Elecs Compilation 1
September 18, 2022 | Author: Anonymous | Category: N/A
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TOPIC: AC/DC CIRCUITS CIRCUITS 1.
Multiplying equation (2) by 3 and equation (3) by 4 and add, then we get, 0 = -90 + 460 Therefore = 5.111 Substituting in equation 3, we have = 0.0115 A = 11.5 mA
Suppose you double the voltage in a simple dc circuit, and cut the resistance in half. The current will become: A. B. C. D.
I
I I
Four times as great. Twice as great. The same as it was before. Half as great.
I
I
I
REFERENCE: REFERENCE: ELECTRICAL AND ELECTRONIC TECHNOLOGY 10TH EDITION BY HUGHES
ANSWER: A Solution:
Therefore,
3.
V=IR 2V=I I = =
Using Nodal analysis, calculate the voltages V1 and V2 in the circuit of Figure below
REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
2.
A. 3/10 V B. 10/3 V
C. 3 D. 10VV
The resistances of the various arms of a bridge are given in Figure below. The battery has an e.m.f. of 2.0 V and a negligible internal resistance. Determine the value and direction of the current in BD, using Kirchhoff’s laws. laws.
ANSWER: B Solution: Solution: At node 1
−=1⇨V = 1 ⇨ eq.1 At node 2
− =
⇨ V = 0 ⇨ eq.2
From equation (2), by multiplying each term by 21, A. B. C. D.
11.5 mA 15.1 mA 51.1 µA 11.5 µA
7
Therefore,
ANSWER: A Solution: Current in BC= Current in DC= Applying Kirchhoff’s second law to the mesh formed by ABC and the battery, we have 2=10 +30 =40 eq.1 Similarly for mesh ABDA, 0 = 10 + 40 - 20 eq. 2 and for mesh BDCB
I I I I
I
I I I I 30I ⇨ I I ⇨
0 = 40 + 15 = -30 + 15 + 85
I I I I I I I I−⇨ 30 eq.3
Rogelyn L. Barbosa| BS ECE
V - V (7 + 3) = 0 7
V = 10V ⇨ V = V
From equation (1), by multiplying each term by 15,
8V 5V = 15 ⇨ 4.5 V = 15 = REFERENCE: ELECTRICAL AND ELECTRONIC REFERENCE: TECHNOLOGY 10TH EDITION BY HUGHES
4.
B.
a defined rise and an instantaneous decay C. a defined rise and a defined decay, and the two are equal D. an instantaneous rise and an instantaneous decay
Which of the following can vary with ac, but not with dc?
A. B. C. D.
Power. Voltage. Frequency. Magnitude.
ANSWER: C EXPLANATION: Sawtooth EXPLANATION: Sawtooth waves can have rise and delay slopes in an infinite number of different combinations. One common example is shown in the figure. In this case, the rise and decay are both finite and equal. This is known as triangular wave.
ANSWER: C EXPLANATION:: In ac, the polarity reverses EXPLANATION reverses at regular intervals. The instantaneous amplitude (that is, the amplitude at any given instant in time) of ac usually varies because of the repeated reversal of polarity. But there are certain cases where the amplitude remains constant, even though the polarity keeps reversing. The rate of change of polarity is the variable that makes ac so much different from the dc. The behavior of an ac wave depends largely on this rate: the frequency.
REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS BY STAN GIBILISCO 4 TH ED. 5.
The period of an ac wave is: A. The same as the frequency. B. Not related to the frequency. C. Equal to 1 divided by the frequency. D. Equal to the amplitude divided by the frequency.
7.
What is the power factor (PF) of a purely resistive ac circuit? circuit? A. B. C. D.
ANSWER: C ANSWER: EXPLANATION:: In a periodic ac wave, the EXPLANATION function of instantaneous amplitude versus time repeats itself over and over, so that the same pattern recurs indefinitely. The length of time between one repetitions of the pattern, or one cycle, and the next is called the period of the wave. This is illustrated below:
0 1 0.707 without values, determined
it
cannot
be
ANSWER: B EXPLANATION: In ac circuits with reactance, EXPLANATION: the real power P in watts is equals ,or VI cos θ, where θ is the phase angle. The real power is the power dissipated as heat in resistance. Cos θ is the power factor factor of the circuit. Multiplying VI by the cosine of the phase angle provides the resistive component for real power equal to . In a purely resistive circuit, all circuit power is dissipated by the resistor(s). Voltage and current are in phase with each other. Therefore the phase angle is zero.
I R
I R
P.F= cos θ
; P.F= cos (0) =
1 REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
6.
REFERENCE: GROB’S BASIC ELECTRONICS 11TH REFERENCE: EDITION 8.
A triangular wave exhibits: A. an instantaneous rise and a defined decay
Rogelyn L. Barbosa| BS ECE
A 10 10Ω Ω R is in parallel with a 15- Ω XL . The applied voltage Is 120 Vac. How much is the apparent power in the circuit? circuit?
A. B. C. D.
ANSWER: D. Solution: Given R= 10 Ω ; VA
Solution: Given:
2.4 kW. 1.44 kVA. 1.44 kW. 1.73 kVA
X =15Ω ; V= 120
==12Ω I = = 8Ω I = II I = 8 8 12
= 14 14.4.422 A P=VI= 12014.42 = . . . REFERENCE : GROB’S BASIC ELECTRONICS 11TH REFERENCE: EDITION 9.
A 15- Ω resistance is in series with 50 Ω of X L and 30 Ω of X C . If the applied voltage equals 50 V, how much real power is dissipated by the circuit? A. B. C. D.
60 W. 100 W. 100 VA. 4.16 W.
ANSWER: A Solution:
Z = R R X X = 15 15 5030 = 25Ω I = VZ = 5025ΩV = 2 A
I = 5A; V = 12 1200 V; θ = 53.13° P=VIcosθ= 120V 120V5A 5A cos cos53.13° 3° = . . REFERENCE: GROB’S BASIC ELECTRONICS 11TH REFERENCE: EDITION 11. In an ac circuit with only parallel inductors, A. lags by 90° B. lags by 90° C. and are in phase D. None of the above
I V V I V I
ANSWER: A EXPLANATION: Inductors do EXPLANATION: do not behave the same as resistors. Whereas resistors simply oppose the flow of electrons through them (by dropping a voltage directly proportional to the current), inductors oppose changes in current through them, by dropping a voltage directly proportional to the rate of change of current. In accordance with Lenz’s Law, Law, this induced voltage is always of such a polarity as to try to maintain current at its present value. That is, if current is increasing in magnitude, the induced voltage will “push against” the electron flow; if current is decreasing, the polarity will reverse and “push with” the electron flow to oppose the decrease. This opposition to current change is called reactance, rather than resistance. resistance.
− − − =.° =arctan = arctan− θ=arctan
P = VI cocosθ = 50 V2A cos53.13° =. REFERENCE: GROB’S BASIC ELECTRONICS 11TH REFERENCE: EDITION 10. A parallel ac circuit with 120 Vac applied has a total current,IT ,of 5 A. If the phase angle of the circuit is -53.13° how much real power is dissipated by the circuit? A. 600 VA. B. 480 W. C. 360 W. D. 3.6 kVA ANSWER: C
Rogelyn L. Barbosa| BS ECE
Remember, the voltage dropped across an inductor is a reaction against the change in current through it. Therefore, the instantaneous whenever the instantaneous voltage current isiszero at a peak (zero
change, or level slope, on the current sine wave), and the instantaneous voltage is at a peak wherever the instantaneous current is at maximum change (the points of steepest slope on the current wave, where it crosses the zero line). This results in a voltage wave that is 90o out of phase with the current wave. Looking at the graph, the voltage wave seems to have a “head start” on the current wave; the voltage “leads” the current, and the current “lags” behind the voltage. voltage.
I = 6 × I = 6× 0.1 = . REFERENCE: GROB’S BASIC ELECTRONICS 11TH REFERENCE: EDITION 14. Determine the primary impedance Z P for the transformer circuit in Figure below. below.
REFERENCE: GROB’S BASIC ELECTRONICS 11TH REFERENCE: EDITION 12. TOPIC: AC/DC CIRCUITS Determine the input impedance to the series network of Fig. 15.23. Draw the impedance diagram.
A. B. C. D.
182 Ω Ω 812 Ω Ω 128 Ω Ω 218Ω 218Ω
ANSWER: C Solution:
A. B. C. D.
6.325 18.43° 7.325 18.43° 8.325 18.43° 9.325 18.43°
ANSWER: A Solution:
Z = Z Z Z = R0°X90°X 90° =
RjX jX
= R j X X = 6 Ωj Ωj10Ω12Ω 10Ω12Ω = 6Ωj2Ω Z =. .°
REFERENCE : BOYLESTAD INTRODUCTORY REFERENCE: CIRCUIT ANALYSIS 10TH EDITION 13. A transformer with a 1:6 turns ratio has 720 V across 7200 Ω in the secondary. Calculate the value of A. 0.5 A B. 0.6 A C. 0.7 A D. 0.8 A
I .
ANSWER: B Solution:
= 0.1 A I = = Ω
Rogelyn L. Barbosa| BS ECE
Z = NN × R = × 8Ω ×8Ω =16 = 128 Ω
REFERENCE: GROB’S BASIC ELECTRONICS 11TH REFERENCE: EDITION 15. Two series coils, each with an L of 250 µH, have a total inductance of 550 µH connected series-aiding and 450 µH series-opposing. How much is the mutual inductance LM between the two coils? A. 10 µH B. 15 µH C. 20 µH D. 25 µH ANSWER: D Solution:
L = − = − = = 25 µH
REFERENCE: GROB’S BASIC ELECTRONICS 11TH REFERENCE: EDITION 16. A current of 1.2 A flows in a coil with an inductance of 0.4 H. How much energy is stored in the magnetic field? field? A. B. C. D.
0.882 J 0.828 J 0.288 J 0.228 J
ANSWER: C Solution:
Energy= = .×. = 0.288 J
Solution: Note that 6 s is twice the RC time of 3 s. Then t/RC =2.
0.868
REFERENCE: GROB’S BASIC ELECTRONICS 11TH REFERENCE: EDITION 17. What is the total Z of a 600-Ω 600-ΩR in parallel with a 300-Ω 300-ΩXL? Assume 600 V for the applied voltage. A. 286 Ω Ω B. 268 Ω Ω C. 862 Ω Ω D. 826 Ω Ω ANSWER: B
A I = Ω = 1 I = Ω = 2 A I = I I = 1 2 = √ 5 I =2.2244 A Z = = =Ω
Solution:
.
REFERENCE: GROB’S BASIC ELECTRONICS 11TH REFERENCE: EDITION 18. A 200-µH coil has a Q of 40 at 0.5 MHz. Find Re. Re.
A. B. C. D.
V =antilog log400.434×2 log400.434×2 =antilog 1.602 = antilog (0.734) = 5.42 V
REFERENCE: GROB’S BASIC ELECTRONICS 11TH REFERENCE: EDITION 20. An RC circuit has an R of 10 kΩ and a C of 0.05 µF. The applied voltage for charging is 36 V. How long will it take C to charge to 24 V? V? A. 0.459 ms B. 0.495 ms C. 0.549 ms D. 0.459 ms ANSWER: C Solution: RC= 10 kΩ x 0.05 µF= 0.5 ms The rises to 24 V while drops from 36 to 12 V. Then
v
log 0.477
v
t = 2.3×0.5×10 RClog RClog − × =2.3
=2.3 ×0.5×10− × = 0.549 x 10− or 0.549 ms
REFERENCE: GROB’S BASIC ELECTRONICS 11TH REFERENCE: EDITION
15.7 Ω Ω 17.5 Ω Ω 57.2 Ω Ω 51.7 Ω Ω
TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
ANSWER: B
π Solution: R = = == π×. ××× = 15.7 Ω
REFERENCE: GROB’S BASIC ELECTRONICS 11TH REFERENCE: EDITION 19. An RC circuit has a time constant of 3 s. The capacitor is charged to 40 V. Then C is discharged. After 6 s of discharge, how much is VR? VR? A. 5.42 V B. 5.24 V C. 4.25 V D. 4.52 V ANSWER: A
Rogelyn L. Barbosa| BS ECE
21. The term “semiconductor” arises from:
A. Resistor-like properties of metal oxides. B. Variable conductive properties of some materials. C. The fact that there’s nothing better to call silicon. D. Insulating properties of silicon and GaAs. ANSWER: B EXPLANATION:: EXPLANATION Various elements, compounds, and mixtures can function as semiconductors. The two most common materials are silicon and a compound of gallium arsenic known as gallium arsenide (often abbreviated as GaAs). In the early years of semiconductor technology, germanium formed the basis for many
semiconductors; today it is seen occasionally but not often. Other substances that work as semiconductors are selenium, cadmium compounds, indium compounds, and the oxides of certain metals. REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: TEACH AND ELECTRONICS BY STAN GIBILISCO 4TH ED. 22. The purpose of doping is to:
A. Make the charge carriers move faster. B. Cause holes to flow. C. Give semiconductor material specific properties. D. Protect devices from damage in case of transients. ANSWER: C EXPLANATION: For a semiconductor EXPLANATION: material to have the properties necessary in order to function as electronic components, impurities are usually added. The impurities cause the material to conduct currents in certain ways. The addition of an impurity to a semiconductor is called doping. Sometimes the impurity is called a dopant. REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
23. A semiconductor material is made into N type by: A. B. C. D.
Adding an acceptor impurity. Adding a donor impurity. Injecting electrons. Taking electrons away
ANSWER: B EXPLANATION: When an impurity contains an EXPLANATION: When excess of electrons, the dopant is called a donor impurity. Adding such a substance causes conduction mainly by means of electron flow, as in an ordinary metal such as copper or aluminum. The excess electrons are passed from atom to atom when a voltage exists across the material. Elements that serve as donor impurities include antimony, arsenic, bismuth, and phosphorus. A material with a donor impurity is called an N-type semiconductor, because electrons have negative (N) charge.
