December 2015
Intermediate to advanced
737 pages
23h 39m
English
Content preview from Network Analysis and Synthesis, 1st Edition by Pearson
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Kirchhoff’s Laws, Mesh and Nodal Analysis 75
Example 2.40 Find the current through 1 Ω resistance in the network shown in Figure 2.76
using nodal analysis.
5 Ω
−j 4 Ω
1 Ω
50∠0
°−
55.46∠33.69°
+
+
j 5 Ω j2 Ω
Figure 2.76
Solution: Let the V
1
and V
2
be the nodal voltages as shown in Figure 2.77.
5 Ω
−j 4 Ω
j 5 Ω j 2 Ω 1Ω
12
50∠0
°−
55.46∠33.69°
= −46.145 −j
30.76
+
+
−
I
1
I
3
I
5
I
2
V
2
V
1
I
4
Figure 2.77
By applying KCL at node 1, I
1
is written as follows:
III
V
j
VVV
j
V
j
j VVV
j
jj
123
11 12
1112
50
5
0
52
50
5
255
10
500
=+
−
=
−
+
−
−
=
+−
− 110 10 25 25
11
12
VVjV
jV
=−
+−
()−+ −=
10 35 25 500
12
jV jV j (2.86)
By applying KCL at node 2, we get the following
III
345
=+
(2.87)
or
VV
j
V
j
Vj
12 22
2
0
4
46 145 30 76
1
−
=
−
−
+
−− ...
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