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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
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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.
After selecting Bias
Point/General Settings in the
simulation profile and
running the simulation,
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
+
-
+ -
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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
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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
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+
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
Simulate the circuit of Figure 6.2.1 and verify that KCL, KVL, and Ohms law
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,
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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
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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.
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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
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+
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
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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.
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 6.3.6. 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
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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
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+ 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.
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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
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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
Simulate the circuit of Figure 6.3.8 and verify that KCL, KVL, and Ohms law
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
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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
Simulate the circuit of Figure 6.4.1 and verify that KCL, KVL, and Ohms law
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
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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.
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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
+
+
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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
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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
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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
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P6.1.21 Determine VOin Figure
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
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P6.2.3 Determine IOin Figure P6.2.3,
Ans. 0.
P6.2.4 Determine VOin Figure
P6.2.4.
Ans. 12.5 V.
P6.2.5 Determine VOin Figure
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
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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 Determine VOin Figure
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
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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
+
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8/13/2019 Circuit Equations
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