EE215: Fundamentals of Electrical Engineering
Midterm Exam 2
February 22, 2006
Time: in class, 50 minutes
Class: _
EE215
_____________
Section:
____________________
Name: ____________________
Student ID:
____________________
Please show all steps in solving the problems. Clearly mark your final answers and don’t forget
the units.
Problem 1 (10+10+5 points)
Suppose you are given an unknown power source that is connected to a potentiometer
R
with a
range from 0 to 1000
Ω
(see diagram).
You measure the power
p
dissipated by
R
and obtain the following curve:
0
100
200
300
400
500
600
700
800
900
1000
0
5
10
15
20
25
R
=1000
Ω
+
–
V
Th
R
Th
???
p
/W
R
/
Ω
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View Full Document(a)
What is the Thévenin equivalent of your power source? Give values for
v
Th
and
R
Th
.
(b)
Give the equation for the power
p
as a function of
R
.
(c)
What value would
p
approach if you could make
R
very large (much larger than
1000
Ω
)? Explain your answer.
Answer:
(a) We recall that maximum power transfer occurs for
R
=
R
Th
. From the graph we see that the
power reaches a maximum of
p
= 25W for
R
= 100
Ω
. Therefore,
R
Th
= 100
Ω
. From
p
=
v
2
/
R
we
can calculate the voltage drop across
R
as
v
=
√
pR
= 50V. Since
R
=
R
Th
,
v
Th
is twice this
number, thus
v
Th
= 100V.
(b) In general, the voltage across
R
is
v
=
v
Th
R
/(
R
+
R
Th
) and the current is
i
=
v
Th
/(
R
+
R
Th
), hence
the power is
p
=
vi
=
v
Th
2
R
/ (
R
+
R
Th
)
2
. In our case,
p
(
R
) = 10000
R
/ (
R
+100)
2
for
R
in Ohms and
p
in Watts.
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 Spring '05
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 Electrical Engineering, Mesh Analysis, Black Box, Vout, Voltage drop, open circuit voltage, maximum gain

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