BASIC ELECTRONICS

Department of Computer Science NUCES-FAST

Spring 2015 Weekly Notes

Week – 4 Mesh/Loop Analysis

Reference: Electric Circuits, Nilsson & Riedel, 9th Edition Basic Engineering Circuit Analysis, J. David Irwin, R. Mark Nelms, 10th Edition

4.5 Introduction to the Mesh-Current Method 99

Substituting Eqs. 4.16 and 4.17 into Eq. 4.15 yields

vb J_ _L i R{

+ R2 + (1 + (3)RE

V cc Vr

Ri 0 + P)RE

Solving Eq. 4.18 for ?;h yields

VccMI + WE + V0R1R2 RtR2 + ( 1 + (3)RE(R] + R2) vb =

(4.18)

(4.19)

Using the node-voltage method to analyze this circuit reduces the prob­lem from manipulating six simultaneous equations (see Problem 2.27) to manipulating three simultaneous equations. You should verify that, when Eq. 4.19 is combined with Eqs. 4.16 and 4.17, the solution for i& is identical to Eq. 2.25. (See Problem 4.30.)

^ A S S E S S M E N T PROBLEMS

Objective 1—Understand and be able to use the node-voltage method

4.4 Use the node-voltage method to find va in the circuit shown.

30 0 -AA/V-

io a

10V

Answer: 24 V.

4.5 Use the node-voltage method to find v in the circuit shown.

Answer: 8 V.

4.6 Use the node-voltage method to find v\ in the circuit shown.

20 ft

' ! ,^40ft 20 iA

Answer: 48 V.

NOTE: Also try Chapter Problems 4.24, 4.26, and 4.27.

4.5 Introduction to the Mesh-Current Method

As stated in Section 4.1, the mesh-current method of circuit analysis enables us to describe a circuit in terms of be — (ne - 1) equations. Recall that a mesh is a loop with no other loops inside it. The circuit in Fig. 4.1 (b) is shown again in Fig. 4.18, with current arrows inside each loop to distinguish it. Recall F1gure 4<18 A The d r c u | t shown in Fig 41^t witn the

also that the mesh-current method is applicable only to planar circuits. The mesh currents defined.

100 Techniques of Circuit Analysis

Figure 4.19 • A circuit used to illustrate development of the mesh-current method of circuit analysis.

circuit in Fig. 4.18 contains seven essential branches where the current is unknown and four essential nodes. Therefore, to solve it via the mesh-current method, we must write four [7 - ( 4 - 1)] mesh-current equations.

A mesh current is the current that exists only in the perimeter of a mesh. On a circuit diagram it appears as either a closed solid line or an almost-closed solid line that follows the perimeter of the appropriate mesh. An arrowhead on the solid line indicates the reference direction for the mesh current. Figure 4.18 shows the four mesh currents that describe the circuit in Fig. 4.1(b). Note that by definition, mesh currents automati­cally satisfy Kirchhoffs current law. That is, at any node in the circuit, a given mesh current both enters and leaves the node.

Figure 4.18 also shows that identifying a mesh current in terms of a branch current is not always possible. For example, the mesh current i2 is not equal to any branch current, whereas mesh currents ^, z3, and /4 can be identified with branch currents. Thus measuring a mesh current is not always possible; note that there is no place where an ammeter can be inserted to measure the mesh current i2. The fact that a mesh current can be a fictitious quantity doesn't mean that it is a useless concept. On the contrary, the mesh-current method of circuit analysis evolves quite natu­rally from the branch-current equations.

We can use the circuit in Fig. 4.19 to show the evolution of the mesh-current technique. We begin by using the branch currents (/j, /2, and i3) to formulate the set of independent equations. For this circuit, be — 3 and ne = 2. We can write only one independent current equation, so we need two independent voltage equations. Applying Kirchhoffs current law to the upper node and Kirchhoffs voltage law around the two meshes gener­ates the following set of equations:

fi = h + *3»

vx --

-v2 z ilR-2 ~ *3"3*

(4.20)

(4.21)

(4.22)

We reduce this set of three equations to a set of two equations by solving Eq. 4.20 for /3 and then substituting this expression into Eqs. 4.21 and 4.22:

vx = /'!(/?! + i?3) - *2#3, (4.23)

•V2 •itR3 + i2(R2 + /½). (4.24)

'Wv 0 'WV

'., < ^ 3 <h

Figure 4.20 A Mesh currents /., and ih.