Rogelyn L. Barbosa| BS ECE
REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
24. Which of the following does not result from adding an acceptor impurity? impurity? A. The material becomes P type. B. Current flows mainly in the form of holes. C. Most of the carriers have positive electric charge. D. The substance acquires an electron surplus. ANSWER: D EXPLANATION: If an impurity has a deficiency EXPLANATION: If of electrons, the dopant is called an acceptor impurity. When a substance such as aluminum, boron, gallium, or indium is added to a semiconductor, the material conducts by means of hole flow. A hole is a missing electron— electron —or more precisely, a place in an atom where an electron should be, but isn’t. A semiconductor with an acceptor impurity is called a P-type semiconductor, because holes have, in effect, a positive (P) charge. REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS BY STAN GIBILISCO 4 TH ED. 25. When a P-N junction does not conduct even though a voltage is applied, the junction is A. Reverse biased at a voltage less than the avalanche voltage B. overdriven C. Biased past the breaker voltage. D. In a state of avalanche effect. ANSWER: A EXPLANATION: When the battery or dc EXPLANATION: power-supply polarity is switched so the Ntype material is positive with respect to the P type, the situation is called reverse bias. Electrons in the N-type material are pulled toward the positive charge pole, away from the P-N junction. In the P-type material, holes are pulled toward the negative charge pole, also away from the P-N junction. The electrons are the majority carriers in the Ntype material, and the holes are the majority carriers in the P-type material. The charge therefore becomes depleted in the vicinity of the P-N junction, and on both sides of it. This zone, where majority carriers are deficient, is
called the depletion region. A shortage of majority carriers in any semiconductor substance means that the substance cannot conduct well. Thus, the depletion region acts like an electrical insulator. This is why a semiconductor diode will not normally conduct when it is reverse-biased. A diode is, in effect, a one-way current gate— gate —usually! REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
Other junctions are slower. The main limiting factor is the capacitance at the P-N junction during conditions of reverse bias. As the junction capacitance of a diode increases, maximum frequency at which it can alternate between the conducting state and the nonconducting state decreases. The junction capacitance of a diode depends on several factors, including the operating voltage, the type of semiconductor material, and the cross-sectional area of the P-N junction.
26. If the reverse bias exceeds the avalanche voltage in a P-N junction:
A. The junction will be destroyed. B. The junction will insulate; no current will flow. C. The junction will conduct current. D. The capacitance will become extremely high. ANSWER: C EXPLANATION: Sometimes, a diode conducts EXPLANATION: when it is reverse-biased. The greater the reverse-bias voltage, the more like an electrical insulator a P-N junction gets— gets—up to a point. But if the reverse bias rises past a specific critical value, the voltage overcomes the ability of the junction to prevent the flow of current, and the junction conducts as if it were forward-biased. This phenomenon is called the avalanche effect because conduction occurs in a sudden and massive way, something like a snow avalanche on a mountainside. REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
27. When a P-N junction is reverse-biased, the capacitance depends on all of the following except: except: A. The frequency. B. The width of the depletion region. C. The cross-sectional area of the junction. D. The type of semiconductor material. ANSWER: A EXPLANATION: Some P-N junctions can EXPLANATION: alternate between conduction (in forward bias) and nonconduction (in reverse bias) millions or billions of times per second.
Rogelyn L. Barbosa| BS ECE
If you examine Figure above, you might get the idea that the depletion region, sandwiched between two semiconducting sections, can play a role similar to that of the dielectric in a capacitor. This is true! In fact, a reverse-biased P-N junction actually is a capacitor. Some semiconductor components, called varactor diodes, are manufactured with this property specifically in mind. The junction capacitance of a diode can be varied by changing the reverse-bias voltage, because this voltage affects the width of the depletion region. The greater the reverse voltage, the wider the depletion region gets, and the smaller the capacitance becomes. REFERENCE : TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
28. Holes flow the opposite way from electrons because:
A. Charge carriers flow continuously. B. Charge carriers are passed from atom to atom. C. They have the same polarity. D. No! Holes flow in the same direction as electrons. ANSWER: B EXPLANATION: EXPLANATION: Charge carriers in semiconductor materials are either electrons, each of which has a unit negative charge, or holes, each of which has a unit positive charge. In any semiconductor substance, some of the current takes the form of electrons passed from atom to atom in a negative-to-positive direction,
and some of the current occurs as holes that move from atom to atom in a positiveto-negative direction.
Sometimes electrons account for most of the current in a semiconductor. This is the case if the material has donor impurities, that is, if it is of the N type. In other cases, holes account for most of the current. This happens when the material has acceptor impurities, and is thus of the P type. The dominating charge carriers (either electrons or holes) are called the majority carriers. The less abundant ones are called the minority carriers. The ratio of majority to minority carriers can vary, depending on the way in which the semiconductor material has been manufactured. Figure above is a simplified illustration of electron flow versus hole flow in a sample of N-type semiconductor material, where the majority carriers are electrons and the minority carriers are holes. The solid black dots represent electrons. Imagine them moving from right to left in this illustration as they are passed from atom to atom. Small open circles represent holes. Imagine them moving from left to right in the illustration. In this particular example, the positive battery or power-supply powersupply terminal (or “source of holes”) would be out of the picture toward the left, and the negative battery or power-supply terminal (or “source of electrons”) would be out of the picture toward the right. REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS BY STAN GIBILISCO 4 TH ED. 29. TOPIC: SEMICONDUCTORS DIODES AND TRANSISTORS
INCLUDING
Avalanche voltage is routinely exceeded when a P-N junction acts as a: A. B. C. D.
voltage drops back below the critical value, the junction behaves normally again.Some components are designed to take advantage of the avalanche effect. In other cases, the avalanche effect limits the performance of a circuit. In a device designed for voltage regulation, called a Zener diode, diode, you’ll hear about the avalanche voltage or Zener voltage specification. This can range from a couple of volts to well over 100 V. Zener diodes are often used in voltage-regulating circuits. REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS BY STAN GIBILISCO 4 TH ED. 30. When a P-N junction is forward biased, conduction will not occur unless A. the applied voltage exceeds the forward breakover voltage B. the applied voltage is less than the forward breakover voltage C. the junction capacitance is high enough
D.
the depletion region is wide enough
ANSWER: A EXPLANATION: It EXPLANATION: It takes a specific, well-defined minimum applied voltage for conduction to occur through a semiconductor diode. This is called the forward breakover voltage. Depending on the type of material, the forward breakover voltage varies from about 0.3 V to 1 V. If the voltage across the junction is not at least as great as the forward breakover voltage, the diode will not conduct, even when it is connected as shown in Figure below. This effect, known as the forward breakover effect or the P-N junction threshold effect can be of use in circuits designed to limit thethatpositive and/or voltages signals can attain.negative The effectpeak can also be used in a device called a threshold detector, in which a signal must be stronger than certain amplitude in order to pass through.
Current rectifier. Variable resistor. Variable capacitor. Voltage regulator.
ANSWER: D
REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
EXPLANATION: The avalanche effect does not EXPLANATION: damage a P-N junction (unless the voltage is extreme). It’s a temporary thing. When the
Rogelyn L. Barbosa| BS ECE
31. A diode is normally operated in
A. B. C. D.
reverse breakdown the forward-bias region the reverse-bias region either B or C
ANSWER: D EXPLANATION: Generally EXPLANATION: Generally the term bias refers to the use of a dc voltage to establish certain operating conditions for an electronic device. In relation to a diode, there are two bias conditions; forward and reverse. Either of these bias conditions is established by connecting a sufficient dc voltage of the proper polarity across the pn junction. REFERENCE: REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
32. Ideally, a diode can be represented by a
forward-biased, it is equivalent to a closed switch in series with a small equivalent voltage source ( ) equal to the barrier potential (0.7V) with the positive side toward the anode. This equivalent voltage source represents the barrier potential that must be exceeded by the bias voltage before the diode will conduct and is not an active source of voltage. When conducting, a voltage drop of 0.7V appears across the diode.
V
REFERENCE: REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED. 34. A certain power-supply filter produces an output with a ripple of 100 mV peak-to-peak and a dc value of 20 V. The ripple factor is A. B. C. D.
0.05 0.005 0.00005 0.02
A. Voltage source
B. switch resistance C. D. all of these ANSWER: C EXPLANATION: The ideal model of a EXPLANATION: diode is the least accurate approximation and can be represented by a simple switch. When the diode is forwardbiased, it ideally acts like a closed (on) switch. When the diode is reverse-biased, it ideally acts like an open (off) switch. Although the barrier potential, the forward dynamic resistance, and the reverse current are neglected, this model is adequate for most troubleshooting when you are trying to determine if the diode is working properly. REFERENCE: REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED. 33. In the practical diode model, A. The barrier potential is taken into account B. The forward dynamic resistance is taken into account C. None of these D. Both A and B ANSWER: A EXPLANATION EXPLANATION: : The practical includes the barrier potential. Whenmodel the diode is
Rogelyn L. Barbosa| BS ECE
ANSWER B. Solution:
0.005 r = = = 0.005
REFERENCE: REFERENCE: ELECTRONIC DEVICES CONVENTIONAL CURRENT VERSION BY THOMAS L. FLOYD 8th ED. 35. A 60 V peak full-wave rectified voltage is applied to a capacitor-input filter. If , the ripple voltage is
120 Hz, Hz,RR = 10 kΩ,and Ω,and C = 10 10µF µF A. B. C. D.
ff =
0.6 V 6 mV 5.0 V 2.88 V
ANSWER: C Solution:
V =60 V ⇨
the unfiltered peak full-wave rectified voltage The frequency of a full-wave rectified voltage is 120 Hz. The approximate peak-to-peak ripple voltage at the output is
V− ≅ fR1C V 1 = 120 1201010 kΩ kΩ10µF 10µF 60 =
REFERENCE: REFERENCE: ELECTRONIC DEVICES CONVENTIONAL CURRENT VERSION BY THOMAS L. FLOYD 8 TH ED. 36. TOPIC: SEMICONDUCTORS DIODES AND TRANSISTORS
INCLUDING
A certain full-wave has a peak output voltage of 30 V. A to50 the µF rectifier. capacitor-input filterthe is connected Calculate peak-to-peak ripple and the dc output voltage developed across a 600Ω 600Ω load resistance. A. B. C. D.
V =8.33 V; V = 25 25.8.8 V V =9.33 V; V = 20 20.0.0 V V =9.80V; V = 30 30.0.0 V V =10.0 V; V = 28 28.0.0 V
ANSWER: A Solution: the unfiltered peak full-wave rectified voltage. The frequency of a full-wave rectified voltage is 120 Hz. The
V = 30 V ⇨
D. Vp= 18.4 V ; PIV= 19.1 V ANSWER: B Solution: The peak output voltage (taking into account the two diode drops) is
V =1.414V = 1.41412 V ≅ 17 V V = V 1diode .4 V =is 17 V 1.4 The V = PIV. rating for each PIV=V 0.7 V = 15 15..6 V 0.7 VV== .. REFERENCE: REFERENCE: ELECTRONIC DEVICES CONVENTIONAL CURRENT VERSION BY THOMAS L. FLOYD 8 TH ED. 38. Determine the peak value of the output voltage in the figure below if the turns ratio is 0.5
approximate ripple voltage at the output peak-to-peak is
V− ≅ fR1C V 1 = 120 120600Ω 600Ω50µF 50µF 30 = .
The approximate dc value of the output voltage is determined as follows:
V =1 2fR1C V 1 =1 240Hz 240Hz600Ω 600Ω50µF 50µF 30V = . . REFERENCE: REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. th FLOYD 8 ED. 37. Determine the peak output voltage for the bridge rectifier in the figure below. Assuming practical model, what PIV rating is required for the diodes? The transformer is specified to have a 12 V rms secondary voltage for the standard 120 V across the primary.
A. Vp= 17 V ; PIV= 17.7 V B. C.
Vp= 15.6 V ; PIV= 16.3 V Vp= 12 V ; PIV= 12.7 V
Rogelyn L. Barbosa| BS ECE
A. B. C. D.
85 V 85.7 V 84.3 V 170 V
ANSWER: C Solution: The peak secondary voltage is
V 1700 V = V = 17 V = n V =0.5170 170 = 85 V =0.5 The rectified peak output voltage is V = V 0.7 V = 85 V 0.7 V = . REFERENCE: REFERENCE: ELECTRONIC DEVICES CONVENTIONAL CURRENT VERSION BY THOMAS L. FLOYD 8 TH ED. 39. What is the average value of half-wave rectified voltage in the figure below?
A. 7.96 V B. 50 V C. 0 V
D.
15.9 V
ANSWER: D Solution:
VG = π = π = .
REFERENCE: REFERENCE: ELECTRONIC DEVICES CONVENTIONAL CURRENT VERSION BY THOMAS L. FLOYD 8 TH ED.
