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8.6 Nodal Analysis
In the previous section we applied Kirchhoff’s voltage law to arrive at loop
currents in a network. In this section we will apply Kirchhoff’s current law
to determine the potential difference (voltage) at any node with respect to
some arbitrary reference point in a network. Once the potentials of all nodes
are known, it is a simple matter to determine other quantities such as current
and power within the network.
The steps used in solving a circuit using nodal analysis are as follows:
1. Arbitrarily assign a reference node within the circuit and indicate this
node as ground. The reference node is usually located at the bottom of
the circuit, although it may be located anywhere.
2. Convert each voltage source in the network to its equivalent current
source. This step, although not absolutely necessary, makes further calcu-
lations easier to understand.
3. Arbitrarily assign voltages (V1, V2, . . . , Vn) to the remaining nodes in
the circuit. (Remember that you have already assigned a reference node,
so these voltages will all be with respect to the chosen reference.)
4. Arbitrarily assign a current direction to each branch in which there is no
current source. Using the assigned current directions, indicate the corre-
sponding polarities of the voltage drops on all resistors.
5. With the exception of the reference node (ground), apply Kirchhoff’s cur-
rent law at each of the nodes. If a circuit has a total of n� 1 nodes (includ-
ing the reference node), there will be n simultaneous linear equations.
6. Rewrite each of the arbitrarily assigned currents in terms of the potential
difference across a known resistance.
7. Solve the resulting simultaneous linear equations for the voltages (V1, V2,
. . . , Vn).
288 Chapter 8 ■ Methods of Analysis
EXAMPLE 8–14 Given the circuit of Figure 8–30, use nodal analysis to
solve for the voltage Vab.
I2 = 50 mA
20 �
R1
30 �R3I3
200 mA
I1
R2
40 �
R3
E
30 �
a
b
6 V
200 mA
FIGURE 8–30
Section 8.6 ■ Nodal Analysis 289
Steps 3 and 4: Arbitrarily assign node voltages and branch currents. Indicate
the voltage polarities across all resistors according to the assumed current
directions.
Step 5: We now apply Kirchhoff’s current law at the nodes labelled as V1 and
V2:
Node V1: Ientering Ileaving
200 mA � 50 mA I1 � I2
Node V2: Ientering Ileaving
200 mA � I2 50 mA � I3
Step 6: The currents are rewritten in terms of the voltages across the resistors
as follows:
I1 !20
V1
�!
I2 !V
4
1
0
"
�
V2!
I3 !30
V2
�!
The nodal equations become
200 mA � 50 mA !20
V1
�!�!
V
4
1
0
"
�
V2!
200 mA �!V
4
1
0
"
�
V2! 50 mA �!
30
V2
�!
FIGURE 8–31
R1
V1 V2
200 mA 200 mA
50 mA
(Reference)
20 � R3 30 �
40 �
I1
I2
! "
!
"
I3
!
"
R2
Solution
Step 1: Select a convenient reference node.
Step 2: Convert the voltage sources into equivalent current sources. The
equivalent circuit is shown in Figure 8–31.
290 Chapter 8 ■ Methods of Analysis
Substituting the voltage expressions into the original nodal equations, we
have the following simultaneous linear equations:
�!20
1
�!�!
40
1
�! V1 " �!40
1
�! V2 250 mA
"�!40
1
�! V1 � �!30
1
�!�!
40
1
�! V2 150 mA
These may be further simplified as
(0.075 S)V1 " (0.025 S)V2 250 mA
"(0.025 S)V1 � (0.0583!)V2 150 mA
Step 7: Use determinants to solve for the nodal voltages as
" "0.250 "0.025
0.150 0.0583!
!0
0
.
.
0
0
0
1
3
8
7
3!5! 4.89 V
(0.250)(0.0583!) " (0.150)(" 0.025)!!!!(0.075)(0.0583!) " ("0.025)("0.025)
V1
" "0.075 "0.025
0.025 0.0583!
If we go back to the original circuit of Figure 8–30, we see that the voltage V2
is the same as the voltage Va, namely
Va 4.67 V 6.0 V � Vab
Therefore, the voltage Vab is simply found as
Vab 4.67 V " 6.0 V "1.33 V
" "0.075 0.250
"0.025 0.150V2
" "
!0
0
.
.
