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31
Mesh-current and
nodal analysis
At the end of this chapter you should be able to:
ž solve d.c. and a.c. networks using mesh-current analysis
ž solve d.c. and a.c. networks using nodal analysis
31.1 Mesh-current
analysis
Mesh-current analysis is merely an extension of the use of Kirchhoff’s
laws, explained in Chapter 30. Figure 31.1 shows a network whose circulating currents I1 , I2 and I3 have been assigned to closed loops in the
circuit rather than to branches. Currents I1 , I2 and I3 are called meshcurrents or loop-currents.
Figure 31.1
In mesh-current analysis the loop-currents are all arranged to circulate in the same direction (in Figure 31.1, shown as clockwise direction).
Kirchhoff’s second law is applied to each of the loops in turn, which
in the circuit of Figure 31.1 produces three equations in three unknowns
which may be solved for I1 , I2 and I3 . The three equations produced from
Figure 31.1 are:
I1 Z1 C Z2 I2 Z2 D E1
I2 Z2 C Z3 C Z4 I1 Z2 I3 Z4 D 0
I3 Z4 C Z5 I2 Z4 D E2
The branch currents are determined by taking the phasor sum of the
mesh currents common to that branch. For example, the current flowing
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in impedance Z2 of Figure 31.1 is given by (I1 I2 ) phasorially. The
method of mesh-current analysis, called Maxwell’s theorem, is demonstrated in the following problems.
Problem 1. Use mesh-current analysis to determine the current
flowing in (a) the 5 resistance, and (b) the 1 resistance of the
d.c. circuit shown in Figure 31.2.
Figure 31.2
The mesh currents I1 , I2 and I3 are shown in Figure 31.2. Using Kirchhoff’s voltage law:
For loop 1, 3 C 5I1 5I2 D 4
1
For loop 2, 4 C 1 C 6 C 5I2 5I1 1I3 D 0
2
For loop 3, 1 C 8I3 1I2 D 5
3
Thus
4D0
10 D0
20 C 9I3 C 5 D 0
30 8I1 5I2
5I1 C 16I2 I3
I2
Using determinants,
5
16
1
I1
0
1
9
D 8
5
0
4 0 5 I2
D 8
0 4 1 0 5
0
9
5 D 8
5
0
I3
5
16
1
4 0 5 1
5
16
1
0 1 9 2
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1
5 9
I1
16
0 4
1
5
D
1 9 1
8 9
D
5
4 0
D
16
8 1
I2
5
0 4
0
5
1 9 I3
8
16 C 5 5
1 1
5
1 C
5
0
9
5 16 1 9 I1
I2
I3
D
D
55 4143
85 445
45 C 5103
1
D
8143 C 545
I2
I3
1
I1
D
D
D
547
140
495
919
Hence I1 D
I3 D
547
140
D 0.595 A, I2 D
D 0.152 A, and
919
919
495
D 0.539 A
919
Thus current in the 5 Z resistance D I1 I2 D 0.595 0.152
D 0.44 A,
and current in the 1 Z resistance D I2 I3 D 0.152 0.539
D 0.69 A
Problem 2. For the a.c. network shown in Figure 31.3 determine,
using mesh-current analysis, (a) the mesh currents I1 and I2 (b) the
current flowing in the capacitor, and (c) the active power delivered
by the 1006 0° V voltage source.
Figure 31.3
(a)
For the first loop 5 j4I1 j4I2 D 1006 0°
(1)
For the second loop 4 C j3 j4I2 j4I1 D 0
(2)
Rewriting equations (1) and (2) gives:
5 j4I1 C j4I2 100 D 0
10 j4I1 C 4 jI2 C 0 D 0
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Thus, using determinants,
I2
I
1
D j4
5 j4
100 4 j
0 j4
1
D 5 j4
j4
100 0
4 j j4
I1
I2
1
D
D
400 j100
j400
32 j21
Hence
I1 D
400 j100
412.316 14.04°
D
32 j21
38.286 33.27°
D 10.776 19.23° A D 10.8 6 −19.2° A,
correct to one decimal place
I2 D
4006
90°
38.286 33.27°
D 10.456 56.73° A
D 10.5 6 −56.7° A,
correct to one decimal place
(b)
Current flowing in capacitor D I1 I2
D 10.776 19.23° 10.456 56.73°
D 4.44 C j12.28 D 13.16 70.12° A,
i.e., the current in the capacitor is 13.1 A
(c)
Source power P D VI cos
D 10010.77 cos 19.23°
D 1016.9 W D 1020 W,
correct to three significant figures.
