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Chapter 6 Circuit Equations

Kirchhoffs laws and Ohms law provide the number of equations necessary to

analyze an electric circuit, but the number of equations can become unwieldy except

in simple circuit configurations. Although the theorems, procedures and techniques

discussed in Chapter 3, 4, and 5 can be invaluable in simplifying the analysis in many

cases of practical interest, there remains cases where these methods are of limited

use. Systematic methods based on Kirchhoffs laws have therefore been developed

to facilitate analysis of more complex circuits and reduce the likelihood of error in

writing the equations that govern the behavior of electric circuits. The most commonly

used node-voltage and mesh-current methods are presented in this chapter,

including some special considerations and generalizations.

6.1 Node-Voltage Method

Concept In the node-voltage method, voltages of essential nod es are

assigned w ith respect to on e of the essential nodes taken as a

reference. This automatical ly sat isf ies KVL in every mesh in the

circui t . Equations based on KCL are then wri t ten direct ly in terms of

Ohms law for each essential node other than the reference node.

The node-voltage method will be illustrated by the circuit of Figure 6.1.1a,which is excited by a current source and in which resistance values are expressed as

conductances. The first step is to select one of the essential nodes as the reference

node. This can be done quite arbitrarily, but it is somewhat convenient to select as a

reference node the node that has the largest number of connections, which is usually

a grounded node. It is also convenient to select as a reference node, one of the node

with respect to which a required voltage is defined, so that only one unknown needs

to be determined. If the

node voltages arerequired with respect to a

node other than the

selected reference node,

this can be done very

simply, as will be shown

later. When the reference

node is not a grounded

node, it is denoted by a

a b c

d

G1

G2

G3

G4

G5

GsrcISRC

Va Vb Vc

(a)

+

+ +

+

+

Vab Vbc

Vca

VcdVdb

+

Vad


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filled or an open arrow, as

in Figure 6.1.1a, where

node d is chosen as the

reference node.

The reference

node is assigned a

voltage of zero, and the

voltages of all the other

essential nodes are

expressed with reference to this node. To verify that KVL is satisfied around every

mesh, consider the upper mesh to begin with. The voltage drops around this mesh

are denoted as Vab, Vbc, and Vcain Figure 6.1.1a. KVL around this mesh will then be:

Vab+ Vbc+ Vca= 0 (6.1.1)

The voltage drops Vab, Vbc, and Vcacan be expressed in terms of the assigned

node voltages as: Vab= VaVb, Vbc= VbVc, and Vca= VcVa. The LHS of

Equation 6.1.1 becomes:

(VaVb) + (VbVc) + (VcVa) = 0

The node voltages cancel out in pairs, so they sum to zero, as required by

KVL. The same is true of the mesh on the RHS of Figure 6.1.1a, for which KVL takes

the form:

Vbc+ Vcd+ Vdb= 0 (6.1.2)

Substituting Vbc= VbVc, Vcd= VcVd, and Vdb= VdVbsatisfies Equation

6.1.2. KVL for the mesh composed of V1, V2, and Gsrcis:

VadVab+ Vdb= 0 (6.1.3)

Substituting Vad= VaVd, Vab= VaVb, and Vdb= VdVbagain satisfies

Equation 6.1.3.

With KVL satisfied, KCL is written for each of the essential nodes in terms of

Ohms law and any source currents entering or leaving the node. For node a, thecurrent leaving the node through the conductances connected to the node is: Iad+ Iac

+ Iab(Figure 6.1.1b), where Iad= GsrcVa, Iac= G5(VaVc) and Iab= G1(VaVb). The

source current entering node a is ISRC. Equating the current leaving the node

through the conductances to the current entering the node from the source and

collecting terms in the node voltages gives KCL for node a as:

(Gsrc+ G1+ G5)Va G1Vb G5Vc = ISRC (6.1.4)

The current leaving node b through the conductances connected to the node

is: Iba+ Ibc+ Ibd, where Iba= G1(VbVa), Ibc= G3(VbVc), and Ibd= G2Vb. There is no

source current entering node b. Collecting terms in the node voltages gives KCL for

Figure 6.1.1

ab c

d

G1

G2

G3

G4

G5

GsrcISRC

Va Vb Vc

(b)

IabIad

Iac

Iba Ibd

Ibc

Icb I

cd

Ica


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node b as:

G1Va (G1+ G2+ G3)Vb G3Vc = 0 (6.1.5)

The current leaving node c through the conductances connected to the node

is: Icb+ Ica+ Icd, where Icb= G3(VcVb), Ica= G5(VcVa), and Icd= G4Vc. There is no

source current entering node c. Collecting terms in the node voltages gives KCL for

node c as:

G5Va G3Vb (G3+ G4+ G5)Vc = 0 (6.1.6)

Comparing Equations 6.1.4 to 6.1.6 reveals a pattern that allows writing the

node-voltage equations by inspection, namely:

1. in the equation for a given node, the coefficient multiplying the voltage of this

node is the sum of all the conductances connected to that node. Thus, in the

equation for node a (Equation 6.1.4) Vais multiplied by (Gsrc+ G1+ G5), the sum

of the three conductances connected to node a. Similarly, in the equation for

node b (Equation 6.1.5), Vbis multiplied by (G1+ G2+ G3), and in the equation

for node c (Equation 6.1.6), Vcis multiplied by (G3+ G4+ G5).These coefficients

are known as the self conductancesof the nodes.

2. In the equation for a given node, the coefficient multiplying the voltage of each of

the other nodes is the conductance that directly connects this node to the given

node, with a minus sign. The minus sign arises from subtracting the voltages of

the other nodes from the voltage of the given node. Thus, the current leaving

node a through G1is (VaVb)G1= G1VaG1Vb, where G1connects node a to

node b. In the equation for node a (Equation 6.1.4) Vbis therefore multiplied by

-G1. The current leaving node a through G5is (VaVc)G5= G5VaG5Vc, where

G5connects node a to node c. In the equation for node a (Equation 6.1.4) Vcis

therefore multiplied byG5, The same is true of the other node equations. These

coefficients are known as the mutual conductancesbetween the nodes. If there

is no conductance that directly connects a certain node with the node in question,

the corresponding mutual conductance is zero.If the coefficients of

the node voltages in

Equations 6.1.4 to 6.1.6

are arranged in an array,

as illustrated in Figure

6.1.2, the array has the

following features, which

provide a useful check on

(Gsrc + G1 + G5) G1 G5

G1 (G1 + G2 + G3) G3

G5 G3 (G3 + G4 + G5)

Figure 6.1.2


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1/12 S 0.25 S

1/3 S

0.1 S

Figure 6.1.3

d

0.5 S

0.25 S

6 A

Va Vb

Vc

the correctness of the node-voltage equations:

1. the self conductances are the diagonal entries in the array.