0
0
0
1
3
7
7
5
5! 4.67 V
(0.075)(0.150) " ("0.025)(0.250)!!!!
0.00375
0.075 0.0255
"0.025 0.0583!
and
Section 8.6 ■ Nodal Analysis 291
EXAMPLE 8–15 Determine the nodal voltages for the circuit shown in
Figure 8–32.
Solution By following the steps outlined, the circuit may be redrawn as
shown in Figure 8–33.
2 A5 �R1
R2 = 3 �
3 A6 �
4 �R3
R46 �
18 V
FIGURE 8–32
FIGURE 8–33
V1 V2
2 A 3 A
(reference)
6 �R3 4 �
R2 = 3 �
I2I4
I3
R1 5 �
I1
Applying Kirchhoff’s current law to the nodes corresponding to V1 and V2,
the following nodal equations are obtained:
Ileaving Ientering
Node V1: I1 � I2 2 A
Node V2: I3 � I4 I2 � 3 A
The currents may once again be written in terms of the voltages across the
resistors:
I1 !5
V
�
1!
I2 !V1
3
"
�
V2!
I3 !4
V
�
2!
I4 !6
V
�
2!
292 Chapter 8 ■ Methods of Analysis
The nodal equations become
Node V1: !5
V
�
1!�!
(V1
3
"
�
V2)! 2 A
Node V2: !4
V
�
2!�!
6
V
�
2! !
(V1
3
"
�
V2)!� 3 A
These equations may now be simplified as
Node V1: �!51
�!�!
3
1
�! V1 " �!3
1
�! V2 2 A
Node V2: "�!31
�! V1 � �!4
1
�!�!
6
1
�!�!
3
1
�! V2 3 A
The solutions for V1 and V2 are found using determinants:
" "2 "0.333
3 0.750V1 !2
0
.
.
5
2
0
8
0
9! 8.65 V
" "0.533 "0.333
"0.333 0.750
" "0.533 2
"0.333 3V2 !
2
0
.
.
2
2
6
8
7
9! 7.85 V
" "0.533 "0.333
"0.333 0.750
In the previous two examples, you may have noticed that the simultaneous
linear equations have a format similar to that developed for mesh analysis.
When we wrote the nodal equation for node V1 the coefficient for the vari-
able V1 was positive, and it had a magnitude given by the summation of the
conductance attached to this node. The coefficient for the variable V2 was
negative and had a magnitude given by the mutual conductance between
nodes V1 and V2.
Format Approach
A simple format approach may be used to write the nodal equations for any
network having n � 1 nodes. Where one of these nodes is denoted as the ref-
erence node, there will be n simultaneous linear equations which will appear
as follows:
G11V1 " G12V2 " G13V3 "… " R1nVn I1
"G21V1 � G22V2 " G23V3 "… " R2nVn I2
"Gn1V1 " Gn2V2 " Gn3V3 "… � RnnVn In
The coefficients (constants) G11, G22, G33, . . . , Gnn represent the sum-
mation of the conductances attached to the particular node. The remaining
coefficients are called the mutual conductance terms. For example, the
mutual conductance G23 is the conductance attached to node V2, which is
common to node V3. If there is no conductance that is common to two nodes,
then this term would be zero. Notice that the terms G11, G22, G33, . . . , Gnn
are positive and that the mutual conductance terms are negative. Further, if
the equations are written correctly, then the terms will be symmetrical about
the principal diagonal, e.g., G23 G32.
The terms V1, V2, . . . , Vn are the unknown node voltages. Each voltage
represents the potential difference between the node in question and the ref-
erence node.
The terms I1, I2, . . . , In are the summation of current sources entering
the node. If a current source has a current such that it is leaving the node,
then the current is simply assigned as negative. If a particular current source
is shared between two nodes, then this current must be included in both
nodal equations.
The method used in applying the format approach of nodal analysis is as
follows:
1. Convert voltage sources into equivalent current sources.
2. Label the reference node as . Label the remaining nodes as V1, V2, . . . , Vn.
3. Write the linear equation for each node using the format outlined.
4. Solve the resulting simultaneous linear equations for V1, V2, . . . , Vn.
The next examples illustrate how the format approach is used to solve
circuit problems.
Section 8.6 ■ Nodal Analysis 293
EXAMPLE 8–16 Determine the nodal voltages for the circuit shown in
Figure 8–34.
