Download EE305 Final Exam

EE305 Final Exam
Name:
May 5, 2011
1. Test time: 1:00 pm till 3:00 pm
2. You are allowed to use the formula collection handed out by the instructor.
3. Only a No 2 pencil is allowed.
4. Besides the items in the two previous points no other material is allowed!
5. No calculator is allowed!
6. ALL cell phones are off!
7. ALL electronic devices are off and in your backpack!
8. ALL caps and hats are off!
9. The exam is 32.5% of the final grade. You can get 32.5 plus 2 bonus points in this exam, ie each
point in this exam is 1% of your final grade.
10. There 69 questions, so each question gives you 0.5 points.
11. You keep the formula collection.
final exam version May 2011 α
1
1. The unit of power is
(a) Coulomb
(b) Watt
(c) Ampere
(d) Volt
2. The voltage 10kV is the same as
(a) 10+2 V
(b) 10+3 V
(c) 10+4 V
(d) 10+5 V
3. During 10 minutes 36C of charge pass a conductor. The current through the conductor is given by
(a) 0.06A
(b) 0.6A
(c) 6A
(d) 60A
4. 300,000 electrons will provide a charge of
(a) qtotal = −4.806 · 10−16 C
(b) qtotal = −4.806 · 10−15 C
(c) qtotal = −4.806 · 10−14 C
(d) qtotal = −4.806 · 10−13 C
5. Which of the following sources is a sinusoidal voltage source:
(a)
(b)
(c)
2
(d)
6.
Considering the labeling and the given signs of the voltages and currents, which of the elements is
actually passive?
−7A
−5V
7A
−5V
5V
(a)
−7A
7A
(b)
−5V
(c)
(d)
7. An ideal current source is defined by the following sentence:
(a) Provides the prescribed power at its terminals regardless of the circuit connected.
(b) Provides the prescribed resistance at its terminals regardless of the circuit connected.
(c) Provides the prescribed voltage at its terminals regardless of the circuit connected.
(d) Provides the prescribed current at its terminals regardless of the circuit connected.
8. Given is the circuit below. The correct KVL is:
(a) −VA + VB − VC + VD + VE = 0
(b) −VA − VB − VC + VD + VE = 0
(c) −VA − VB − VC + VD − VE = 0
(d) −VA + VB − VC + VD − VE = 0
9. The current I is equal to
(a) 16A
(b) 12A
(c) 2A
2A
3A
(d) −2A
−7A
3
I
4A
10. A toaster draws 0.25A from a 120V line. The resistance of the toaster is:
(a) 0.96kΩ
(b) 1.8kΩ
(c) 0.48kΩ
(d) 3.6kΩ
11. We have two series resistors, R1 and R2 . The equivalent resistance is
(a) Req = R1 + R2
1
1
+
(b) Req =
R1
R2
R1 + R2
(c) Req =
R1 R2
R1 R2
(d) Req =
R1 + R2
12. Given is the following circuit. The voltage V2 can be described as:
R1
(a) V2 = Vs
R1
Vs
R1 + R2
R2
Vs
(c) V2 =
R1 + R2
R1 R2
(d) V2 =
Vs
R1 + R2
(b) V2 =
Vs
R2
V2
13. Given is the following circuit. The current I2 in terms of R1 , R2 , and Is is given by
R2
Is
R1 + R2
R1
Is
(b) I2 =
R1 + R2
R1 R2
(c) I2 =
Is
R1 + R2
1
(d) I2 =
Is
R1 + R2
(a) I2 =
Is
I2
R1
4
R2
14. Given is the following circuit. Find the equivalent resistance with respect to the ideal voltage source.
R4
(a) Req = (R1 + R2 ) || (R3 + R4 )
(b) Req = ((R1 + R2 ) ||R3 ) + R4
R1
(c) Req = R1 ||R2 ||R3 ||R4
R3
Vs
(d) Req = (R1 + R2 ) ||R3 ||R4
R2
15. For the given circuit, which equation describes V3 correctly?
(a) V3 = +IA (R1 + R2 )
IA
(b) V3 = −IA (R1 + R2 )
R1
(c) V3 = +Is R3
(d) V3 = −Is R3
+
Is
R3
R2
V3
−
16. The voltage drop across a 1.1kW hairdryer that draws 5A is
(a) 120V.
(b) 125V.
(c) 220V.
(d) 240V.
