω1Ω

2Ω 2Ω

Fig. P2-39. Circuit for Problem P2-39.

P2-40 Find all of the element voltages and currents inthe circuit of Figure P2-40, by using the mesh method.Be sure to identify the variables clearly.

s(t)

is(t)

1Ω 1Ω

Fig. P2-40. Circuit for Problem P2-40.

P2-41

(a) What is the minimal number of KVL equations thatneed to be written if the mesh method is to be usedto analyze the circuit in Figure P2-41?

s(t)

i1(t)

3i1(t)iα

is(t)

10Ω

Fig. P2-41. Circuit for Problem P2-41.

(b) Write a KVL equation over the indicated mesh.Express your answer in terms of the mesh currentiα(t), i1(t), vs(t), and is(t).

(c) Express i1(t) as a function of iα(t).(d) Write i1(t) as a function of is(t) and vs(t).

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100 CHAPTER 2 WRITING CIRCUIT EQUATIONS

P2-42 Solve for v(t) in the circuit in Figure P2-42.

i(t)

2i(t)is(t)+

(t)–

2Ω 2Ω

Fig. P2-42. Circuit for Problem P2-42.

P2-43 Consider the circuit with a CCVS shown inFigure P2-43.

(a) Redraw the circuit with the CCVS replaced byan equivalent CCVS that depends upon the meshcurrents in the two meshes iα(t) and iβ(t).

(b) Write two KVL equations over paths defined by themeshes in terms of the variables iα(t) and iβ(t).

(c) Solve for i(t).

s(t)

i(t)

10i(t)2Ω

2Ω10Ω

Fig. P2-43. Circuit for Problem P2-43.

P2-44

(a) Which method, the mesh method or the nodemethod, will result in fewer equations to solve inorder to find v(t) in the circuit in Figure P2-44?

(b) Compute v(t), using the method that you selectedin (a).

is1(t) is2

(t)

+(t)

100Ω 200Ω

50Ω

Fig. P2-44. Circuit for Problem P2-44.

P2-45

(a) For the circuit in Figure P2-45, express the meshcurrent in the lower right mesh in terms of iα(t) andiβ(t).

s(t) 3(t)+(t)–

2Ω 2Ω

Fig. P2-45. Circuit for Problem P2-45.

(b) Redraw the circuit with the VCCS replaced byan equivalent CCCS that depends upon the meshcurrents iα(t) and iβ(t). iα(t) is the mesh current inthe upper mesh of the circuit, and iβ(t) is the meshcurrent associated with the lower left mesh.

(c) Write KVL equations over the upper mesh and thesupermesh in terms of the variables iα(t) and iβ(t).

(d) Solve for iα(t) and iβ(t).

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PROBLEMS 101

(e) What is v(t)?

P2-46 Find v(t) in the circuit in Figure P2-46 by anymethod that you choose.

s(t) is(t)+(t)–

1Ω 2Ω

Fig. P2-46. Circuit for Problem P2-46.

P2-47 In the circuit in Figure P2-47, express v(t) interms of vs(t) and is(t).

s(t) 3i(t)

is(t)

i(t)

+(t)–

2Ω 2Ω

Fig. P2-47. Circuit for Problem P2-47.

P2-48 The node method and the mesh method arenot the only systematic methods for finding the solutionof a circuit, although they are the most popular. As anexample of a different approach, consider the circuit inFigure P2-48.

is(t)

ie(t)

ia(t) ib(t)ic(t) id(t)

Fig. P2-48. Circuit for Problem P2-48.

(a) Show that all of the currents in the circuit (and,therefore, all of the voltages) can be expressed interms of ia(t), ib(t), and is(t); that is, express ic(t),id(t), and ie(t) in terms of ia(t), ib(t), and is(t).

(b) Write a set of two independent KVL equations overappropriate paths, using only ia(t), ib(t), and is(t)as variables.

(c) Express your equations in matrix–vector form byfilling in the missing entries in the equation below:

ia(t)ib(t)

=

is(t).

P2-49 Find the power delivered by each source in thecircuit in Figure P2-49.

100Ω 200Ω

200Ω 40Ω

0.5A 20V

Fig. P2-49. Circuit for Problem P2-49.

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102 CHAPTER 2 WRITING CIRCUIT EQUATIONS

2-8-3 Advanced Problems

P2-50 Solve for i(t) in the circuit in Figure P2-50.

s1(t)

s2(t)

is(t)

i(t)

Fig. P2-50. Circuit for Problem P2-50.

P2-51 In the circuit in Figure P2-51,

(a) Compute the power supplied by the voltage source.(Note: The power supplied is the negative of thepower absorbed.)

(b) Compute the power supplied by the current source.

s(t)

is(t)

3Ω6Ω2Ω

Fig. P2-51. Circuit for Problem P2-51.

P2-52 Our derivations of the simplified exhaustivemethod and the mesh method were limited to planarcircuits. (The node method had no such restriction). Inthis problem, we extend the simplified exhaustive methodto nonplanar circuits. An example of a nonplanar circuit isshown in Figure P2-52. All of the resistors in that circuithave resistance R.

s(t)

Fig. P2-52. Nonplanar circuit for Problems P2-52 andP2-53.

The procedure is summarized in the following steps

1. Create a supernode around each voltage source.

2. Write a v–i relation for each element and a KCLequation at all but one node/supernode, as before.

3. Remove sufficient branches to eliminate all branchcrossings, to obtain a planar network, N .

4. Write a KVL equation at each mesh or supermesh ofthe resulting circuit that does not contain an exteriorcurrent source.

5. For each branch that was removed, form a closedpath consisting of that branch and some subset ofbranches fromN . Write a KVL equation on each ofthese paths.