Solution The circuit has a total of three nodes: the reference node (at a
potential of zero volts) and two other nodes, V1 and V2.
By applying the format approach for writing the nodal equations, we get
two equations:
Node V1: �!31
�!�!
5
1
�! V1 " �!5
1
�! V2 "6 A � 1 A
Node V2: "�!51
�! V1 � �!5
1
�!�!
4
1
�! V2 "1 A " 2 A
R1
V1 V2I2
I1 6 A I3 2 A
1 A
3 � R3 4 �
R2 = 5 �
FIGURE 8–34
294 Chapter 8 ■ Methods of Analysis
EXAMPLE 8–17 Use nodal analysis to find the nodal voltages for the cir-
cuit of Figure 8–35. Use the answers to solve for the current through R1.
Solution In order to apply nodal analysis, we must first convert the voltage
source into its equivalent current source. The resulting circuit is shown in
Figure 8–36.
2 mA 5 k�
3 mA
3 k�5 k�
10 V
2 k�R4
2 mA
I4R2
R3 = 4 k�
R1
I2
FIGURE 8–35
On the right-hand sides of the above, those currents that are leaving the nodes
are given a negative sign.
These equations may be rewritten as
Node V1: (0.533 S)V1 " (0.200 S)V2 "5 A
Node V2: "(0.200 S)V1 � (0.450 S)V2 "3 A
Using determinants to solve these equations, we have
" ""5 "0.200
"3 0.450V1 !
"
0.
2
2
.
0
8
0
5! "14.3 V
" "0.533 "0.200
"0.200 0.450
" "0.533 "5
"0.200 "3V2 !
"
0.
2
2
.
0
6
0
0! "13.0 V
" "0.533 "0.200
"0.200 0.450
Section 8.6 ■ Nodal Analysis 295
R4
V2V1
2 k�R2
R3 = 4 k�
3 k�5 k�
I4 2 mAI2 3 mA
2 mA
FIGURE 8–36
Labelling the nodes and writing the nodal equations, we obtain the following:
Node V1: �!5 k
1
�!�!
3 k
1
�!�!
4 k
1
�! V1 " �!4 k
1
�! V2 2 mA " 3 mA
Node V2: "�!4 k
1
�! V1 � �!4 k
1
�!�!
2 k
1
�! V2 2 mA
Because it is inconvenient to use kilohms and milliamps throughout our cal-
culations, we may eliminate these units in our calculations. You have already
seen that any voltage obtained by using these quantities will result in the units
being “volts.” Therefore the nodal equations may be simplified as
Node V1: (0.7833)V1 " (0.2500)V2 "1
Node V2: "(0.2500)V1 � (0.750)V2 2
The solutions are as follows:
" ""1 "0.250
2 0.750V1 !
"
0
0
.5
.2
2
5
5
0! "0.476 V
" "0.7833 "0.250
"0.2503 0.750
" "0.7833 "1
"0.2503 2V2 !
1
0
.3
.5
1
2
6
5
7! 2.51 V
" "0.7833 "0.250
"0.2503 0.750
Using the values derived for the nodal voltages, it is now possible to solve for
any other quantities in the circuit. To determine the current through resistor
R1 5 k�, we first reassemble the circuit as it appeared originally. Since the
node voltage V1 is the same in both circuits, we use it in determining the
desired current. The resistor may be isolated as shown in Figure 8–37.
V1 = "0.476 V
10 V
5 k�R1I
FIGURE 8–37
A common mistake is that the
current is determined by using
the equivalent circuit rather than
the original circuit. You must
remember that the circuits are
only equivalent external to the
conversion.
NOTES...
Answers: V1 3.00 V, V2 6.00 V, V3 "2.00 V
296 Chapter 8 ■ Methods of Analysis
The current is easily found as
I 10 V "!("0
5
.4
k
7
�
6 V)! 2.10 mA (upward)
PRACTICE
PROBLEMS 4
Use nodal analysis to determine the node voltages for the circuit of Figure 8–38.
1 �
9 V
1 �2 �
8 �
4 A
3 A
V2
V3V1
FIGURE 8–38
60 �
10 �
90 �
15 �
30 �30 V
30 �
I
FIGURE 8–39
8.7 Delta-Wye (Pi-Tee) Conversion
Delta-Wye Conversion
You have previously examined resistor networks involving series, parallel,
and series-parallel combinations. We will next examine networks which
cannot be placed into any of the above categories. While these circuits
may be analyzed using techniques developed earlier in this chapter, there
is an easier approach. For example, consider the circuit shown in Figure
8–39.
