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Circle your class time: 4-5:30PM, TuTh 1-2:30PM, MW 8:30-10AM, TuTh
DO NOT OPEN THIS BOOKLET
UNTIL INSTRUCTED TO DO SO
ECE 2300 -- EXAM #1
February 22, 2003
1. This exam is closed book, closed notes. You may use one 8.5” x 11” crib
sheet, or its equivalent. You may use any calculator. Turn all cell phones or other
communications devices off.
2. Show all work on these pages. Show all work necessary to complete the
problem. If you go on to another page, indicate clearly where your work can
be found. A solution without the appropriate work shown will receive no credit.
Clearly indicate your answer (for example, by enclosing it in a box).
3. Show all units in solutions, intermediate results, and figures. Units in the exam
will be included between square brackets.
4. Do not use red ink. Do not use red pencil.
5. You will have 90 minutes to work on this exam.
6. Be sure to sign the statement below before you turn in your exam. If you fail to
sign this statement, your exam will not be graded and you will earn no credit.
1. ________________/1
2. ________________/24
3. ________________/28
4. ________________/24
5. ________________/34
Total = 111
Statement: I understand the Academic Honesty Policy of the University of
Houston. I certify that I will not communicate with any other person concerning
the material on the exam while I am taking this exam.
ECE 2300 Exam 1 – February 22, 2003 – Page 2
Signature: ___________________________________ Date:
________________
ECE 2300 Exam 1 – February 22, 2003 – Page 3
1) {1 Point} On the front page, circle your class time.
Room for extra work
ECE 2300 Exam 1 – February 22, 2003 – Page 4
2) {24 Points} Show that the total power delivered by the components in the
circuit below equals the total power absorbed by the components in this circuit. In
other words, find the sum of the powers delivered by each of the elements that
deliver positive power, and compare that with the sum of the powers absorbed by
the elements that absorb positive power.
+ vY
+
vS1 =
100[V]
vS2= vY
+
iS1= - 0.4[S]vX
R1 =
10[]
R2 =
20[]
-
+
vX
iS2 =
10[A]
-
ECE 2300 Exam 1 – February 22, 2003 – Page 5
Room for extra work
ECE 2300 Exam 1 – February 22, 2003 – Page 6
3) {28 Points} A device can be modeled by an ideal voltage source in series with
a resistance. The device was connected to a 7.8[mA] ideal current source, as
shown in Figure 1, and a voltage v1 of – 215.2[V] resulted. The same device was
then connected to a 5.6[mA] ideal current source, as shown in Figure 2, and a
voltage v2 of 39.4[V] resulted.
Find the power absorbed by the device if a 50[V] ideal voltage source is
connected to the same device, as shown in Figure 3.
a
a
+
v2
+
v1
7.8[mA]
Device
b
b
Figure 2
Figure 1
a
+
50[V]
Device
-
b
Figure 3
5.6[mA]
Device
ECE 2300 Exam 1 – February 22, 2003 – Page 7
Room for extra work
ECE 2300 Exam 1 – February 22, 2003 – Page 8
4) {24 Points} The circuit shown below has two ammeters, two voltmeters, and
an unknown resistor RX.
Each ammeter has a full-scale current of 10[mA], and a meter resistance of 7[].
Each voltmeter has a full-scale voltage of 100[V], and a meter resistance of
75[k].
The values of resistors R2 and R3 are always in a ratio of 1:10.
Ammeter #2 reads a current, i2 = 3[mA].
Voltmeter #2 reads a voltage, v2 = 13.1[V].
Find the numerical value of RX.