40. In many cases, a PNP transistor can be replaced with an NPN device and the circuit will do the same thing, provided that A. the power supply or battery polarity is reversed B. the collector and the emitter is interchanged C. the arrow is pointing inward D. Forget it! A PNP transistor can never be replaced with an NPN transistor ANSWER: A EXPLANATION: It’s easy to tell whether a EXPLANATION: bipolar transistor in a diagram is NPN or PNP. If the device is NPN, the arrow at the emitter points outward. If the device is PNP, the arrow at the emitter points inward. Generally, PNP and NPN transistors can perform the same functions. The differences are the polarities of the voltages and the directions of the resulting currents. In most applications, an NPN device can be replaced with a PNP device or vice versa, the power-supply polarity can be reversed, and the circuit will work in the same way— way—as long as the new device has the appropriate specifications. REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS BY STAN GIBILISCO 4 TH ED. 41. TOPIC: SEM SEM For operation as an amplifier, the base of an npn transistor must be A. B. C. D.
Positive with respect to the emitter Negative with respect to the emitter Positive with respect to the collector 0V
in Figure 4 – –4. 4. These free electrons easily diffuse through the forward based BE junction into the lightly doped and very thin p-type base region, as indicated by the wide arrow. The base has a low density of holes, which are the majority carriers, as represented by the white circles. A small percentage of the total number of free electrons injected into the base regionelectrons recombine with holes and region move as valence through the base and into the emitter region as hole current, indicated by the red arrows.
When the electrons that have recombined with holes as valence electrons leave the crystalline structure of the base, they become free electrons in the metallic base lead and produce the external base current. Most of the free electrons that have entered the base do not recombine with holes because the base is very thin. As the free electrons move toward the reverse-biased BC junction, they are swept across into the collector region by the attraction of the positive collector supply voltage. The free electrons move through the collector region, into the external circuit, and then return into the emitter region along with the base current, as indicated. The emitter current is slightly greater than the collector current because of the small base current that splits off from the total current injected into the base region from the emitter. REFERENCE: REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED. 42. TOPIC: SEMICONDUCTORS DIODES AND TRANSISTORS TRANSISTORS
INCLUDING
When operated in cutoff and saturation, the transistor acts like a A. B. C. D.
Linear amplifier Switch Variable capacitor Variable resistor
ANSWER: B
ANSWER: A The heavily doped n-type
EXPLANATION: A transistor can be operated EXPLANATION:
emitter region has a very high density of conduction-band (free) electrons, as indicated
as an electronic switch in cutoff and saturation. In cutoff, both pn junctions are
EXPLANATION: EXPLANATION:
Rogelyn L. Barbosa| BS ECE
reverse-biased and there is no collector current. The transistor ideally behaves likes an open switch between collector and emitter. In saturation, both pn junctions are forwardbiased and the collector current is maximum. The transistor behaves like an open switch between collector and emitter.
REFERENCE: REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. th FLOYD 8 ED. 43. TOPIC: SEMICONDUCTORS DIODES AND TRANSISTORS TRANSISTORS
INCLUDING
A JFET always operates with A. The gate-to-source pn junction reversed biased B. The gate-to-source pn junction forward- biased C. The drain connected to ground D. The gate connected to the source ANSWER: A EXPLANATION: To illustrate the operation of a EXPLANATION: To JFET, Figure 8 – –2 2 shows dc bias voltages applied to an n-channel device. VDD provides a drain-to-source voltage and supplies current from
A. 6.72 V B. 2.72 V C. 10. 72 V D. 4 V ANSWER: C Solution: Since , . The minimum value of for the JFET to be in its constant-current region is
VGS = 4 V V = 4 V VS VS = V = 4 V In the constant-current region with VGS = 0 V I = ISS = 12 mA V = IR = 12 mA560Ω = 6.7722 V V = VS V = 4 V 6.72 V = .
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED. 45. Determine the drain-to-source voltage in the circuit of Figure below. The MOSFET datasheet gives
VGS = 8 V anandd ISS =
12 mA
drain to source. VGG sets the reverse-bias voltage between the gate and the source, as shown. The JFET is always operated with the gate-source pn junction reverse-biased. Reversebiasing of the gate-source junction with a negative gate voltage produces a depletion region along the pn junction, which extends into the n channel and thus increases its resistance by restricting the channel width. REFERENCE : REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
VGS =
44. For the JFET in Figure below, Determine the minimum value of VDD required to put the device in the constant-current region of operation when VGS = 0 V.
4 V and ISS = 12 mA, mA,
A. B. C. D.
18 V 10.6 V 0V None of the above
ANSWER: B Solution:
V = V ISSR = 18 V 12 mA mA620Ω 620ΩS= . or 10.6 V REFERENCE : REFERENCE: ELECTRONIC DEVICES CONVENTIONAL CURRENT VERSION BY THOMAS L. FLOYD 8 TH ED. 46. A certain transistor is to be operated with . If its maximum power rating is 250 mW, what is the most collector current that it can handle?
V = 6 V
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A. B. C. D.
V ≅ R RR V = 15.5.66kΩkΩ 10 V = 3.5599 V V = V V = 3.5599 V 0.77V= V = 2.89 8 9 V = 5.1166 mA ; I = = .Ω I ≅ I =
41.7 mA 41.7 nA 41.7 µA 0.47 A
ANSWER: A Solution:
I = P V = 2506 mW V = .. REFERENCE: ELECTRONIC REFERENCE: DEVICES CONVENTIONAL CURRENT VERSION BY THOMAS L. FLOYD 8 TH ED. 47. Determine the voltage gain and the ac output voltage in the given figure if
r′ =
50Ω
.
V = V IR = 10 V 5.16 mA 1.0 kΩ = 4.8844 V V = V V = 4.84 V 2.89 8 9 V = . REFERENCE: ELECTRONIC DEVICES REFERENCE: CONVENTIONAL CURRENT VERSION BY THOMAS L. FLOYD 8 TH ED. ED.
49. Determine the dc input resistance looking in at the base of the transistor in the figure below. And
β =125
A. B. C. D.
V = 4 V
A = 20 ; V = 2 V rms rms A = 20 ; V = 4 V rms rms A = 20 ; V = 5 V rms rms A = 20 ; V = 10 V rms rms
ANSWER: A Solution:
Ω = A ≅ = .Ω V = AVb = 2020100mV 100mV = REFERENCE: ELECTRONIC REFERENCE: DEVICES CONVENTIONAL CURRENT VERSION BY THOMAS L. FLOYD 8 TH ED.
V I β =100
48. Determine the and in the stiff voltagedivider biased transistor circuit of the figure shown if
A. B. C. D.
132 kΩ kΩ 142 kΩ kΩ 152 kΩ kΩ 162 kΩ kΩ
ANSWER: C Solution:
I = −. −. = .= βΩI V= 3.=31254V Ω Ω RNS 3.mmA3AmA = REFERENCE: ELECTRONIC REFERENCE: DEVICES CONVENTIONAL CURRENT VERSION BY THOMAS L. FLOYD 8 TH ED.
A. B. C. D.
I = 6.4411 mA ; V = 3.5599 V I = 5.1166 mA ; V = 1.9955 V I = 5.1166 mA; V =2.89 V I = 6.4411 mA; mA; V =4.84 V
ANSWER: B Solution:
Rogelyn L. Barbosa| BS ECE
50. Find VDS and VGS in Figure below. For the particular JFET in this circuit, the parameter values such as gm, VGS(off), and IDSS are such that a drain current (ID) of approximately 5 mA is produced. Another JFET, even of the same type, may not produce the same results when connected in this circuit due to the variations in parameter values.
B.
cannot be controlled by gate current. C. increases as the gate current increases. D. none of the above ANSWER: A EXPLANATION: A. B. C. D.
VVSS == 10 13 13.9.9V V; V;GSVGS==1.11.1 1V.1 V VS = 9 V ; VGS = 1.1 V VS = 8.9 V ; VGS = 1.1 1.1 V
ANSWER: D Solution:
VS = IRS = 5 mA220Ω = 1.1 V V = V IR = 15 V 5 mA 1.0 kΩ = 15 V 5 V = 10 V VS = V VS = 10 V 1.1 V = . V, Since V = 0 V, VGS = VG VS =G0 V 1.1 V = . REFERENCE: REFERENCE: ELECTRONIC DEVICES CONVENTIONAL CURRENT VERSION BY THOMAS L. FLOYD 8 TH ED.
RS
51. Determine the value of required to selfbias a p-channel JFET with datasheet values of 25 mA and 15 V. is to be 5 V.
ISS = VGS A. B. C. D.
VGSO =
250Ω 250Ω 350Ω 350Ω 450Ω 450Ω 550Ω 550Ω
ANSWER: C Solution:
=
I ≅ ISS 1 25 mA 11.1.1 mA mA mA 11 = 11 RS = | | = . =Ω REFERENCE: REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8 th ED. TOPIC: INDUSTRIAL AND POWER ELECTRONICS ELECTRONICS 52. The forward breakover voltage of an SCR A. decreases increases.
as
Rogelyn L. Barbosa| BS ECE
the
gate
The Figure above shows how the level of the gate current, IG , can control the forward breakover voltage, VBRF . The maximum forward breakover voltage, VBRF, occurs when the gate current, I G , equals zero. When the gate-cathode junction is forward biased, the SCR will fi re at a lower anode-cathode voltage. Notice in the figure, that as the gate current, IG , is increased, the value of V BRF is decreased. As the value of gate current, IG , is increased, the SCR functions much like an ordinary rectifier diode. An important characteristic of an SCR is that once it is turned on by gate current, the gate loses all control. The only way to turn off the SCR is to reduce the anode current below the level of holding current, IH . Not even a negative gate voltage will turn the SCR off in this case. In most cases, the anode supply voltage is an alternating voltage. This means that the SCR will automatically turn off when the anode voltage drops to zero or goes negative. Of course, when the anode voltage is negative, the SCR is reverse-biased. The process of turning off an SCR is called commutation. REFERENCE : GROB’S BASIC ELECTRONICS 11 TH REFERENCE: EDITION 53. An RC phase-shift network is used in SCR and triac circuits to
current A. Control the conduction angle of the thyristor
B. Handle some of the load current C. Vary the holding current D. None of the above ANSWER: A EXPLANATION: A. 20.1 µA B. 19.1 mA C. 20.1mA D. 19.1 µA SCRs are frequently used to control the amount of power that is delivered to a load. The figure above shows how the conduction angle of an SCR can be controlled over the range of 0°to 180° by using an RC phase-shifting network. Recall from basic ac circuit theory that the capacitor and resistor voltage in a series RC circuit are always 90° out of phase. In figure above, the voltage across the capacitor is applied to the anode side of the diode, D1. The cathode lead of the diode connects to the gate of the SCR. Again, the purpose of using the diode is to ensure that the negative alternation of the input voltage cannot apply excessive reverse-bias voltage to the SCR’s gate cathode junction. When R is increased to nearly its maximum value, the phase angle θ between V in and the capacitor voltage, V C , is approximately 90°. This means it will take longer for the voltage across C to reach the voltage required to f re the SCR. Since the RC network provides a “delay,” the SCR can be triggered in the 90°to 180° portion of the input cycle, resulting in smoother control of the load current. REFERENCE: GROB’S BASIC ELECTRONICS REFERENCE: ELECTRONICS 11TH EDITION 54. Determine the value of anode current in Figure below when the device is on. 10 V. Assume the forward voltage drop is 0.9 V.
ANSWER: B Solution:
V =0.9 V ⇨ voltage at the anode VS = VSV V 19. = 210VV 0.9 VV== 19.1 V S .. I = RS = 1.0 kΩ = REFERENCE: ELECTRONIC REFERENCE: DEVICES CONVENTIONAL CURRENT VERSION BY THOMAS L. FLOYD 8 TH ED. 55. Determine the gate trigger current and the anode current when the switch, SW1, is momentarily closed in Figure below. Assume
VK = 0.2 VV,V,VGK = 0.7 V and I = 5 mA
A. B. C. D.
IG = 41 4100 mA mA ; I = 44 4488 µµAA IG = 45 4500 µµAA ; I= 44 4488 mA mA IG = 40 4000 mA mA ; I = 44 4488 µµAA
IG = 41 4100 µµAA ; I = 44 4488 mA mA 0.7 V = µµ IG = VGRG VGK = 3 V 5. 6 kΩ I = − = −. Ω = ANSWER: D Solution:
REFERENCE: ELECTRONIC REFERENCE: DEVICES CONVENTIONAL CURRENT VERSION BY THOMAS L. FLOYD 8 TH ED. 56. A half-wave rectifier circuit employing an SCR is adjusted to have a gate current of 1mA. The forward breakdown voltage of SCR is 100 V for Ig = 1mA. If a sinusoidal voltage of 200 V peak is applied, find the firing angle.
Rogelyn L. Barbosa| BS ECE
A. B. C. D.
I = I = 1 A; α =30° π = 3.36 A I = +
30° 45° 60° 90°
ANSWER: A Solution: v = Vm sin θ θ Here, v = 100 V, Vm = 200 V
REFERENCE: PRINCIPLES OF ELECTRONICS REFERENCE: BY V.K MEHTA
= 200 sin=θ θ30° θ =100 sin−1 (0.5) 30° Firing angle, α = θ = 30°
∴
REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
59. In figure below, the switch is closed. A diac with a breakover voltage is connected in the circuit. If the triac has a trigger voltage of 1 V and a trigger current of 10 mA, what is the capacitor voltage that triggers the triac?
VO = 30 V
OF
57. An a.c. voltage v = 240 sin314 t is applied to an SCR half-wave rectifier. If the SCR has a forward breakdown voltage of 180 V, find the time during which SCR remains off. A. B. C. D.
