CHAPTER 3 - METHODS OF ANALYSIS

List of topics for this chapter :Solving Systems of EquationsNodal AnalysisNodal Analysis with Voltage SourcesMesh AnalysisMesh Analysis with Current SourcesNodal and Mesh Analysis by InspectionCircuit Analysis with PSpice

SOLVING SYSTEMS OF EQUATIONS

Problem 3.1 Invert a general nn matrix.

The inverse of a nonsingular nn matrix

=

nn2n1n

n22221

n11211

aaa

aaa

aaa

A

is given by

==

nnn2n1

2n2212

1n2111T

1-

ccc

ccc

ccc

1CA

where is the determinant of the matrix A and ijc is the cofactor of ija in .

The value of the determinant, , can be obtained by expanding along the ith row

ininni

2i2i2i

1i1i1i

nn2n1n

n22221

n11211

ma)1-(ma)1-(ma)1-(

aaa

aaa

aaa

+++ +++==

or the jth column

njnjjn

j2j2j2

j1j1j1

nn2n1n

n22221

n11211

ma)1-(ma)1-(ma)1-(

aaa

aaa

aaa

+++ +++==

where ijm , the minor of ija in , is a )1n()1n( determinant of the submatrix of Aobtained by removing the ith row and the jth column.

The cofactor of ija in is ijji

ij m)1-(c += .

The transpose of the cofactor matrix is also known as the adjoint of the matrix; i.e., TCAadj = .Therefore, =1-A

====

TCAdetAadj

Problem 3.2 Solve a general system of simultaneous equations using Cramer's rule.

Given a system of simultaneous equations having the form

nnnn22n11n

2nn2222121

1nn1212111

bxaxaxa

bxaxaxabxaxaxa

=+++

=+++

=+++

where there are n unknowns, 1x , 2x , , nx , to be determined.

The matrix representation of the system of simultaneous equations is

=

n

2

1

n

2

1

nn2n1n

n22221

n11211

b

bb

x

x

x

aaa

aaa

aaa

or BAX =

where

=

nn2n1n

n22221

n11211

aaa

aaa

aaa

A

,

=

n

2

1

x

x

x

X ,

=

n

2

1

b

bb

B

Note that A is a square ( nn ) matrix, while X and B are ( 1n ) column matrices.

Cramer's rule states that the solution to the system of simultaneous equations, BAX = , is

=1x 1

, =2x 2

, , =nx n

where the 's are the determinants given by

nn2n1n

n22221

n11211

aaa

aaa

aaa

=

nn2nn

n2222

n1121

1

aab

aabaab

=

nnn1n

n2221

n1111

2

aba

abaaba

=

n2n1n

22221

11211

n

baa

baabaa

=

Notice that is the determinant of matrix A and k is with its kth column replaced withmatrix B.

Obviously, Cramer's rule only applies when 0 . In the case that 0= , the set of equationshas no unique solution because the equations are linearly dependent.

See Problem 3.1 to find out how to calculate the value of a determinant of a matrix.

NODAL ANALYSIS

Problem 3.3 [3.3] Find the currents 1i through 4i and the voltage ov in Figure 3.1.

Figure 3.1

20

i2

10 10 A

vo

i1

2 A30

i3

60

i4

Applying KCL to the non-reference node,

60v

230v

20v

10v

10 oooo ++++=

= ov12480 =ov V40

Thus,

10v

i o1 = 20v

i o2 = 30v

i o3 = 60v

i o4 =

=1i A4 =2i A2 =3i A3333.1 =4i mA7.666

Problem 3.4 Given the circuit in Figure 3.1, find xI .

Figure 3.1

=xI A2

NODAL ANALYSIS WITH VOLTAGE SOURCES

Problem 3.5 Given the circuit in Figure 3.1, solve for xV .

Figure 3.1

3 A10 20

10

2 A

Ix

9 A20

20

+

20 V+

Vx

10 V

+

5

10 10

Carefully DEFINE the problem.Each component is labeled completely. The problem is clear.

