BARBOSA, ROGELYN L.
1.
TOPIC: AC/DC CIRCUITS
Suppose you double the voltage in a simple dc circuit, and cut the resistance in half. The current will become:
A.
B.
C.
D.
Four times as great.
Twice as great.
The same as it was before.
Half as great.
ANSWER: A
V = IR
Solution:
R
2V = I ( )
2V
I=1
2
Therefore,
2
(R)
𝐈 = 𝟒 𝐭𝐢𝐦𝐞𝐬 𝐚𝐬 𝐠𝐫𝐞𝐚𝐭
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
2.
TOPIC: AC/DC CIRCUITS
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.
A.
B.
C.
D.
11.5 mA
15.1 mA
51.1 µA
11.5 µA
ANSWER: A
Solution:
Current in BC=I1 − I3
Current in DC=I2 + I3
Applying Kirchhoff’s second law to the mesh formed by ABC and the battery, we have
2=10I1 +30I1 − I3
=40I1 − 30I3
⇨ eq.1
Similarly for mesh ABDA,
0 = 10I1 + 40I3 - 20I2
⇨ eq. 2
and for mesh BDCB
0 = 40I3 + 15I2 + I3 − 30I1 − I3
= -30I1 + 15I2 + 85I3
⇨ eq.3
Multiplying equation (2) by 3 and equation (3) by 4 and add, then we get,
0 = -90I1 + 460I3
Therefore
I1 = 5.111I3
Substituting I1 in equation 3, we have
I3 = 0.0115 A = 11.5 Ma
REFERENCE: ELECTRICAL AND ELECTRONIC TECHNOLOGY 10TH EDITION BY HUGHES
3.
TOPIC: AC/DC CIRCUITS
Using Nodal analysis, calculate the voltages V1 and V2 in the circuit of Figure below
A.
B.
C.
D.
3/10 V
10/3 V
3V
10 V
ANSWER: B
Solution: At node 1
At node 2
V1
5
+(
V1 −V2
3
V1 −V2
3
=
V2
7
1
1
V2
5
3
3
) = 1 ⇨ V1 ( + ) −
⇨
V1
3
1
1
3
7
=1
− V2 ( + ) = 0
⇨ eq.1
⇨ eq.2
From equation (2), by multiplying each term by 21,
7V1 - V2 (7 + 3) = 0
Therefore,
7V1 = 10V2
⇨ V2 =
7
10
V1
From equation (1), by multiplying each term by 15,
8V1 − 5V2 = 15 ⇨ 4.5 V1 = 15
𝐕𝟏 =
𝟏𝟎
𝟑
𝐕
REFERENCE: ELECTRICAL AND ELECTRONIC TECHNOLOGY 10TH EDITION BY HUGHES
4.
TOPIC: AC/DC CIRCUITS
Which of the following can vary with ac, but not with dc?
A.
B.
C.
D.
Power.
Voltage.
Frequency.
Magnitude.
ANSWER: C
EXPLANATION: In ac, the polarity 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 AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
5.
TOPIC: AC/DC CIRCUITS
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.
ANSWER: C
EXPLANATION: In a periodic ac wave, the 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:
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4TH ED.
6.
TOPIC: AC/DC CIRCUITS
A triangular wave exhibits:
A.
B.
C.
D.
an instantaneous rise and a defined decay
a defined rise and an instantaneous decay
a defined rise and a defined decay, and the two are equal
an instantaneous rise and an instantaneous decay
ANSWER: C
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.
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
7.
TOPIC: AC/DC CIRCUITS
What is the power factor (PF) of a purely resistive ac circuit?
A.
B.
C.
D.
0
1
0.707
without values, it cannot be determined
ANSWER: B
EXPLANATION: In ac circuits with reactance, the real power P in watts is equals I2 R,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 of the circuit. Multiplying
VI by the cosine of the phase angle provides the resistive component for real power equal to I2 R. 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.
P.F= cos θ
; P.F= cos (0) = 1
REFERENCE: GROB’S BASIC ELECTRONICS 11TH EDITION
8.
TOPIC: AC/DC CIRCUITS
A 10Ω R is in parallel with a 15- Ω XL . The applied voltage Is 120 Vac. How much is the apparent power in the circuit?
A.
B.
C.
D.
2.4 kW.
1.44 kVA.
1.44 kW.
1.73 kVA
ANSWER: D.