2.4 millisecond 2.5 millisecond 2.6 millisecond 2.7 millisecond
ANSWER: D Solution: v = Vm sin 314 t Here v = 180 V; Vm = 240 V 180 = 240 sin (314 t) 314 t = sin−1 (0.75) = 48.6° = 0.848 radian t = 0.848 0.0027 sec 314 = 2.7 millisecond
∴
∴
REFERENCE: PRINCIPLES OF ELECTRONICS REFERENCE: BY V.K MEHTA 58. In an SCR half-wave rectifier circuit, what peak-load current will occur if we measure an average (d.c.) load current of 1A at a firing angle of 30° ? A. B. C. D.
3.26 A 3.36 A 3.46 A 3.56 A
ANSWER: B
I be the peak load I = 1cos α = 1cos α
Solution: Let current
π
Q I =
(
π
π I = + α
Rogelyn L. Barbosa| BS ECE
A. B. C. D.
13 V 21 V 31 V 41 V
ANSWER: C Solution: When switch is closed, the capacitor starts charging and voltage at point A increases. When voltage at point A becomes equal to of diac plus gate triggering voltage of the triac, the triac is fired into conduction. Therefore,
VO
V
VG V = VO VG = 30 V 1V =
REFERENCE: PRINCIPLES OF ELECTRONICS BY REFERENCE: V.K MEHTA 60. A unijunction transistor has 10 V between the bases. If the intrinsic standoff ratio is 0.65, What will be the peak voltage if the forward voltage drop in the pn junction is 0.7 V? A. 5.2 V B. 6.2 V C. 7.2 V D. 8.2 V ANSWER: C Solution:
V = 10 V; Ƞ = 0.65; V = 0.7 V = ȠV = 0.65 × 10 = 6.5 V V = ȠV V = 6.5 0.7 = .
Standoff voltage Peak-point voltage,
REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
OF
TOPIC: MICROELECTRONICS INCLUDING OP AMPS AMPS
which is its slew rate. Figure 33 – 33 –9 9 illustrates this concept. Here the op amp’s output output waveform should be an amplified version of the sinusoidal input, V id . In this case, waveform A would be the expected output. However, if the slope of the output sine wave exceeds the S R
61. The input stage of every op amp is a A. Differential amplifier B. Push-pull amplifier C. Common-base amplifier D. None of the above ANSWER: A EXPLANATION: EXPLANATION: A typical op-amp is made up of three types of amplifier circuits: a differential amplifier, a voltage amplifier, and a push-pull amplifier. The differential amplifier is the input stage for the op-amp. It provides amplification of the difference voltage between the two points. The second stage is usually a class A amplifier that provides additional gain. Some op-amps may have more than one voltage amplifier stage. A push-pull class B amplifier is typically used for the output stage
REFERENCE: ELECTRONIC REFERENCE: DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
rating of triangular. the op amp, the waveform appears Therefore, slewrate distortion of a sine wave produces a triangular wave, such as waveform B. REFERENCE: REFERENCE: ELECTRONIC DEVICES CONVENTIONAL CURRENT VERSION BY THOMAS L. FLOYD 8 TH ED. 63. The input impedance of an inverting amplifier is approximately equal to
R zero
A. B. C. Infinity
D.
R ANSWER: A EXPLANATION:
62. The slew-rate specification of an op amp is the A. Maximum value of positive or negative output voltage B. Maximum rate at which its output voltage can change C. Attenuation against a commonmode signal D. Frequency where the voltage gain is is one or unity ANSWER: B EXPLANATION:
Another very important op-amp specification is its slew rate, usually designated S R . The slew-rate specification of an op amp tells how fast the output voltage can change in volts per microsecond, or V/µs. For a 741 op amp, the S R is 0.5 V/µs. This means that no matter how fast the input voltage to a 741 op amp changes, the output voltage can change only as fast as 0.5 V/µs,
Rogelyn L. Barbosa| BS ECE
The circuit is called an inverting amplifier because the input and output signals are 180_ out of phase. The 180° inversion occurs because V in phase is applied to the inverting ( –) –) input terminal of the op amp. Resistors R F and R i provide the negative feedback, which in turn controls the circuit’s overall voltage gain. The output signal is fed back to the inverting input through resistors R F and R i . The voltage between the inverting input and ground is the differential input voltage, designated V id . The exact value of V id is determined by the values A VOL and V out . Even with negative feedback, the output voltage of an op amp can be found from Vout = AVOL X Vid
For all practical purposes, V id is so small that it can be considered zero in most cases. This introduces little or no error in circuit analysis. Because V id is so small (practically zero), the inverting input terminal of the op amp is said to be at virtual ground. This means that the voltage at op potential amp’s inverting inputyet is at thethe same as ground, it can sink no current. Because the inverting input of the op amp is at virtual ground, the voltage source,V in , sees an input impedance equal to R i . Therefore, Zin Ri. The inverting input of the op amp has extremely high input impedance, but its value is not the input impedance of the circuit.
≅
A. 1.59 KHz B. 1.95 KHz C. 5.91 KHz D. 5.19 KHz ANSWER: D Solution:
f = R 1.42R4 C 2.7 kΩ =8.4kΩ R 2R = 3 kΩ 22. f = . ..× = . Ω
REFERENCE: REFERENCE: GROB’S TH ELECTRONICS 11 EDITION
BASIC
REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
64. In LM317 voltage regulator shown in the figure below, is adjusted to 2.4 kΩ. If the value of is 240 Ω, determine the regulated d.c. output voltage for the circuit.
R R
OF
66. A differential amplifier has an open-circuit voltage gain of 100. This amplifier has a common input signal of 3.2 V to both terminals. This results in an output signal of 26 mV. Determine the common mode voltage gain and CMRR in dB. A.
A =0.0081 and CMRR = 80.8 A =0.0081 and CMRR =81.8 dB A =0.0081 and CMRR =82.8 dB A =0.0081 and CMRR = 83.8 dB
B. A. B. C. D.
C.
17.53 V 15.37 V 13.57 V 13.75 V
D.
dB
ANSWER: Solution: B
ANSWER: D Solution:
V =1.25RR 1 =1.25 . ΩΩ 1 = .
REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
OF
65. Determine the frequency of the circuit shown in the figure below. Given that Ω Ω µ
3 k ; R = 2.7 k and and C= C = 0.003333 F
Rogelyn L. Barbosa| BS ECE
R =
v 2 V ; v = 26 mV =3. A = = =. common . mode voltage gain 100 =20log CMRR = 20 lo AA 0.0081 =. REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
OF
67. For the circuit shown below, find the common mode voltage gain and the CMRR in dB
. S f = π () = π = π ⇨ (0.5 V/µs= 500 kHz) = 9.95 kHz
A = 0.5 ; CMRR CMRR = 45 45.0.099 dB A46.09=dB0. 25 25 ; CM CMRR RR = A = 0.5 ; CMRR CMRR = 48 48.0.099 dB A = 0.25 25 ; CM CMRR RR = 47.09 dB
A. B.
C. D.
tail
current
Id.c emitter = I=current I⁄2 =in56.each 5 µAtransistor ⁄2 = 28 28.2.255 µA ⇨ r′ = = . =884.96 ⇨ a.c emitter resistance
Ω =56.5 ⇨ Differential = = . voltage gain CMMR=20log A
5 =20log 56. = .. 0.25
REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.KMEHTA
OF
68. Determine the maximum operating frequency for the circuit shown. The slew rate is 0.5 V/µs
A. B. C. D.
8.95 kHz 9.95 kHz 7.95 kHz 6.95 kHz
ANSWER: B Solution: the maximum peak output voltage is approximately 8 V.
V
Rogelyn L. Barbosa| BS ECE
OF
69. For the noninverting amplifier circuit shown , find peak-to-peak output voltage
A. B. C. D.
ANSWER: D Solution:
Ω =. A = = Ω Common mode voltage gain = − = −.Ω =56.5µA ⇨
REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
6V 12 V 18 V 24 V
ANSWER: B Solution: the input signal is 2 V peakto-peak
A = 1 =1 ΩΩ = 6 ⇨ Voltage gain Axv = 6 x 2 = peak-to-peak output voltage
REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
OF
70. The most popular form of IC package is A. B. C. D.
DIL TO-5 Flatpack None of the above
ANSWER: B EXPLANATION: In order EXPLANATION: order to protect ICs from an external environment and to provide mechanical protection, various forms of encapsulation are used for integrated circuits.
The figure above shows TO-5 package which resembles a small signal transistor in both appearance and but differs in thatThe it has 8, 10 orsize 12 pigtail-type leads. close
leads spacing and the difficulty of removal from a printed circuit board has diminished the popularity of this package with the users. REFERENCE : REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
OF
TOPIC: INSTRUMENTATION AND MEASUREMENTS MEASUREMENTS 71. An ammeter is connected in __________ with the circuit element whose current we wish to measure. A. B. C. D.
Series Parallel Series-parallel None of the above
ANSWER: A EXPLANATION: Ammeter measures EXPLANATION: value of current flowing in circuit, so current should flow inside ammeter to give proper result. And it has very low resistance to ensure the correct measurement of current in the circuit. If it is connected in parallel across any load then all current in circuit will choose lower resistive path (i.e ammeter) to cause its circuit to be damaged. Hence it is used in series. REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
OF
72. A Voltmeter is connected in ____with the circuit component across which potential difference is to be measured. A. B. C. D.
Series Parallel Series-parallel None of the above
ANSWER: B EXPLANATION: Voltmeter has very high EXPLANATION: resistance to ensure that it's connection do not alter flow of current in the circuit. Now if it is connected in series then no current will be there in the circuit due to its high resistance. Hence it is connected in parallel to the load across which potential difference is to be measured. REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
Rogelyn L. Barbosa| BS ECE
OF
73. An ideal ammeter has _____resistance. A. B. C. D.
Low Infinite Zero High
ANSWER: C EXPLANATION: Just like voltmeters, EXPLANATION: ammeters tend to influence the amount of current in the circuits they’re connected to. However, unlike the ideal voltmeter, the ideal ammeter has zero internal resistance, so as to drop as little voltage as possible as electrons flow through it. Note that this ideal resistance value is exactly opposite as that of a voltmeter. With voltmeters, we want as little current to be drawn as possible from the circuit under test. With ammeters, we want as little voltage to be dropped as possible while conducting current REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
OF
74. When an ammeter is inserted in the circuit, the circuit current will A. B. C. D.
Increase Decrease Remains the same None of the above
ANSWER: B EXPLANATION: When an ammeter is placed in series with a circuit, it ideally drops no voltage as current goes through it. In other words, it acts very much like a piece of wire, with very little resistance from one test probe to the other. Consequently, an ammeter will act as a short circuit if placed in parallel (across the terminals of) a substantial source of voltage. If this is done, a surge in current will result, potentially damaging the meter: REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA 75. The resistance of an ideal voltmeter is A. Low B. C.
Infinite Zero
OF
rectifier circuit are of silicon. Calculate the value of multiplier resistor required.
D. High
RS
ANSWER: B EXPLANATION: Since voltmeters are EXPLANATION: always connected in parallel with the component or components under test, any current through the voltmeter will contribute to the overall current in the tested circuit, potentially affecting the voltage being measured. A perfect voltmeter has infinite resistance, so that it draws no current from the circuit under test REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
OF
76. In the circuit shown, it is desired to measure the voltage across a 10 kΩ resistance. If a multimeter of sensitivity 4 kΩ/volt and range 0-10 V is used for the purpose, what will be the reading?
A. B. C. D.
890.7 kΩ kΩ 790.7 kΩ kΩ 690.7 k Ω Ω 590.7 k Ω Ω
ANSWER: A Solution: 100 µA
⇨ average current I.. =100 µA ⇨ F.S.D current of meter
R = RS R = RS 1000Ω ⇨ total circuit resistance
V = √ 2 V.. = √ 2 x 100 100 V= V = 141.4 141.4 V ⇨ peak value of applied voltage Total rectifier drop = 2V = 2 x 0.7V = 1.4 V = . −. . + µ
RS = . . Ω REFERENCE: PRINCIPLES OF ELECTRONICS REFERENCE: PRINCIPLES BY V.K MEHTA
A. B. C. D.
78. An a.c. voltmeter uses a bridge rectifier with silicon diodes and a PMMC instrument with f.s.d. current of 75 µA. if meter coil resistance is 900 Ω and the multiplier resistor is 708 kΩ, calculate the applied r.m.s. voltage when the meter reads f.s.d.
6.88 V 7.88 V 8.88 V 9.88 V
A. B. C. D.
ANSWER: C Solution: Resistance of meter = 4 k Ω x 10= 40 k Ω Total circuit resistance = 40 k Ω||10 k Ω+10 k Ω Ω
ANSWER: C Solution:
1 + 100 = 8 10 = 18 kΩ 1 1 mA mA Circuit current= Ω = 1.11
Voltage read by multimeter= 8 k Ω x1.11 8.88 V mA=8.88 mA= OF
77. A PMMC instrument with a full-scale deflection (f.s.d) current of 100 µA and Ω is to be used as a voltmeter of range 0100 V (r.m.s). The diodes used in the bridge
R =
1k
Rogelyn L. Barbosa| BS ECE
µ . Now peak f.s.d current − = µ = √ ..−. Or V... = + Peak f.s.d. meter current=
=
REFERENCE : REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
40 V 50 V 60 V 70 V
REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
OF
79. A multimeter has full scale deflection current of 1 mA. Determine its sensitivity.
A. B. C. D.
10 Ω/V Ω/V 100 Ω/V Ω/V 1000 Ω/V Ω/V 10, 000 Ω/V Ω/V
Negative-AND gate is not the same as a NAND gate. Its truth table, actually, is identical to a NOR gate:
ANSWER: A Solution: Full scale deflection current,
I = 1 mA Multimeter sensitivity = = = 1000 Ω/V Ω/V
REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
OF
80. The deflection sensitivity of a CRT is 0.01 mm/V. Find the shift produced in the spot when 400 V are applied to the vertical plates. A. B. C. D.