PRESENT everything you know about the problem.The goal of the problem is to find xV . Letting the lower node be the reference node, we needto find the voltage to the right of the 10-V voltage source.

A supernode is formed by enclosing a (dependent or independent) voltage source connectedbetween two non-reference nodes and any elements connected in parallel with it. Hence, it isclear that the two nodes on either side of the 10-V voltage source form a supernode.

A supermesh results when two meshes have a (dependent or independent) current source incommon. Hence, the two meshes on the right half of the circuit create a supermesh, with the9-A current source in common.

Establish a set of ALTERNATIVE solutions and determine the one that promises thegreatest likelihood of success.The two methods of analysis of simple circuits, such as the one above, are nodal analysis andmesh analysis. Because the circuit contains three nodes in addition to the reference node,nodal analysis produces a set of three equations and three unknowns. Yet, the supernodechanges the set to two equations and a constraint equation. On the other hand, the circuit hasfour loops. Thus, mesh analysis produces four equations and four unknowns. The supermeshchanges the set to three equations and a constraint equation.

Because the goal of the problem is to find a voltage, the obvious choice is to use nodalanalysis to find the voltages at each node of the circuit. If mesh analysis were used to findthe mesh currents, Ohm's law would also be needed to find the voltage across the resistor.

ATTEMPT a problem solution.Begin the problem solution by identifying the nodes, including the supernode.

Clearly, 2x vV =

Use nodal analysis to find 1v , 2v , and 3v .At the supernode (nodes 1 & 2): At node 3:

05

vv

100v

200v

2020v 32211

=

+

+

+ 0

100v9

5vv 323

=

+

9 A20

20

+

20 V+

Vx

10 V

+

5

10 10

v1 v3v2

Simplifying,0)vv)(4(v2v)20v( 32211 =+++ 90v)vv)(2( 323 =+

20v4v6v2 321 =+ 90v3v2- 32 =+10v2v3v 321 =+

This is a set of two equations and three unknowns. Thus, we must find a constraint equation.The supernode will provide the constraint equation.

10vv 12 += or 10vv 21 =

Substitute the constraint equation into the simplified equation from the supernode. Then, thisequation plus two times the simplified equation from node 3 will isolate 3v .

10v2v3)10v( 322 =+]90v3v2-[)2( 32 =+

]20v2v4[ 32 = + ]180v6v4-[ 32 =+ = ]200v4[ 3 =50v3 = volts

The equation at node 3 can be written,90v3v2 32 =

30)60)(21()90150)(21(]90)50)(3()[21(v2 ==== volts

The constraint equation gives20103010vv 21 === volts

Therefore, 30vV 2x == volts

EVALUATE the solution and check for accuracy.This circuit can be analyzed using mesh analysis to verify the solution. This would providepractice analyzing a circuit with a supermesh. Mesh analysis will be discussed later in thischapter. So, we will check our solution using KCL at each node.

For the supernode,

0431055030

1030

2020

202020

5vv

10v

20v

2020v 32211

=++=

+++

=

+++

For node 3,

05941050

953050

10v

95vv 323

=+=+

=+

KCL is not violated. Thus, our check for accuracy was successful.

Has the problem been solved SATISFACTORILY? If so, present the solution; if not,then return to ALTERNATIVE solutions and continue through the process again.This problem has been solved satisfactorily.

=xV V30

Problem 3.6 [3.7] Using nodal analysis, find oV in Figure 3.1.

Figure 3.1

05V4v

13v

5v

0iii o111321 =

+

+=++

15V4v7 o1 =

But 1o v52

V = or o1 V25

v =

So that 27)15)(2(

V15V4V25)7( ooo ==

=oV V1111.1

Problem 3.7 Given the circuit in Figure 3.1, solve for xV using matrix inversion.