(Check: power in 5 resistor D I21 5 D 10.772 5 D 579.97 W
and power in 4 resistor D I22 4 D 10.452 4 D 436.81 W
Thus total power dissipated D 579.97 C 436.81
D 1016.8 W D 1020 W, correct
to three significant figures.)
Problem 3. A balanced star-connected 3-phase load is shown in
Figure 31.4. Determine the value of the line currents IR , IY and IB
using mesh-current analysis.
Figure 31.4
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Two mesh currents I1 and I2 are chosen as shown in Figure 31.4.
From loop 1, I1 3 C j4 C I1 3 C j4 I2 3 C j4 D 4156 120°
i.e., 6 C j8I1 3 C j4I2 4156 120° D 0
1
From loop 2, I2 3 C j4 I1 3 C j4 C I2 3 C j4 D 4156 0°
i.e., 3 C j4I1 C 6 C j8I2 4156 0° D 0
2
Solving equations (1) and (2) using determinants gives:
3 C j4
6 C j8
I1
I2
D 6 C j8
3 C j4
6
°
415 0
4156
120° 4156 120° 4156 0° 1
D 6 C j8
3 C j4
3 C j4 6 C j8 I2
I1
D
20756 53.13° C 41506 173.13°
41506 53.13° 20756 173.13°
D
1006
1
106.26° 256 106.26°
I1
I2
1
D
D
35946 143.13°
35946 83.13°
756 106.26°
Hence I1 D
35946 143.13°
D 47.96 36.87° A
756 106.26°
I2 D
35946 83.13°
D 47.96 23.13° A
756 106.26°
and
Thus line current IR D I1 D 47.96 36.87° A
IB D I2 D 47.96 23.23° A
D 47.96 156.87° A
and
IY D I2 I1 D 47.96 23.13° 47.96 36.87°
D 47.96 −83.13° A
Further problems on mesh-current analysis may be found in Section 31.3,
problems 1 to 9, page 559.
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31.2
Figure 31.5
Nodal analysis
A node of a network is defined as a point where two or more branches
are joined. If three or more branches join at a node, then that node is
called a principal node or junction. In Figure 31.5, points 1, 2, 3, 4 and
5 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 is
chosen as the reference node then V13 is assumed to mean the voltage
at 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 with
respect to a particular chosen reference node, the notation V1 for V13 and
V2 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 find
voltages 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 which
may be simultaneously solved using determinants.
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
V1 V2
V1 Vx
C
C
D0
ZA
ZD
ZB
1
Similarly, for node 2, assuming all branch currents are leaving the node
as shown in Figure 31.7,
V2
V2 C VY
V2 V1
C
C
D0
ZB
ZE
ZC
Figure 31.7
2
In equations (1) and (2), the currents are all assumed to be leaving the
node. In fact, any selection in the direction of the branch currents may
be made — the resulting equations will be identical. (For example, if for
node 1 the current flowing in ZB is considered as flowing towards node 1
instead of away, then the equation for node 1 becomes
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V1
V2 V1
V1 Vx
C
D
ZA
ZD
ZB
which if rearranged is seen to be exactly the same as equation (1).)
Rearranging equations (1) and (2) gives:
1
1
1
C
C
ZA
ZB
ZD
1
ZB
V1 C
V1 1
ZB
1
1
1
C
C
ZB
ZC
ZE
V2 V2 C
1
ZA
1
ZC
Vx D 0
3
VY D 0
4
Equations (3) and (4) may be rewritten in terms of admittances (where
admittance Y D l/Z ):
YA C YB C YD V1 YB V2 YA Vx D 0
5
YB V1 C YB C YC C YE V2 C YC VY D 0
6
Equations (5) and (6) may be solved for V1 and V2 by using determinants.