This circuit could be analyzed using mesh analysis. However, you
see that the analysis would involve solving four simultaneous linear
equations, since there are four separate loops in the circuit. If we were
to use nodal analysis, the solution would require determining three node
voltages, since there are three nodes in addition to a reference node. Unless
a computer is used, both techniques are very time-consuming and prone to
error.
As you have already seen, it is occasionally easier to examine a circuit
after it has been converted to some equivalent form. We will now develop a
technique for converting a circuit from a delta (or pi) into an equivalent wye
(or tee) circuit. Consider the circuits shown in Figure 8–40. We start by
making the assumption that the networks shown in Figure 8–40(a) are equiv-
alent to those shown in Figure 8–40(b). Then, using this assumption, we will
determine the mathematical relationships between the various resistors in the
equivalent circuits.
The circuit of Figure 8–40(a) can be equivalent to the circuit of Figure
8–40(b) only if the resistance “seen” between any two terminals is exactly
550 Electrical Circuit Theory and Technology
31.2 Nodal analysis A node of a network is defined as a point where two or more branchesare joined. If three or more branches join at a node, then that node iscalled a principal node or junction. In Figure 31.5, points 1, 2, 3, 4 and5 are nodes, and points 1, 2 and 3 are principal nodes.A node voltage is the voltage of a particular node with respect to a
node called the reference node. If in Figure 31.5, for example, node 3 ischosen as the reference node then V13 is assumed to mean the voltageat node 1 with respect to node 3 (as distinct from V31). Similarly, V23
would be assumed to mean the voltage at node 2 with respect to node 3,and so on. However, since the node voltage is always determined withrespect to a particular chosen reference node, the notation V1 for V13 andV2 for V23 would always be used in this instance.The object of nodal analysis is to determine the values of voltages at
all the principal nodes with respect to the reference node, e.g., to findvoltages V1 and V2 in Figure 31.5. When such voltages are determined,the currents flowing in each branch can be found.Kirchhoff’s current law is applied to nodes 1 and 2 in turn in
Figure 31.5 and two equations in unknowns V1 and V2 are obtained whichmay be simultaneously solved using determinants.
Figure 31.5 Figure 31.6
The branches leading to node 1 are shown separately in Figure 31.6.Let us assume that all branch currents are leaving the node as shown.Since the sum of currents at a junction is zero,
V1 Vx
ZACV1
ZDCV1 V2
ZBD 0 ⊲1⊳
Similarly, for node 2, assuming all branch currents are leaving the nodeas shown in Figure 31.7,
V2 V1
ZBCV2
ZECV2 C VY
ZCD 0 ⊲2⊳
In equations (1) and (2), the currents are all assumed to be leaving thenode. In fact, any selection in the direction of the branch currents maybe made— the resulting equations will be identical. (For example, if fornode 1 the current flowing in ZB is considered as flowing towards node 1instead of away, then the equation for node 1 becomesFigure 31.7
Mesh-current and nodal analysis 551
V1 Vx
ZACV1
ZDDV2 V1
ZB
which if rearranged is seen to be exactly the same as equation (1).)Rearranging equations (1) and (2) gives:
(
1
ZAC
1
ZBC
1
ZD
)
V1
(
1
ZB
)
V2
(
1
ZA
)
Vx D 0 ⊲3⊳
(
1
ZB
)
V1 C
(
1
ZBC
1
ZCC
1
ZE
)
V2 C
(
1
ZC
)
VY D 0 ⊲4⊳
Equations (3) and (4) may be rewritten in terms of admittances (whereadmittance Y D l/Z ):
⊲YA C YB C YD⊳V1 YBV2 YAVx D 0 ⊲5⊳
YBV1 C ⊲YB C YC C YE⊳V2 C YCVY D 0 ⊲6⊳
Equations (5) and (6) may be solved for V1 and V2 by using determinants.Thus
V1∣
∣
∣
∣
YB YA⊲YB C YC C YE⊳ YC
∣
∣
∣
∣
D V2
∣
∣
∣
∣
⊲YA C YB C YD⊳ YA YB YC
∣
∣
∣
∣
D1
∣
∣
∣
∣
⊲YA C YB C YD⊳ YB YB ⊲YB C YC C YE⊳
∣
∣
∣
∣
Current equations, and hence voltage equations, may be written at eachprincipal node of a network with the exception of a reference node. Thenumber of equations necessary to produce a solution for a circuit is, infact, always one less than the number of principal nodes.Whether mesh-current analysis or nodal analysis is used to determine
currents in circuits depends on the number of loops and nodes the circuitcontains, Basically, the method that requires the least number of equationsis used. The method of nodal analysis is demonstrated in the followingproblems.
