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Transcript
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
vS2=
5[W] iX
vS1=
3[V]
B
A
+
-
+
-
-
+
+
vA
R1=
1[W]
vB
R2 =
2[W]
vS3=
10 vX
R3 =
3[W]
vS4=
3[V]
D
+ vC
-
vX
R4 =
4[W]
vE + +
+
+ vD
-
-
+
-
+
-
-
vS5=
15[W] iX
E
-
C
R6 =
6[W]
R8=
8[W]
iX
R5 =
5[W]
+ vF
R7 =
7[W]
-
iS1=
7[A]
F
We can write
v v
v v v
v v
v
A+B: A D  A S 1 C  B E  B  0.
R2
R1
R4
R3
The constraint equation is
A+B: vB  v A  vS 2 .
Similarly, for node C we have two voltage sources connected. Again, the vS1
source has the R1 resistor in series, but the vS3 is a voltage source between two nonreference essential nodes. We might think that we would have a supernode here,
but if we look closely, we see that we have yet another voltage source connected to
node D. Since this one is connected to the reference node, this vS4 voltage source
determines the voltage at node D, and therefore all we will need is a constraint
equationf or node C. Let’s write the equations, starting with node C where we
have a constraint equation,
C+D: vD - vC  vS 3.
The voltage at node D is determined by the vS4 voltage source between it and the
reference node. We write
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