ECE3050 Homework Set 1
1. For
V
=18V
,
R
1
=39k
Ω
,
R
2
=43k
Ω
,and
R
3
=11k
Ω
,useOhm
’sLaw
,vo
ltaged
iv
is
ion
,
and current division to solve for
V
1
,
V
2
,
I
1
,
I
2
I
3
.
V
1
=18
39k
Ω
39 k
Ω
+43k
Ω
k
11k
Ω
=14
.
7V
V
2
43k
Ω
k
Ω
Ω
Ω
k
Ω
=3
.
30V
I
1
=
18
Ω
Ω
k
Ω
= 376
.
8
µ
A
I
2
=
Ω
43 k
Ω
+11k
Ω
I
1
=76
.
8
µ
A
I
3
=
Ω
Ω
Ω
I
1
= 300
µ
A
2. For
I
= 250
µ
A
,
R
1
= 100k
Ω
,
R
2
=68k
Ω
R
3
=82k
Ω
, use Ohm’s Law, voltage division,
and current division to solve for
I
1
,
I
2
V
3
.
I
1
= 250
µ
A
68k
Ω
+82k
Ω
100k
Ω
+68k
Ω
Ω
= 150
µ
A
I
2
= 250
µ
A
100 k
Ω
Ω
Ω
Ω
= 100
µ
A
V
3
= 100
µ
A
×
82k
Ω
=8
.
2V
3. It is given that
R
1
=1k
Ω
,
A
v
=10
−
4
,
A
i
=50
R
2
=40k
Ω
.
(a) With
i
o
=0
, use superposition to write the equations for
i
1
and
v
o
(
oc
)
. Solve the equations
for
v
o
(
oc
)
as a function of
v
s
.
i
1
=
v
s
−
A
v
v
o
R
1
v
o
(
oc
)
=
−
A
i
i
1
R
2
1
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o
(
oc
)
=
−
A
i
R
2
R
1
1
−
A
i
A
v
R
2
R
1
v
s
=
−
2500
v
s
(b) With
v
o
=0
, use superposition to write the equations for
i
1
and
i
o
(
sc
)
. Solve the equations
for
i
o
(
sc
)
as a function of
v
s
.
i
o
(
sc
)
=
−
A
i
v
s
R
1
=
−
0
.
05
v
s
(c) Use the solutions for
v
o
(
oc
)
and
i
o
(
sc
)
to show that
r
out
=50k
Ω
.
(d) Show that the Thévenin equivalent circuit seen looking into the output is a voltage source
v
o
(
oc
)
=
−
2500
v
s
in series with a resistance
r
out
Ω
.
(e) Show that the Norton equivalent circuit seen looking into the output is a current source
i
o
(
sc
)
=
−
0
.
05
v
s
in parallel with a resistance
r
out
Ω
.
(f) If a load resistor
R
L
=20k
Ω
is connected from the output node to ground, use both the
Thévenin and the Norton equivalent circuits to show that
v
o
=
−
714
.
3
v
s
for both.
4. It is given that
R
1
=10
Ω
,
G
m
.
5S
,
R
2
=5
Ω
,
R
3
=50
Ω
,and
A
v
=4
.
(a) With
i
o
, use superposition to solve for
v
o
(
oc
)
as a function of
v
s
.
v
o
(
oc
)
=
R
3
R
2
+
R
3
v
s
+
A
v
v
s
R
2
R
2
+
R
3
=
R
3
+
A
v
R
2
R
2
+
R
3
v
s
=1
.
273
v
s
(b) With
v
o
, use superposition to solve for
i
o
(
sc
)
as a function of
v
s
.
i
o
(
sc
)
=
v
s
R
2
+
A
v
v
s
R
3
=
v
s
μ
1
R
2
+
A
v
R
3
¶
.
28
v
s
(c) Solve for
r
out
.
r
out
=
v
o
(
oc
)
i
o
(
sc
)
.
545
Ω
(d) Show that the Thévenin equivalent circuit seen looking into the output is a voltage source
v
o
(
oc
)
.
273
v
s
in series with a resistance
r
out
.
545
Ω
.
(e) Show that the Norton equivalent circuit seen looking into the output is a current source
i
o
(
sc
)
.
28
v
s
in parallel with a resistance
r
out
.
545
Ω
.
(f) If a load resistor
R
L
Ω
is connected from the output node to ground, show
v
o
.
667
v
s
.
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 Summer '08
 HOLLIS
 Volt, Electrical impedance, Thévenin's theorem, Norton's theorem, Current Source, kΩ

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