Problem 4. For the network shown in Figure 31.8, determine thevoltage VAB, by using nodal analysis.
Figure 31.8
552 Electrical Circuit Theory and Technology
Figure 31.8 contains two principal nodes (at 1 and B) and thus only one
nodal equation is required. B is taken as the reference node and the equa-
tion for node 1 is obtained as follows. Applying Kirchhoff’s current law
to node 1 gives:
IX C IY D I
i.e.,V1
16C
V1
⊲4C j3⊳D 20 6 0°
Thus V1
(
1
16C
1
4C j3
)
D 20
V1
(
0.0625C4 j3
42 C 32
)
D 20
V1⊲0.0625C 0.16 j0.12⊳ D 20
V1⊲0.2225 j0.12⊳ D 20
from which, V1 D20
⊲0.2225 j0.12⊳D
20
0.2528 6 28.34°
i.e., voltage V1 D 79.1 6 28.34° V
The current through the ⊲4C j3⊳� branch, Iy D V1/⊲4C j3⊳
Hence the voltage drop between points A and B, i.e., across the 4 �
resistance, is given by:
VAB D ⊲Iy⊳⊲4⊳ DV1⊲4⊳
⊲4C j3⊳D
79.1 6 28.34°
5 6 36.87°⊲4⊳ D 63.36 6 −8.53° V
Problem 5. Determine the value of voltage VXY shown in thecircuit of Figure 31.9.
Figure 31.9The circuit contains no principal nodes. However, if point Y is chosen as
the reference node then an equation may be written for node X assuming
that current leaves point X by both branches.
ThusVX 8 6 0°
⊲5C 4⊳CVx 8 6 90°
⊲3C j6⊳D 0
from which, VX
(
1
9C
1
3C j6
)
D8
9C
j8
3C j6
VX
(
1
9C
3 j6
32 C 62
)
D8
9Cj8⊲3 j6⊳
32 C 62
Mesh-current and nodal analysis 553
VX⊲0.1778 j0.1333⊳ D 0.8889C48C j24
45
VX⊲0.2222 6 36.86°⊳ D 1.9556C j0.5333
D 2.027 6 15.25°
Since point Y is the reference node,
voltage VX D VXY D2.027 6 15.25°
0.2222 6 36.86°D 9.12 6 6 52.11° V
Problem 6. Use nodal analysis to determine the current flowingin each branch of the network shown in Figure 31.10.
Figure 31.10This is the same problem as problem 1 of Chapter 30, page 536, whichwas solved using Kirchhoff’s laws. A comparison of methods canbe made.There are only two principal nodes in Figure 31.10 so only one nodal
equation is required. Node 2 is taken as the reference node.
The equation at node 1 is I1 C I2 C I3 D 0
i.e.,V1 100 6 0°
25CV1
20CV1 50 6 90°
10D 0
i.e.,
(
1
25C
1
20C
1
10
)
V1 100 6 0°
25
50 6 90°
10D 0
0.19 V1 D 4C j5
Thus the voltage at node 1, V1 D4C j5
0.19D 33.70 6 51.34° V
or ⊲21.05C j26.32⊳V
Hence the current in the 25 � resistance,
I1 DV1 100 6 0°
25D
21.05C j26.32 100
25
D 78.95C j26.32
25
D 3.33 6 6 161.56° A flowing away from node 1
⊲or 3.33 6 ⊲161.56° 180°⊳A D 3.33 6 6 −18.44° A flowing towardnode 1)
The current in the 20 � resistance,
I2 DV1
20D
33.70 6 51.34°
20D 1.69 6 6 51.34° A
flowing from node 1 to node 2
554 Electrical Circuit Theory and Technology
The current in the 10 � resistor,
I3 DV1 50 6 90°
10D
21.05C j26.32 j50
10D
21.05 j23.68
10
D 3.17 6 6 −48.36° A away from node 1
⊲or 3.17 6 ⊲ 48.36° 180°⊳ D 3.17 6 228.36° A D 3.17 6 6 131.64° A
toward node 1)
Problem 7. In the network of Figure 31.11 use nodal analysis todetermine (a) the voltage at nodes 1 and 2, (b) the current in thej4 � inductance, (c) the current in the 5 � resistance, and (d) themagnitude of the active power dissipated in the 2.5 � resistance.