Figure 3.1

+ Vx

80 V10 10

10

+

10

2 A

4 Vo2

5 3

+

i1

i2

+

Vo

+

1

3 V

v1 i3

Clearly, 2x1 vVv += or 21x vvV =

Use nodal analysis to find 1v and 2v .At node 1 : At node 2 :

010

vv210

0v 211=

+ 0

10vv

100v

1080v 1222

=

+

+

Simplifying,0vv20v 211 =+ 0vvv80v 1222 =++

20vv2 21 = 80v3v- 21 =+

The system of simultaneous equations is

=

8020

v

v

31-1-2

2

1

=

=

+

+=

=

3628

180140

51

160208060

51

8020

2113

161

v

v

2

1

Therefore, 3628vvV 21x ===xV V8-

Problem 3.8 Solve Problem 3.5 using Cramer's rule.

=xV V30

This answer is the same as that found in Problem 3.5.

+ Vx

80 V10 10

10

+

v1 v210

2 A

MESH ANALYSIS

Problem 3.9 Given the circuit in Figure 3.1, solve for the loop currents, 1i and 2i , usingmesh analysis.

Figure 3.1

For loop 1 : For loop 2 :0)ii)(20(i1070- 211 =++ 0100i20)ii)(20( 212 =++

70i20i30 21 = 100-i40i20- 21 =+7i2i3 21 = 10-i4i2- 21 =+

Thus, the system of simultaneous equations is

=

10-7

ii

42-2-3

2

1 or

=

=

=

2-1

16-8

81

10-7

3224

4121

ii

2

1

Therefore, =1i A1 and =2i A2-

Problem 3.10 [3.33] Apply mesh analysis to find i in Figure 3.1.

Figure 3.1

100 V20

10

+

+

70 V i1

20

i2

4

1

10

+

8 V

i

i2

6 V

+

2

i3

i1

5

For loop 1 : 2121 ii63i2i126 == (1)For loop 2 : 321312 ii7i28ii2i78- +== (2)For loop 3 : 2323 ii620ii668- ==++ (3)

Putting (1), (2), and (3) in matrix form,

=

283

iii

61-017-201-6

3

2

1

-23461-017-201-6== 240

620182036

2 == -3821-087-231-6

3 ==

Clearly,

234-24038-

iii 2323

=

==

=i A188.1

MESH ANALYSIS WITH CURRENT SOURCES

Problem 3.11 Given the circuit shown in Figure 3.1, find xI using mesh analysis.

Figure 3.1

Carefully DEFINE the problem.Each component is labeled completely. The problem is clear.

PRESENT everything you know about the problem.A supermesh results when two meshes have a (dependent or independent) current source incommon. Hence, the leftmost mesh and the middle mesh create a supermesh, with the 1-Acurrent source in common.

1 A 20

10

+

25 V

Ix

50 V+

20

Establish a set of ALTERNATIVE solutions and determine the one that promises thegreatest likelihood of success.The problem clearly states to use mesh analysis. This makes sense because the goal of theproblem is to find a current, xI , and mesh analysis produces the currents in each mesh, orloop, of a circuit.

ATTEMPT a problem solution.Begin the problem solution by identifying the meshes, including the supermesh.

Clearly, 3x iI =

Use mesh analysis to find 1i , 2i , and 3i .For the supermesh (loops 1 & 2) : 0)ii)(20(i1025- 321 =++where 12 ii1 = or 1ii 21 = (constraint equation)For loop 3 : 0i20)ii)(20(50 323 =++

Substitute the constraint equation into the equation for the supermesh and simplify,0i20i2010i1025- 322 =++

0i20i20i2050 323 =++

Simplifying further,35i20i30 32 = or 7i4i6 32 =

50-i40i20- 32 =+ 5-i4i2- 32 =+

The system of simultaneous equations is

=

5-7

ii

42-4-6

3

2

=

=

=

1-5.0

16-8

161

5-7

6244

8241

ii

3

2

and 5.0-15.01ii 21 ===

Therefore, 1-iI 3x == amp

1 A 20

10

+

25 V

Ix

50 V+

20

i1 i3i2

EVALUATE the solution and check for accuracy.This circuit has only one unknown node after identifying the lower node as the referencenode. Hence, it can easily be analyzed using nodal analysis.