1 mm 2 mm 3 mm 4 mm
ANSWER: D Solution: Spot shift= deflection applied voltage = 0.01 x 400= 4 mm
sensitivity
REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
x
82. Suppose an AND gate is followed by an inverter. Under what conditions is the output of the resulting black box low? A. If and only if both inputs are high B. If and only if both inputs are low C. If and only if one input is high and the other is low D. Under no conditions (the output is always high) ANSWER: A
OF
TOPIC: DIGITAL ELECTRONICS ELECTRONICS 81. TOPIC: DIGITAL ELECTRONICS ELECTRONICS Suppose inverters are placed in series with both inputs of an AND gate. Under what conditions is the output of the resulting black box high? A. If and only if both inverter inputs are high B. If and only if both inverter inputs are low C. If and only if one inverter input is high and the other is low D. Under no conditions (the output is always low) ANSWER: B EXPLANATION:: A Negative-AND gate EXPLANATION functions the same as an AND gate with all its inputs inverted (connected through NOT gates). In keeping with standard gate symbol convention, these inverted inputs are signified by bubbles. Contrary to most peoples’ first instinct, the logical behavior of a
Rogelyn L. Barbosa| BS ECE
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4TH ED.
EXPLANATION: A variation on the idea of EXPLANATION: the AND gate is called the NAND gate. The word “NAND” is a verbal contraction of the words NOT and AND. Essentially, a NAND gate behaves the same as an AND gate with a NOT (inverter) gate connected to the output terminal. To symbolize this output signal inversion, the NAND gate symbol has a bubble on the output line. The truth table for a NAND gate is as one might expect, exactly opposite as that of an AND gate:
As with AND gates, NAND gates are made with more than two inputs. In such cases, the same general principle applies: the output will be “low” (0) if and only if all inputs are “high” (1). If any input is “low” (0), the output will go “high” (1). (1).
REFERENCE : REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4TH ED. 83. In Boolean algebra, addition represents A. The logical NOT Operation B. The logical AND operation C. The logical OR operation D. The logical NAND operation ANSWER: C EXPLANATION: Boolean algebra is a EXPLANATION: system of mathematical logic using the numbers 0 and 1 with the operations (addition), and AND (multiplication), OR (addition), NOT (negation). Combinations of these operations are NAND (NOT AND) and NOR (NOT OR). This peculiar form of mathematical logic, which gets its name from the nineteenth-century British mathematician George Boole, is used in the design of digital logic circuits.
D. (-X)+(-Y) is equivalent to – to –(X*Y) (X*Y) ANSWER: D EXPLANATION:: Statements on either EXPLANATION side of the equals sign in each case are logically equivalent. When two statements are logically equivalent, it means that one is true if and only if (iff ) the other is true. For example, the statement X = Y means that X implies Y, and also that Y implies X. Logical equivalence is sometimes symbolized by a double arrow with one or two shafts (↔or). Boolean theorems are used to analyze and simplify complicated logic functions. This makes it possible to build a circuit to perform a specific digital function, using the smallest possible number of logic switches.
REFERENCE: REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4TH ED. 84. If the output to a logical inverter is low, it means that A. B. C. D.
Both of the two inputs are high Both of the two inputs are low The single input is high The single output is low
REFERENCE: REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4TH ED. 86. Convert the decimal number 37 to its equivalent binary number.
ANSWER: C
A.
EXPLANATION: An inverter or NOT gate EXPLANATION: An has one input and one output. It
C.
B.
reverses the state of the input.
REFERENCE: REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4TH ED. 85. DeMorgan’s Theorem states that, for all logical statements X and Y, A. –(X*Y) –(X*Y) is equivalent to X+Y
B. C.
X*Y is equivalent to –(X+Y) to – (X+Y) (-X)+(-Y) is equivalent to X*Y
Rogelyn L. Barbosa| BS ECE
D.
101001 100101 100010 101010
ANSWER: B Solution: Division 37/2=18 18/2=9 9/2=4 4/2=2 2/2=1 1/2=0 Therefore,
Remainder 1 0 1 0 0 1
37 =
REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
OF
87. Convert the decimal number 76 to octal equivalent.
Y= AB + AC + B + BC Y= AB + AC + B (1+C)
114 141 411 414
A. B. C. D.
Y= AB + AC + B * 1 Y= AB + AC + B Y= B + (A+1) + AC
ANSWER: Solution: A Division 76/8=9 9/8=1 1/8=0
Y= B * 1 + AC
Remainder 4 1 1
Y= B + AC REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
76 =
Therefore,
REFERENCE: REFERENCE: PRINCIPLES ELECTRONICS BY V.K MEHTA
OF
OF
90. Simplify the given Boolean expression: Y= (A + B + C) * (A + B)
88. Convert decimal number hexadecimal number
A. B. C. D.
423
to
1C7 1B7 1D7 1A7
D. Y= ABC ANSWER: B Solution: Y= (A + B + C) * (A + B)
ANSWER: D D Solution: Division 423/16=26 26/16=1 1/16=0 Therefore,
A. Y= A + B B. Y= AB C. Y= A + B + C
= A*A + A * B + B * A + B * B + C * A + C * B Y = A + AB + AB + B + AC + BC
Remainder 7 10 1
= A + AB + B + AC + BC BC = A+ B + AC + BC
423 =
= A (1 + C) + B (1 + C)
REFERENCE: PRINCIPLES OF ELECTRONICS BY REFERENCE: V.K MEHTA 89. Using Boolean techniques, following expression Y= AB + A (B+C) +B (B+C) A. B. C. D.
= A * 1+ B * 1 Y=A+B
simplify the
Y= A+BC Y=C+AB Y=B+AC Y= A+B+C
REFERENCE: PRINCIPLES OF ELECTRONICS BY REFERENCE: V.K MEHTA
TOPIC: OTHERS OTHERS 91. In the case of line regulation,
ANSWER: C Solution:
Y= AB + A (B+C) +B (B+C) Y= AB + AB + AC + BB + BC Y= AB + AB + AC + B + BC
Rogelyn L. Barbosa| BS ECE
A. When the temperature varies, the output voltage stays constant B. When the output voltage changes, the load current stays constant C. When the input voltage changes, the output voltage stays constant
D. When the load changes, the output voltage stays constant ANSWER: C EXPLANATION: Two basic categories of EXPLANATION: voltage regulation are line regulation and load regulation. The purpose of line regulation is to maintain a nearly constant output voltage when the input voltage varies. The purposely of load regulation is to maintain a nearly constant output voltage when the load varies. REFERENCE : REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED. 92. The basic difference between a series regulator and a shunt regulator is A. The amount of current that can be handled B. The position of the control element C. The type of sample circuit D. The type of error detector
switching type of voltage regulator than with the linear types because the transistor is not always conducting. Switching regulator efficiencies can be greater than 90 %. Therefore, switching regulators can provide greater load currents at low voltage than linear regulators because the control transistor doesn’t dissipate as much power. REFERENCE: ELECTRONIC REFERENCE: DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED. 94. When the input to a particular voltage regulator decreases by 5 V, the output decreases by 0.25 V. The nominal output is 15 V. Determine the line regulation in % V. A. B. C. D.
0. 222% V 0. 333% V 0. 444% V 0. 555% V
ANSWER: B Solution:
ANSWER: B EXPLANATION: A basic voltage consists EXPLANATION: of a reference voltage source, an error detector, a sampling element and a control device. Protection circuitry is also found in most regulators. Two basic categories of voltage regulators are linear and switching. Two basic types of linear regulators are series and shunt. In a linear series regulator, the control element is a transistor in series with the load. In a linear shunt regulator, the control element is a transistors in
∆⁄% line regulation= ∆ = . ⁄ = .% .% REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED. 95. Determine the output voltage for the regulator in the given figure.
parallel with the load. REFERENCE: ELECTRONIC REFERENCE: DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED. 93. In a switching regulator, transistor is conducting
the
control
A. Part of the time B. All of the time C. Only when the input voltage exceeds a set limit D. Only when there is an overload ANSWER: A EXPLANATION: EXPLANATION: efficiency can
Rogelyn L. Barbosa| BS ECE
A much be realized
greater with a
A. B. C. D.
6.2 V 7.2 V 9.2 V 10.2 V
ANSWER: D Solution:
V =5.1 ⇨the zener voltage R VOU =1 R V =1 10 kΩΩ 5.1 V = . 10 k
REFERENCE: REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
96. Determine the maximum current that the regulator in the figure below can provide to a load.
secondary voltages are 11 000 V and 400 V respectively and the core loss is 2.9 kW, assuming the power factor of the load to be 0.8. Calculate the efficiency on full load. A. 98.3% B. 93.8% C. 89.3% D. 83.9% ANSWER: A Solution: Full-load secondary current is,
× =1250A
Full-load primary current is,
A. B. C. D.
45.5.5 A ≃ 500×1000 11000 = 45 Therefore secondary I R loss on full load is, 1250 × 0.00019 019== 29 2969 69 W and primary I2R loss on full load is, 45.5 ×0.42=870W
0.7 A 0.8 A 0.9 A 0.10 A
ANSWER: A Solution:
0.7 V 0.7 V I = R = 1.0 Ω = .
Total I2R loss on full load
REFERENCE: ELECTRONIC REFERENCE: DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
R
97. In figure below, what power rating must have if the maximum input voltage is 12.5 V?
= 3839 W = 3.84 kW and Total loss on full load = 3.84 + 2.9 = 6.74 kW Output power on full load = 500 × 0.8 = 400 kW Input power on full load = 400 + 6.74 = 406.74 kW Efficiency on full load is,
Ƞ = 1 inputlosses power 6.74 = 0.998383 uuninitt 11 406. 74
= 98.3%
A. B. C. D.
5.10 W W 6.10 7.10 W 8.10 W
ANSWER: C Solution:
V = VN VOU = 12 12.5.5 V V 12. 5 V P = R = 22Ω = . REFERENCE: ELECTRONIC DEVICES CONVENTIONAL CURRENT VERSION BY THOMAS L. FLOYD 8 TH ED.
REFERENCE: ELECTRICAL ELECTRONIC TECHNOLOGY EDITION BY HUGHES
99. A 100 kVA transformer has 400 turns on the primary and 80 turns on the secondary. The primary and secondary resistances are 0.3 Ω and 0.01 Ω respectively, and the corresponding leakage reactances are 1.1 Ωand 0.035 Ωrespectively. The supply voltage is 2200 V. Calculate the equivalent impedance referred to the primary circuit.
98. The primary and secondary windings of a 500
A. 2.50 Ω Ω B. 2.05 Ω Ω C. 5.02 Ω Ω
kVA transformer have resistances of 0.42 Ω and 0.0019 Ω respectively. The primary and
D. 5.20 Ω Ω
Rogelyn L. Barbosa| BS ECE
AND 10TH
ANSWER: B Solution:
= = ⇨
single resistance in the primary circuit equivalent to the primary and secondary resistances of the actual transformer
=0.30.01 =0.55Ω = = ⇨single reactance in the primary circuit
REFERENCE: ELECTRICAL AND ELECTRONIC TECHNOLOGY 10TH EDITION BY HUGHES TOPIC: AC AND DC CIRCUITS 101. Suppose you double the voltage in a simple dc circuit, and cut the resistance in half. The current will become:
A. Four times as great. B. Twice as great. C. The same as it was before. D. Half as great ANSWER: A Solution:
=1.10.035400 80 =1.975Ω
I = V/R I = 2V/0.5R
= ⇨ equivalent impedance of the primary and secondary windingsreferred to the primary circuit
= 0.0.55 1.975 =.Ω REFERENCE: ELECTRICAL ELECTRONIC TECHNOLOGY EDITION BY HUGHES
AND 10TH
100. A single-phase transformer has 1000 turns on the primary and 200 turns on the secondary. The no-load current is 3 A at a power factor 0.2 lagging when the secondary current is 280 A at a power factor of 0.8 lagging. Calculate the power factor. Assume the voltage drop in the windings to be negligible. A. 0.78 lagging B. 0.78 leading C. 0.87 lagging D. 0.87 leading E. ANSWER: A Solution:
I ×1000=280×200 ⇨ I = 56 A Cos∅ =0.8 ⇨ sin∅ =0.6 Cos∅ =0.2 ⇨ sin∅ =0.98 I cos∅ = I cos∅ I cos∅ 3×0.2 = 45 = 56×0.8 45.4.4 A I sin∅ = I sin∅ I sin∅ = 36 = 56×0.6 .3×0.98 36.5.544 A Tan∅ = . =0.805 ∅ =38°50 cos∅ =cos38°50 =. Rogelyn L. Barbosa| BS ECE
I = 4 V/R V/R REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 102. A wiring diagram would most likely be found in: A. An engineer’s general circuit idea notebook. B. An advertisement for an electrical device. C. The service/repair manual for a radio receiver. D. A procedural flowchart. ANSWER: C EXPLANATION: Wiring diagrams are especially useful and necessary when you must service or repair an electronic device because it has the information needed like the value of the component used. REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 103. A circuit has a total resistance of 473,332 ohmand draws 4.4 mA. The best expression for the voltage of the source is: A. 2082 V. B. 110 kV. C. 2.1 kV. D. 2.08266 kV. ANSWER: C Solution: V = IR V = (4.4 mA)(473332 ohm)
kV V = 2.1 kV
ANSWER: C Solution:
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO
W P = IV = (8mA)(250V) = 2 W
104. Good engineering practice usually requires that a series-parallel resistive network be made:
REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO
A. B. C.
From resistors resistors that that are are all all the verysame. rugged. From From a series combination of resistors in parallel. D. From a parallel combination of resistors in series. ANSWER: B EXPLANATION:: By doing this, the total power EXPLANATION handling capacity of the resistance can be greatly increased over that of a single resistor REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 105. You have an unlimited supply of 1-W, 1000 ohmresistors, and you need a 500 ohmresistance rated at 7 W or more. This can be done by assembling: A. Four sets of two 1000 ohm resistors in series, and connecting these four setsin parallel. B. Four sets of two 1000 ohm resistors in parallel, and connecting these four setsin series. C. A 3 _ 3 series-parallel matrix of 1000-_ resistors. D. Something other than any of the above. ANSWER: A Solution: Power rating = 1W (no. of resistors) = 1W W (4x2) = 8 W
107. Four resistors are connected in series across a 6.0-V battery. The values are R1 = 10 ohm, R2 = 20 ohm, R3 = 50 ohm, and R4 = 100 ohmas shown in Figure below. The voltage across R2 is:
A. 0.18 V. B. 33 mV. C. 5.6 mV. D. 670 mV. ANSWER: D Solution: RT = 10+20+50+100 = 180 IT = 60/180 =3.33 mA VR2 = 3.33mA (20) = 670 Mv REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 108. For question number 7, the voltage across R1+R3 is A. 10V B. 20V C. 25V D. Is not important. ANSWER: B Solution:
Resistance = 1000(2)// 1000(2)// ohm 1000(2)// 1000(2) = 500 ohm REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 106. The voltage is 250 V and the current is 8.0 mA. The power dissipated by thepotentiometer is: A. 31 mW. B. 31 W. C. 2.0 W. D. 2.0 mW.