Figure 31.11
(a) At node 1,V1 25 6 0°
2C
V1
j4CV1 V2
5D 0
Rearranging gives:
(
1
2C
1
j4C
1
5
)
V1
(
1
5
)
V2 25 6 0°
2D 0
i.e., ⊲0.7C j0.25⊳V1 0.2V2 12.5 D 0 ⊲1⊳
At node 2,V2 25 6 90°
2.5CV2
j4CV2 V1
5D 0
Rearranging gives:
(
1
5
)
V1 C
(
1
2.5C
1
j4C
1
5
)
V2 25 6 90°
2.5D 0
i.e., 0.2V1 C ⊲0.6 j0.25⊳V2 j10 D 0 ⊲2⊳
Thus two simultaneous equations have been formed with twounknowns, V1 and V2. Using determinants, if
⊲0.7C j0.25⊳V1 0.2V2 12.5 D 0 ⊲1⊳
Mesh-current and nodal analysis 555
and 0.2V1 C ⊲0.6 j0.25⊳V2 j10 D 0 ⊲2⊳
thenV1
∣
∣
∣
∣
∣
0.2 12.5
⊲0.6 j0.25⊳ j10
∣
∣
∣
∣
∣
D V2
∣
∣
∣
∣
∣
⊲0.7C j0.25⊳ 12.5
0.2 j10
∣
∣
∣
∣
∣
D1
∣
∣
∣
∣
∣
⊲0.7C j0.25⊳ 0.2
0.2 ⊲0.6 j0.25⊳
∣
∣
∣
∣
∣
i.e.,V1
⊲j2C 7.5 j3.125⊳D
V2
⊲ j7C 2.5 2.5⊳
D1
⊲0.42 j0.175C j0.15C 0.0625 0.04⊳
andV1
7.584 6 8.53°D
V2
7 6 90°D
1
0.443 6 3.23°
Thus voltage, V1 D7.584 6 8.53°
0.443 6 3.23°D 17.12 6 5.30° V
D 17.1 6 6 −5.3° V, correct to one decimal place,
and voltage, V2 D7 6 90°
0.443 6 3.23°D 15.80 6 93.23° V
D 15.8 6 6 93.2° V, correct to one decimal place.
(b) The current in the j4 � inductance is given by:
V2
j4D
15.80 6 93.23°
4 6 90°D 3.95 6 6 3.23° A flowing away from node 2
(c) The current in the 5 � resistance is given by:
I5 DV1 V2
5D
17.12 6 5.30° 15.80 6 93.23°
5
i.e., I5 D⊲17.05 j1.58⊳ ⊲ 0.89C j15.77⊳
5
D17.94 j17.35
5D
24.96 6 44.04°
5
D 4.99 6 6 −44.04° A flowing from node 1 to node 2
(d) The active power dissipated in the 2.5 � resistor is given by
P2.5 D ⊲I2.5⊳2⊲2.5⊳ D
(
V2 25 6 90°
2.5
)2
⊲2.5⊳
D⊲0.89C j15.77 j25⊳2
2.5D⊲9.273 6 95.51°⊳2
2.5
556 Electrical Circuit Theory and Technology
D85.99 6 191.02°
2.5by de Moivre’s theorem
D 34.4 6 169° W
Thus the magnitude of the active power dissipated in the 2.5 Z
resistance is 34.4 W
Problem 8. In the network shown in Figure 31.12 determine thevoltage VXY using nodal analysis.
Figure 31.12
Node 3 is taken as the reference node.