2. ii) The array is symmetrical with respect to the diagonal, as illustrated by the

coefficients pointed to by the arrows in Figure 6.1.2. This symmetry is because

the conductance is independent of the direction of current. For example, in the

expression Iab= G1(VaVb), G1is the same as in the expression Iba= G1(VbVa),

although Iab= -Iba. It will be shown in the next section that this symmetry is

destroyed when the dependency relations of dependent sources are taken into

account.

3. All the mutual conductances have a negative sign, as explained previously.

4. In any row or column, the mutual conductances are part of the self conductances

in that row or column. Thus, in the first row or first column in the array of Figure

6.1.2, G1and G5are included in the self conductance term (Gsrc+ G1+ G5).

The procedure for writing the node-voltage equation for a given essential

node n can be summarized as follows:

1. The voltage of node n is multiplied by the sum of all the conductances

connected directly to this node, so that node n constitutes one terminal for each

of these conductances.

2. The voltage of every other node is multiplied by the conductance connected

directly between this node and node n, with a negative sign. If there is no such

conductance, the coefficient is zero.

3. The LHS of the node-voltage equation for node n is the sum of the terms from

the preceding steps, ordered as the unknown node voltages. This sum is the total

current leaving node n through the conductances connected to this node.

4. The RHS of the equation is equal to the algebraic sum of source currents

entering node n. Thus, a source current entering node n will have a positive

sign, whereas a source current leaving node n will have a negative sign.

Example 6.1.1

It is required to analyze the

circuit of Figure 6.1.3 using the

node-voltage method.

Solution:Taking node d as a

reference node, and following the

aforementioned standard procedure,

the node voltage equations are:


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Node a: (0.5 + 1/3 + 0.1)Va(1/3)Vb0.1Vc= 6 (6.1.7)

Node b: (1/3)Va+ (1/3 + 0.25 + 1/12)Vb0.25Vc= 0 (6.1.8)

Node c: 0.1Va0.25Vb+ (0.25 + 0.25 + 0.1)Vc= 0 (6.1.9)

Equations 6.1.7 to 6.1.9 can be simplified to:

(14/15)Va (1/3)Vb 0.1Vc= 6 (6.1.10)

(1/3)Va+ (2/3)Vb0.25Vc= 0 (6.1.11)

0.1Va 0.25Vb+ (3/5)Vc= 0 (6.1.12)

Equations 6.1.10 to 6.1.12 can be solved by any of the usual methods for

solving linear simultaneous equations, or by using appropriate calculators, to give: Va

= 9 V, Vb= 6 V, and Vc= 4 V.

Linear simultaneous equations can be conveniently solved using MATLAB.

To do so, the conductance coefficients on the LHS of the node-voltage equations are

entered as a square matrix, and the source currents on the RHS of the node-voltage

equations are entered as a column matrix. In the example under consideration, the

matrix of coefficients are entered in MATLAB as follows:

C = [14/15,-1/3,-0.1;-1/3,2/3,-0.25;-0.1,-0.25,3/5]

In MATLAB, a matrix is entered between square brackets. Elements in a row

are separated by commas, whereas rows are separated by semicolons. The matrix of

source currents is entered as:

S = [6;0;0]

The command:

C\S

is equivalent to [inv(C)]*S and gives a column matrix of the node voltages in the order

Va, Vb, and Vc. MATLAB displays the solution to the simultaneous Equations 6.1.6 as:

9.000

6.000

4.000

Simulation:The circuit isentered as in Figure 6.1.4.

After selecting Bias

Point/General Settings in

the simulation profile and

running the simulation,

pressing the I and V buttons

displays the currents and voltages, respectively, indicated in Figure 6.1.4.

3 4

4122

0

6Adc

6.000V

500.0mA1.000A

1.000A500.0mA

4.500A

9.000V

Figure 6.1.4

10

500.0mA

+

-

4.000V


6/32



Example 6.1.2

It is required to analyze

the circuit of Figure 6.1.5 using

the node-voltage method. The

circuit is of the same form as

that in Figure 6.1.3, but with

different circuit parameters

and with the uppermost

conductance replaced by a

current source. One of the

mutual conductances is now

zero, and there is algebraic summation of source currents at one of the nodes.

Solution:In writing the node-voltage equations for a given node as KCL equations,

the LHS of the equations represents current leaving the node through conductances,

and the RHS represents source current entering the node. Hence in writing the node-

voltage equation for node a, a source current leaving the node must be entered on

the RHS with a negative sign. Alternatively, it may be considered that the net source

current entering node a is (12 6) A. The equation for node a is:

Node a: (0.5 + 0.25)Va0.25Vb0Vc= 126 (6.1.13)

Note that since there is no conductance that directly connects node a tonode c, the mutual conductance between these two node is zero, which means that

Vcno longer appears in the node-voltage equation for node a, nor does Vaappear in

the node-voltage equation for node c. The remaining node voltage equation are:

Node b: -0.25Va+ (1/3 + 0.25 + 0.5)Vb0.5Vc= 0 (6.1.14)

Node c: 0Va0.5Vb+ (0.25 + 0.5)Vc= 6 (6.1.15)

The solution to these equations gives Va= 11 V, Vb= 9 V, and Vc= 14 V.

Note that if the 0 coefficient is removed, the matrix is no longer square and the array

of coefficients is no longer

symmetrical about the

diagonal.

Simulation:The circuit is

entered as in Figure 6.1.6.