6. Solve the resulting set of equations, using anyappropriate method.

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PROBLEMS 103

(a) On a drawing of the circuit, define as variables thenine resistor currents. Using those currents, writeKCL equations at all but one of the nodes of thenetwork. (There should be five of them.)

(b) Using the above procedure, write four KVLequations in terms of the current variables.

(c) Solve your equations for the resistor currents.

P2-53 Problem P2-52 outlined a procedure for extendingthe simplified exhaustive method to nonplanar circuits.A similar approach can be used to extend the meshmethod. A mesh current is defined for each of the meshes,supermeshes and/or loops identified in steps 3 and 4.KVL equations are then written in terms of those meshcurrents over this set of loops. Since the variables aremesh currents, there is no need to write KCL equations.

(a) Use the nonplanar extension of the mesh method tosolve the circuit in Figure P2-52.

(b) Solve that same circuit by using the node method.Compute the resistor currents from the nodepotentials and show that the resulting values are thesame as those derivable from the mesh currents thatyou computed in part (a).

P2-54 In our derivation of the mesh method, we stressedits duality with the node method: the similarity of the twomethods if the roles of voltages and currents and of nodesand meshes are reversed. This problem lets you explorethis issue further. Consider the network in Figure P2-54.

(a) Use the node method to construct the set ofequations that must be solved to find the circuitsolution. Let the ground node be the node that isomitted in writing your KCL equations. Expressthese equations in the form

Cv(t) = s1vs1(t)+ s2vs2(t),

s1(t) s2

(t)

2Ω 2Ω

2Ω 2Ω

Fig. P2-54. Circuit for Problem P2-54.

where v(t) is a vector of node potentials, s1 and s2

are column vectors of constants, and C is a constantmatrix.

(b) Now design a different network containing twocurrent sources with currents is1(t) and is2(t), suchthat the set of mesh equations that need to be solvedto find the complete solution is

Ci(t) = s1is1(t)+ s2is2(t),

where i(t) is the vector of mesh currents and whereC, s1, and s2 are the same as in your solution inpart (a).

(c) Solve your equations from part (b).

P2-55 In the circuit in Figure P2-55, express v(t) interms of is(t).

is(t) 2(t)

+(t)

Fig. P2-55. Circuit for Problem P2-55.

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104 CHAPTER 2 WRITING CIRCUIT EQUATIONS

P2-56 A thermistor is a sensor whose resistance RTvaries as a function of temperature. A plot of RTvs. T for a particular (idealized) device is shown inFigure P2-56(a). A Wheatstone bridge, such as the onein Figure P2-56(b), is a circuit that can be used to converta variable resistance into a variable voltage. Produce aplot of voltage v versus temperature T as T goes from 0to 60.

RT300Ω

T (˚C)0˚ 30˚ 60˚

(a)

RT

+ –

200Ω200Ω

200Ω

10V

(b)

Fig. P2-56. (a) Plot of resistance vs. temperature for aparticular thermistor. (b) Wheatstone bridge circuit forProblem P2-56.

P2-57 Section 2-6 demonstrated that the conservationof power was implied by KCL and KVL. Thatdemonstration was done by using node potentials for

the specific circuit in Figure 2.23. For the same circuit,produce a similar derivation using mesh currents.

2-8-4 Design Problems

P2-58 The circuit in Figure P2-58 contains a light-emitting diode, which is modelled as a 2 resistor. Thediode will light if the current is greater than 10 mA, butwill burn out if its current is greater than 15 mA. Choosethe nominal value of the of the resistance R, if R canchange by ±10% with changes in temperature.

100Ω

500Ω6

R

Fig. P2-58. Circuit for Problem P2-58.

P2-59 The circuit in Figure P2-59 has two sources, eachof which is constant as a function of time. The voltagesource has a value of 16V, and the current source hasa value of imA. Find the value of i that will cause thevoltage v, indicated in the circuit, to have a value of 4V.

+

–3kΩ2kΩ

4kΩ 1kΩ

16 i

Fig. P2-59. Circuit with two sources for Problem P2-59.

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PROBLEMS 105

P2-60 The circuit in Figure P2-60 represents a modelof an amplifier that is driving a loudspeaker, which ismodelled by the 16 resistor. The input to the amplifieris the signal vs(t) = 0.2 cos(1000t).

(a) Compute the value of the resistance R such thatv(t) = 20 cos(1000t).

(b) The power in the signal x(t) = A cos(ωt) is A2/2.Compute the ratio of the power in the signal v(t)delivered to the loudspeaker to the power in the inputsignal vs(t).

s(t)

+

ab

150ab

+

(t)

300Ω

500kΩ 16Ω

R

Fig. P2-60. Circuit for Problem P2-60.

P2-61 A distraught engineering student accidentallythrew away a circuit design that he had been working onfor some time. Rifling through his trash, he came upon thepreliminary circuit in Figure P2-61 and the intermediatematrix–vector equation shown below.

15 0 0 1

21 1 0−1− 1

5 1 13 0

0 1−1 0

v1(t)

v2(t)

v3(t)

v4(t)

=

000−1

vs(t)+

10−1

0

is(t).

He has no memory of the resistor values, which voltagesare which, or what their reference directions should be.There are a few things that you should know about him:(1) he hates to do laundry, (2) he is not particularlysystematic, and (3) he does not call home nearly as oftenas he should.

(a) He vaguely remembers that he used the simplifiedexhaustive method to write the circuit equations.Assuming this to be the case, how many KCL andKVL equations should he have written?

(b) What are the KCL equations for the circuit?

(c) What are the KVL equations for the circuit?

(d) On the circuit drawing, label the four voltages andtheir reference directions.

(e) On the circuit drawing, label the four resistancevalues.

Fig. P2-61. Circuit for Problem P2-61.