17. Resistors of the E24 series (5% tolerance) have the following values within the decade between 10
and 100
(a) 10, 11, 12, 13, 15, 16, 18, 20, 22, 24, 27, 30, 33, 36, 39, 43, 47, 51, 56, 62, 68, 75, 82, 91
(b) 10, 12, 15, 18, 22, 27, 33, 39, 47, 56, 68, 82
(c) 10, 15, 22, 33, 47, 68
(d) 10, 22, 47
18. The maximum current through a R = 6kΩ resistor is given by I = 5mA. Which minimum power
rating is needed for the resistor?
(a) 1W
(b) 0.5W
(c) 0.25W
(d) 0.125W
5
19. Given is the following circuit, which is a balanced Wheatstone bridge if:
α
R1
R2
(a)
=
R3
R4
R1
R4
R1
R2
(b)
=
R2
R3
R1
R2
β
δ
(c)
=
Vs
R4
R3
R1
R4
(d)
=
R3
R4
R3
R2
γ
20. A practical current source is modeled by the following circuit:
(a)
(b)
(c)
(d)
21. A practical voltmeter is described by
(a) an ideal voltmeter with an internal resistance in series.
(b) an ideal voltmeter with an internal resistance in parallel.
(c) an ideal voltmeter with an ideal ohmmeter in series.
(d) an ideal voltmeter with an ideal ohmmeter in parallel.
22. To measure the voltage across a circuit element
(a) we have to disconnect the element from the circuit and connect the voltmeter to its terminals.
(b) the voltmeter must be put in parallel to this element, while the element is still in the circuit.
(c) the voltmeter must be put in series to this element, while the element is still in the circuit.
(d) the voltmeter must be put in series to the circuit’s source which provides the voltage for the
circuit.
6
23. Given is the circuit below. The reference node has a known voltage of 0V. All elements in the
circuit are known. We apply the general version of node voltage analysis to the circuit. For which
nodes do we have to calculate unknown voltages?
1
(a) nodes 1, 2, 3
(b) nodes 1, 2
(c) nodes 2, 3
R2
(d) nodes 1, 3
R5
R4
2
R1
3
R3
Is
reference node
24. We want to apply General Node Voltage Analysis to the circuit below. The node voltages at nodes
a, b and c are to be determined, the reference node has a voltage of 0V. The KCL for node b is
correctly displayed as:
R2
R1
c
Va − Vb
Vb
Vc − Vb
(a)
+
−
=0
R5
R3
R6
IS1
R3
Vc
Vb
Va − Vb
+
−
=0
(b)
R4
R5
R4
R6
Va − Vb
Vc − Vb
Vb
(c)
+
−
=0
R5
R3
R6
a
Vc − Vb
Vb
Va − Vb − Vc
ref erence
b
+
−
=0
(d)
R5
R6
R5
R3
R6
R7
IS2
7
25. Given is one small part of a circuit as seen below. Node i and j are connected by the resistors R2 ,
R3 and R4 and there are no other connections between these two nodes. Using the dummy version
of the node voltage analysis, you find Gi,j to be
1
1
1
−
−
R2
R3
R4
1
1
=−
−
R2
R3
1
=−
R4
1
= − R2 R3
R2 +R3 + R4
(a) Gi,j = −
(b) Gi,j
(c) Gi,j
(d) Gi,j
R2
R4
i
j
R3
26. Using the General Version of Mesh Current Analysis, the KVL equation for mesh 2 in the following
circuit is given by:
R2
R1
(a) (R2 + R4 ) · Im2 + R6 · Im2 + R3 · Im2 = 0
(b) (R2 − R4 ) · Im2 + R6 · (Im2 + Im3 ) + R3 · (Im2 + Im1 ) = 0
(c) (R2 − R4 ) · Im2 + R6 · (Im2 + Im3 ) + R3 · (Im2 − Im1 ) = 0
(d) (R2 + R4 ) · Im2 + R6 · (Im2 − Im3 ) + R3 · (Im2 − Im1 ) = 0
R3
Im1
Im2
R5
R7
R6
Im3
VS
8
R4
27. Using the Dummy Version of Mesh Current Analysis, R1,2 in the following circuit is given by:
R2
R1
(a) R1,2 = −R3 − R2 − R4 − R6 − R5 − R1
(b) R1,2 = −R1 − R2
(c) R1,2 = −R3
R3
(d) R1,2 = −R3 − R5 − R6
Im1
Im2
R5
R4
R6
Im3
R7
VS
28. Given is the following circuit:
Vs
R1
R2
Is
R3
If we zero out the voltage source the following circuit displays the resulting circuit:
R1
Is
R1
R2
Is
Is
R3
R3
R2
(a)
Is
R1
R3
R2
(b)
(c)
9
(d)
29. To zero out an ideal current source, it must be replaced by
(a) a voltage source.
