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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Network Theorems and Applications 205
∆=
−−
−−
−−
=×−+−−
−+
=
11 52
12520
15053
11 25 53 1000 55320250 25
3
()()
()
0060
∆=
−−
−
−
=+ +−−
=+
=
1
05 2
02520
65053
050 120 20 150
600 300
900
()
()
∆=
−
−−
−
=++− −−
=+
=
2
11 02
10 20
16 53
11 0 120 02
60
1320 12
1332
()
()
i
i
1
1
2
2
900
3060
0 294
1332
3060
0 435
=
∆
∆
==
=
∆
∆
==
.
.
A
A
Current through 5 Ω resistance due to 12 V source = i
2
− i
1
(from right to left) = 0.435 − 0.294
= 0.141 A.
By superposition theorem, we can say the following:
Current through 5 Ω resistances = Due to 9 A source + Due to 12 V source
= 2.9612 + (0.141)
= 3.1022 A.
Using Thevenin’s theorem, the following steps are to be followed:
Step 1: Open-circuit the branch containing 5 Ω resistance, ...
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