A. Current and voltage,…power,… current and voltage b. Current or voltage,…power,… current or voltage
C. power or voltage,… current,… power or voltage
D. power or current,…voltage,… power or current
E. power and current,… energy,…voltage
Solution:
Proportionality applies when the input and output are current or voltage. Output power is not linearly related to the input current or voltage.
Answer: B.
49. Practically transients occur during period if time, which is equal to
A. TC B. 2TC C. 3TC D.4TC E.5TC
Answer: E.
50. The unit of current is
A. V B. J C. A D. F E. C
Solution:
Current is defined as dq/dt and is a measure of the flow of electrical charge. The unit of current is Ampere (A). 1 A = 1 Coulomb/second.
Answer: C.
51. Three resistors R1, R2, R3 are in series with a constant voltage V. The voltage across R1is 20V, the power in R2 is 25W, and R3=2 Ω. Find the total voltage V if the current is 5A.
A. 25V B. 30V C. 35V D.40V E.45V
Solution:
The voltage across R2 is V2=P/I=25/5=5V.
The voltage across R3 is V3=IR=5*2=10V.
The total voltage is V=V1+V2+V3=20+5+10=35V.
Answer: C.
52. The net positive charge flowing through a device varies as q(t)=2t2C. Find the current through the device at t=1s.
A. 2 A B. 3A. C. 4A D. 5A E. 6A
Solution:
i(t)= dq/dt =d(2t2)/dt =4t;
i(1)= 4A.
Answer: C.
53. The circuit consists of a 10 Ω resistor in series with a parallel combination of two resistors with value of 15 Ω and 5 Ω. Current through 5 Ω resistor equals 6A. What total power absorbs by the circuit?
A.500W B.880W C. 560W D.740W E. 960W
Solution:
C
urrent
in lower branch is three times less than in upper one,
it equals 6/3=2A. P=I2R
Pupper=I2R=36*5=180W
Plower= I2R=4*15=60W
Pmid=(6+2)2*10=640W
Ptotal=180+60+640=880W
Answer: B.
54. The unit of conductance is
A.C B.A C.J D. Ω E.S
Solution:
Conductance G=1/R, has the unit Siemens, S.
Answer: E.
55. The circuit, which is called a Wheatstone bridge, consists of ______ resistors and a source.
A. 2 B.3 C. 4 D.5 E.6
Solution:
Answer: D.
56. The first-order step response for RC circuit can be represented as ____ for t≥0
A. vC(t)=[vC(0) – vC(∞)]e-t/Tc B. vC(t)=[vC(∞) – vC(0)]e-t/Tc+ vC(∞)
C. vC(t)=[vC(∞) – vC(0)]e-t/Tc D. vC(t)=[vC(∞) – vC(0)]e-t/Tc- vC(∞)
E. vC(t)=[vC(0) – vC(∞)]e-t/Tc + vC(∞)
Solution:
1)
2)
Answer: E.
57. The natural response has its origin in _____ of the circuit and ______ the form of the input.
A. the physical characteristic …. doesn’t depend on
B. characteristic … doesn’t depend on
C. the physical characteristic …depends on
D. characteristic … depends on
E. set of elements … depends on
Solution:
The natural response has its origin in the physical characteristic of the circuit and doesn’t depend on the form of the input.
Answer: A.
58. The first-order step response for RC circuit can be represented as ____ for t≥0
A. vC(t)=[vC(0) – vC(∞)]e-t/Tc B. vC(t)=[vC(∞) – vC(0)]e-t/Tc+ vC(∞)
C. vC(t)=[vC(∞) – vC(0)]e-t/Tc D. vC(t)=[vC(∞) – vC(0)]e-t/Tc- vC(∞)
E. vC(t)=[vC(0) – vC(∞)]e-t/Tc + vC(∞)
Solution:
1)
2)
Answer: E.
59. The real part of impedance is called ______ and imaginary part is called _____. The real part of admittance is called ______ and imaginary part is called _____.
A. resistivity,…conductivity,… susceptance,… conductance
B. reactance,…resistance,… susceptance,… conductance
C. resistance,…reactance,… susceptance,.. conductance
D. reactance,…resistance,…conductance,…susceptance
E. resistance,…reactance,…conductance,…susceptance
Solution:
Z = R ± jX
Y = G ± jB
The real part of impedance is called resistance and imaginary part is called reactance. The real part of admittance is called conductance and imaginary part is called susceptance.