We can solve Eqs. 4.23 and 4.24 for /2 and i2 to replace the solution of three simultaneous equations with the solution of two simultaneous equations. We derived Eqs. 4.23 and 4.24 by substituting the ne — 1 current equations into the be - (ne - 1) voltage equations. The value of the mesh-current method is that, by defining mesh currents, we automatically eliminate the ne - 1 current equations. Thus the mesh-current method is equivalent to a systematic substitution of the ne — 1 current equations into the be ~ (ne ~ 1) voltage equations. The mesh currents in Fig. 4.19 that are equivalent to eliminating the branch current /3 from Eqs. 4.21 and 4.22 are shown in Fig. 4.20. We now apply Kirchhoffs voltage law around the two meshes, expressing all voltages across resistors in terms of the mesh cur­rents, to get the equations

vl = /ai?! + (/, - ib)R3, (4.25)

-V2 = O'b " /'a)^3 + kRl- (4-26)

Collecting the coefficients of z'a and /b in Eqs. 4.25 and 4.26 gives

th = k(Ri + Ri) ~ M & (4.27)

-v2 = -LR3 + UR2 + R3). (4.28)

4.5 Introduction to the Mesh-Current Method 101

Note that Eqs. 4.27 and 4.28 and Eqs. 4.23 and 4.24 are identical in form, with the mesh currents 4 and 4 replacing the branch currents /, and ij. Note also that the branch currents shown in Fig. 4.19 can be expressed in terms of the mesh currents shown in Fig. 4.20, or

h — 4 ~ 4

(4.29)

(4.30)

(4.31)

The ability to write Eqs. 4.29-4.31 by inspection is crucial to the mesh-current method of circuit analysis. Once you know the mesh currents, you also know the branch currents. And once you know the branch currents, you can compute any voltages or powers of interest.

Example 4.4 illustrates how the mesh-current method is used to find source powers and a branch voltage.

Example 4.4 Using the Mesh-Current Method

a) Use the mesh-current method to determine the power associated with each voltage source in the circuit shown in Fig. 4.21.

b) Calculate the voltage va across the 8 (1 resistor.

Solution

a) To calculate the power associated with each source, we need to know the current in each source. The circuit indicates that these source currents will be identical to mesh currents. Also, note that the circuit has seven branches where

40V Snkv„ | 6 Q 20V

Figure 4.21 • The circuit for Example 4.4.

the current is unknown and five nodes. Therefore we need three [b - (n - 1 ) - 7 - (5 - 1)] mesh-current equations to describe the circuit. Figure 4.22 shows the three mesh currents used to describe the circuit in Fig. 4.21. If we assume that the voltage drops are positive, the three mesh equations are

- 4 0 + 2*a 4- 8(/a - ih) = 0,

8(4 - 4) + 6/b + 6(4 - /c) = 0,

6(4 - 4) + 4/c + 20 = 0. (4.32)

Your calculator can probably solve these equa­tions, or you can use a computer tool. Cramer's method is a useful tool when solving three or more simultaneous equations by hand. You can review this important tool in Appendix A. Reorganizing Eqs. 4.32 in anticipation of using your calculator, a computer program, or Cramer's method gives

10/a - 8/b + 0/c = 40;

- 8 4 + 20/b - 64 = 0;

04 - 64 + IO4 = -20 . (4.33)

The three mesh currents are

4 = 5.6 A,

4 = 2.0 A,

L = -0.80 A.

40 V 20 V

Figure 4.22 • The three mesh currents used to analyze the circuit shown in Fig. 4.21.

The mesh current 4 is identical with the branch current in the 40 V source, so the power associ­ated with this source is

P40V = - 4 0 4 = -224 W.

102 Techniques of Circuit Analysis

The minus sign means that this source is deliver­ing power to the network. The current in the 20 V source is identical to the mesh current /c; therefore

p2W = 20ic = - 1 6 W.

The 20 V source also is delivering power to the network.

b) The branch current in the 8 il resistor in the direction of the voltage drop v0 is /a - /b. Therefore

v(> = 8(4 - 4) = 8(3.6) = 28.8 V.

^/ASSESSMENT PROBLEM

Objective 2—Understand and be able to use the mesh-current method

4.7 Use the mesh-current method to find (a) the power delivered by the 80 V source to the cir­cuit shown and (b) the power dissipated in the 8 O resistor.

Answer: (a) 400 W;

(b)50W.

NOTE: Also try Chapter Problems 4.33 and 4.34.

80 V sa

4.6 The Mesh-Current Method and Dependent Sources

If the circuit contains dependent sources, the mesh-current equations must be supplemented by the appropriate constraint equations. Example 4.5 illustrates the application of the mesh-current method when the circuit includes a dependent source.

