EE302, Intro to Elec & Computer Engineering – Quiz 3
 Solutions
There were two versions of each problem with a slight difference.
Yours should be shown below.
Problem 1
(25 points)
Determine voltages
v
1
through
v
3
in the circuit below.
Show your work below.
Put your answers on the last page in the slots provided.
Note that the S (Siemens) in the diagram refers
to conductances which are reciprocals of resistances.
Chapter 3, Solution 16.
At the supernode,
2 = v
1
+ 2 (v
1
 v
3
) + 8(v
2
– v
3
) + 4v
2
, which leads to
2 = 3v
1
+ 12v
2
 10v
3
(1)
But
v
1
= v
2
+ 2v
0
and v
0
= v
2
.
Hence
v
1
= 3v
2
(2)
v
3
= 13V
(3)
Substituting (2) and (3) with (1) gives,
v
1
=
18.858 V
, v
2
=
6.286 V
,
v
3
=
13 V
2 A
v
3
v
2
v
1
8 S
4 S
1 S
i
0
–
+
13 V
2 S
+
v
0
–
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View Full DocumentEE302, Intro to Elec & Computer Engineering – Quiz 3
 Solutions
There were two versions of each problem with a slight difference.
Yours should be shown below.
Problem 1
(25 points)
Determine voltages
v
1
through
v
3
in the circuit below.
Show your work below.
Put your answers on the last page in the slots provided.
Note that the S (Siemens) in the diagram refers
to conductances which are reciprocals of resistances.
Chapter 3, Solution 16. (modified)
At the supernode,
3 = v
1
+ 2 (v
1
 v
3
) + 8(v
2
– v
3
) + 4v
2
, which leads to
3 = 3v
1
+ 12v
2
 10v
3
(1)
But
v
1
= v
2
+ 2v
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 Spring '06
 MCCANN
 Operational amplifier applications, Thévenin's theorem, slight difference, Engineering – Quiz

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