Solution: Given R= 10 Ω ; XL = 15Ω ; V= 120 VA
V
120
IL = A =
= 8Ω
IR =
XL
VA
R
=
IT = √(IL
15
120
10
)2 +
= 12Ω
(IR )2 = √(8)2 + (12)2 = 14.42 A
P = VI = (120)(14.42) = 𝟏𝟕𝟑𝟎. 𝟔𝟔 𝐕𝐀 𝐨𝐫 𝟏. 𝟕𝟑𝐤𝐕𝐀
REFERENCE: GROB’S BASIC ELECTRONICS 11TH EDITION
9.
TOPIC: AC/DC CIRCUITS
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:
ZT = √(R)2 + (XL − XC )2 = √(15)2 + (50 − 30)2 = 25Ω
IT =
VT
ZT
=
50 V
25Ω
=2A
X
XL −XC
R
R
θ = arctan−1 = arctan−1
= arctan−1
20
15
= 53.13°
P = VI cosθ = (50 V)(2A) cos(53.13°) = 𝟔𝟎. 𝟎𝟎𝐖
REFERENCE: GROB’S BASIC ELECTRONICS 11TH EDITION
10. TOPIC: AC/DC CIRCUITS
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.
B.
C.
D.
600 VA.
480 W.
360 W.
3.6 kVA
ANSWER: C
Solution:
Given: IT = 5A; Vac = 120 V; θ = −53.13°
P = VI cos θ = (120V)(5A) cos(−53.13°) = 𝟑𝟔𝟎. 𝟎𝟎 𝐖
REFERENCE: GROB’S BASIC ELECTRONICS 11TH EDITION
11. TOPIC: AC/DC CIRCUITS
In an ac circuit with only parallel inductors,
A. IT lags VA by 90°
B. VA lags IT by 90°
C. VA and IT are in phase
D. None of the above
ANSWER: A
EXPLANATION: Inductors 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, 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.
Remember, the voltage dropped across an inductor is a reaction against the change in current through it. Therefore,
the instantaneous voltage is zero whenever the instantaneous current is 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.
REFERENCE: GROB’S BASIC ELECTRONICS 11TH 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.
6.325 18.43°
7.325 18.43°
8.325 18.43°
9.325 18.43°
ANSWER: A
ZT = Z1 + Z2 + Z3
Solution:
= R0° + XL 90° + XC  − 90°
= R + jXL − jXC
= R + j(XL − XC ) = 6 Ω + j(10Ω − 12Ω) = 6Ω − j2Ω
ZT = 𝟔. 𝟑𝟐𝟓𝟏𝟖. 𝟒𝟑°
REFERENCE: BOYLESTAD INTRODUCTORY CIRCUIT ANALYSIS 10TH EDITION
13. TOPIC: AC/DC CIRCUITS
A transformer with a 1:6 turns ratio has 720 V across 7200 Ω in the secondary. Calculate the value of Ip .
A.
B.
C.
D.
0.5 A
0.6 A
0.7 A
0.8 A
ANSWER: B
Solution: Is =
Vs
RL
=
720 V
7200Ω
= 0.1 A
Ip = 6 × Is = 6 × 0.1 = 𝟎. 𝟔 𝐀
REFERENCE: GROB’S BASIC ELECTRONICS 11TH EDITION
14. TOPIC: AC/DC CIRCUITS
Determine the primary impedance Z P for the transformer circuit in Figure below.
A.
B.
C.
D.
182 Ω
812 Ω
128 Ω
218Ω
ANSWER: C
Np 2
Solution: Zp = ( ) × R L
Ns
4 2
= ( ) × 8Ω
1
= 16 × 8Ω
= 128 Ω
REFERENCE: GROB’S BASIC ELECTRONICS 11TH EDITION
15. TOPIC: AC/DC CIRCUITS
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.
B.
C.
D.
10 µH
15 µH
20 µH
25 µH
ANSWER: D
Solution: LM =
LTs−LTo
4
=
550−450
4
=
100
4
= 25 µH
REFERENCE: GROB’S BASIC ELECTRONICS 11TH EDITION
16. TOPIC: AC/DC CIRCUITS
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?
A.
B.
C.
D.
0.882 J
0.828 J
0.288 J
0.228 J
ANSWER: C
LI2
0.4×(1.2)2
Solution: Energy =
=
2
= 0.288 J
2
REFERENCE: GROB’S BASIC ELECTRONICS 11TH EDITION
17. TOPIC: AC/DC CIRCUITS
What is the total Z of a 600-ΩR in parallel with a 300-ΩXL? Assume 600 V for the applied voltage.