Thus
V
1
Y
B
YB C YC C YE V
2
D YA C YB C YD YB
YA YC YA YC 1
Y
C
Y
C
Y
YB
B
D
A
YB
YB C YC C YE D Current equations, and hence voltage equations, may be written at each
principal node of a network with the exception of a reference node. The
number of equations necessary to produce a solution for a circuit is, in
fact, 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 circuit
contains, Basically, the method that requires the least number of equations
is used. The method of nodal analysis is demonstrated in the following
problems.
Problem 4. For the network shown in Figure 31.8, determine the
voltage VAB , by using nodal analysis.
Figure 31.8
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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 equation for node 1 is obtained as follows. Applying Kirchhoff’s current law
to node 1 gives:
IX C IY D I
V1
V1
C
D 206 0°
16
4 C j3
i.e.,
V1
Thus
1
1
C
16 4 C j3
V1 0.0625 C
4 j3
42 C 32
D 20
D 20
V1 0.0625 C 0.16 j0.12 D 20
V1 0.2225 j0.12 D 20
20
20
D
0.2225 j0.12
0.25286 28.34°
from which, V1 D
i.e., voltage V1 D 79.16 28.34° V
The current through the 4 C j3 branch, Iy D V1 /4 C j3
Hence the voltage drop between points A and B, i.e., across the 4 resistance, is given by:
VAB D Iy 4 D
V1 4
79.16 28.34°
4 D 63.3 6 −8.53° V
D
4 C j3
56 36.87°
Problem 5. Determine the value of voltage VXY shown in the
circuit of Figure 31.9.
Figure 31.9
The 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.
VX 86 0°
Vx 86 90°
C
D0
5 C 4
3 C j6
Thus
from which,
VX
VX
1
1
C
9 3 C j6
1
3 j6
C
9 32 C 62
D
8
j8
C
9 3 C j6
D
8 j83 j6
C
9
32 C 62
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VX 0.1778 j0.1333 D 0.8889 C
48 C j24
45
VX 0.22226 36.86° D 1.9556 C j0.5333
D 2.0276 15.25°
Since point Y is the reference node,
voltage VX D VXY D
2.0276 15.25°
D 9.12 6 52.11° V
0.22226 36.86°
Problem 6. Use nodal analysis to determine the current flowing
in each branch of the network shown in Figure 31.10.
Figure 31.10
This is the same problem as problem 1 of Chapter 30, page 536, which
was solved using Kirchhoff’s laws. A comparison of methods can
be 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
V1
V1 506 90°
V1 1006 0°
C
C
D0
25
20
10
i.e.,
i.e.,
1
1006 0°
1
1
506 90°
V1 C
C
D0
25 20 10
25
10
0.19 V1 D 4 C j5
Thus the voltage at node 1, V1 D
4 C j5
D 33.706 51.34° V
0.19
or 21.05 C j26.32V
Hence the current in the 25 resistance,
I1 D
V1 1006 0°
21.05 C j26.32 100
D
25
25
D
78.95 C j26.32
25
D 3.336 161.56° A flowing away from node 1
or 3.336 161.56° 180° A D 3.33 6 −18.44° A flowing toward
node 1)
The current in the 20 resistance,
I2 D
V1
33.706 51.34°
D
D 1.696 51.34° A
20
20
flowing from node 1 to node 2
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The current in the 10 resistor,
I3 D
V1 506 90°
21.05 C j26.32 j50
21.05 j23.68
D
D
10
10
10
D 3.176 −48.36° A away from node 1
or 3.176 48.36° 180° D 3.176 228.36° A D 3.176 131.64° A
toward node 1)
Problem 7. In the network of Figure 31.11 use nodal analysis to
determine (a) the voltage at nodes 1 and 2, (b) the current in the
j4 inductance, (c) the current in the 5 resistance, and (d) the
magnitude of the active power dissipated in the 2.5 resistance.