Rogelyn L. Barbosa| BS ECE
RT = 10+20+50+100 = 180 IT = 60/180 =3.33 mA VR1+R3 = 3.33mA (10+50) = 20V 20V REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 109. The maximum voltage output from a voltage divider: A. Is a fraction of the power supply voltage. B. C.
Depends on the total resistance. Is equal to the supply voltage.
D. Depends on the ratio of resistances. ANSWER: D Solution: Vb = V(Rb)/Ra+Rb Vb = V(Rb)/Ra+Rb The voltage output depends
on the ratio of +.
REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 110. Bleeder resistors: A. Are connected across the capacitor in a power supply. B. Keep a transistor from drawing too much current. C. Prevent an amplifier from being overdriven. D. Optimize the efficiency of an amplifier. ANSWER: A EXPLANATION:: It is connected across the EXPLANATION capacitor to drain the stored voltage so that servicing the supply is not dangerous. REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 111. Suppose a 1-K ohmresistor will dissipate 1.05 W, and you have many 1-W resistors of all common values. If there’s room for 20percent resistance error, the cheapest solution is to use: A. Four 1 K ohm, 1-W resistors in seriesparallel. B. Two 2.2 K ohm, 1-W resistors in parallel. C. Three 3.3 K ohm, 1-W resistors in parallel. D. One 1 K ohm, 1-W resistor, since manufacturers allow for a 10-percent margin of safety. ANSWER: B Solution: Power = 1W x 2 = 2W 20% = 1.6W to 2.4W Resistance = 2.2k ohm // 2.2 K ohm = 1.1k ohm 20% = 880 ohm to 1320 ohm Four 1 K ohm, 1-W resistors in series-parallel is the cheapest
B.
Has a slow rise time and a fast decay time. C. Has equal rise and decay rates. D. Rises and falls abruptly. ANSWER: C EXPLANATION: The positive-going slope is the same as the negative-going slope. This is calletriangular wave. wave. REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 113. In a 117-V utility circuit, the peak voltage is: A. 82.7 V. B. 165 V. C. 234 V. D. 331 V. ANSWER: B
Solution: Utility voltage = rms = 0.707 Vp Vp = 165 V REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 114. If a 175-V dc source were connected in series with the utility mains from astandard wall outlet, the result would be: A. Smooth dc. B. Smooth ac. C. Ac with one peak greater than the other. D. Pulsating dc ANSWER: D EXPLANATION: Ripples were caused by the utility mains from a standard wall outlet. REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO GIBILISCO
115. If a wave has a frequency of 440 Hz, how long does it take for 10 degrees of phase? A. 0.00273 second. B. 0.000273 second. C. 0.0000631 second. D. 0.00000631 second. ANSWER: C Solution: f(360°/10°) = 1/t t = 1/(440Hz)(36°) = 0.0000631 second
112. A triangular wave: A. Has a fast rise time and a slow decay time.
REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO
±
±
Rogelyn L. Barbosa| BS ECE
116. A wave has a frequency of 300 kHz. One complete cycle takes: A. 1⁄300 second second B. 0.00333 second C. 1⁄3,000 second second D. 0.00000333 second ANSWER: D Solution: f = 1/t t= 1/ 300000 Hz = 0.00000333 second REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: TEACH AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 117. An RC circuit has a capacitance of 0.015 microF. The resistance is 52 ohm. What is the phase angle at 90 kHz? A. –24 –24 degrees. B. –0.017 –0.017 degrees. C. –66 –66 degrees. D. None of the above.
Pf = cos = R/Z = (50/88.4) x100 = 56.6% REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 120. A circuit has apparent power= 100 kVa, reactive power 60kVar. What is the quality factor? A. 0.6 B. 1 C. 0.75 D. It can’t be calculated from this data. data. ANSWER: C Solution:
√100 60
R= = 80 W Qf = tan = Q/R= 60/80 = 0.75 REFERENCE: REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO
ANSWER: C
TOPIC: SEMICONDUCTORS INCLUDING DIODES AND
Solution:
TRANSISTORS
Xc=1/2fC = 1/ 2 (90 kHz)(0.015 F) = 117.89
= tan-1 (Xc/R) = tan -1 (117.89/52) = 66 degrees = -66 degrees because it is capacitive. capacitive. REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: TEACH AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 118. A coil shows an inductive reactance of 200 Ω at 500 Hz. What is its inductance? A. 0.637 H. B. 628 H. C. 63.7 mH. D. 628 mH. ANSWER: C Solution: 63.7mH XL = 2fL; L= 200/2 (500) = 63.7mH REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: TEACH AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 119. A series circuit has Z = 88.4 ohm, with R = 50.0 ohm. What is PF? A. 99.9 percent. B. 56.6 percent. C. 60.5 percent. D. 29.5 percent. ANSWER: B Solution:
Rogelyn L. Barbosa| BS ECE
121. The amount of time between the creation of a hole and its disappearance is called A. Doping B. Lifetime C. Recombination D. Valence ANSWER: B EXPLANATION: It is the duration of the creation of hole until it accept electron. electron. REFERENCE: ELECTRONIC REFERENCE: ELECTRONIC DEVICES AND CIRCUITS THEORY 11 THEDITION BY ROBERT L. BOYLESTAD 122. When a voltage is applied to a semiconductor, holes will flow A. Away from the negative potential B. Toward the positive potential C. In the external circuit D. None of the above ANSWER: D EXPLANATION: Hole do not move, electrons move. REFERENCE: ELECTRONIC DEVICES AND REFERENCE: ELECTRONIC CIRCUITS THEORY 11 THEDITION BY ROBERT L. BOYLESTAD 123. The absence of bias across a semiconductor diode, the net flow charge is A. zero B. high
C. infinite D. None of the above ANSWER: A EXPLANATION: Under no bias condition, the electron will not flow, therefore the net flow charge is zero.
A. 22 V B. -0.7 V C. -23.3 V D. 0.7 V ANSWER: C Solution:
REFERENCE: ELECTRONIC REFERENCE:
AND
Ideally, a negative dc value equal to the
CIRCUITS THEORY 11 EDITION BY ROBERT L. BOYLESTAD
input peak less the diode is inserted by the clamping circuit.
124. Determine the ripple factor for the filtered bridge rectifier with a load as indicated in the figure. A. 0.25 B. 0.025 C. 0.079 D. 0.7 ANSWER: C
Vdc = -(Vp(in) – 0.7 – 0.7 V) Vdc = -(24V – -(24V – 0.7 0.7 V) V Vdc = -23.3 V REFERENCE: ELECTRONIC DEVICES 7TH REFERENCE: EDITION BY THOMAS L. FLOYD
DEVICES
TH
126. When a 60 Hz sinusoidal voltage is applied to the input of a half-wave rectifier, the output frequency is A. 60 Hz B. 120 Hz C. 0 Hz D.
n= 0.1
REFERENCE: ELECTRONIC DEVICES AND REFERENCE: CIRCUITS THEORY 11TH EDITION BY ROBERT L. BOYLESTAD
Vp(primary) = 1.414Vrms = 1.414(115V) = 163 V Vp(secondary) = nVp(primary) = 0.1(163V) = 16.3 V Vp(rec) = Vp(secondary) – 1.4 – 1.4 V = 14.9 V
V
.ℎ (14.9 V) = 1.13 V )V =( =(1 1
p(rec)
VDC
30 Hz
ANSWER: A Solution: In have wave rectifier f iinn = f out 60Hz out = 60Hz
Solution:
Vr(pp) =
=
p(rec)
)(14.9 V) = 14.3 V . ℎ = . = 0.079 r= . 0.079 REFERENCE: ELECTRONIC REFERENCE: DEVICES EDITION BY THOMAS L. FLOYD
7TH
125. What is the output voltage that you would expect to observe across RL in the circuit shown below. Assume that RC is large enough to prevent significant capacitor discharge.
127. When the peak output voltage is 100 V, the PIV for each diode in a center-tapped fullwave rectifier is (neglecting the diode drop) A. B. C. D.
100 V 141 V 200 V 50 V
ANSWER: C Solution: In center tapped full wave rectifier PIV = 2Vp =2(100V) = 200V REFERENCE: ELECTRONIC DEVICES AND REFERENCE: CIRCUITS THEORY 11THEDITION BY ROBERT L. BOYLESTAD 128. The depletion region is created by A. B. C. D.
Rogelyn L. Barbosa| BS ECE
ionization diffusion recombination all of the above
ANSWER: D EXPLANATION: The region of uncovered positive and negative ions, diffusion and recombination causes depletion region.
REFERENCE: ELECTRONIC DEVICES AND REFERENCE: CIRCUITS THEORY 11THEDITION BY ROBERT L. BOYLESTAD 132. If the current gain is 200 and the collector current is 100 mA, the base current is
REFERENCE: ELECTRONIC DEVICES AND REFERENCE: CIRCUITS THEORY 11TH EDITION BY
A. 0.5 mA
ROBERT L. BOYLESTAD 129. A certain Zener diode has a maximum power rating of 400 mW at 50°C and a derating factor of 3.2mW/°C. Determine the maximum power the Zener can dissipate at a temperature of 90°C. A. 272 mW B. 528 mW C. 250 mW D. 400 mW ANSWER: A Solution: PD(derated) = PD(max) – (mW/°C) – (mW/°C)T PD(derated) = 400mW(3.2 mW/°C)(90°C - 50°C)
REFERENCE: ELECTRONIC DEVICES 7 TH REFERENCE: EDITION BY THOMAS L. FLOYD 130. What is the PIV across each diode of a bridge rectifier with a secondary voltage of 20 V rms? A. 14.1 V B. 20 V C. 28.3 V D. 34 V ANSWER: C Solution: Vrms=PIV/0.707 PIV= 20(0.707) = 28.3 V REFERENCE: ELECTRONIC DEVICES AND REFERENCE: CIRCUITS THEORY 11THEDITION BY ROBERT L. BOYLESTAD
ANSWER: C EXPLANATION:
It
commonly
is
abundant
used
and
cheaper other semiconductor has good than temperature characteristic. and characteristic.
Rogelyn L. Barbosa| BS ECE
ANSWER: C Solution: Current gain = Ic/Ib Ib = 200(100mA) = 2A REFERENCE: ELECTRONIC DEVICES AND REFERENCE: ELECTRONIC CIRCUITS THEORY 11 THEDITION BY ROBERT L. BOYLESTAD 133. Line regulation is determined by A. zener current and load current B.
PD(derated) = 400 mW – mW – 128 128 mW PD(derated) = 272 mW
131. What is the most semiconductor? A. GaAs B. germanium C. silicon D. gold
B. 2 2 mA C. A D. 20 A
changes voltage in load resistance and output C. load current D. changes in output voltage and input voltage ANSWER: D EXPLANATION: Line regulation the change in output voltage for a given change in input (line) voltage, normally expressed as a percentage. percentage. REFERENCE: ELECTRONIC DEVICES AND REFERENCE: CIRCUITS THEORY 11TH EDITION BY ROBERT L. BOYLESTAD 134. For a certain 12 V Zener diode, a 10 mA change in Zener current produces a 0.1 V changes in Zener voltage. The Zener impedance for this current ranges is A. 0.1 Ω Ω B. 100 Ω C. 10 Ω Ω D. 1 Ω Ω ANSWER: C Solution: Ω Z = V/I = 0.1V/10mA = 10 Ω REFERENCE: ELECTRONIC DEVICES AND REFERENCE: CIRCUITS THEORY 11TH EDITION BY ROBERT L. BOYLESTAD 135. The βDC of a transistor is its
A. internal resistance B. power gain C. voltage gain D. current gain ANSWER: D EXPLANATION: It is the ratio of collector current to base current. current. REFERENCE: ELECTRONIC DEVICES REFERENCE: EDITION BY THOMAS L. FLOYD
7 TH
136. When operated in cutoff and saturation, the transistor acts like A. B. C. D.
a switch a linear amplifier a variable capacitor a variable resistor
ANSWER: A EXPLANATION: It is because in saturation and cut off, the output is on and off, therefore acts as a switch.