A. 286 Ω
B. 268 Ω
C. 862 Ω
D. 826 Ω
ANSWER: B
600 V
Solution: IR =
=1A
600Ω
IL =
600 V
300 Ω
=2A
IT = √IR 2 + IL 2 = √12 + 22 = √5
IT = 2.24 A
V
ZEQ = A =
IT
600 V
2.24 A
= 𝟐𝟔𝟖Ω
REFERENCE: GROB’S BASIC ELECTRONICS 11TH EDITION
18. TOPIC: AC/DC CIRCUITS
A 200-µH coil has a Q of 40 at 0.5 MHz. Find Re.
A.
B.
C.
D.
15.7 Ω
17.5 Ω
57.2 Ω
51.7 Ω
ANSWER: B
Solution: R e =
XL
Q
=
2πfL
=
Q
2π×0.5×106 ×200×10−6
628
40
=
40
= 15.7 Ω
REFERENCE: GROB’S BASIC ELECTRONICS 11TH EDITION
19. TOPIC: AC/DC CIRCUITS
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?
A. 5.42 V
B. 5.24 V
C. 4.25 V
D. 4.52 V
ANSWER: A
Solution: Note that 6 s is twice the RC time of 3 s. Then t/RC =2.
VR = antilog (log 40 − 0.434 × 2)
= antilog (1.602 − 0.868)
= antilog (0.734)
= 5.42 V
REFERENCE: GROB’S BASIC ELECTRONICS 11TH EDITION
20. TOPIC: AC/DC CIRCUITS
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?
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 vc rises to 24 V while vR drops from 36 to 12 V. Then
V
t = 2.3 RC log
v
36
= 2.3 × 0.5 × 10−3 × log
12
= 2.3 × 0.5 × 10−3 × 0.477
= 0.549 x 10−3 or 0.549 ms
REFERENCE: GROB’S BASIC ELECTRONICS 11TH EDITION
21. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
The term “semiconductor” arises from:
A.
B.
C.
D.
Resistor-like properties of metal oxides.
Variable conductive properties of some materials.
The fact that there’s nothing better to call silicon.
Insulating properties of silicon and GaAs.
ANSWER: B
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 AND ELECTRONICS BY STAN GIBILISCO 4TH ED.
22. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
The purpose of doping is to:
A.
B.
C.
D.
Make the charge carriers move faster.
Cause holes to flow.
Give semiconductor material specific properties.
Protect devices from damage in case of transients.
ANSWER: C
EXPLANATION: For a semiconductor 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. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
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 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 Ntype semiconductor, because electrons have negative (N) charge.
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
24. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
Which of the following does not result from adding an acceptor impurity?
A.
B.
C.
D.
The material becomes P type.
Current flows mainly in the form of holes.
Most of the carriers have positive electric charge.
The substance acquires an electron surplus.
ANSWER: D
EXPLANATION: If an impurity has a deficiency 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—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 AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
25. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
When a P-N junction does not conduct even though a voltage is applied, the junction is
A.
B.
C.
D.
Reverse biased at a voltage less than the avalanche voltage
overdriven
Biased past the breaker voltage.
In a state of avalanche effect.
ANSWER: A
EXPLANATION: When the battery or dc power-supply polarity is switched so the N-type 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 N-type 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—usually!
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4TH ED.
26. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
If the reverse bias exceeds the avalanche voltage in a P-N junction:
A.
B.
C.
D.
The junction will be destroyed.
The junction will insulate; no current will flow.
The junction will conduct current.
The capacitance will become extremely high.
ANSWER: C
EXPLANATION: Sometimes, a diode conducts when it is reverse-biased. The greater the reverse-bias voltage, the
more like an electrical insulator a P-N junction 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 AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
27. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
When a P-N junction is reverse-biased, the capacitance depends on all of the following except:
A.
B.
C.
D.
The frequency.
The width of the depletion region.
The cross-sectional area of the junction.
The type of semiconductor material.
ANSWER: A
EXPLANATION: Some P-N junctions can alternate between conduction (in forward bias) and nonconduction (in
reverse bias) millions or billions of times per second. 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.
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 reversebiased 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 AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
28. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
Holes flow the opposite way from electrons because:
A.
B.
C.
D.
Charge carriers flow continuously.
Charge carriers are passed from atom to atom.
They have the same polarity.
No! Holes flow in the same direction as electrons.
ANSWER: B
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 positive-to-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 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 AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
29. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
Avalanche voltage is routinely exceeded when a P-N junction acts as a:
A.
B.
C.
D.
Current rectifier.
Variable resistor.
Variable capacitor.
Voltage regulator.
ANSWER: D
EXPLANATION: The avalanche effect does not damage a P-N junction (unless the voltage is extreme). It’s a temporary
thing. When the 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, 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 AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
30. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
When a P-N junction is forward biased, conduction will not occur unless
A.