Figure 31.11
(a)
At node 1,
V1
V1 V2
V1 256 0°
C
C
D0
2
j4
5
Rearranging gives:
1
1
1
V1 C
C
2 j4 5
1
256 0°
V2 D0
5
2
0.7 C j0.25V1 0.2V2 12.5 D 0
i.e.,
At node 2,
1
V2
V2 V1
V2 256 90°
C
C
D0
2.5
j4
5
Rearranging gives:
i.e.,
1
V1 C
5
1
256 90°
1
1
V2 C
C
D0
2.5 j4 5
2.5
0.2V1 C 0.6 j0.25V2 j10 D 0
2
Thus two simultaneous equations have been formed with two
unknowns, V1 and V2 . Using determinants, if
0.7 C j0.25V1 0.2V2 12.5 D 0
1
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and 0.2V1 C 0.6 j0.25V2 j10 D 0
then
V
1
0.2
0.6 j0.25
2
V
2
D 12.5 0.7 C j0.25
j10
0.2
12.5 j10 1
D 0.7 C j0.25
0.2
0.2
0.6 j0.25 i.e.,
V2
V1
D
j2 C 7.5 j3.125
j7 C 2.5 2.5
D
and
7.5846
1
0.42 j0.175 C j0.15 C 0.0625 0.04
V2
1
V1
D
D
6
6
8.53°
7 90°
0.443 3.23°
Thus voltage, V1 D
7.5846 8.53°
D 17.126 5.30° V
0.4436 3.23°
D 17.1 6 −5.3° V, correct to one decimal place,
and voltage, V2 D
76 90°
D 15.806 93.23° V
0.4436 3.23°
D 15.8 6 93.2° V, correct to one decimal place.
(b)
The current in the j4 inductance is given by:
V2
15.806 93.23°
D 3.95 6 3.23° A flowing away from node 2
D
j4
46 90°
(c)
The current in the 5 resistance is given by:
V1 V2
17.126 5.30° 15.806 93.23°
D
5
5
17.05 j1.58 0.89 C j15.77
I5 D
5
17.94 j17.35
24.966 44.04°
D
D
5
5
D 4.99 6 −44.04° A flowing from node 1 to node 2
I5 D
i.e.,
(d)
The active power dissipated in the 2.5 resistor is given by
P2.5 D I2.5 2 2.5 D
D
V2 256 90°
2.5
2
2.5
0.89 C j15.77 j252
9.2736 95.51° 2
D
2.5
2.5
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85.996 191.02°
by de Moivre’s theorem
2.5
D 34.46 169° W
D
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 the
voltage VXY using nodal analysis.
Figure 31.12
Node 3 is taken as the reference node.
256 0° D
At node 1,
i.e.,
4 j3 1
1
V1 V2 25 D 0
C
25
5
5
0.3796 18.43° V1 0.2V2 25 D 0
or
or
i.e.,
1
V2
V2
V2 V1
C
C
D0
j10 j20
5
At node 2,
i.e.,
V1
V1 V2
C
4 C j3
5
0.2V1 C
1
1
1
V2 D 0
C
C
j10 j20 5
0.2V1 C j0.1 j0.05 C 0.2V2 D 0
0.2V1 C 0.256 36.87° V2 C 0 D 0
2
Simultaneous equations (1) and (2) may be solved for V1 and V2 by using
determinants. Thus,
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V1
0.2
0.256 36.87°
V
2
D 6
25 0.379 18.43°
0.2
0
D i.e.,
0 1
0.3796
25 18.43°
0.2
6
°
0.25 36.87 0.2
V1
V2
1
D
D
6.256 36.87°
5
0.094756 55.30° 0.04
D
1
0.0796 79.85°
Hence voltage, V1 D
6.256 36.87°
D 79.116 42.98° V
0.0796 79.85°
and voltage, V2 D
5
D 63.296 79.85° V
0.0796 79.85°
The current flowing in the 4 C j3 branch is V1 /4 C j3. Hence the
voltage between point X and node 3 is:
V1
79.116 42.98° 36 90° j3 D
4 C j3
56 36.87°
D 47.476 96.11° V
Thus the voltage
VXY D VX VY D VX V2 D 47.476 96.11° 63.296 79.85°
D 16.21 j15.10 D 22.156 −137° V
Problem 9. Use nodal analysis to determine the voltages at nodes
2 and 3 in Figure 31.13 and hence determine the current flowing
in the 2 resistor and the power dissipated in the 3 resistor.
This is the same problem as Problem 2 of Chapter 30, page 537, which
was solved using Kirchhoff’s laws.
In Figure 31.13, the reference node is shown at point A.