(b) a short circuit.
(c) an open circuit.
(d) a resistor.
30. You have a circuit with two sources. To find the total voltage across one specific resistor using the
principle of superposition you need to find the voltage across this resistor
(a) once.
(b) twice and add up the results.
(c) three times and add up the results.
(d) four times and add up the results.
31. A one–port network has
(a) four terminals.
(b) three terminals.
(c) two terminals.
(d) one terminal.
32. To compute the Norton Current
(a) we have to set all independent current sources to zero.
(b) we have to set all independent voltage sources to zero.
(c) we have to set all independent voltage and current sources to zero.
(d) we do not have to change anything about the sources.
Vs2
R1
33. What is the Thévenin Resistance of the circuit?
R1 R2
+ R3
R1 + R2
(b) RT = R1 + R2 + R3
(a) RT =
Vs1
R2
RLoad
(c) RT = R2 + R3
(d) RT = R2
R3
10
34. Integrating a sinusoidal signal over a full period yields
(a) 2π
(b) π
2
(c)
π
(d) 0
35. An AC voltage source vs (t) generates a rectangular voltage. What is the period of the signal?
vs (t)
V
(a) T = 5s
(b) T = 4s
+5.0
(c) T = 3s
+2.5
(d) T = 2s
1
2
3
4
5
−2.5
t
s
−5.0
36. An AC voltage source vs (t) generates the given voltage. The average value of the signal is
(a) 0V
vs (t)
V
(b) positive
+5.0
(c) negative
2.5
(d) √ V
2
+2.5
1
2
3
4
5
−2.5
−5.0
π
37. Given is the following complex number: z = 23ej 2 The conjugate complex is given by
π
(a) z ∗ = 23ej 2
π
(b) z ∗ = −23ej 2
π
(c) z ∗ = 23e−j 2
π
(d) z ∗ = −23e−j 2
11
t
s
38. Given is the following voltage:
π
vs (t) = 10V cos ωt − with f = 60Hz.
2
What is the RMS value of the signal?
(a) 0V
(b) 10V
(c) −10V
10
(d) √ V
2
39. Given is the complex number
z1 = −1 + j1
Which point in the complex plane describes z1 ?
j
j
j
j
1
1
1
1
−1
−1
1
−1
−1
1
−1
(a)
1
−1
(b)
(c)
−1
1
−1
(d)
3π
40. The peak–value phasor corresponding to the following signal i(t) = 15Acos ωt −
is given by
4
(a) I = 15ej
3π
4
(b) I = −15Aej
3π
4
(c) i(t) = 15Aej
3π
4
ejωt
−j 3π
4
(d) I = 15Ae
41. A capacitor is a device which stores
(a) potential energy.
(b) kinetic energy.
(c) electric energy.
(d) magnetic energy.
12
42. An inductor consists of
(a) a coil usually wound around a core material.
(b) two plates (usually of identical shape) facing each other at a distance and carrying equal but
opposite charges.
(c) a cable enclosed by some magnetic material.
(d) a plate (usually of square shape) connected to two terminals carrying equal but opposite
charges.
43. An ideal capacitor in a DC circuit can be replaced by
(a) a resistor.
(b) a short circuit.
(c) an open circuit.
(d) a voltage source.
44. We have two parallel inductors L1 and L2 . The equivalent inductance is given by
1
1
+
L1
L2
= L1 + L2
L1 + L2
=
L1 L2
L1 L2
=
L1 + L2
(a) Leq =
(b) Leq
(c) Leq
(d) Leq
45. Given is a long cylindrical coil. It has an inductance L1 . For a second inductor L2 we use a core
material, which has only half the cross sectional area of L1 . All other parameters stay the same.