B.
C.
D.
the applied voltage exceeds the forward breakover voltage
the applied voltage is less than the forward breakover voltage
the junction capacitance is high enough
the depletion region is wide enough
ANSWER: A
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 the positive and/or negative peak voltages that signals can attain. The effect 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.
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
31. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
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 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: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
32. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
Ideally, a diode can be represented by a
A.
B.
C.
D.
Voltage source
resistance
switch
all of these
ANSWER: C
EXPLANATION: The ideal model of a diode is the least accurate approximation and can be represented by a simple
switch. When the diode is forward-biased, 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: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
33. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
In the practical diode model,
A.
B.
C.
D.
The barrier potential is taken into account
The forward dynamic resistance is taken into account
None of these
Both A and B
ANSWER: A
EXPLANATION: The practical model includes the barrier potential. When the diode is forward-biased, it is equivalent
to a closed switch in series with a small equivalent voltage source ( VF ) 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.
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
34. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
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
ANSWER B.
Solution: r =
Vr(p−p)
VDC
=
100 mV
20 V
= 0.005
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
35. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
A 60 V peak full-wave rectified voltage is applied to a capacitor-input filter. If f = 120 Hz, R L = 10 kΩ, and C =
10µF, the ripple voltage is
A.
B.
C.
D.
0.6 V
6 mV
5.0 V
2.88 V
ANSWER: C
Solution: Vp(rect) = 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
1
1
Vr(p−p) ≅ (
)V
=(
) (60) = 𝟓 𝐕
(120)(10 kΩ)(10µF)
fR L C p(rect)
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
36. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
A certain full-wave has a peak output voltage of 30 V. A 50 µF capacitor-input filter is connected to the rectifier.
Calculate the peak-to-peak ripple and the dc output voltage developed across a 600Ω load resistance.
A.
B.
C.
D.
Vr
Vr
Vr
Vr
= 8.33 V; VDC = 25.8 V
= 9.33 V; VDC = 20.0 V
= 9.80V; VDC = 30.0 V
= 10.0 V; VDC = 28.0 V
ANSWER: A
Solution: Vp(rect) = 30 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
1
1
Vr(p−p) ≅ (
) Vp(rect) = (
) (30) = 𝟖. 𝟑𝟑 𝐕
(120)(600Ω)(50µF)
fR L C
The approximate dc value of the output voltage is determined as follows:
VDC = (1 −
1
1
) Vp(rect) = (1 −
) (30V) = 𝟐𝟓. 𝟖𝟑 𝐕
(240Hz)(600Ω)(50µF)
2fR L C
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
37. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
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.
B.
C.
D.
Vp= 17 V ; PIV= 17.7 V
Vp= 15.6 V ; PIV= 16.3 V
Vp= 12 V ; PIV= 12.7 V
Vp= 18.4 V ; PIV= 19.1 V
ANSWER: B
Solution: The peak output voltage (taking into account the two diode drops) is
Vp(sec) = 1.414Vrms = 1.414(12 V) ≅ 17 V
Vp(out) = Vp(sec) − 1.4 V = 17 V − 1.4 V = 𝟏𝟓. 𝟔 𝐕
The PIV rating for each diode is
PIV = Vp(out) + 0.7 V = 15.6 V + 0.7 V = 𝟏𝟔. 𝟑𝐕
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
38. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
Determine the peak value of the output voltage in the figure below if the turns ratio is 0.5
A.
B.
C.
D.
85 V
85.7 V
84.3 V
170 V
ANSWER: C
Solution: Vp(pri) = Vp(in) = 170 V
The peak secondary voltage is
Vp(sec) = n Vp(pri) = 0.5(170) = 85 V
The rectified peak output voltage is
Vp(out) = Vp(sec) − 0.7 V = 85 V − 0.7 V = 84.3 V
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
39. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
What is the average value of half-wave rectified voltage in the figure below?
A.
B.
C.
D.
7.96 V
50 V
0V
15.9 V
ANSWER: D
Solution: VAVG =
Vp
π
=
50
π
= 𝟏𝟓. 𝟗 𝐕
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
40. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
In many cases, a PNP transistor can be replaced with an NPN device and the circuit will do the same thing, provided
that
A.
B.
C.
D.
the power supply or battery polarity is reversed
the collector and the emitter is interchanged
the arrow is pointing inward
Forget it! A PNP transistor can never be replaced with an NPN transistor
ANSWER: A
EXPLANATION: It’s easy to tell whether a 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 powersupply polarity can be reversed, and the circuit will work in the same way—as long as the new device has the
appropriate specifications.