Answer: E.
60. The general view of exponential function is
A. v(t)=[VAe-t] B. v(t)=[VAe-t] u(t) C. v(t)=[VAe-t/Tc] D. v(t)=[VAe-t/Tc]u(t)
E. v(t)=[VAet/Tc]u(t)
Answer: D.
61. Two circuits are said to be equivalent if they have
A. the same view after reduction process
B. identical i-v-characteristics after reduction process
C. identical i-v-characteristics at a specified pair of terminals
D. identical i-v-characteristics
E. identical parameters of elements
Answer: C.
62. To apply superposition principle we must ______ in turn all sources, except one. When the current source is _________, we replace it by _________.
A. “turn on”, turned off, short circuit. B. “turn off”, turned on, open circuit.
C. “turn off”, turned on, short circuit. D. “turn off”, turned off, open circuit.
E. “turn off”, turned off, short circuit.
Solution:
To apply superposition principle we must “turn off” all sources in turn. When the current source is turned off, we replace it be open circuit.
Answer: D.
63. To apply superposition principle we must ______ in turn all sources, except one. When the voltage source is _________, we replace it by _________.
A. “turn on”, turned off, short circuit B. “turn off”, turned on, open circuit
C. “turn off”, turned on, short circuit D. “turn off”, turned off, open circuit
E. “turn off”, turned off, short circuit
Solution:
To apply superposition principle we must “turn off” all sources in turn. When the voltage source is turned off, we replace it by short circuit.
Answer: E.
64. Voltage v is defined as
A. dw/dt B. dw/dq C.dq/dt D. dq/dv E. dp/dt
Solution:
Voltage is a measure of the energy required to move a small charge from one point to another and is defined as dw/dq.
Answer: B.
65. Voltage across the capacitor for RC circuit is equal to vC(t)=30e-200t-20 V t≥0. What is the initial capacitor’s voltage at t=0?
A. 10 V B. 20V C. –10 V D. –20 V E. 30 V
Solution:
vC(t)=[vC(0) – vC(∞)]e-t/Tc+ vC(∞)
vC(∞)= –20V
vC(0) – vC(∞)=30
vC(0) – (–20)= 30
vC(0)=30 – 20=10V
Answer: A.
66. Voltage across the capacitor for RC circuit is equal to vC(t)=20e-200t-10 V t≥0. What is the amplitude of the input step function?
A. 10 V B. 20V C. –10 V D. –20 V E. 30 V
Solution:
vC(t)=[vC(0) – vC(∞)]e-t/Tc+ vC(∞)
vA = vC(∞)= –10V
Answer: C
67. Voltage across the capacitor for RC circuit is equal to vC(t)=40e-200t-30 V t≥0. What is the initial capacitor’s voltage at t=0?
A. 10 V B. 20V C. –10 V D. –20 V E. 30 V
Solution:
vC(t)=[vC(0) – vC(∞)]e-t/Tc + vC(∞)
vC(∞)= –30 V
vC(0) – vC(∞) = 40 V
vC(0) – (–30) = 40
vC(0)= 40 – 30 = 10 V
Answer: A.
68. Unit step function is defined as
A.u(t)=1 for t>0 B. u(t)=0 for t<0 C. u(t)=1 for t<0 D. u(t)=0 for t>0
E. u(t)=1 for t>0 and u(t)=0 for t<0
Solution:
Unit step function is defined as
69. What statement is correct?
A. Electric energy is measured by watts.
B. An electrical circuit is an interconnection of electrical devices.
C. When current and voltage have the same algebraic signs, the device is delivering power.
D. Voltage is a measure of the rate at which energy is being transferred.
E. Unit of charge is ampere.
Solution:
A – wrong. Electric energy is measured in Joules (J); watt (W) is a unit of power.
B – correct.
C – wrong. Under passive sign convention, the current reference arrow is directed toward the terminal with the positive voltage reference mark. Under this convention, the device power is positive when it absorbs power and is negative when it delivers power.
When current and voltage have the same algebraic signs, p will be positive, so the device will absorb power.
D – wrong. Voltage is a measure of the energy required to move a small charge from one point to another. Power is a measure of the rate at which energy is being transferred.
E – wrong. Unit of charge is Coulomb (C). Unit of current is Ampere (A).