Example 4.5 Using the Mesh-Current Method with Dependent Sources

Use the mesh-current method of circuit analysis to determine the power dissipated in the 4 fl resistor in the circuit shown in Fig. 4.23.

in

5ft 4 f t

50 VI r 20 ft 15 4

Figure 4.23 A The circuit for Example 4.5.

Solution

This circuit has six branches where the current is unknown and four nodes. Therefore we need three mesh currents to describe the circuit. They are

defined on the circuit shown in Fig. 4.24. The three mesh-current equations are

50 = 5(/, - i2) + 20(/, - /3),

0 = 5(/2 - /,) + l/2 + 4(/2 - /3),

0 = 20(/3 - h) + 4(/3 - /2) + 15iV (4.34)

We now express the branch current controlling the dependent voltage source in terms of the mesh currents as

l4> ~ h ~ z3' (4.35)

4.7 The Mesh-Current Method: Some Special Cases 103

which is the supplemental equation imposed by the presence of the dependent source. Substituting Eq. 4.35 into Eqs. 4.34 and collecting the coeffi­cients of / j , /2, and /3 in each equation generates

50 = 25i*! - 5/2 - 20i3,

0 = - 5 / , + 10j2 - 4/3,

0 = — 5/j - 4/2 + 9/3.

1 n - A W

51) 4 a

50 V' 20 a 15/,,,

Figure 4.24 • The circuit shown in Fig. 4.23 with the three mesh currents.

Because we are calculating the power dissipated in the 4 O resistor, we compute the mesh currents i2

and /3:

i2 = 26 A,

/3 = 28 A.

The current in the 4 H resistor oriented from left to right is /3 — i2 , or 2 A. Therefore the power dissipated is

PAH = (¾ - <2)2(4) = (2)2(4) = 16 W.

What if you had not been told to use the mesh-current method? Would you have chosen the node-voltage method? It reduces the problem to finding one unknown node voltage because of the presence of two voltage sources between essential nodes. We present more about making such choices later.

^ A S S E S S M E N T PROBLEMS

Objective 2—Understand and be able to use the mesh-current method

4.8 a) Determine the number of mesh-current equations needed to solve the circuit shown.

b) Use the mesh-current method to find how much power is being delivered to the dependent voltage source.

Answer: (a) 3;

(b) -36 W.

4.9 Use the mesh-current method to find va in the circuit shown.

Answer: 16 V.

NOTE: Also try Chapter Problems 4.38 and 4.39.

4.7 The Mesh-Current Method: Some Special Cases

When a branch includes a current source, the mesh-current method requires some additional manipulations. The circuit shown in Fig. 4.25 depicts the nature of the problem.

We have defined the mesh currents i.d, i^ and /c, as well as the voltage across the 5 A current source, to aid the discussion. Note that the circuit contains five essential branches where the current is unknown and four essential nodes. Hence we need to write two [ 5 - ( 4

100 V

Figure 4.25 • A circuit illustrating mesh 1)] mesh-current a branch contains an independent current

50 V

analysis when source.

106 Techniques of Circuit Analysis

I /ASSESSMENT PROBLEMS

Objective 2—Understand and be able to use the mesh-current method

4.10 Use the mesh-current method to find the power dissipated in the 2 ft resistor in the circuit shown.

30 V

Answer: 72 W.

4.11 Use the mesh-current method to find the mesh current /a in the circuit shown.

75 VI

Answer: 15 A.

4.12 Use the mesh-current method to find the power dissipated in the 1 ft resistor in the cir­cuit shown.

16 A

^vw 4 Wv 4

10 A e 2 0

'a <\t, 1 5 n

in •A^V-

2V*

2A e iovT* j 2 a

Answer: 36 W.

NOTE: Also try Chapter Problems 4.42, 4.44, 4.48, and 4.51.

6V

4,8 The Node-Voltage Method Versus the Mesh-Current Method

The greatest advantage of both the node-voltage and mesh-current meth­ods is that they reduce the number of simultaneous equations that must be manipulated. They also require the analyst to be quite systematic in terms of organizing and writing these equations. It is natural to ask, then, "When is the node-voltage method preferred to the mesh-current method and vice versa?" As you might suspect, there is no clear-cut answer. Asking a number of questions, however, may help you identify the more efficient method before plunging into the solution process:

• Does one of the methods result in fewer simultaneous equations to solve?

• Does the circuit contain supernodes? If so, using the node-voltage method will permit you to reduce the number of equations to be solved.