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
41. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
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
ANSWER: A
EXPLANATION: The heavily doped n-type emitter region has a very high density of conduction-band (free) electrons,
as indicated in Figure 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 region recombine with holes and move as valence electrons through the base region 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: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
42. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
When operated in cutoff and saturation, the transistor acts like a
A.
B.
C.
D.
Linear amplifier
Switch
Variable capacitor
Variable resistor
ANSWER: B
EXPLANATION: A transistor can be operated as an electronic switch in cutoff and saturation. In cutoff, both pn
junctions are 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 forward-biased and the collector current is
maximum. The transistor behaves like an open switch between collector and emitter.
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
43. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
A JFET always operates with
A.
B.
C.
D.
The gate-to-source pn junction reversed biased
The gate-to-source pn junction forward- biased
The drain connected to ground
The gate connected to the source
ANSWER: A
EXPLANATION: To illustrate the operation of a JFET, Figure 8–2 shows dc bias voltages applied to an n-channel device.
VDD provides a drain-to-source voltage and supplies current from
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: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
44. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
For the JFET in Figure below, VGS(off) = −4 V and IDSS = 12 mA, Determine the minimum value of VDD required to
put the device in the constant-current region of operation when VGS = 0 V.
A.
B.
C.
D.
6.72 V
2.72 V
10. 72 V
4V
ANSWER: C
Solution: Since VGS(off) = −4 V, Vp = 4 V. The minimum value of VDS for the JFET to be in its constant-current region
is
VDS = Vp = 4 V
In the constant-current region with VGS = 0 V
ID = IDSS = 12 mA
VRD = ID R D = (12 mA)(560Ω) = 6.72 V
VDD = VDS + VRD = 4 V + 6.72 V = 𝟏𝟎. 𝟕𝟐 𝐕
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
45. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
Determine the drain-to-source voltage in the circuit of Figure below. The MOSFET datasheet gives VGS = −8 V and
IDSS = 12 mA
A.
B.
C.
D.
18 V
10.6 V
0V
None of the above
ANSWER: B
Solution:
VDS = VDD − IDSS R D = 18 V − (12 mA)(620Ω) = 𝟏𝟎. 𝟓𝟔 𝐕 or 10.6 V
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
46. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
A certain transistor is to be operated withVCE = 6 V. If its maximum power rating is 250 mW, what is the most
collector current that it can handle?
A.
B.
C.
41.7 mA
41.7 nA
41.7 µA
D.
0.47 A
ANSWER: A
Solution: IC =
PD(max)
=
VCE
250 mW
6V
= 𝟒𝟏. 𝟔𝟕 𝐦𝐀
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8 th ED.
47. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
Determine the voltage gain and the ac output voltage in the given figure if r′e = 50Ω
A.
B.
C.
D.
Av
Av
Av
Av
= 20 ;
= 20 ;
= 20 ;
= 20 ;
ANSWER: A
Solution: Av ≅
RC
r′e
=
Vout
Vout
Vout
Vout
= 2 V rms
= 4 V rms
= 5 V rms
= 10 V rms
1.o kΩ
50Ω
Vout = Av Vb = (20)(100mV) = 𝟐 𝐕 𝐫𝐦𝐬
= 𝟐𝟎
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
48. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
Determine the VCE and IC in the stiff voltage-divider biased transistor circuit of the figure shown if βDC = 100
A.
B.
C.
D.
IC
IC
IC
IC
= 6.41 mA ; VCE = 3.59 V
= 5.16 mA ; VCE = 1.95 V
= 5.16 mA; VCE = 2.89 V
= 6.41 mA; VCE = 4.84 V
ANSWER: B
Solution: VB ≅ (
R2
R1 +R2
) VCC = (
5.6 kΩ
15.6 kΩ
) (10 V) = 3.59 V
VE = VB − VBE = 3.59 V − 0.7V = 2.89 V
IE =
VB
RE
=
2.89 V
560Ω
= 5.16 mA ; IC ≅ IE = 𝟓. 𝟏𝟔 𝐦𝐀
VC = VCC − IC R C = 10 V − (5.16 mA)(1.0 kΩ) = 4.84 V
VCE = VC − VE = 4.84 V − 2.89 V = 𝟏. 𝟗𝟓 𝐕
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
49. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
Determine the dc input resistance looking in at the base of the transistor in the figure below. βDC = 125 And VB =
4V
A.
B.
C.
D.
