Page 1 of 18 A E E 220 Final Exam Wednesday December 20, 2006 110 Minutes Name: ____________________________________ This will be a closed book, closed notes exam. You may use the equation sheet attached to the back of the exam as well as a calculator. Be sure to circle your answers and fill them in on the scantron form. There will be zero tolerance to continuing beyond the announced end time. No one should get up to leave during the last 10 minutes of the exam. 1 2 3 5 4 6 Shade in Class Section
Transcript
Page 1 of 18
AE E 220 Final Exam
Wednesday December 20, 2006110 Minutes
Name: ____________________________________
This will be a closed book, closed notes exam. You may use the equation sheet attached
to the back of the exam as well as a calculator. Be sure to circle your answers and fill them in
on the scantron form. There will be zero tolerance to continuing beyond the announced end
time. No one should get up to leave during the last 10 minutes of the exam.
1 2 3 54 6
Shade inClass Section
Page 2 of 18
20 k
10 k
20 k
10 k
10 k
20 k
10 k
10 k
10 k10 k
1) The equivalent resistance for the resistive network shown below is most nearly:
a) 10 k b) 19 k c) 20 k d) 22 k e) 26 k
f) 29 k g) 32 k h) 36 k i) 40 k j) None of the above
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8 k1 mA
5 k
30 k
2 k
RL
4 k
6 k 8 k
2) For the circuit shown below, the value for the load resistor RL that will maximize the
power to this load is most nearly:
a) 10 k b) 15 k c) 17 k d) 21 k e) 28 k
f) 35 k g) 37 k h) 55 k i) 63 k j) None of the above
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3 & 4) Determine VD and I3 in the diagram shown below:
3) VD is most nearly:
a) -10 V b) -8 V c) -2 V d) 2 V e) 4 V
f) 8 V g) 10 V h) 14 V i) 18 V j) None of the above
4) I3 is most nearly:
a) -3 mA b) -2 mA c) -1 mA d) 1 mA e) 2 mA
f) 3 mA g) 4 mA h) 7 mA i) 7 mA j) None of the above
X
Y
I1
I2
I3
I425 V
+
- 6 mA
+ VB -
VA
+
-
3 V
+
-
+ 7 V - + 10 V -
VC
+
-
- VD +
2 mA
6 mA
3 mA
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5) The power supplied by the current source is most nearly:
a) -20 mW b) -15 mW c) -5 mW d) 5 mW e) 10 mW
f) 15 mW g) 16.7 mW h) 20 mW i) 26.7 mW j) None of the above
+-
4 k
8 k
20 k
0.5 mA 30 V
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200 FCEQ =>
100 F
200 F
200 F
100 F
6) The equivalent capacitance for the capacitor network shown below is most nearly:
a) 100 F b) 140 F c) 167 F d) 200 F e) 245 F
f) 267 F g) 325 F h) 400 F i) 650 F j) None of the above
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40 mH
90 mH
LEQ =>
40 mH
30 mH
20 mH
7) The equivalent inductance for the inductor network shown below is most nearly:
a) 7 mH b) 12 mH c) 15 mH d) 18 mH e) 26 mH
f) 36 mH g) 45 mH h) 72 mH i) 90 mH j) None of the above
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t = 0
+-
10
300 F
40 mH
10 V 30
8) For the circuit shown below, determine the stored energy when the circuit has achieved
DC steady-state conditions. (There is no stored energy before the switch closes.)
a) 8.5 mJ b) 10 mJ c) 15 mJ d) 17 mJ e) 20 mJ
f) 25 mJ g) 27 mJ h) 30 mJ i) 50 mJ j) None of the above
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9) A circuit consists of the series connection of a 10 resistor, a 36 mH inductor, and a
0.5 mF capacitor. If the source voltage applied to this load is v(t) = 7 cos (500t + 10E) V, then
the magnitude of the impedance, |Z| seen by the source will be most nearly:
a) 5.5 b) 10 c) 10.7 d) 11.7 e) 14.1
f) 17.2 g) 20.6 h) 24 i) 28 j) None of the above
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30 k40 k+- 15 k
22 V
12 k
Vout
+
-
12 k
10) For the circuit shown below, Vout is most nearly:
a) 3.0 V b) 4.5 V c) 6.5 V d) 6.9 V e) 8.7 V
f) 9.8 V g) 11 V h) 12.2 V i) 15.7 V j) None of the above
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11 & 12) For the circuit shown below, determine ZL of maximum power transfer.
VS = (20 p 20E) V ZS = (800 p -30E) Ω ZL = ?