At node 1, 25 6 0° DV1
4C j3CV1 V2
5
i.e.,
(
4 j3
25C
1
5
)
V1 1
5V2 25 D 0
or ⊲0.379 6 18.43°⊳V1 0.2V2 25 D 0 ⊲1⊳
At node 2,V2
j10CV2
j20CV2 V1
5D 0
i.e., 0.2V1 C
(
1
j10C
1
j20C
1
5
)
V2 D 0
or 0.2V1 C ⊲ j0.1 j0.05C 0.2⊳V2 D 0
i.e., 0.2V1 C ⊲0.25 6 36.87°⊳V2 C 0 D 0 ⊲2⊳
Simultaneous equations (1) and (2) may be solved for V1 and V2 by using
determinants. Thus,
Mesh-current and nodal analysis 557
V1∣
∣
∣
∣
∣
0.2 25
0.25 6 36.87° 0
∣
∣
∣
∣
∣
D V2
∣
∣
∣
∣
∣
0.379 6 18.43° 25
0.2 0
∣
∣
∣
∣
∣
D1
∣
∣
∣
∣
∣
0.379 6 18.43° 0.2
0.2 0.25 6 36.87°
∣
∣
∣
∣
∣
i.e.,V1
6.25 6 36.87°D V2
5D
1
0.09475 6 55.30° 0.04
D1
0.079 6 79.85°
Hence voltage, V1 D6.25 6 36.87°
0.079 6 79.85°D 79.11 6 6 42.98° V
and voltage, V2 D5
0.079 6 79.85°D 63.29 6 6 79.85° V
The current flowing in the ⊲4C j3⊳� branch is V1/⊲4C j3⊳. Hence thevoltage between point X and node 3 is:
V1
⊲4C j3⊳⊲j3⊳ D
⊲79.116 42.98°⊳⊲3 6 90°⊳
5 6 36.87°
D 47.47 6 96.11° V
Thus the voltage
VXY D VX VY D VX V2 D 47.47 6 96.11° 63.29 6 79.85°
D 16.21 j15.10 D 22.156 6 −137° V
Problem 9. Use nodal analysis to determine the voltages at nodes2 and 3 in Figure 31.13 and hence determine the current flowingin the 2 � resistor and the power dissipated in the 3 � resistor.
This is the same problem as Problem 2 of Chapter 30, page 537, whichwas solved using Kirchhoff’s laws.In Figure 31.13, the reference node is shown at point A.
At node 1,V1 V2
1CV1
6CV1 8 V3
5D 0
i.e., 1.367V1 V2 0.2V3 1.6 D 0 ⊲1⊳
At node 2,V2
2CV2 V1
1CV2 V3
3D 0
Figure 31.13
558 Electrical Circuit Theory and Technology
i.e., V1 C 1.833V2 0.333V3 C 0 D 0 ⊲2⊳
At node 3,V3
4CV3 V2
3CV3 C 8 V1
5D 0
i.e., 0.2V1 0.333V2 C 0.783V3 C 1.6 D 0 ⊲3⊳
Equations (1) to (3) can be solved for V1, V2 and V3 by usingdeterminants. Hence
V1∣
∣
∣
∣
∣
∣
∣
1 0.2 1.6
1.833 0.333 0
0.333 0.783 1.6
∣
∣
∣
∣
∣
∣
∣
D V2
∣
∣
∣
∣
∣
∣
∣
1.367 0.2 1.6
1 0.333 0
0.2 0.783 1.6
∣
∣
∣
∣
∣
∣
∣
DV3
∣
∣
∣
∣
∣
∣
∣
1.367 1 1.6
1 1.833 0
0.2 0.333 1.6
∣
∣
∣
∣
∣
∣
∣
D 1
∣
∣
∣
∣
∣
∣
∣
1.367 1 0.2
1 1.833 0.333
0.2 0.333 0.783
∣
∣
∣
∣
∣
∣
∣
Solving for V2 gives: V2
1.6⊲ 0.8496⊳C 1.6⊲ 0.6552⊳
D 1
1.367⊲1.3244⊳C 1⊲ 0.8496⊳ 0.2⊲0.6996⊳
hence V2
0.31104D
1
0.82093from which, voltage,V 2 D
0.31104
0.82093
D 0.3789 V
Thus the current in the 2 Z resistor DV2
2D
0.3789
2D 0.19 A,
flowing from node 2 to node A.
Solving for V3 gives:V3
1.6⊲0.6996⊳C 1.6⊲1.5057⊳D
1
0.82093
henceV3
1.2898D
1
0.82093from which, voltage,V3 D
1.2898
0.82093
D −1.571 V
Power in the 3 Z resistor D ⊲I3⊳2⊲3⊳ D
(
V2 V3
3
)2
⊲3⊳
D⊲0.3789 ⊲ 1.571⊳⊳2
3D 1.27 W
Further problems on nodal analysis may be found in Section 31.3following, problems 10 to 15, page 560.