Point/General Settings in the

simulation profile and


1/3 S 0.25 S

0.25 S

Figure 6.1.5

d

0.5 S

0.5 S

12 A

Va Vb

Vc

6 A

4 2

432

0

12Adc

6Adc

9.000V2.500A

500.0mA

14.00V

3.500A3.000A

5.500A

11.00V

Figure 6.1.6

+

-

+ -


7/32



pressing the I and V buttons displays the currents and voltages, respectively,

indicated in Figure 6.1.6.

Exercise 6.1.1

Verify that KCL and Ohms law are satisfied in Figure 6.1.5.

Exercise 6.1.2

Reverse the direction of the 6 A source in the circuit of Figure 6.1.5, derive

the node-voltage equations and determine Va, Vb, and Vc. Simulate the circuit and

verify the values of the node voltages, KCL, and Ohms law.

Ans. Va= 25 V, Vb= 3 V, and Vc= -6 V.

Change of Reference Node

Consider Figure 6.1.1a. Any branch voltage is the difference between two

node voltages. For example, Vab= VaVb. If the same quantity is added to both Va

and Vb, it will cancel out from the RHS, leaving Vabthe same. In Example 6.1.1, node

d was taken as a reference and the node voltages were found to be: Va= 9 V, Vb, =

6 V, Vc= 4 V, with Vd= 0 because it is the reference node. Suppose after finding the

node voltages with respect to node d as reference, we wish to determine the node

voltages with respect to another node, say node b as reference, which means that

Vbmust be zero. To make Vb= 0 without changing the branch voltages, we simply

subtract 6 V from all the node voltages, which gives, Va= 3 V, Vb, = 0 V, Vc= 1 V,

and Vd= -6 V. The branch voltages are evidently the same. With node d as

reference, Vab= 96 = 3 V, and with node b as reference, Vab= 30 = 3 V. If the

branch voltages remain the same, then the branch currents and KCL will also remain

the same.

Concept In any circui t , the branch voltages and cur rents are independent

of the cho ice of reference nod e.

Exercise 6.1.3

Redo Example 6.1.2 with node b as reference and verify the new values of

node voltages.

Non-Transform able Voltage Source*

When the node-voltage method is to be used in a circuit that has an ideal

voltage source in series with a resistor, the combination is conveniently transformed

to a current source in parallel with the same resistor. But when the ideal voltage


8/32



source does not have a

resistor in series with it, it

cannot be transformed to a

current source and must be

left unaltered. The circuit of

Figure 6.1.7, for example, is

the same as that of Figure

6.1.5 but with the 6 A source

replaced by a 3 V source that

cannot be transformed to a current source. In applying the node-voltage method, an

unknown current Ixis assigned an arbitrary direction through the voltage source and

the standard procedure followed, treating Ixlike a source current, in accordance with

the substitution theorem. The node-voltage equations become:

Node a: (0.5 + 0.25)Va0.25Vb0Vc= 12Ix (6.1.16)

Node b: 0.25Va+ (1/3 + 0.25 + 0.5)Vb0.5Vc= 0 (6.1.17)

Node c: 0Va0.5Vb+ (0.25 + 0.5)Vc= Ix (6.1.18)

Ixcan be eliminated by adding together Equations 6.1.16 and 6.1.18 for the

two nodes between which the voltage source is connected. The resulting equation is:

0.75Va0.75Vb+ 0.75Vc= 12 (6.1.19)

Equation 6.1.19 is sometimes referred to as the equation of a supernode thatresults from combining nodes a and c. The node-voltage equation for the

supernode can be written following the usual procedure, without having to introduce

an unknown source current. However, introducing such a current is more

fundamental and is less likely to cause an error.

Adding two node-voltage equations to eliminate the unknown source current

reduces the number of independent voltage equations by one. But an additional

equation in the node voltages is provided by the relation between the node voltages

and the source voltage. In Figure 6.1.7,

VcVa= 3 (6.1.20)

Equations 6.1.17, 6.1.19 and 6.1.20 are three independent equations that can

be solved to give: Va= 11 V, Vb= 9 V, and Vc= 14 V. These values are the same as

in Example 6.1.2, because in this example VcVa= 3, as for the voltage source

between nodes a and c.

Exercise 6.1.4

Simulate the circuit of Figure 6.1.7 and verify that KCL, KVL, and Ohms law

1/3 S 0.25 S

0.25 S

Figure 6.1.7

d

0.5 S

0.5 S

12 A

Va Vb

Vc

Ix

3 V

+

Ix


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are satisfied.

6.2 Dependent Sources in Node-Voltage Method

Dependent current

sources are treated in exactly

the same manner as

independent current sources.

Consider, for example, the

circuit of Figure 6.2.1, which is

the same as that of Figure 6.1.5

but with the 6 A independent

source replaced by a dependent

current source. The node-

voltage equations are written in the usual way as:

Node a: (0.5 + 0.25)Va0.25Vb0Vc= 122Ib (6.2.1)

Node b: 0.25Va+ (1/3 + 0.25 + 0.5)Vb0.5Vc= 0 (6.2.2)

Node c: 0Va0.5Vb+ (0.25 + 0.5)Vc= 2Ib (6.2.3)

Note that in these equations, the net current entering node a is 12 2Iband

the current entering node c is 2Ib. Leaving the 0 coefficient in the equations

maintains the symmetry in the array of coefficients.In order to solve the node-voltage equations the controlling variable, Ibin this

case, should be expressed in terms of the node voltages. In the circuit of Figure

6.2.1, Ib= Vb/3 A. Substituting and moving the term in Vbto the LHS, Equations 6.1.8

to 6.1.10 become:

Node a: (0.5 + 0.25)Va(0.252/3)Vb0Vc= 12 (6.2.4)

Node b: 0.25Va+ (1/3 + 0.25 + 0.5)Vb0.5Vc= 0 (6.2.5)

Node c: 0Va (0.5 + 2/3)Vb+ (0.25 + 0.5)Vc= 0 (6.2.6)

Solving these equations gives Va= 11 V, Vb= 9 V, and Vc= 14 V, the same

as in Example 6.1.2, because in this example Ib= 3 A, and 2Ib= 6 A, the same as the

independent current. Note that the array of coefficients is symmetrical with respect to

the diagonal in Equations 6.2.1 to 6.2.3, when 2Ibis on the RHS, but the symmetry is

destroyed in Equations 6.2.4 to 6.2.5 when 2Ibis substituted for in terms of Vband

moved to the LHS.