11) Magnitude (3 points)
a) 4 b) 8 c) 20 d) 40 e) 80
f) 800 g) 1.6 k h) 8 k i) 64 k j) None of the above
12) Phase (2 points)
a) -50E b) -30E c) -20E d) -10E e) -3E
f) 3E g) 10E h) 20E i) 30E j) 50E
10 : 1I1 I2
ZL
+
-VL
+-
ZS
VS ~
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13 & 14) For the circuit shown below, answer the questions the follow:
VS = (10 p -25E) V
13) The reactive power for the load is most nearly: (4 points)
a) 1.5 b) 1.87 c) 2 d) 2.5 e) 3
f) 3.75 g) 4 h) 5 i) 6.25 j) None of the above
14) What are the proper units for this answer? (1 point)
a) mW b) W c) kW d) mVA e) VA
f) kVA g) mVAR h) VAR i) kVAR j) unit-less
+-~
3
j4
5
VS Load
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15 & 16) Given the following transformer, the load current, I2 is most nearly:
VS = (12 p 0) kVRMS ZS . 0 Ω ZL = 10 Ω
15) Amplitude (4 points)
a) 1.2 b) 2.4 c) 6.0 d) 12 e) 24
f) 60 g) 120 h) 140 i) 600 j) None of the above
16) Units (1 point)
a) ARMS b) mARMS c) ARMS d) kARMS e) MARMS
f) VRMS g) mVRMS h) VRMS i) kVRMS j) MVRMS
50 : 1I1 I2
ZL
+
-VL
+-
ZS
VS ~
~
~~
~
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19 & 20) For the following circuit, the switch is closed at t = 0. For t < 0, vC (t) = 9 V.
Determine the time t > 0, when vR(t) = – 2 V.
19) Number (4 points)
a) 0.21 b) 0.24 c) 0.56 d) 1.0 e) 2.0
f) 3.2 g) 4.0 h) 5.6 i) 6.3 j) None of the above
20) Units (1 points)
a) ns b) s c) ms d) s e) light-years
f) eons g) Ashlee h) Ivan i) Tim j) EE
25
0.5 mFt = 0
+ vR(t) -+
vC(t)-
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23 & 24) For the balanced three phase circuit shown below, the line current IbB is most nearly:
(Positive phase sequence)
Van = (480 p 30E) VRMS ZL = (60 p -20E) Ω
23) Amplitude (3 points)
a) 4.0 ARMS b) 6.9 ARMS c) 8.0 ARMS d) 9.8 ARMS e) 13.6 ARMS
f) 19.6 ARMS g) 24 ARMS h) 30 ARMS i) 36 ARMS j) None of the above
24) Phase (2 points)
a) -160E b) -130E c) -70E d) -40E e) -10E
f) 50E g) 80E h) 110E i) 170E j) None of the above
(n)
+-
c Vbn
+
-
Vcn
Van
b
a
C B
A
ZL
ZL ZL
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25 & 26) For the balanced three phase circuit shown below, the voltage Vbn is most nearly:
(Positive phase sequence)
VAB = (207 p -90E) VRMS ZL = (20 p 10E) Ω
25) Amplitude (3 points)
a) 60 VRMS b) 90 VRMS c) 120 VRMS d) 135 VRMS e) 170 VRMS
f) 207 VRMS g) 240 VRMS h) 360 VRMS i) 480 VRMS j) None of the above
26) Phase (2 points)
a) -170E b) -120E c) -60E d) 60E e) 110E
f) 120E g) 130E h) 170E i) 180E j) None of the above
(n)
+-
c Vbn
+
-
Vcn
Van
b
a
C B
A
ZL
ZL ZL
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27 & 28) For the circuit shown below, the amplitude and phase of the complex power
delivered to load Z3 is most nearly:
V1 = (120 p 30E) VRMS
V2 = (120 p 90E) VRMS
Z1 = (8 p 10E)
Z2 = (10 p -15E)
Z3 = (15 p 20E)
27) (3 points)
a) 0.96 kVA b) 1.44 kVA c) 1.92 kVA d) 2.88 kVA e) 3.84 kVA
f) 21.6 kVA g) 43 kVA h) 325 kVA i) 650 kVA j) None of the above
28) (2 points)
a) -80E b) -60E c) -40E d) -20E e) 0E
f) 20E g) 40E h) 60E i) 70E j) 80E
+-
+-
Z2
Z1
Z3
~
V2
+
-~
V1
+
-
~
~
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29 & 30) Given VAB = (207 p -90E) VRMS, IBC = (10 p 130E) ARMS, determine the
complex power supplied by each source. Please write answer. (Sorry)