Mesh-current and nodal analysis 559
31.3 Further problemson mesh-current and
nodal analysis
Mesh-current analysis
1 Repeat problems 1 to 10, page 542, of Chapter 30 using mesh-
current analysis.
2 For the network shown in Figure 31.14, use mesh-current analysis
to determine the value of current I and the active power output of
the voltage source. [6.966 49.94° A; 644 W]
3 Use mesh-current analysis to determine currents I1, I2 and I3 for the
network shown in Figure 31.15.
[I1 D 8.73 6 1.37° A, I2 D 7.02 6 17.25° A,
I3 D 3.05 6 48.67° A]Figure 31.14
Figure 31.15
4 For the network shown in Figure 31.16, determine the current
flowing in the ⊲4C j3⊳� impedance. [0]
Figure 31.16
5 For the network shown in Figure 31.17, use mesh-current analysis
to determine (a) the current in the capacitor, IC, (b) the current in
the inductance, IL, (c) the p.d. across the 4 � resistance, and (d) the
total active circuit power.
[(a) 14.5 A (b) 11.5 A (c) 71.8 V (d) 2499 W]
Figure 31.17
6 Determine the value of the currents IR, IY and IB in the network
shown in Figure 31.18 by using mesh-current analysis.
[IR D 7.84 6 71.19° AI IY D 9.04 6 37.50° A;
IB D 9.89 6 168.81° A]
7 In the network of Figure 31.19, use mesh-current analysis to
determine (a) the current in the capacitor, (b) the current in the 5 �
resistance, (c) the active power output of the 15 6 0° V source, and
(d) the magnitude of the p.d. across the j2 � inductance.
[(a) 1.03 A (b) 1.48 A(c) 16.28 W (d) 3.47 V]
8 A balanced 3-phase delta-connected load is shown in Figure 31.20.
Use mesh-current analysis to determine the values of mesh currents
560 Electrical Circuit Theory and Technology
Figure 31.18
I1, I2 and I3 shown and hence find the line currents IR, IYand IB.
[I1 D 83 6 173.13° A, I2 D 83 6 53.13° A,I3 D 83 6 66.87° A IR D 143.86 143.13° A,IY D 143.8 6 23.13° A, IB D 143.8 6 96.87° A]
9 Use mesh-circuit analysis to determine the value of currents IA to IEin the circuit shown in Figure 31.21.
[IA D 2.40 6 52.52° A; IB D 1.02 6 46.19° A;IC D 1.39 6 57.17° A; ID D 0.67 6 15.57° A;
IE D 0.996 6 83.74° A]
Figure 31.19 Figure 31.20
Figure 31.21 Figure 31.22
Nodal analysis
10 Repeat problems 1, 2, 5, 8 and 10 on page 542 of Chapter 30, andproblems 2, 3, 5, and 9 above, using nodal analysis.
Mesh-current and nodal analysis 561
Figure 31.23
11 Determine for the network shown in Figure 31.22 the voltage atnode 1 and the voltage VAB
[V1 D 59.0 6 28.92° V; VAB D 45.3 6 10.89° V]
12 Determine the voltage VPQ in the network shown in Figure 31.23.[VPQ D 55.87 6 50.60° V]
13 Use nodal analysis to determine the currents IA, IB and IC shown inthe network of Figure 31.24.
[IA D 1.21 6 150.96° AI IB D 1.06 6 56.32° A;IC D 0.55 6 32.01° A]
Figure 31.24
14 For the network shown in Figure 31.25 determine (a) the voltages atnodes 1 and 2, (b) the current in the 40 � resistance, (c) the currentin the 20 � resistance, and (d) the magnitude of the active powerdissipated in the 10 � resistance
[(a) V1 D 88.12 6 33.86° V, V2 D 58.72 6 72.28° V(b) 2.20 6 33.86° A, away from node 1,(c) 2.80 6 118.65° A, away from node 1, (d) 223 W]
Figure 31.25 Figure 31.26
15 Determine the voltage VAB in the network of Figure 31.26, usingnodal analysis. [VAB D 54.23 6 102.52° V]
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