082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

8.12 Analysis of electrical networks 309
2 V I 4 V
a
3 V
+
– I b
3 V
I 1
1 V
I c
5 V I 2 2 V I 3
3 V
I d – I e +
2 V 4 V
+ 4 V –
I f
Figure8.3
Theelectrical network ofFigure 8.2
with values forthe source voltagesand
resistorsadded.
Solution
We have already obtained the equations for this network. Substituting actual values
for the resistorsand voltage sources gives
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
3 10 −3 −1 I 1
⎝−2−4⎠ = ⎝−3 14 −2⎠
⎝I 2
⎠
4 −1−2 6 I 3
This is now in the formV =RI ′ . We shall solve these equations by Gaussian elimination.
Forming the augmented matrix, we have
⎛
⎞
10−3−1 3
⎝−3 14 −2 −6⎠
−1−2 6 4
Then
and
Hence,
R 1
R 2
→ 10R 2
+3R 1
R 3
→ 10R 3
+R 1
R 1
R 2
R 3
→ 131R 3
+23R 2
Similarly,
I 3
= 4460
7200 = 0.619 A
I 2
=
Finally,
I 1
=
−51 +23(0.619)
131
3 +0.619 +3(−0.281)
10
⎛
⎞
10 −3 −1 3
⎝0 131 −23 −51⎠
0 −23 59 43
⎛
⎞
10 −3 −1 3
⎝0 131 −23 −51⎠
0 0 7200 4460
= −0.281 A
= 0.278 A
➔