139. For a common-collector amplifier, RE = 100 Ω, Ic= 2.5 mA, and βAC = 150. re = ? ? A. 10.33 Ω Ω B. 10.33kΩ C. 10 Ω D. 11 Ω ANSWER: A Solution:
re=26mV/Ie βAC = Ic/Ib Ic/Ib 150 = 2.5 mA/Ib Ib = 16.67 microA. Ie = Ic + Ib = 2.5 mA = 16.67 microA = 2.51667mA re=26mV/2.51667 mA re= 10.33 ohms
REFERENCE: ELECTRONIC DEVICES AND REFERENCE: CIRCUITS THEORY 11TH EDITION BY ROBERT L. BOYLESTAD 140. A certain JFET data sheet gives VGS(off) = - 4 V. The pinch-off voltage, Vp,
TH
REFERENCE: ELECTRONIC DEVICES REFERENCE: EDITION BY THOMAS L. FLOYD
7
137. Find the value of emitter current if collector current = 3mA, alpha = 0.98. A. B. C. D.
3.06 mA 2.98 mA 6 mA requires fewer components than all the other methods ANSWER: A Solution: Alpha = collector current / emitter current Emitter current = 3mA/0.98 = 3.06 Ma REFERENCE: ELECTRONIC DEVICES AND REFERENCE: CIRCUITS THEORY 11TH EDITION BY ROBERT L. BOYLESTAD 138. If the dc emitter current in a certain transistor amplifier is 3 mA, the approximate value of re is A. B. C. D.
3Ω 3 kΩ kΩ 0.33 kΩ kΩ 8.33 Ω Ω
A. B. C. D.
ANSWER: D Solution: Vp = - Vgs(off) = -(-4) = 4V REFERENCE: ELECTRONIC DEVICES 7TH REFERENCE: EDITION BY THOMAS L. FLOYD 141. A MOSFET differs from a JFET mainly because A. of the power rating B. C.
the JFET has a pn junction MOSFETs do not have a physical channel D. the MOSFET has two gates ANSWER: B EXPLANATION: MOSFET has no pn junction instead a SG junction and they are not the same. same. REFERENCE: ELECTRONIC DEVICES 7 TH REFERENCE: EDITION BY THOMAS L. FLOYD 142. A certain D-MOSFET is biased at VGS = 0 V. Its data sheet specifies IDSS = 20 mA and VGS(off) = - 5 V. The value of the drain
ANSWER: D TH
REFERENCE: ELECTRONIC DEVICES REFERENCE: EDITION BY THOMAS L. FLOYD
Rogelyn L. Barbosa| BS ECE
cannot be determined is – is –4 4V depends on VGS is + 4 V
7
current
A. B. C. D.
EXPLANATION: D MOSFET with positive Vgs is operating in E- mode because it is to the right of vertical axis(Vgs=0). axis(Vgs=0).
is 0 A is 10 mA is 20 mA cannot be determined
REFERENCE: ELECTRONIC DEVICES REFERENCE: EDITION BY THOMAS L. FLOYD
ANSWER: A Solution: Id = Idss(1-Vgs/Vp)2
146. In a certain common-source (CS) amplifier, VDS = 3.2 V rms and VGS = 280 mV rms. The voltage gain is
Id = 20mA (1-1) Id = 0A 0A REFERENCE: ELECTRONIC DEVICES REFERENCE: EDITION BY THOMAS L. FLOYD
7 TH
143. Determine the magnitude of gm for a JFET with IDSS = 8mA and Vp = -4V at Vgs = -0.5V. A. B. C. D.
3mS 3.5mS 2mS Tomorrow
ANSWER: B Solution: gmo = 2IDSS/|Vp|= 2(8mA)/4 = 4mS gm = gmo(1-Vgs/Vp) gm = 4mS (1 – (1 – 0.5/4) 0.5/4) 3.5mS gm = 3.5mS REFERENCE: ELECTRONIC DEVICES 7 TH REFERENCE: EDITION BY THOMAS L. FLOYD 144. Determine the magnitude of gm for a JFET with IDSS = 8mA and Vp = -4V at Vgs = -1.5V.
A. 1 B. 11.4 C. 8.75 D. 3.2 ANSWER: B Solution: Gain = VDS / VGS Gain = 3.2V/280mV = 11.4 11.4 REFERENCE: ELECTRONIC DEVICES 7TH EDITION BY THOMAS L. FLOYD 147. A certain common-drain (CD) amplifier with RS 1.0voltage kΩ hasgain a transconductance of 6000 μS. =The is A. B. C. D.
1 0.86 0.98 6
ANSWER: B Solution: AV =
A. 2.57mS B. 3mS C. 4mS D. 1mS ANSWER: A Solution: gmo = 2IDSS/|Vp|= 2(8mA)/4 = 4mS gm = gmo(1-Vgs/Vp) gm = 4mS (1 – (1 – 1.5/4) 1.5/4) 2.57mS gm = 2.57mS REFERENCE: ELECTRONIC DEVICES 7 TH REFERENCE: EDITION BY THOMAS L. FLOYD 145. An n-channel D-MOSFET with positive VGS is operating in A. B. C. D.
the depletion mode cutoff the enhancement mode saturation
ANSWER: C
Rogelyn L. Barbosa| BS ECE
7 TH
ℎ + = +ℎ = 0.86
REFERENCE: ELECTRONIC DEVICES 7TH REFERENCE: EDITION BY THOMAS L. FLOYD 148. Determine the magnitude of gm for a JFET with IDSS = 8mA and Vp = -4V at Vgs = -0.5V. A. 3mS B. 2.5mS C. 2mS D. 1.5mS ANSWER: D Solution: gmo = 2IDSS/|Vp|= 2(8mA)/4 = 4mS gm = gmo(1-Vgs/Vp) gm = 4mS (1 – (1 – 2.5/4) 2.5/4) ; gm = 1.5mS REFERENCE: ELECTRONIC DEVICES 7TH REFERENCE: EDITION BY THOMAS L. FLOYD 149. There is a _____ degrees of inversion between gate and drain voltages of an FET. A. Zero
B. 90 C. 180 D. 270 ANSWER: C EXPLANATION: Means that input is equal to negative output output REFERENCE: ELECTRONIC DEVICES 7TH EDITION BY THOMAS L. FLOYD 150. Two FET amplifier are cascaded. The first stage has a voltage gain of 5 and the second stage has a voltage gain of 7. The overall voltage gain is A. 35 B. 12 C. dependent on the second stage loading D. 15
153. The usual way to protect a load from excessive supply voltage is with a A. Crowbar B. Zener diode C. Four-layer diode D. Thyristor ANSWER: A EXPLANATION: Crowbar is a simple overvoltage protection circuit. It uses a Zener diode that monitor the DC output voltage. voltage. REFERENCE: ELECTRONIC DEVICES 7TH REFERENCE: EDITION BY THOMAS L. FLOYD
154. Determine the gate current when the switch, SW1 is momentarily closed in figure below. Assume VAK = 0.8V, VGK= 0.7V and IH = 20mA.
ANSWER: A Solution: AT = A1 A2 AT = 5x7 35 AT = 35 REFERENCE: ELECTRONIC REFERENCE: DEVICES EDITION BY THOMAS L. FLOYD
7TH A. B. C. D.
TOPIC: INDUSTRIAL AND POWER ELECTRONICS 151. Athyristor has A. B. C. D.
two pn junctions three pn junctions four pn junctions only two terminals
ANSWER: B EXPLANATION:
It
is
4 mA 4.1 mA 4.2 mA 4.3 mA
ANSWER: B Solution: Solution:
= 4.1 mA = − = −. ℎ a
four
layer
semiconductor therefore three junctions were formed. formed. REFERENCE : REFERENCE: ELECTRONIC DEVICES EDITION BY THOMAS L. FLOYD
7 TH
152. The unijunction transistor acts as a A. Four-layer diode B. Diac C. Triac D. Latch ANSWER: D EXPLANATION: It has only one pn junction therefore act as a latch. REFERENCE: ELECTRONIC REFERENCE: DEVICES EDITION BY THOMAS L. FLOYD
7TH
REFERENCE: REFERENCE:
ELECTRONIC
DEVICES
EDITION BY THOMAS L. FLOYD 155. For question no. 5, determine the anode current. A. 2 mA B. 3 mA C. 23.2 mA D. 24.2 mA ANSWER: C Solution:
23.2mA = − = −. ℎ = 23.2mA REFERENCE: ELECTRONIC DEVICES 7TH REFERENCE: EDITION BY THOMAS L. FLOYD
Rogelyn L. Barbosa| BS ECE
7TH
156. Determine the value of R1 in the figure that will ensure power turn-on and turn-off the UJT. The characteristic of the UJT exhibits the following values:
SCR. REFERENCE: ELECTRONIC DEVICES 7 TH REFERENCE: EDITION BY THOMAS L. FLOYD 159. Determine the value of the anode current in the figure when rhe device is on. V BR(F) = 100V. Assume Vbe = 0.7V and V CE(sat) = 0.1V for the internal. A. B. C. D.
=0.5, Vv=1
V,
Iv=10 mA, Ip=20 A and Vp= 14 V. V. A. 800K 800K ,2.9K ,2.9K B. 2.9K 2.9K , 800K 800K C. 3.1K 3.1K , 800K 800K , 3.1K D. 800K 800K 3.1K ANSWER: A
ANSWER: B Solution: Va=Vbe-VCE(sat) = 0.7 +0.1 = 0.8V VRs = Vbias – Vbias – Va Va = 110V-0.8V = 109.2V IA = VRs/Rs = 109.2/1kohm = 109.2mA
Solution:
> > 3014 20 > > 301 10 > >2.9
REFERENCE: ELECTRONIC DEVICES 7TH REFERENCE: EDITION BY THOMAS L. FLOYD
160. A 4-layervoltage diode turns on when the anode to cathode exceeds
REFERENCE: ELECTRONIC DEVICES REFERENCE: EDITION BY THOMAS L. FLOYD
7TH
157. The data sheet of a certain UJT gives =0.6. determine the peak point emitter voltage Vp if Vbb=20V. A. 11V B. 12V C. 10V D. 12.7V ANSWER: D Solution: Vp = Vbb+Vd = 0.6(20V) + 0.7V = 12.7V REFERENCE: ELECTRONIC DEVICES 7TH REFERENCE: EDITION BY THOMAS L. FLOYD 158. An SCR can be turned off by A. B. C. D.
forced commutation a negative pulse on the gate anode current interruption answer a, b and c
ANSWER: D EXPLANATION: A,B and C stops the conduction therefore turning off the
Rogelyn L. Barbosa| BS ECE
10mA 109.2mA 102.9mA 100Ma
A. 0.7 V B. the gate voltage C. the forward-breakover voltage D. the forward-blocking voltage ANSWER: C EXPLANATION: It is like the threshold voltage in diode where the electron cannot flow without overcoming it. REFERENCE: ELECTRONIC DEVICES 7 TH REFERENCE: EDITION BY THOMAS L. FLOYD TOPIC: MICROELECTRONICS INCLUDING OP AMPS AMPS 161. The open-loop gain of an op-amp is given at 225,000. Find the gain in dB. A. 1070 B. 107 C. 214 D. 10.7 ANSWER: B Solution: Solution: 107 AdB = 20 log (225000) = 107 REFERENCE: ELECTRONIC DEVICES 7 TH EDITION REFERENCE: BY THOMAS L. FLOYD 162. Determine the open loop gain of an Op-am whose cut-off frequency is 100 Hz with a midrange voltage gain of 100,000. A. 7,071 B. 707.1 C. 70,710
REFERENCE: ELECTRONIC DEVICES REFERENCE: EDITION BY THOMAS L. FLOYD
D. 70.71 ANSWER: C Solution:
7TH
166. The output of a particular op-amp increases 8 V in 12 μs. The slew rate is A. 96 V/μs V/μs B. 0.67 V/ μs μs C. 1.5 V/μs V/μs D. none of the above
AOL =
= 100000/ 1 = 70,710 + +
REFERENCE: ELECTRONIC REFERENCE: DEVICES EDITION BY THOMAS L. FLOYD 163.
7TH
If the voltage gain for each input of a summing amplifier with a 4.8 kilo-ohm feedback resistor is unity, the input resistors must have a value of _____ kilo-ohm. A. 4.8 times the number of inputs B. 48 C. 4.8 divided by the number of inputs D. 4.8 ANSWER: D A = Rf/Rin Rin = 1(4.8 k ohm) = 4.8 k-ohm REFERENCE: ELECTRONIC REFERENCE: DEVICES EDITION BY THOMAS L. FLOYD FLOYD
7TH
7TH
165. If Av(d) = 3500 and Acm = 0.35, the CMRR is 1225 10,000 80 dB answers b and c
ANSWER: D CMRR = Av/Ac = 3500/0.35 = 10,000 CMRRdB = 20 log(3500/0.35) = 80 dB
Rogelyn L. Barbosa| BS ECE
A. 100,000 B. 1000 C. 101 D. 100 ANSWER: B Solution: A=Rf/Ri = 100/0.1 = 1000 REFERENCE: ELECTRONIC DEVICES REFERENCE: EDITION BY THOMAS L. FLOYD
7TH
A. 1 kHz B. 9 kHz C. 10 kHz
ANSWER: D EXPLANATION: Measure of op amp ability to reject common mode signals. signals.