In the case of dependent voltage sources, if the source is in series with a

resistance, the dependent voltage source is transformed to a dependent current

source in parallel with this resistance, and the standard procedure followed. If the

1/3 S 0.25 S

0.25 S

Figure 6.2.1

d

0.5 S

0.5 S

12 A

Va Vb

Vc

Ib

2 Ib


10/32



+

I3

d

I2

b

I3

I3I2

I3I1

R3

R2 R4

R5

(a)

I2

c

I1

a I2

I1 I1I2 I1I3

VSRC

R1Rsrc

dependent voltage source cannot be transformed to a current source, an unknown

current is assigned to the voltage source, and the procedure explained in connection

with Figure 6.1.7 is followed. Several examples of this type are included in problems

at the end of the chapter.

It will be observed from the preceding that the effect of the dependent source

in Figure 6.2.1 is to modify some of the conductance coefficients in the node-voltage

equations, leaving only the values of independent source on the RHS of the

equations. This is in accordance with the discussion of Section 4.1, Chapter 4, that

dependent sources alone cannot excite a circuit, and with discussion in connection

with Equation 5.1.5, Chapter 5.

Exercise 6.2.1


are satisfied.

6.3 Mesh-Current Method

Concept In the mesh-current m ethod, the unkn own m esh currents are

assigned in s uch a manner that KCL is automatically sat isf ied at

every essential nod e. Equations based on KVL are then wri t ten for

each mesh directly in terms of Ohms law.

The mesh-current method will be illustrated by the circuit of Figure 6.3.1a.

The first step is to assign a current to each mesh, conventionally in the clockwise

direction, as illustrated by I1, I2, and I3in the figure. Since the same mesh current

enters and leaves any given node, KCL is automatically satisfied at each of the

essential nodes a, b, c, and d. To verify this, consider node a. The mesh current

I1enters node a through Rsrcand leaves through R1. The mesh current I2enters

node a through R1

and leaves throughR5. The current

leaving node a

through R1can be

considered as I1I2.

Equating the total

current entering node

a to that leaving it: I1

= I1I2+ I2= I1,


11/32



which satisfies KCL.

At node b,

the current entering

the node through R1

is I1I2. The current

leaving node b

through R2is I1I3

and that leaving it

through R3is I3I2.

Equating the total

current entering node b to that leaving it: I1I2= I1I3+ I3I2= I1I2, thereby

satisfying KCL. At node c, the current entering the node through R5is I2and that

entering through R3is I3I2. The current leaving node c through R4is I3. Equating

the total current entering node c to that leaving it: I2+ I3I2= I3, thereby satisfying

KCL. At node d, the current entering node d through R4is I3and that entering it

through R2is I1I3. The current leaving node d through thesource is I1. Equating

the total current entering node d to that leaving it: I3+ I1I3= I1, thereby satisfying

KCL.

The next step is to write KVL around each mesh. Figure 6.3.1b indicates the

voltage drop in each resistor due to the mesh currents flowing through the resistor.

Considering R1, for example, the net current through R1is (I1I2) in the direction of

I1, and the voltage drop in R1in the direction of I1is R1(I1I2). Similarly, the voltage

drop in R2is R2(I1I3) in the direction of I1. The total voltage drop in the direction of I1

due to the resistors in the mesh is therefore RsrcI1+ R1(I1I2) + R2(I1I3). According

to KVL, the total voltage drop due to the resistors in the mesh must be equal to the

voltage rise due any sources in the mesh, which is VSRCin the direction of I1.

Equating the voltage drop to the voltage rise and collecting terms in the mesh

currents gives:(Rsrc+ R1+ R2)I1 R1I2 R2I3 = VSRC (6.3.1)

The term (Rsrc+ R1+ R2)I1is the voltage drop in mesh 1 due to I1alone. The

negative signs in Equation 6.3.1 can be usefully interpreted in terms of voltage rises.

Thus, in the term R1(I1I2), R1I2is a voltage rise in mesh 1, in the direction of I1, due

to I2flowing in R1(Figure 6.3.1b). Since the LHS of Equation 6.3.1 is the voltage drop

in the direction of I1due to the resistors in mesh 1, the term R1I2is a voltage rise and

will have a negative sign on the LHS of Equation 6.3.1. Similarly, since R2I3is a

voltage rise in mesh 1, in the direction of I1, due to I3flowing in R2, the term R2I3will

have a negative sign on the LHS of Equation 6.3.1.

+

VSRC

I2

I3

(b)

I1

+ R1I1+

R1I2 +

RsrcI1

+ R5I2

+

R2I1

+

R2I3

R3I2 +

R3I3+ +

R4I3

Figure 6.3.1


12/32



In mesh 2, the net current in R1in the direction of I2is (I2I1) and the voltage

drop in R1in the direction of I2is R1(I2I1). Similarly, the net current in R3in the

direction of I2is (I2I3) and the voltage drop in R3in the direction of I2is R3(I2I3).

The total voltage drop in the direction of I2due to the resistors mesh 2 is: R1(I2I1) +

R5I2+ R3(I2I3). As there are no sources in mesh 2, this total voltage drop must be

equal to zero. Collecting terms in the mesh currents, gives the mesh current equation

for mesh 2 as:

R1I2 (R1+ R3+ R5)I1 R3I3 = 0 (6.3.2)

In mesh 3, the net current in R2in the direction of I3is (I3I1) and the voltage

drop in R2in the direction of I3is R2(I3I1). Similarly, the net current in R3in the

direction of I3is (I3I2) and the voltage drop in R3in the direction of I3is R3(I3I2).