132 kΩ
142 kΩ
152 kΩ
162 kΩ
ANSWER: C
Solution:
IE =
VB −0.7V
RE
R IN(BASE) =
=
4 V−0.7 V
= 3.3 mA
1.0 kΩ
βDC VB
125(4V)
IE
=
3.3 mA
= 𝟏𝟓𝟐 𝐤Ω
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
50. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
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.
A.
B.
C.
D.
VDS
VDS
VDS
VDS
ANSWER: D
Solution:
= 10 V ; VGS = 1.1 V
= 13.9 V ; VGS = −1.1 V
= 9 V ; VGS = 1.1 V
= 8.9 V ; VGS = −1.1 V
VS = ID R S = (5 mA)(220Ω) = 1.1 V
VD = VDD − ID R D = 15 V − (5 mA)(1.0 kΩ) = 15 V − 5 V = 10 V
VDS = VD − VS = 10 V − 1.1 V = 𝟖. 𝟗 𝐕
Since VG = 0 V,
VGS = VG − VS = 0 V − 1.1 V = −𝟏. 𝟏 𝐕
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
51. TOPIC: SEMICONDUCTORS INCLUDING DIODES AND TRANSISTORS
Determine the value of R S required to self-bias a p-channel JFET with datasheet values of IDSS =25 mA and
VGS(OFF) =15 V. VGS is to be 5 V.
A.
B.
C.
D.
250Ω
350Ω
450Ω
550Ω
ANSWER: C
ID ≅ IDSS (1 −
Solution:
VGS
VGS(OFF)
VGS
RS = |
ID
2
) = (25 mA) (1 −
|=
5V
11.1 mA
5V 2
15 V
) = 11.1 mA
= 𝟒𝟓𝟎Ω
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
52. TOPIC: INDUSTRIAL AND POWER ELECTRONICS
The forward breakover voltage of an SCR
A.
B.
C.
D.
decreases as the gate current increases.
cannot be controlled by gate current.
increases as the gate current increases.
none of the above
ANSWER: A
EXPLANATION:
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 11TH EDITION
53. TOPIC: INDUSTRIAL AND POWER ELECTRONICS
An RC phase-shift network is used in SCR and triac circuits to
A.
B.
C.
D.
Control the conduction angle of the thyristor
Handle some of the load current
Vary the holding current
None of the above
ANSWER: A
EXPLANATION:
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 11TH EDITION
54. TOPIC: INDUSTRIAL AND POWER ELECTRONICS
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.
A.
B.
C.
D.
20.1 µA
19.1 mA
20.1mA
19.1 µA
ANSWER: B
Solution: VA = 0.9 V ⇨ voltage at the anode
VRS = VBIAS − VA = 20 V − 0.9 V = 19.1 V
V
19.1 V
IA = RS =
= 𝟏𝟗. 𝟏 𝐦𝐀
RS
1.0 kΩ
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
55. TOPIC: INDUSTRIAL AND POWER ELECTRONICS
Determine the gate trigger current and the anode current when the switch, SW1, is momentarily closed in Figure
below. Assume VAK = 0.2 V, VGK = 0.7 V and IH = 5 mA
IG
A.
B.
C.
D.
IG
IG
IG
IG
= 410 mA ; IA = 448 µA
= 450 µA ; IA = 448 mA
= 400 mA ; IA = 448 µA
= 410 µA ; IA = 448 mA
ANSWER: D
V
−V
3 V−0.7 V
Solution: IG = TRIG GK =
= 𝟒𝟏𝟎 µ𝐀
RG
5.6 kΩ
IA =
VA −VK
RA
=
15 V−0.2 V
33 Ω
= 𝟒𝟒𝟖 𝐦𝐀
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
56. TOPIC: INDUSTRIAL AND POWER ELECTRONICS
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.
A.
B.
C.
D.
30°
45°
60°
90°
ANSWER: A
Solution:
v = Vm sin θ
Here, v = 100 V, Vm = 200 V
∴ 100 = 200 sin θ
θ = sin−1 (0.5) = 30°
Firing angle, α = θ = 30°
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
57. TOPIC: INDUSTRIAL AND POWER ELECTRONICS
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 BY V.K MEHTA
58. TOPIC: INDUSTRIAL AND POWER ELECTRONICS
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
Solution: Let Im be the peak load current
V
Iav = m (1 + cos α)
2πRL
=
Im =
Im
2π
(1 + cos α)
2πIav
(Q Im =
Vm
RL
)
(1+cos α)
Iav = Idc = 1 A; α = 30°
Im =
2π(1)
(1+cos 30)
= 3.36 A
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
59. TOPIC: INDUSTRIAL AND POWER ELECTRONICS
In figure below, the switch is closed. A diac with a breakover voltage VBO = 30 V 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?
