This preview shows pages 1–3. Sign up to view the full content.
Instructor: Alexander Ganago
Problem 6
(Suggested by SingRong Li; modified)
The circuit on this diagram includes a voltage
source and a potentiometer; it is used as a sensor
for the angular position (angle
α
) of a robotic arm.
At
α
= 0, the potentiometer’s tap is in its middle
position so that
R
X
= R
0
= R
P
/2
At an arbitrary, nonzero angle
α
(in degrees) the
resistance
R
X
equals
R
X
= R
P
/2 + K
⋅
α
where K is constant measured in k
Ω
/degrees.
Assume
R
P
= 100 k
Ω
and determine the maximal value of
K
such that the angles
measured with this sensor cover the range from – 40 (negative 40) to +40 degrees.
The maximal value of
K
equals …
A.
0.625 k
Ω
/degree
B.
1.25 k
Ω
/degree
C.
2. 5 k
Ω
/degree
D.
5 k
Ω
/degree
E.
Depends on the source voltage, which is not given.
Solution
As the angle becomes negative and changes from 0 to 40 (negative 40) degrees, the
resistance decreases from
R
X
= R
P
/2 = 50 k
Ω
to
R
X
= R
P
/2  K
⋅
40.
For the sensor to work properly,
R
X
= R
P
/2 + K
⋅
α
should remain nonnegative.
Thus, at the maximal value of
K,
at the limit position of – 40 (negative 40) degrees the
resistance should equal zero; thus 50 k
Ω
+( 40 degrees)
⋅
K (k
Ω
/degrees) = 0,
or K = 1.25 k
Ω
/degree.
At positive angle
α
that increases from 0 to +40 degrees, the resistance
R
X
grows from
50 k
Ω
to
R
X
= R
P
/2 + K
⋅
40;
at the maximal value of
K,
at the limit position of
+40 degrees
R
X
should be 100 k
Ω
.
Thus 50 k
Ω
+( 40 degrees)
⋅
K (k
Ω
/degrees) = 100 k
Ω
, or K = 1.25 k
Ω
/degree.
Answer: B
EECS 314
Old exam problems with solutions, for practice
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Problem 7
In the circuit shown on this diagram, the current
source absorbs …
A.
– 36 W
B.
– 24 W
C.
– 12 W
D.
+12 W
E.
+36 W
Solution
To solve the problem, we need reference marks, as
shown on this diagram:
Note that the polarity of voltage
V
X
across the
current source is chosen in agreement with the
passive sign convention, so that the current enters
the positive terminal of the source.
The voltage
V
R
across the resistor equals,
according to Ohm’s law,
(3
Ω
)
⋅
(2 A) = 6 V
The
V
R
polarity is such that the current enters the more positive terminal of the resistor.
Also, note that the same 2 A current enters the positive terminal of the voltage source.
KVL equation around the loop is:
12 V –
V
R
–
V
X
= 0
or
12 V – 6 V –
V
X
= 0
Thus
V
X
= – 18 V
According to the passive sign convention, the current source absorbs
(2 A)
⋅
V
X
= – 36 W
To doublecheck the answer, verify the power balance:
the voltage source absorbs (12 V)
⋅
(2 A) = +24 W
the resistor absorbs (3
Ω
)
⋅
(2 A)
⋅
(2 A) = +12 W
Thus the total power in the circuit equals (– 36 W) +24 W +12 W = 0, as expected.
This is the end of the preview. Sign up
to
access the rest of the document.
 Winter '07
 Ganago
 Volt

Click to edit the document details