The total voltage drop in the direction of I3due to the resistors mesh 3 is: R2(I3I1) +

R4I3+ R3(I3I2). As there are no sources in mesh 3, this total voltage drop must be

equal to zero. Collecting terms in the mesh currents, gives the mesh current equation

for mesh 3 as:

R2I1 R3I2+ (R2+ R3+ R4)I3 = 0 (6.3.3)

Comparing Equations 6.3.1 to 6.3.3 reveals a pattern that allows writing the

mesh-voltage equations by inspection, namely:

1. In the equation for a given mesh, the coefficient multiplying the mesh current is

the sum of all the resistances in the mesh. Thus, in the equation for mesh 1

(Equation 6.3.1) I1is multiplied by (Rsrc+ R1+ R2). Similarly, in the equation for

mesh 2 (Equation 6.3.2), I2is multiplied by (R1+ R3+ R5), and in the equation for

mesh 3 (Equation 6.3.3), I3is multiplied by (R2+ R3+ R4).These coefficients are

known as the self resistancesof the meshes.

2. In the equation for a given mesh, the coefficient multiplying the current of each of

the other meshes is the resistance that is common to the two meshes, with a

minus sign. Thus, in the equation for mesh 1 (Equation 6.3.1) I2is multiplied by

-R1, where R1is the resistance that is common between meshes 1 and 2, and I3is multiplied by -R2, where R2is the resistance that is common between meshes 1

and 3. The same is true of the other mesh equations. These coefficients are

known as the mutual resistancesbetween the meshes. If there is no resistance

that is common between a certain mesh with the mesh in question, the

corresponding mutual resistance is zero.

If the coefficients of the mesh currents in Equations 6.3.1 to 6.3.3 are

arranged in an array, as illustrated in Figure 6.3.2, the array has the following

features, which are a useful check on the correctness of the mesh-current equations:

1. The self resistances are the diagonal entries in the array.


13/32



2. The array is

symmetrical with

respect to the diagonal,

as illustrated by the

coefficients pointed to

by the arrows in Figure

6.3.2. This symmetry is

because the same

resistance that is

common between two meshes appears in the equation of each of the two

meshes as a multiplier of the current in the other mesh. It will be shown in the

next section that this symmetry is destroyed when the dependency relations of

dependent sources are taken into account.

3. All the mutual resistances have negative signs when mesh currents are in the

same sense, that is clockwise or anticlockwise, because the voltage across a

common resistance is a voltage drop when this resistance is part of the self

resistance of a mesh, but is a voltage rise if the resistance is a mutual resistance,

as explained in connection with Equation 6.3.1.

4. In any row or column, the mutual resistances are part of the self resistance in that

row or column.

The procedure for writing the mesh-current equation for a given mesh n can

be summarized as follows:

1. The current of mesh n is multiplied by the sum of all the resistances in the mesh.

2. The current of every other mesh is multiplied by the resistance that is common

between this mesh and mesh n, with a negative sign. If there is no such

resistance, the coefficient is zero.

3. The LHS of the mesh-current equation for mesh n is the sum of the terms from

the preceding steps, ordered as the unknown mesh currents. This sum is the totalvoltage drop, in the direction of the mesh current, due to all the resistances in the

mesh.

4. The RHS of the equation is equal to the algebraic sum of source voltages in

mesh n. A source voltage that is a voltage rise in the direction of the mesh

current will have a positive sign, whereas a source voltage that is a voltage drop

in the direction of the mesh current will have a negative sign.

(Rsrc + R1 + R2) R1 R2

R1 (R1 + R2 + R5)

R3

R2 R3 (R2 + R3 + R4)

Figure 6.3.2


14/32



+

4

12 4

3

10

Figure 6.3.3

12 V

2

I2

I3I1

Example 6.3.1

It is required to analyze the circuit of Figure 6.3.3 using the mesh current

method.

Solution:The circuit is redrawn in Figure 6.3.3 showing the mesh currents.

Following the standard procedure, the mesh current equations are written as:

Mesh 1: (2 + 3 +12)I13I212I3= 12 (6.3.4)

Mesh 2: -3I1+ (3 + 4 + 10)I24I3= 0 (6.3.5)

Mesh 3: -12I14I2+ (4 + 4 +12)I3= 0 (6.3.6)

These equations reduce to:

17I1 3I212I3= 12 (6.3.7)

-3I1+17I2 4I3= 0 (6.3.8)

-12I14I2+ 20I3= 0 (6.3.9)

The solution to these equations

gives: I1= 1.5 A, I2= 0.5 A, and I3

= 1 A.

Simulation:The circuit is entered as in

Figure 6.3.4. After

selecting Bias

Point/General Settings

in the simulation profile

and running the

simulation, pressing the

I and V buttons displays

the currents and

voltages, respectively,

indicated in Figure

6.3.4.

Example 6.3.2

It is required to analyze the circuit of Figure 6.3.5 using the mesh-current

method. The circuit is of the same form as that in Figure 6.3.3, but with different

circuit parameters, and with the 3 resistor replaced by an 8 V source. One of the

mutual resistances is now zero, and there is algebraic summation of source voltages

in mesh 1.

Solution:Following the standard procedure, and starting with mesh 1, the self

resistance of this mesh is (2 + 8) , the mutual resistance with mesh 2 is zero, and

2

10

12

4

4

0

12Vdc+

-

6.000V 500.0mA

3.000V

9.000V

1.500A 1.000mA

500.0mA

1.000mA500.0mA

Figure 6.3.4

3


15/32



the mutual resistance with mesh

3 is 8 . The LHS of the mesh

current equation, which

accounts for the total voltage

drop in the direction of I1due to

the resistances in mesh 1 is:(2 +

8)I1+ 0I28I3. The source

voltage of the 12 V source is a

voltage rise in the direction of I1,

and the source voltage of the 8 V source is a voltage drop in the direction of I1. The

net voltage rise in the direction of I1due to the voltage sources in mesh 1 is (128)

V. The mesh-current equation for mesh 1 is therefore:

Mesh 1: (2 + 8)I10I28I3= 128 (6.3.10)

The mesh-current equations for the other two meshes, in accordance with the

standard procedure are:

Mesh 2: 0I1+ (8 + 8)I28I3= 8 (6.3.11)

Mesh 3: -8I18I2+ 20I3= 0 (6.3.12)

These equations reduce to:

10I1 0I28I3= 4 (6.3.13)

0I1+ 16I28I3= 8 (6.3.14)

-8I18I2+ 20I3= 0 (6.3.15)

The solution to these equations gives: I1= 1 A, I2= 0.875 A, and I3= 0.75 A.