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 VA at
point A becomes equal to VBO of diac plus gate triggering voltage VGT of the triac, the triac is fired into conduction.
Therefore,
VA = VBO + VGT = 30 V + 1V = 𝟑𝟏 𝐕
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
60. TOPIC: INDUSTRIAL AND POWER ELECTRONICS
A unijunction transistor has 10 V between the bases. If the intrinsic stand off ratio is 0.65, What will be the peak
voltage if the forward voltage drop in the pn junction is 0.7 V?
A.
B.
C.
D.
5.2 V
6.2 V
7.2 V
8.2 V
ANSWER: C
Solution: VBB = 10 V; Ƞ = 0.65; VD = 0.7 V
Stand off voltage= ȠVBB = 0.65 × 10 = 6.5 V
Peak-point voltage, VP = ȠVBB + VD = 6.5 + 0.7 = 𝟕. 𝟐 𝐕
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
61. TOPIC: MICROELECTRONICS INCLUDING OP AMP
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: 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 DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
62. TOPIC: MICROELECTRONICS INCLUDING OP AMP
The slew-rate specification of an op amp is the
A.
B.
C.
D.
Maximum value of positive or negative output voltage
Maximum rate at which its output voltage can change
Attenuation against a common-mode signal
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, which is its slew rate. Figure 33–9 illustrates this concept. Here the op amp’s 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 rating of the op amp, the waveform
appears triangular. Therefore, slew-rate distortion of a sine wave produces a triangular wave, such as waveform B.
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
63. TOPIC: MICROELECTRONICS INCLUDING OP AMP
The input impedance of an inverting amplifier is approximately equal to
A.
B.
C.
D.
Ri
zero
Infinity
RF
ANSWER: A
EXPLANATION:
The circuit is called an inverting amplifier because the input and output signals are 180_ out of phase. The 180° phase
inversion occurs because V in 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 the op amp’s inverting input is at the same potential as ground, yet 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.
REFERENCE: GROB’S BASIC ELECTRONICS 11TH EDITION
64. TOPIC: MICROELECTRONICS INCLUDING OP AMP
In LM317 voltage regulator shown in the figure below, R 2 is adjusted to 2.4 kΩ. If the value of R1 is 240 Ω,
determine the regulated d.c. output voltage for the circuit.
A.
B.
17.53 V
15.37 V
C.
D.
13.57 V
13.75 V
ANSWER: D
Vout = 1.25 (
Solution:
= 1.25 (
R2
R1
2.4 kΩ
240 Ω
+ 1)
+ 1) = 𝟏𝟑. 𝟕𝟓 𝐕
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
65. TOPIC: MICROELECTRONICS INCLUDING OP AMP
Determine the frequency of the circuit shown in the figure below. Given that R1 = 3 kΩ; R 2 = 2.7 kΩ and C =
0.033µF
A.
B.
C.
D.
1.59 KHz
1.95 KHz
5.91 KHz
5.19 KHz
ANSWER: D
Solution:
f = (R
1.44
1 +2R2 )C
R1 + 2R 2 = 3 kΩ + (2)(2.7 kΩ) = 8.4kΩ
f = (8.4
1.44
kΩ)(0.033×10−6 )
= 𝟓. 𝟏𝟗 𝐊𝐇𝐳
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
66. TOPIC: MICROELECTRONICS INCLUDING OP AMP
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.
B.
C.
D.
ACM
ACM
ACM
ACM
ANSWER: B
Solution:
=0.0081 and CMRR dB = 80.8 dB
=0.0081 and CMRR dB =81.8 dB
=0.0081 and CMRR dB =82.8 dB
=0.0081 and CMRR dB = 83.8 dB
vin(CM) = 3.2 V ; vout = 26 mV
ACM =
vout
vin(CM)
=
CMRR dB = 20
26 mV
= 𝟎. 𝟎𝟎𝟖𝟏 ⇨ common mode voltage gain
3.2
A
lo10 DM
A
CM
= 20log
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
100
0.0081
= 𝟖𝟏. 𝟖 𝐝𝐁
67. TOPIC: MICROELECTRONICS INCLUDING OP AMP
For the circuit shown below, find the common mode voltage gain and the CMRR in dB
A.
ACM = 0.5 ; CMRR dB = 45.09 dB
B.
ACM = 0.25 ; CMRR dB = 46.09 dB
C.
ACM = 0.5 ; CMRR dB = 48.09 dB
D.