126 C H A P T E R 3 N O D A L A N D L O O P A N A LY S I S T E C H N I Q U E S

E3.13 Use mesh equations to find in the circuit in Fig. E3.13.Vo

Learning AssessmentANSWER: Vo =

33

5 V.

±–

+-4 k�

2 k�

2 k�

6 k� Vo

3 V

6 V

+

-

2 k�

10 V8 V

12 V4 k�

3 k�

3 k�

4 k�

6 k�

6 k� Vo

+

–

+– +

–

-+

Figure E3.13

Let us find both and in the circuit in Fig. 3.22.

Although it appears that there are two unknown mesh currents, the current goes directlythrough the current source and, therefore, is constrained to be 2 mA. Hence, only thecurrent is unknown. KVL for the rightmost mesh is

And, of course,

I1 = 2 * 10-3

2k( I2 - I1) - 2 + 6k I2 = 0

I2

I1

I1

V1Vo

SOLUTION

E3.14 Find Vo in Fig. E3.14 using mesh analysis. ANSWER: Vo = 8.96 V.

Figure E3.14

CIRCUITS CONTAINING INDEPENDENT CURRENT SOURCES Just as the pres-ence of a voltage source in a network simplified the nodal analysis, the presence of a currentsource simplifies a loop analysis. The following examples illustrate the point.

EXAMPLE

3.14

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S E C T I O N 3 . 2 L O O P A N A LY S I S 127

These equations can be written as

Solving these equation for I2 yields I2 = 3/4kA and thus

To obtain we apply KVL around any closed path. If we use the outer loop, the KVLequation is

And therefore,

Note that since the current is known, the resistor did not enter the equation infinding However, it appears in every loop containing the current source and, thus, isused in finding V1 .

Vo .4-k�I1

V1 =

21

2 V

-V1 + 4kI1 - 2 + 6kI2 = 0

V1

Vo = 6kI2 =

9

2 V

I1 = 2�k

- 2kI1 + 8kI2 = 2

2 mA

2 V

2 k�

4 k�

6 k�

V1

I1 I2 Vo

+

-

+-Figure 3.22

Circuit used inExample 3.14.

SOLUTION

EXAMPLE

3.15We wish to find in the network in Fig. 3.23.

Since the currents and pass directly through a current source, two of the three requiredequations are

The third equation is KVL for the mesh containing the voltage source; that is,

These equations yield

and hence,

Vo = 6kI3 - 3 =

-3

2 V

I3 =

1

4 mA

4kAI3 - I2B + 2kAI3 - I1B + 6kI3 - 3 = 0

I2 = -2 * 10-3

I1 = 4 * 10-3

I2I1

Vo

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128 C H A P T E R 3 N O D A L A N D L O O P A N A LY S I S T E C H N I Q U E S

-+

4 mA

2 mA 3 V

2 k�

4 k�

6 k�

4 k�Vo

+

-

I3

I1

I2

Figure 3.23

Circuit used inExample 3.15.

What we have demonstrated in the previous example is the general approach for dealingwith independent current sources when writing KVL equations; that is, use one loop througheach current source. The number of “window panes” in the network tells us how many equa-tions we need. Additional KVL equations are written to cover the remaining circuit elementsin the network. The following example illustrates this approach.

Let us find in the network in Fig. 3.24a.

First, we select two loop currents and such that passes directly through the 2-mAsource, and passes directly through the 4-mA source, as shown in Fig. 3.24b. Therefore,two of our three linearly independent equations are

The remaining loop current must pass through the circuit elements not covered by the twoprevious equations and cannot, of course, pass through the current sources. The path for thisremaining loop current can be obtained by open-circuiting the current sources, as shown inFig. 3.24c. When all currents are labeled on the original circuit, the KVL equation for thislast loop, as shown in Fig. 3.24d, is

Solving the equations yields

and therefore,

Next consider the supermesh technique. In this case the three mesh currents are specifiedas shown in Fig. 3.24e, and since the voltage across the 4-mA current source is unknown,it is assumed to be The mesh currents constrained by the current sources are

The KVL equations for meshes 2 and 3, respectively, are

-6 + 1kI3 + Vx + 1kAI3 - I1B = 0

2kI2 + 2kAI2 - I1B - Vx = 0

I2 - I3 = 4 * 10-3

I1 = 2 * 10-3

Vx .

Io = I1 - I2 - I3 =

-4

3 mA

I3 =

-2

3 mA

-6 + 1kI3 + 2kAI2 + I3B + 2kAI3 + I2 - I1B + 1kAI3 - I1B = 0

I3

I2 = 4 * 10-3

I1 = 2 * 10-3

I2

I1I2I1

Io

SOLUTION

In this case the 4-mA currentsource is located on theboundary between two mesh-es. Thus, we will demonstratetwo techniques for dealingwith this type of situtation.One is a special loop tech-nique, and the other is knownas the supermesh approach.