Note the symmetry of the coefficients with respect to the diagonal.

Simulation::The circuit is

entered as in Figure 6.3.6.


Point/General Settings in the

simulation profile and


pressing the I and V buttons

displays the currents and

voltages, respectively,

indicated in Figure 6.3.6. I1is

the same as the current in

the 12 V source, I2is the same as the current in the upper 8 resistor, and I3is the

same as the current in the 4 resistor.

2

8

8

8

4

0

8Vdc12Vdc

+

-

+ -

2.000V125.0mA

3.000V

10.00V

1.000A 125.0mA

875.0mA

750.0mA250.0mA

Figure 6.3.6

+

8

8 4

8

12 V

2

I2

I3I1

+

8 V

Figure 6.3.5


16/32



Exercise 6.3.1

Verify that KVL and Ohms law are satisfied in Figure 6.3.5.

Exercise 6.3.2

Reverse the polarity of the 8 V source in the circuit of Figure 6.3.5, derive the

mesh-current equations and determine I1, I2, and I3. Simulate the circuit and verify the

values of the mesh currents, KVL, and Ohms law.

Ans. I1= 3 A, I2= 0.125 A, and I3= 1.25 A.

General izat ion of Mesh-Current Method*

It is sometimes convenient not to take all mesh currents in the same sense, or

to consider as a variable

a loop current rather than

a mesh current.

Compared to Figure

6.3.3, for example, I1in

Figure 6.3.7 is the

current in the outer loop,

I2is the current in the

same mesh 2, and I3is

the current in the same

mesh 3 but in the counterclockwise sense. How can the equations relating I1, I2, and

I3be written in the same form as the previously described mesh-current equations?

Doing so enhances the understanding of how KVL is applied in a more general

context.

Considering loop 1, the self resistances in the loop are the (2 + 10 + 4) .

The voltage drop in the direction of I1due to the self resistance of the loop is (2 + 10+ 4)I1. The mutual resistance between loop 1 and mesh 2 is the 10 resistance. But

I1and I2flow in the same direction in this resistor, so that the voltage across this

resistor due to I2is a voltage drop in loop 1 in the direction of I1, just like the voltage

in the 10 resistor due to I1. The effect of I2in the 10 resistor is to adda voltage

drop 10I2in loop 1. The mutual resistance between loop 1 and mesh 3 is the 4

resistance. But with I1and I3flowing in opposite directions in this resistor, I3produces

a voltage rise 4I3in loop 1, which subtractsfrom the total voltage drop in the direction

of I1in loop 1. The net voltage drop in the direction of I1in loop 1 is therefore: (2 + 10

+

4

12 4

3

10

Figure 6.3.7

12 V

2

I2

I3

I1


17/32



+ 4)I1+ 10I24I3= 16I1+ 10I24I3. The voltage rise due to source voltages in loop 1

is 12 V. Hence the loop-current equation for this loop is:

16I1+ 10I24I3= 12 (6.3.16)

Considering mesh 2, the voltage drop in the direction of I2due to the self

resistance in the mesh is (10 + 4 + 3)I2. I1flowing in the 10 resistor, and I3flowing

in the 4 resistor, both add to the voltage drop in mesh 2. The total voltage drop in

the direction of I2in mesh 2 is therefore: (10 + 4 + 3)I2+ 10I1+ 4I3= 17I2+ 10I1+ 4I3.

With no sources in mesh 2, the mesh-current equation for mesh 2 is:

10I1+ 17I2+ 4I3= 0 (6.3.17)

Considering mesh 3, the voltage drop in the direction of I3due to the self

resistances in the mesh is (12 + 4 + 4) I3. I1flowing in the 4 resistor produces a

voltage rise in mesh 3, whereas I2flowing in the 4

resistor produces a voltage dropin mesh 3. The net voltage drop in the direction of I3in mesh 3 is therefore: (12 + 4 +

4)I34I1+ 4I2= 20I24I1+ 4I2. With no sources in mesh 3, the mesh-current

equation for mesh 3 is:

-4I1+ 4I2+ 20I3= 0 (6.3.18)

Equations 6.3.16 to 6.3.18 are the three independent equations that can be

solved to give:I1= 1.5 A, I2= -1 A, and I3= 0.5 A. These values are in agreement

with those derived for the same circuit in Figure 6.3.3. The current in the 12 V source

is 1.5 A, the current in the 10 resistor is I1+ I2= 5 A, and the current in the 4

resistor on the side is I1I3= 1 A, as found from the mesh current equation in Figure

6.3.3.

When using loop currents, with or without mesh currents, the following should

be noted:

1. The only modification from the standard procedure is that if the loop or mesh

currents flow in the same direction in a mutual resistance, this resistance is

written with a positive sign in the KVL equations.

2. The number of independent equations is the same as the number of meshes in

the circuit.

2. The array of coefficients is symmetrical with respect to the diagonal, in the

absence of dependent sources, as in Equations 6.3.16 to 6.3.80. The symmetry

also applies in the presence of dependent sources but before the dependency

relations are taken into account, as discussed in connection with Equations 6.2.4

to 6.2.5.


18/32



Exercise 6.3.3

Redo Example 6.3.2 using the same loop and mesh currents as in Figure

6.3.6.

Non-Transform able Current Source*

When the mesh-current method is to be used in a circuit that has an ideal

current source in parallel with a resistor, the combination is conveniently transformed

to a voltage source in series with the same resistance. But when the ideal current

source does not have a resistor in parallel with it, it cannot be transformed to a

current source and must be left unaltered. The circuit of Figure 6.3.8, for example, is

the same as that of Figure 6.3.5, except that the 8 V source has been replaced by a

current source that cannot be transformed to a voltage source. An unknown voltage

Vxof arbitrary polarity is assumed across the current source and is treated like a

source voltage. The mesh

current equations become:

Mesh 1: (2 + 8)I10I28I3

= 12Vx (6.3.19)

Mesh 2: 0I1+ (8 + 8)I28I3

= Vx (6.3.20)

Mesh 3: -8I18I2+ (8 + 8 +4)I3= 0 (6.3.21)

Vxcan be eliminated

by adding together Equations

6.3.19 and 6.3.20 for the two meshes between which the current source is

connected. The resulting equation is:

10I1+ 16I216I3= 12 (6.3.22)

Equation 6.3.22 is sometimes referred to as the equation of a supermesh

that results from combining meshes 1 and 2. The mesh-current equation for the

supermesh can be written following the usual procedure, without having to introduce

an unknown source voltage. However, introducing such a voltage is more

fundamental and is less likely to cause an error.