At node 1,
i.e.,
Figure 31.13
At node 2,
V1 V2
V1
V1 8 V3
C
C
D0
1
6
5
1.367V1 V2 0.2V3 1.6 D 0
1
V2
V2 V1
V2 V3
C
C
D0
2
1
3
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i.e.,
V1 C 1.833V2 0.333V3 C 0 D 0
At node 3,
V3 V2
V3 C 8 V1
V3
C
C
D0
4
3
5
2
0.2V1 0.333V2 C 0.783V3 C 1.6 D 0
i.e.,
3
Equations (1) to (3) can be solved for V1 , V2 and V3 by using
determinants. Hence
1
1.833
0.333
V1
0.2
0.333
0.783
D 1.367
1
0.2
1.6 0 1.6 V2
0.2
0.333
0.783
1.6 0 1.6 V3
1
D 1.367
1.367
1
1.6
1
0.2 0 1.833
1.833 0.333 1
1
0.2 0.333
0.2 0.333
1.6 0.783 D Solving for V2 gives:
V2
1.60.8496 C 1.60.6552
D
hence
1
1.3671.3244 C 10.8496 0.20.6996
0.31104
V2
1
D
from which, voltage, V 2 D
0.31104
0.82093
0.82093
D 0.3789 V
0.3789
V2
D
D 0.19 A,
Thus the current in the 2 Z resistor D
2
2
flowing from node 2 to node A.
Solving for V3 gives:
hence
V3
1
D
1.60.6996 C 1.61.5057
0.82093
V3
1.2898
1
D
from which, voltage,V3 D
1.2898
0.82093
0.82093
D −1.571 V
Power in the 3 Z resistor D I3 2 3 D
D
V2 V3
3
2
3
0.3789 1.5712
D 1.27 W
3
Further problems on nodal analysis may be found in Section 31.3
following, problems 10 to 15, page 560.
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31.3 Further problems
on mesh-current and
nodal analysis
Mesh-current analysis
1
Repeat problems 1 to 10, page 542, of Chapter 30 using meshcurrent 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 8.736 1.37° A, I2 D 7.026 17.25° A,
I3 D 3.056 48.67° A]
4
For the network shown in Figure 31.16, determine the current
flowing in the 4 C j3 impedance.
[0]
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 31.14
Figure 31.15
Figure 31.16
Figure 31.17
6
7
8
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 7.846 71.19° AI IY D 9.046 37.50° A;
IB D 9.896 168.81° A]
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 156 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]
A balanced 3-phase delta-connected load is shown in Figure 31.20.
Use mesh-current analysis to determine the values of mesh currents
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I1 , I2 and I3 shown and hence find the line currents IR , IY
and IB .
[I1 D 836 173.13° A, I2 D 836 53.13° A,
I3 D 836 66.87° A IR D 143.86 143.13° A,
IY D 143.86 23.13° A, IB D 143.86 96.87° A]
9
Use mesh-circuit analysis to determine the value of currents IA to IE
in the circuit shown in Figure 31.21.
[IA D 2.406 52.52° A; IB D 1.026 46.19° A;
IC D 1.396 57.17° A; ID D 0.676 15.57° A;
IE D 0.9966 83.74° A]
Figure 31.18
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, and
problems 2, 3, 5, and 9 above, using nodal analysis.
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11
Determine for the network shown in Figure 31.22 the voltage at
node 1 and the voltage VAB
[V1 D 59.06 28.92° V; VAB D 45.36 10.89° V]
12
Determine the voltage VPQ in the network shown in Figure 31.23.
[VPQ D 55.876 50.60° V]
13
Use nodal analysis to determine the currents IA , IB and IC shown in
the network of Figure 31.24.
[IA D 1.216 150.96° AI IB D 1.066 56.32° A;
IC D 0.556 32.01° A]
14
For the network shown in Figure 31.25 determine (a) the voltages at
nodes 1 and 2, (b) the current in the 40 resistance, (c) the current
in the 20 resistance, and (d) the magnitude of the active power
dissipated in the 10 resistance
[(a) V1 D 88.126 33.86° V, V2 D 58.726 72.28° V
(b) 2.206 33.86° A, away from node 1,
(c) 2.806 118.65° A, away from node 1, (d) 223 W]
Figure 31.23
Figure 31.24
Figure 31.25
Figure 31.26
15
Determine the voltage VAB in the network of Figure 31.26, using
nodal analysis.
[VAB D 54.236 102.52° V]
17