[ h i n t ]

EXAMPLE

3.16

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130 C H A P T E R 3 N O D A L A N D L O O P A N A LY S I S T E C H N I Q U E S

E3.15 Find in the network in Fig. E3.15.Vo

Learning AssessmentsANSWER: Vo =

33

5 V.

E3.16 Find in the network in Fig. E3.16.Vo ANSWER: Vo =

32

5 V.

±–2 k� 4 k�

6 k�

4 mA

5 V

Vo+ -

Figure E3.15

±–

1 k�

4 k� Vo

2 k�

2 mA

4 mA

4 V

+

-Figure E3.16

E3.17 Find in Fig. E3.17 using loop analysis.Vo ANSWER: = 9.71 V.Vo

Figure E3.17

E3.18 Find in Fig. E3.17 using mesh analysis.Vo ANSWER: = 9.71 V.Vo

2 k�

10 V8 V

12 V2 mA

3 k�

3 k�

4 k�

6 k�

6 k� Vo

+

–

+– +

–

-+

irwin03_101-155hr.qxd 30-06-2010 13:12 Page 130

132 C H A P T E R 3 N O D A L A N D L O O P A N A LY S I S T E C H N I Q U E S

Let us find in the circuit in Fig. 3.26, which contains a voltage-controlled current source.

The currents and are drawn through the current sources. Therefore, two of the equationsneeded are

The KVL equation for the third mesh is

where

Combining these equations yields

The equations can be expressed in matrix form as IR = V, where

-1 2 0 i1 0R = 0 1 0 I = i2 and V = 0.002

-2000 0 8000 i3 3

Performing the operation I = R-1V, produces the currents

i1 = 4.0 mAi2 = 2.0 mAi3 = 1.375 mA

And hence, . Vo = 8.25 V

- 2kI2 + 8kI3 = 3

I2 = 2�k

- I1 + 2I2 = 0

Vx = 4k (I1 - I2)

- 3 + 2k(I3 - I1) + 6kI3 = 0

I2 = 2 * 10 - 3

I1 =

Vx

2000

I2I1

Vo

±–

Vx

I1

I34 k�

6 k�

2 mA

2 k�

3 V

Vx2000—

Vo

+

-

I2

+-

Figure 3.26

Circuit used inExample 3.18.

The network in Fig. 3.27 contains both a current-controlled voltage source and a voltage-controlled current source. Let us use MATLAB to determine the loop currents.

The equations for the loop currents shown in the figure are

1kAI4 - I3B + 1kAI4 - I2B + 12 = 0

-1kIx + 2kAI3 - I1B + 1kAI3 - I4B = 0

I2 =

Vx

2k

I1 =

4

k

SOLUTION

SOLUTION

EXAMPLE

3.18

EXAMPLE

3.19

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S E C T I O N 3 . 2 L O O P A N A LY S I S 133

where

Combining these equations yields

In matrix form the equations are

=

The equations are in the form RI = V, and the solution to I = R-1V is

i1 = 4.0 mA

i2 = 6.0 mA

i3 = -2.0 mA

i4 = -1.0 mA

E4

k

0

8

12

UDI1

I2

I3

I4

TD1

1

0

0

0

1

1k

1k

0

-1

3k

1k

0

0

-2k

-2k

T

1kI2 + 1kI3 - 2kI4 = 12

1kI2 + 3kI3 - 2kI4 = 8

I1 + I2 - I3 = 0

I1 =

4

k

Ix = I4 - I2

Vx = 2kAI3 - I1B

±–

±–

4 mA

2 k�

1 k�

2 k� 1 k�

12 V

IxVx

I41kIx

I1

I3

I2Vx2k—

+ -

Figure 3.27

Circuit used inExample 3.19.

At this point we will again examine the circuit in Example 3.10 and analyze it using loopequations. Recall that because the network has two voltage sources, the nodal analysis wassomewhat simplified. In a similar manner, the presence of the current sources shouldsimplify a loop analysis.

Clearly, the network has four loops, and thus four linearly independent equations arerequired to determine the loop currents. The network is redrawn in Fig. 3.28 where the loop cur-rents are specified. Note that we have drawn one current through each of the independent cur-rent sources. This choice of currents simplifies the analysis since two of the four equations are

I3 = -2�k I1 = 2�k

EXAMPLE

3.20

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