A. B. C. D.
167. A certain inverting amplifier has anRi of 0.1 kΩ and an Rf of 100 kΩ. The closed loop gain is
168. The bandwidth of an AC amplifier having a lower critical frequency of 1 kHz and an upper critical frequency of 10 kHz is
164. CMRR stands for A. Common mode Reduction Ratio B. Common mode Reflection Ratio C. Common mode Restarting Ratio D. Common mode Rejection Ratio
REFERENCE: ELECTRONIC REFERENCE: DEVICES EDITION BY THOMAS L. FLOYD
ANSWER: B Solution: V/μs Slew rate = V/μs V/μs = 8/12 = 0.67 V/μs REFERENCE: ELECTRONIC DEVICES 7TH REFERENCE: EDITION BY THOMAS L. FLOYD
D.
11 kHz
ANSWER: B Solution: B= USF-LCF =10kHz-1kHz B = 9kHz 9kHz REFERENCE: ELECTRONIC DEVICES 7TH REFERENCE: EDITION BY THOMAS L. FLOYD 169. With zero volts on both inputs, an op-amp ideally should have an output equal to A. B. C. D.
the positive supply voltage the negative supply voltage zero the CMRR
ANSWER: C Solution:
Vo = Vi (Rf/Ri) Vo = 0 (Rf/Ri) = zero REFERENCE: ELECTRONIC DEVICES REFERENCE: EDITION BY THOMAS L. FLOYD
7TH
170. The bandwidth of a dc amplifier having an upper critical frequency of 100 kHz is A. B. C. D.
REFERENCE: REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 7TH
TOPIC: INSTRUMENTATION AND MEASUREMENTS 171. The meter movement in an illumination meter measures: A. B. C. D.
Current. Voltage. Power. Energy.
ANSWER: A EXPLANATION: Also called light meter is measured in terms of current because as the light strikes the photovoltaic cell, current is produced. REFERENCE: REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 172. The change in the direction of a compass needle, when a current-carrying wirebrought near, is: A. Electromagnetic deflection. deflection. B. Electrostatic force. C. Magnetic force. D. Electroscopic force. ANSWER: A EXPLANATION:: Also called magnetic EXPLANATION deflection, whereas compass works by using magnetic characteristic of the earth. REFERENCE: REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 173. Suppose a certain current in a galvanometer causes the needle to deflect 20degrees, and
Rogelyn L. Barbosa| BS ECE
ANSWER: C EXPLANATION: Galvanometer measures the current so the deflection of the needle is directly proportional to the increase in current.
100 kHz unknown infinity 0 kHz
ANSWER: A Solution: B = cut-off frequency = 100kHz 100kHz REFERENCE: ELECTRONIC REFERENCE: DEVICES EDITION BY THOMAS L. FLOYD
then this current is doubled. The needle deflection: A. Will decrease. B. Will stay the same. C. Will increase. D. Will reverse direction.
174. One important advantage of an electrostatic meter is that: A. It measures very small currents. B. It will handle large currents. C. It can detect ac voltages. D. It draws a large current from the source. ANSWER: C EXPLANATION: Electrostatic voltmeter utilizes the attraction force between two charged surfaces to create a deflection of a pointer directly calibrated in volts. in volts. Since Since the attraction force is the same regardless of the polarity of the charged surfaces (as long as the charge is opposite), the electrostatic voltmeter can measure both direct both direct current and and alternating alternating current. REFERENCE: REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 175. An ammeter shunt is useful because: A. It increases meter sensitivity. B. It makes a meter more physically rugged. C. It allows for measurement of a wide range of currents. D. It prevents overheating of the meter. ANSWER: C EXPLANATION: An ammeter An ammeter shunt allows the measurement of current current values too large to be directly measured by a particular ammeter. REFERENCE: REFERENCE:
TEACH
YOURSELF
ELECTRICITY AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO
176. Voltmeters should generally have: A. Large internal resistance. resistance. B. Low internal resistance. C. Maximum possible sensitivity. D. Ability to withstand large currents. ANSWER: A EXPLANATION: If you make the internal resistance very large, then the current flowing through the voltmeter will be negligible, and the voltage drop across the resistor will not change. REFERENCE: REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 177. A utility meter’s readout indicates: indicates: A. Voltage. B. Power. C. Current. D. Energy. ANSWER: D EXPLANATION: Utility meter is used for metering devices used on utility mains. REFERENCE: REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 178. A VU meter is never used for measurement of: A. Sound. B. Decibels. C. Power. D. Energy. Energy. ANSWER: D EXPLANATION: VU meter device displaying a representation of the signal the signal level iin n audioequipment audioequipment and not energy. REFERENCE: REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 179. A measuring device which will result to full scale reading at shorted terminal condition A. Voltmeter B. Ammeter C. Electrostatic meter D. Ohmmeter ANSWER: D EXPLANATION:
Ohmmeter
measures
resistance which is the opposition to current. The scale start to deflect when
Rogelyn L. Barbosa| BS ECE
the resistance goes down, and at short circuit resistance is zero therefore results in full scale reading. REFERENCE: ELECTRONIC DEVICES 7TH REFERENCE: EDITION BY THOMAS L. FLOYD FLOYD 180. The most accurate type of test equipment used for measuring high power is the _____. A. Bolometer B. Wavemeter C. Calorimeter D. Wattmeter ANSWER: A EXPLANATION: A bolometer consists of an absorptive element, such as a thin layer of metal, connected to a thermal reservoir (a body of constant temperature) through a thermal link. REFERENCE: ELECTRONIC DEVICES 7TH REFERENCE: EDITION BY THOMAS L. FLOYD TOPIC: DIGITAL ELECTRONICS 181. The bit storage capacity of a Read Only Memory(ROM) with a 512 x 8 organization is _____ bits. A. 1024 B. 8192 C. 4096 D. 2048 ANSWER: C C Solution: Bit storage capacity = 512x8 = 4096 REFERENCE: ELECTRONIC DEVICES AND REFERENCE: CIRCUITS THEORY 11TH EDITION BY ROBERT L. BOYLESTAD 182. The symbols 1, 2 and 3 through 9 are what type of numerals? A. Boolean B. Roman C. Binary D. Arabic ANSWER: D EXPLANATION: Arabic Numerals is the EXPLANATION: most common system for the symbolic representation of numbers numbers in the world today. REFERENCE: ELECTRONIC DEVICES AND REFERENCE: CIRCUITS THEORY 11 TH EDITION BY ROBERT L. BOYLESTAD 183. The value of the decimal number 23 in binary form is: A. 1011. B. C.
110111. 10111. 10111.
D. 11100. ANSWER: C Solution: 23/2=11 11/2 =5 5/2 = 2 2/2=1 1/2=.5
1 1 1 0 1
Answer = 10111 REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 184. The binary number 110001 represents the digital number: A. 49. 49. B. 25. C. 21. D. 13. ANSWER: A Solution: 25 + 24+ 20 = 32+16+1=49 32+16+1=49 REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 185. The fifth digit from the right in a binary number carries a decimal value of: A. 64. B. 32. C. 24. D. 16. ANSWER: D Solution: Solution: 11111 24+23+ 22+21 + 20 16 24 = 16 REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 186. The largest possible decimal number that can be represented by six binary digits (bits) is: A. 256. B. 128. C. 64. D. 63. ANSWER: D Solution: 111111 5 4 3 63 2 + 2 +2 + 22+21 + 20= 63 REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO
Rogelyn L. Barbosa| BS ECE
187. Data sent along a single line, one bit after another, is called: A. Serial. B. Synchronous. C. Parallel. D. Asynchronous. ANSWER: A EXPLANATION: Serial communication is the process of sending data sending data one one bit bit at a time, sequentially, over a communication a communication channel or computer or computer bus. bus. REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 188. An advantage of a J-K over an R-S flip-flop is that: A. The J-K flip-flop is faster. B. The J-K can attain more states. C. The J-K always has predictable outputs. D. No! An R-S flip-flop is superior to a J-K. ANSWER: C C EXPLANATION: The main difference between a JK flip-flop and an R-S flip-flop is that in the JK flip-flop, both inputs can be HIGH. When both the J and K inputs are HIGH, the Q output is toggled , which means that the output alternates between HIGH and LOW. Thereby the invalid condition which occurs in the R-S flip flop is eliminated. REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 189. The inputs of an R-S flip-flop are known as: A. Low and high. B. Asynchronous. C. Synchronous. D. Set and reset. ANSWER: D EXPLANATION: The set is the high state and it holds the value until reset to low by a signal at the reset input. REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 190. A frequency synthesizer makes use of A. An OR gate. B. A divider. C. The octal numbering system. D. The hexadecimal numbering system. ANSWER: B
EXPLANATION: Divider is used tomake the frequency synthesizer programmable. programmable. REFERENCE: TEACH REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO
EXPLANATION: Revolving joint is generally known as V – – Joint. Here, the output link axis is perpendicular to the rotational axis, and the input link is parallel to the rotational axes. As like twisting joint, the output link spins about the input link.
OTHERS REFERENCE: REFERENCE: 191. The computer worm that infected computer systems world-wide in 10 minutes, making it the fastest computer virus ever known is called _____. A. Claw Hammer B. Sledge Hammer C. SQL Hammer D. SARS Virus ANSWER: C EXPLANATION: SQL Slammer is a computer worm that caused a a denial of service on some Internet some Internet hosts and dramatically slowed down general Internet traffic, general traffic, starting at 05:30 UTC 05:30 UTC on January 25, 2003. It spread rapidly, infecting most of its 75,000 victims within ten minutes. REFERENCE: "ISS SECURITY BRIEF: MICROSOFT SQL REFERENCE: "ISS SLAMMER WORM PROPAGATION". PROPAGATION". ISSFORUM. 25 JANUARY 2003. RETRIEVED 2008-11-29. 192. In a robotic system, how many degrees of freedom are there? A. Four B. Six C. Three D. Two ANSWER: B EXPLANATION: 6 degrees of freedom is the minimum needed to reach a volume of space from every angle. The longer the arm, the greater the volume that can be reached. More than 6 joints and the redundant – robot becomes kinematically redundant – it can reach the same spot at the same angle in more than one way. For example, you can pinch your nose and wiggle your elbow at the same time. time. REFERENCE: REFERENCE: https://www.marginallyclever.com/2014 /02/what-are-the-six-degrees-offreedom-in-a-robot-arm/ 193. In a robots, he axis that allows rotation are often referred to as _____ joints A. Revolving B. Revolute C. Resolute D. Rotary
ANSWER: A A
Rogelyn L. Barbosa| BS ECE
http://www.roboticsbible.com/robot-linksand-joints.html and-joints.html 194. An android takes the form of: A. An insect. B. A human body. C. A simple robot arm. D. Binocular vision. ANSWER: B EXPLANATION: It looks like a human body because android is labeled as a boy. REFERENCE: ELECTRONIC DEVICES 7TH EDITION BY THOMAS L. FLOYD 195. According to Asimov’s three laws, under what circumstances is it all right for a robot to injure a human being? A. Never. B. When the human being specifically requests it. C. In case of an accident. D. In case the robot controller is infected with a computer virus. ANSWER: A EXPLANATION: It is said in all 3 rules that a robot cannot harm a human no matter what the circumstances is. REFERENCE: ELECTRONIC DEVICES REFERENCE: EDITION BY THOMAS L. FLOYD 196. A manipulator is also known as a:
7TH
A. Track drive. B. Robot arm. C. Vision system. D. Robot controller. ANSWER: B EXPLANATION: Robot arm is a mechanical arm, usually programmable which is analogous to human arm and also called manipulator. REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 197. An android is well suited for operation in: A. Extreme weather conditions. B. C.
Total darkness. An assembly line.
REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: TEACH AND ELECTRONICS THIRD EDITION BY STAN
C. Three dimensions. D. Four dimensions. ANSWER: C EXPLANATION: Spherical coordinates (r, θ, φ) as commonly used in physics (ISO convention): radial distance r, polar angle θ (theta), (theta), and azimuthal angle φ (phi). (phi). The symbol ρ (rho) (rho)
GIBILISCO
is often used instead of r.
D. An environment with children. ANSWER: D EXPLANATION: Android is a robot a robot with its body shape built to resemble the human the human body. body.
198. What is the peak download and peak upload of LTE- advanced? A. 1000Mbps and 500Mbps B. 5000Mbps and 1000Mbps C. 500Mbps and 250Mbps D. None of the above ANSWER: A EXPLANATION: It’sCoordinated It’sCoordinated Multi-point Transmission will also allow more system capacity to help handle the enhanced data speeds REFERENCE: TEACH REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 199. Rodney Brooks is best known for his work with: A. Epipolar navigation. B. Binocular vision. C. Range sensing and plotting. D. Insect robots ANSWER: D EXPLANATION: He is an Australian roboticist, Australian roboticist, Fellow of the Australian Academy of Science, author, Science, author, and robotics entrepreneur, most known for popularizing the actionist the approach to robotics. He was a Panasonic Professor of Robotics at the the Massachusetts Institute of Technology and former director of the MIT the MIT Computer Science and Artificial Intelligence Laboratory. He is a founder and former Laboratory. Chief Technical Officer of iRobot and coFounder, Chairman and Chief Technical Officer of Rethink Robotics (formerly Heartland Robotics). REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: TEACH AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO 200. Spherical coordinates can uniquely define the position of a point inup to: A. One dimension. B. Two dimensions.
Rogelyn L. Barbosa| BS ECE
REFERENCE: TEACH YOURSELF ELECTRICITY REFERENCE: AND ELECTRONICS THIRD EDITION BY STAN GIBILISCO
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