Adding two mesh-current equations reduces the number of independent

mesh-current equations by one. But an additional equation is provided by the relation

between the mesh currents and the source current. From Figure 6.3.8,

I1I2= 0.125 (6.3.23)

Equations 6.3.21 to 6.3.23 are three independent equations that can be

+

8

8

4

8

12 V

2

I2

I3I1

0.125 A

Figure 6.3.8

+ Vx


19/32



solved to give I1= 1 A, I2= 0.875 A, and I3= 0.75 A. These values are the same as in

Example 6.3.2, because I1I2= 0.125 A, as for the current source between meshes

1 and 2.

Exercise 6.3.3


are satisfied.

6.4 Dependent Sources in Mesh-Current Method

Dependent voltage sources are treated in exactly the same manner as

independent voltage sources. Consider, for example, the circuit of Figure 6.4.1,

which is the same as that of Figure 6.3.4 but with the 8 V independent source

replaced by a dependent voltage source. The node voltage equations are written in

the usual way as:

Mesh 1: (2 + 8)I10I28I3= 1232Ib (6.4.1)

Mesh 2: 0I1+ (8 + 8)I2

8I3= 32Ib (6.4.2)

Mesh 3: -8I18I2+ (8 + 8

+ 4)I3= 0 (6.4.3)

Note that in theseequations, the net voltage

rise in mesh 1 is 1232Ib

and the voltage rise in mesh

2 is 32Ib. Leaving the 0

coefficient in the equations

maintains the symmetry in the array of coefficients.

In order to solve the mesh-current equations the controlling variable, Ibin this

case, should be expressed in terms of the mesh currents voltages. In the circuit of

Figure 6.4.1, Ib= I1I2. Substituting and moving the term in Ibto the LHS, Equations

6.4.1 to 6.4.3 become:

Mesh 1: (10 +32)I10I240I3= 12 (6.4.4)

Mesh 2: 32I1+ (8 + 8)I2+ 24I3= 0 (6.4.5)

Mesh 3: -8I18I2+ (8 + 8 + 4)Vc= 0 (6.4.6)

Solving these equations gives I1= 1 A, I2= 0.875 A, and I3= 0.75 A, the same

as in Example 6.3.2, because in this example Ib= 0.25 A, and 32Ib= 8 V, the same

as the independent voltage. Note that the array of coefficients is symmetrical with

+

8

8 4

8

12 V

2

I2

I3I1

Figure 6.4.1

+

Ib

32Ib


20/32



respect to the diagonal in Equations 6.4.1 to 6.4.3, when 32 Ibis on the RHS, but the

symmetry is destroyed in equations 6.4.4 to 6.4.6 when 32Ibis substituted for in

terms of I1and I3and moved to the LHS. Again, the effect of the dependent source is

to modify some of the resistance coefficients in the mesh-current equations.

In the case of dependent current sources, if the source is in parallel with a

resistance, the dependent current source is transformed to a dependent voltage

source, and the standard procedure followed. If the dependent current source cannot

be transformed to a current source, an unknown voltage is assigned across the

current source, and the procedure explained in connection with Figure 6.3.7 is

followed. Several examples of this type are included in problems at the end of the

chapter.

Exercise 6.4.1


are satisfied.

Summary of Main Concepts and Results

In the node-voltage method, voltages of essential nodes are assigned with

respect to one of the essential nodes taken as a reference. This automatically

satisfies KVL in every mesh in the circuit. Equations based on KCL are then

written directly in terms of Ohms law for each essential node other than the

reference node.

In any circuit, the branch voltages and currents are independent of the choice of

reference node.

In the mesh-current method, the unknown mesh currents are assigned in such a

manner that KCL is automatically satisfied at every essential node. Equations

based on KVL are then written for each mesh directly in terms of Ohms law.

In writing the node-voltage and mesh-current methods, dependent sources are

treated in exactly the same way as independent sources.

Problem-Solving Tips

1. The solution to any circuit problem can be checked by making sure that KCL is

satisfied at every node and KVL is satisfied around every mesh.

2. A useful check on the node-voltage and mesh-current equations is that the array

of coefficients on the left-hand side of the equations should be symmetrical about

the diagonal. For the purpose of this check, zero coefficients must be included

and dependent sources should appear as sources on the right-hand side of the


21/32



equations.

3. An additional check is that in any row or column, the mutual conductances

(resistances) are part of the self conductances (resistances) in that row or

column.


22/32


23/32

23-6

P6.1.5 (a) Determine Vain Figure

P6.1.5 by transforming the

voltage sources to current

sources and writing the node-

voltage equation for node a.

(b) Write the same node-

voltage equation based on KCL, without transforming the sources.

Ans. 10 V.

P6.1.6 Given that Va = 25 V and Vb = 12 V, with

node c grounded. Determine Vaif node

b is grounded instead of node c.

Ans. 13 V.

P6.1.7Determine VLand IAin

Figure P6.1.7.

Ans. 12.1 V, 0.35 A.

P6.1.8 Determine the

node voltagesin Figure

P6.1.8.

Ans. Va= 36.43 V,

Vb= 23.57 V,

Vc= 37.86 V

a

b c

Figure P6.1.6

+

0.01 S10 A

0.025 S

0.02 S

cVc

bVb

aIA

0.5 S

0.025 S

0.02 S

Va

VL

+

Figure P6.1.7

20 mA

4 kc

b

a

2 k

Figure P6.1.8

5 mA

10 mA2 k

4 k

1 k

10 V

4 6

6 20 V

a

Figure P6.1.5

+

+


24/32

24-6

P6.1.9 Determine the power delivered

or absorbed by the current

sources in Figure P6.1.9.