1[k]
Voltmeter
#1
Ammeter
#1
Ammeter
#2
iS =
5[mA]
R2 =
10[k]
-
Voltmeter
#2
R1 =
12[k]
i2
RY
RX
v2
+
R3 =
100[k]
ECE 2300 Exam 1 – February 22, 2003 – Page 9
Room for extra work
ECE 2300 Exam 1 – February 22, 2003 – Page 10
5) {34 Points} For the circuit shown, use the node-voltage method to write a
complete set of independent equations that could be used to solve this circuit. Do
not attempt to solve the equations. Do not attempt to simplify the circuit.
iZ
R4
vS6 = iZ
-
+
+
+
vS5
-
+
-
iS1
iX
vY
-
R6
iS2
-
R3
R2
vS4
R5
-
+
+
-
R1
+
vS3 = vY
vS2 = riX
+
vS1
R7
R9
R8
ECE 2300 Exam 1 – February 22, 2003 – Page 11
Room for extra work
ECE 2300 Exam 1 – February 22, 2003 – Page 12
Solutions:
2) {24 Points} Show that the total power delivered by the components in the
circuit below equals the total power absorbed by the components in this circuit. In
other words, find the sum of the powers delivered by each of the elements that
deliver positive power, and compare that with the sum of the powers absorbed by
the elements that absorb positive power.
+ vY
+
vS1 =
100[V]
vS2= vY
+
iS1= - 0.4[S]vX
R1 =
10[]
R2 =
20[]
-
+
vX
iS2 =
10[A]
-
ECE 2300 Exam 1 – February 22, 2003 – Page 13
ECE 2300 Exam 1 – February 22, 2003 – Page 14
ECE 2300 Exam 1 – February 22, 2003 – Page 15
3) {28 Points} A device can be modeled by an ideal voltage source in series with
a resistance. The device was connected to a 7.8[mA] ideal current source, as
shown in Figure 1, and a voltage v1 of – 215.2[V] resulted. The same device was
then connected to a 5.6[mA] ideal current source, as shown in Figure 2, and a
voltage v2 of 39.4[V] resulted.
Find the power absorbed by the device if a 50[V] ideal voltage source is
connected to the same device, as shown in Figure 3.
a
a
+
v2
+
v1
7.8[mA]
Device
b
b
Figure 2
Figure 1
a
+
50[V]
Device
-
b
Figure 3
5.6[mA]
Device
ECE 2300 Exam 1 – February 22, 2003 – Page 16
ECE 2300 Exam 1 – February 22, 2003 – Page 17
4) {24 Points} The circuit shown below has two ammeters, two voltmeters, and
an unknown resistor RX.
Each ammeter has a full-scale current of 10[mA], and a meter resistance of 7[].
Each voltmeter has a full-scale voltage of 100[V], and a meter resistance of
75[k].
The values of resistors R2 and R3 are always in a ratio of 1:10.
Ammeter #2 reads a current, i2 = 3[mA].
Voltmeter #2 reads a voltage, v2 = 13.1[V].
Find the numerical value of RX.
1[k]
Ammeter
#1
Voltmeter
#1
Ammeter
#2
iS =
5[mA]
-
Voltmeter
#2
R1 =
12[k]
i2
RY
R2 =
10[k]
RY+7[]
RX
v2
+
R3 =
100[k]
ECE 2300 Exam 1 – February 22, 2003 – Page 18
ECE 2300 Exam 1 – February 22, 2003 – Page 19
5) {34 Points} For the circuit shown, use the node-voltage method to write a
complete set of independent equations that could be used to solve this circuit. Do
not attempt to solve the equations. Do not attempt to simplify the circuit.
ECE 2300 Exam 1 – February 22, 2003 – Page 20
iZ
R4
vS6 = iZ
-
+
+
+
vS5
-
+
-
iS1
iX
vY
-
R6
iS2
-
R3
R2
vS4
R5
-
+
+
-
R1
+
vS3 = vY
vS2 = riX
+
vS1
R7
R9
R8
ECE 2300 Exam 1 – February 22, 2003 – Page 21
ECE 2300 Exam 1 – February 22, 2003 – Page 22
ECE 2300 Exam 1 – February 22, 2003 – Page 23