Thevenin Probs, 7/24/01

P6.1

@R.A. Decarlo & P. M. Lin
PROBLEM SOLUTIONS CHAPTER 6
S
OLUTION
6.1
.
(a)
V
oc
is found by removing
R
L
and doing voltage division.
V
V
V
OC
28
100
600
600
300
)
700

600
(
)
700

600
(
63
=
+
×
+
=
R
TH
is found by setting the source to zero and by calculating the equivalent resistance seen looking back
between the A and B terminal.
Ω
=
+
=
200
600

]
100
)
600

300
[(
TH
R
(b)
Using
L
L
I
R
P
⋅
=
the power for each resistance may be found by substituting the appropriate
R
L
in the following equation.
L
L
TH
OC
R
R
R
V
P
⋅
+
=
2
For 50
Ω
, 200
Ω
, and 800
Ω
, the power obtained is 627.2 mW, 980.0 mW, and 627.2 mW respectively.
The use of Thevenin equivalent does reduce the effort in obtaining the answer.
S
OLUTION
6.2
.
(a)
To find
R
TH
, open circuit the current source and shortcircuit the voltage source. The
resulting resistance seen from terminal AB is 1 k
Ω
. Using superposition, the contribution of the current
and voltage source at the open circuit output may be summed as 30 V (2 k/4 k) + 10 mA (2 k/4 k) (2 k
Ω
).
V
OC
is then 25 V and
I
SC
=
V
OC
/
R
TH
is 25 mA.
(b)
Following is a plot of
L
L
TH
OC
R
R
R
V
P
⋅
+
=
2
for
R
L
from 100
Ω
to 4 k
Ω
.
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Thevenin Probs, 7/24/01

P6.2

@R.A. Decarlo & P. M. Lin
S
OLUTION
6.3
.
(a)
Turning off the two sources
R
TH
=
(60
+
60)40
=
30
Ω
,
and using superposition
V
OC
=
6
V
40
40
+
60
+
60
+
0.1
A
60
40
+
60
+
60
40
=
3 V.
I
SC
is the obtained as 100 mA.
(b)
Using
L
L
TH
OC
R
R
R
V
P
⋅
+
=
2
a load of 90
Ω
will absorb 56.25 mW.
(c)
It absorbs 75 mW;
hence the 30
Ω
resistor absorbs more power.
S
OLUTION
6.4
.
As both resistor divider ration are the same (3/6), the voltage at A and B is the same
resulting in a
V
OC
of 0 V.
R
TH
=
(3
K
6
K
)
+
(9
K
18
K
)
=
8 k
Ω
The relation
V
OC
/
I
SC
cannot be used in this situation.
S
OLUTION
6.5
.
Using superposition
V
OC
=
20
20
×
10
3
20
×
10
3
+
5
×
10
3

10
5
×
10
3
20
×
10
3
+
5
×
10
3
+
20
⋅
20sin(50
t
)
=
14
+
400sin(50
t
)
V,
and