Ans. 15 A source absorbs 19.5 W, 30 A

source delivers 66 W.

P6.1.10 Determine the branch voltages in

Figure P6.1.10.

Ans. VL= 18.2 V, 31.8 V rise across the 5 A

source, 26.4 V rise across the

dependent source,

P6.1.11 Determine VOin Figure P6.1.11.

Ans. 55/6 V.

P6.1.12 Determine VOin Figure P6.1.12,Ans. 20 V.

15 A

10 S

20 S

10 S

30 A

15 S

Figure P6.1.9

0.4 S

0.2VL

VL

+

Figure P6.1.10

5 A 0.1 S 0.1 S

10 V

2

2 4

5IxA

4

Ix

+

VO

+

Figure P6.1.11

10 V

10

20

40

80

20 V

+

VO

+

+

Figure P6.1.12


25/32

25-6

P6.1.13 Determine VOin

Figure P6.1.13.

Ans. 15.51 V.

P6.1.14 Determine IOin

Figure P6.1.14.

Ans. 15.5 A.

P6.1.15 Determine IOin

Figure P6.1.15.

Ans. -10/3 A.

P6.1.16 Determine VOin Figure

P6.1.16.

Ans. 30 V.

10 V

4

10 A

8

Ix

4Ix

+

VO

2

2

a b

Figure P6.1.13

+

10 V

+

4 S10 A 8 S

Vx4Vx

2 S

2 S

+

IO

Figure P6.1.14

+

4 S

10 A 4 S

5Vx +

2 S

2 S

+

Vx

IO

Figure P6.1.15

2 A 4 A

2

4 4

2

5

+

VO

+

3VO

Figure P6.1.16


26/32

26-6

4 S

2 V

5 S

IO

4 S

3IO

4 V2 S 2 S

a b

Figure P6.1.18

+

+

ISRC1 ISRC2

P6.1.17 Determine VOin

Figure P6.1.17.

Ans. 0.

P61.18 Determine

ISRC1

and ISRC2

in

Figure

P6.1.18.

Ans. ISRC1= 1 A, ISRC2

= 95 A.

P6.1.19 Determine IOin Figure P6.1.19.

Ans. -22 A.

P6.1.20 Determine VO

in Figure

P6.1.20.

Ans. 18.5 V.

20 V

4

10 A+

VO4

2

Figure P6.1.17

2

4

+

+ Vx

2Vx

+

10 V

4 S

2 S 2 S

4 S

2 S 2 A

IO

2Ix

Ix

Figure P6.1.19

+

0.1V

1 S 0.5 S

0.5Vx

+ +

Vx V

0.2 S +

+

0.2 S 0.1 S+VO

20 V

Figure P6.1.20


27/32

27-6


P6.1.21.

Ans. 1.82 V.

P6.1.22 Determine VOin Figure P6.1.22,

assuming that all resistances are

2 .

Ans. 2.56 V.

6.2 Mesh-Current Method

Use the mesh -current m ethod in Pr ob lems P6.2.1.to P6.2.20.

P6.2.1 Determine ISRC1and ISRC2in

Figure P6.2.1.

Ans. ISRC1= 0, ISRC2= 5/3 A.

P6.2.2 Determine ISRC1and ISRC2in

Figure P6.2.2.

Ans. ISRC1= -1.3 A, ISRC2= 2.2 A.

10 V

1

2

2

2

2

2

2

4 4

+

VO

Figure P6.1.21

+

0.5Ix

+0.5Iy

Ix Iy

10 V5 A

+

+VO

Figure P6.1.22

Figure P6.2.1

ISRC1 ISRC2

10 V

4 6

6 20 V+

+

15 V 20

10

15

10

ISRC1 ISRC2+

+

30 V

Figure P6.2.2


28/32

28-6

P6.2.3 Determine IOin Figure P6.2.3,

Ans. 0.


P6.2.4.

Ans. 12.5 V.


P6.2.5.

Ans. 12.1 V.

P6.2.6 Determine IOin Figure P6.2.6

Ans. 0.85 A

10 V

10

20

40

80

20 V

+

+

Figure P6.2.3

IO

20

20

40

40

20

100 +

80 V+ VO

Figure P6.2.4

40

50

VL

+

50 100

40

20 V

2

+

Figure P6.2.5

10

10

10 20

30

20

+

+

+

IO

Figure P6.2.6

11 V

66 V

33 V


29/32

29-6

P6.2.7 Determine Ixin

Figure P6.2.7.

Ans. -1.86 A

P6.2.8 Determine Ixin figure P6.2.8.

Ans. 1.38 A.


P6.2.9.

Ans. 20 V.

P6.2.10 Determine VOinFigure P6.2.10

Ans. 40 V.

4 V

4

10 V4

1

8 V

5

+

Figure P6.2.7

2

+

+

Ix

10

Ix

5 IL+

+

Figure P6.2.8

50 V

25

100

IL

40 V

20

IO

0.5IO

+

VO10 10

20 V

+

+

Figure P6.2.9

20 A 1

2

VO

+

Figure P6.2.10

4

2

4

+

40 V


30/32

30-6

P6.2.11 Determine IOin Figure

P6.2.11.

Ans. 15.51 A.

P6.2.12 Determine IOin Figure

P6.2.12.

Ans. -10/3 A.

P6.2.13 Determine the power

delivered or absorbed by each

current source.

Ans. 2A source delivers 2 W and 4 A

source delivers 380 W.

P6.2.14 Determine IOin Figure P6.2.12.

Ans. -22 A

10 V

+

4 S10 A 8 S

Vx4Vx

2 S

2 S

+

IO

Figure P6.2.11

+

4 S

10 A 4 S

5Vx +

2 S

2 S

+

Vx

IO

Figure P6.2.12

2 A 4 A

2

4 4

2

5

+

VO

+

3VO

Figure P6.2.13

10 V

4 S

2 S 2 S

4 S

2 S 2 A

IO

2Ix

Ix

Figure P6.2.14

+


31/32


32/32

circuit equations

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