020

50v20v

110

25v xxx=

++

120v4 x =30vx = volts

Using Ohm's law,

-120

503020

50vI xx =

=

= amp

This matches the answer that was obtained using mesh analysis. Our check for accuracy wassuccessful.

Has the problem been solved SATISFACTORILY? If so, present the solution; if not,then return to ALTERNATIVE solutions and continue through the process again.This problem has been solved satisfactorily.

=xI A1-

Problem 3.12 [3.35] Use mesh analysis to obtain oi in the circuit of Figure 3.1.

Figure 3.1

12 V

2 1

+

io

6 V

+

3 A5

4

1 A 20

10

+

25 V

Ix

50 V+

20 vx

Loops 1 and 2 form a supermesh. For the supermesh,012i5i4i6 321 =++ (1)

For loop 3,0i4ii76 213 =+ (2)

Also, 12 i3i += (3)

Putting (1), (2), and (3) into matrix form,

=

3612-

iii

011-7-415-46

3

2

1

Solving this system of equations yields-3.067i1 = amps-0.06667i2 = amps-1.3333i3 = amps

Therefore,)3333.1-(-3.067iii 31o ==

=oi A1.7333-

12 V

2

1 +

io

6 V

+

5

4

i1

3 A

i2i3

i2i1

Problem 3.13 Given the circuit as shown in Figure 3.1, solve for xI using mesh analysis.

Figure 3.1

=xI A1

NODAL AND MESH ANALYSIS BY INSPECTION

Problem 3.14 [3.51] Obtain the node-voltage equations for the circuit shown inFigure 3.1 by inspection. Determine the node voltages 1v and 2v .

Figure 3.1

25.141

11

G11 =+= 2112 G-1G == 5.121

11

G 22 =+=

336i1 == -165i2 ==

2 A20

30

+

30 V 24

Ix

6 A

2 4

1

3 A

v1 v2

5 A

Hence, we have

====

1-3

v

v

1.51-1-25.1

2

1

=

25.1115.11

1.51-1-25.1 -1

where 0.875-1)(-1)()5.1)(25.1( ==

=

+

+=

=

24

)4286.1)(1-()1429.1)(3()1429.1)(1-()7143.1)(3(

1-3

4286.11429.11429.17143.1

v

v

2

1

Clearly, =1v V4 and =2v V2

CIRCUIT ANALYSIS WITH PSPICE

Problem 3.15 Solve Problem 3.13 using PSpice.

Clearly, xI is the current flowing through R3 and the current probe reads =xI A0.1 .

This answer is the same as the answer obtained in Problem 3.13.

SearchHelpEWB Help PageWe want your feedbacke-Text Main MenuTextbook Table of ContentsProblem Solving WorkbookWeb LinksTextbook WebsiteOLC Student Center WebsiteMcGraw-Hill Website

PrefaceChapter 1 Basic ConceptsChapter 2 Basic LawsChapter 3 Methods of AnalysisSolving Systems of EquationsNodal AnalysisNodal Analysis with Voltage SourcesMesh AnalysisMesh Analysis with Current SourcesNodal and Mesh Analysis by InspectionCircuit Analysis with PSpice

Chapter 4 Circuit TheoremsChapter 5 Operational AmplifierChapter 6 Capacitors and InductorsChapter 7 First-Order CircuitsChapter 8 Second-Order CircuitsChapter 9 Sinusoids and PhasorsChapter 10 Sinusoidal Steady-State AnalysisChapter 11 AC Power AnalysisChapter 12 Three-Phase CircuitsChapter 13 Magnetically Coupled CircuitsChapter 14 Frequency ResponseChapter 15 Laplace TransformChapter 16 Fourier SeriesChapter 17 Fourier TransformChapter 18 Two-Port Networks

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