The new inductance L2 is given by
(a) L2 = 4L1
(b) L2 = 2L1
L1
(c) L2 =
2
L1
(d) L2 =
4
13
46. The unit of relative permittivity is is
(a) [ε] = 1
Vs
(b) [ε] =
Am
As
(c) [ε] =
Vm
Vm
(d) [ε] =
As
47. The impedance of an inductor is given by
(a) ZL = L
1
(b) ZL =
L
(c) ZL = jωL
1
(d) ZL =
jωL
48. The admittance of a resistor is given by
(a) YR = R
1
(b) YR =
R
(c) YR = jωR
1
(d) YR =
jωR
49. Given is one small part of a circuit as seen below. Node i and j are connected by the resistor R2 and
the capacitor C4 and there are no other connections between those two nodes. Using the dummy
version of the node voltage analysis, you find Yi,j to be
1
R2 + jωC4
1
= − R2 +
jωC4
1
=−
+ jωC4
R2
1
=−
1
R2 + jωC
4
(a) Yi,j = −
(b) Yi,j
(c) Yi,j
(d) Yi,j
C4
i
R2
j
14
50. The impedance of a resistor and an inductor in series is given by
(a) ZRLseries = R + L
(b) ZRLseries = jωR + jωL
(c) ZRLseries = R + jωL
(d) ZRLseries = jωRL
51. The admittance of an inductor in parallel with a capacitor is given by
1
+ jωC
jωL
1
1
= +
L C
1
= +C
L
1
= jωL +
jωC
(a) YL||C =
(b) YL||C
(c) YL||C
(d) YL||C
52. A quantity that contains all the power information in a given load is the
(a) apparent power.
(b) complex power.
(c) average power.
(d) reactive power.
53. What is the apparent power for the following complex power:
S = 225VA∠π/2
(a) The apparent power is 225VA.
(b) The apparent power is 0VA.
(c) The apparent power is 225VAR.
(d) The apparent power is 0W.
54. What is the average power for the following complex power:
S = 225VA∠π/2
(a) The average power is 225VA.
(b) The average power is 0VA.
(c) The average power is 225VAR.
(d) The average power is 0W.
15
55. The two input terminals of an operational amplifier are labeled as:
(a)
(b)
(c)
(d)
high and low.
inverting and noninverting.
top and bottom.
differential and non–differential.
56. For an ideal operational amplifier as given in the figure usually the following assumption is made:
i1
(a) i1 6= 0 and i2 = 0
(b) i1 = 0 and i2 = 0
(c) i1 6= 0 and i2 6= 0
i2
(d) i1 = 0 and i2 6= 0
57. In a circuit with an operational amplifier the feedback loop usually connects the output terminal
with the inverting terminal. This is called
(a)
(b)
(c)
(d)
negative feedback.
inverting feedback.
connecting feedback.
absolute feedback.
58. In the given circuit, the phase of V out will be shifted with respect to V S by
RF
(a) 0◦
(b) 90◦
CS RS
(c) −90◦
(d) 180◦
+
VS
V out
−
RF
59. The following circuit is
(a)
(b)
(c)
(d)
an active low–pass filter
an active high–pass filter
a passive low–pass filter
a passive high–pass filter
CS
RS
+
VS
V out
−
16
60. A band pass filter is designed to
(a) stop all frequencies within a band of frequencies, ω1 < ω < ω2 .
(b) only pass frequencies from DC (0Hz) up to the cutoff frequency ωc .
(c) pass all frequencies above the cutoff frequency ωc .
(d) pass all frequencies within a band of frequencies, ω1 < ω < ω2 .
61. Convert the decimal number 17 into a binary number. The result is
(a) 10001
(b) 11011
(c) 11010
(d) 10011
62. Add the following two binary numbers:
10101
The result is
01110
(a) 111011
(b) 100011
(c) 101111
(d) 11111
63. Subtract the following two binary numbers:
1011
The result is
0101
(a) 010
(b) 100
(c) 110
(d) 111
64. The twos complement of an n-bit binary number is obtained by
(a) subtracting the number itself from n2 − 1
(b) subtracting the number itself from n2
(c) subtracting the number itself from 2n − 1
(d) subtracting the number itself from 2n
17
65. An ideal diode in reverse bias can be replaced by
(a) a short circuit.
(b) an open circuit.
(c) a 0.7V battery.
(d) a resistor.
66. An offset model diode in forward bias can be replaced by
(a) a short circuit.
(b) an open circuit.
(c) a 0.7V battery.
(d) a resistor.
67. A BJT has a reverse biased BE junction and a reverse biased BC junction. The transistor works
in the
(a) Cutoff Region
(b) Active Region
(c) Saturation Region
(d) Breakdown Region
68. An npn BJT (arrow points toward emitter) has vEB = 0.7V.
(a) The BE junction is forward biased.
(b) The BC junction is forward biased.
(c) The BE junction is reversed biased.
(d) The BC junction is reverse biased.
18
69. Which Karnaugh map is equivalent to the following truth table?
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
F
0
1
0
0
0
0
1
1
C
0
1
0
1
0
1
0
1
A
C
1
0
1
1
A
1
0
1
C
1
0
1
0
B
1
(b)
A
C
0
1
1
0
B
(a)
0
0
1
0
1
1
A
0
C
0
B
0
0
1
1
0
1
B
(c)
(d)
19
0
0