ACM = 0.25 ; CMRR dB = 47.09 dB
ANSWER: D
ACM =
Solution:
RC
2RE
=
100 kΩ
2x200 kΩ
𝐕𝐄𝐄 −𝐕𝐁𝐄
𝐈𝐄 =
r′e =
25 mV
IE1
=
25 mV
28.25 µA
= 𝟎. 𝟐𝟓 ⇨ common mode voltage gain
𝟏𝟐 𝐕−𝟎.𝟕 𝐕
=
= 56.5µA ⇨ tail current
𝟐𝟎𝟎 𝐤Ω
IE⁄
56.5 µA⁄
IE1 = IE2 = 2 =
2 = 28.25 µA ⇨ d.c emitter current in each transistor
= 884.96 ⇨ a.c emitter resistance
𝐑𝐄
𝐀𝐃𝐌 =
𝐑𝐂
=
𝟐r′ e
𝟏𝟎𝟎 𝐤Ω
= 56.5 ⇨ Differential voltage gain
𝟐𝐱𝟖𝟖𝟒.𝟗𝟔
𝐀 𝐃𝐌
CMMR dB = 20 log
ACM
= 20 log
56.5
0.25
= 𝟒𝟕. 𝟎𝟗 𝐝𝐁
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
68. TOPIC: MICROELECTRONICS INCLUDING OP AMP
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 (Vpk ) is approximately 8 V.
fmax =
Slew rate
=
V
µs
0.5
2π (Vpk)
2π x 8
500 kHz
=
2π x 8
= 9.95 kHz
⇨ (0.5 V/µs= 500 kHz)
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
69. TOPIC: MICROELECTRONICS INCLUDING OP AMP
For the noninverting amplifier circuit shown , find peak-to-peak output voltage
A.
B.
C.
D.
6V
12 V
18 V
24 V
ANSWER: B
Solution:
the input signal is 2 V peak-to-peak
ACL = 1 +
Rf
Ri
=1+
5 kΩ
1 kΩ
= 6 ⇨ Voltage gain
ACL xvinp = 6 x 2 = 𝟏𝟐 𝐕 ⇨peak-to-peak output voltage
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
70. TOPIC: MICROELECTRONICS INCLUDING OP AMP
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 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 size but
differs in that it has 8, 10 or 12 pigtail-type leads. The close leads spacing and the difficulty of removal from a printed
circuit board has diminished the popularity of this package with the users.
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
71. TOPIC: INSTRUMENTATION AND MEASUREMENTS
An ammeter is connected in __________ with the circuit element whose current we wish to measure.
A.
B.
Series
Parallel
C.
D.
Series-parallel
None of the above
ANSWER: A
EXPLANATION: Ammeter measures 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: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
72. TOPIC: INSTRUMENTATION AND MEASUREMENTS
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 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: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
73. TOPIC: INSTRUMENTATION AND MEASUREMENTS
An ideal ammeter has _____resistance.
A.
B.
C.
D.
Low
Infinite
Zero
High
ANSWER: C
EXPLANATION: Just like voltmeters, 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: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
74. TOPIC: INSTRUMENTATION AND MEASUREMENTS
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: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
75. TOPIC: INSTRUMENTATION AND MEASUREMENTS
The resistance of an ideal voltmeter is
A.
B.
C.
D.
Low
Infinite
Zero
High
ANSWER: B
EXPLANATION: Since voltmeters are 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: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
76. TOPIC: INSTRUMENTATION AND MEASUREMENTS
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.
6.88 V
7.88 V
8.88 V
9.88 V
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
40 x 10
=
+ 10 = 8 + 10 = 18 kΩ
40+10
20 V
Circuit current=
= 1.11 mA
18 kΩ
Voltage read by multimeter= 8 k Ω x1.11 mA=8.88 V
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
77. TOPIC: INSTRUMENTATION AND MEASUREMENTS
A PMMC instrument with a full-scale deflection (f.s.d) current of 100 µA and R m = 1 kΩ is to be used as a voltmeter
of range 0-100 V (r.m.s). The diodes used in the bridge rectifier circuit are of silicon. Calculate the value of multiplier
resistor R S required.
A.
B.
C.
D.
890.7 kΩ
790.7 kΩ
690.7 k Ω
590.7 k Ω
ANSWER: A
Solution: 100 µA ⇨ average current
Im(f.s.d) = 100 µA ⇨ F.S.D current of meter
R T = R S + R m = (R S + 1000Ω)
⇨ total circuit resistance
Vm = √2 Vr.m.s = √2 x 100 V = 141.4 V ⇨ peak value of applied voltage
Total rectifier drop = 2VF = 2 x 0.7V = 1.4 V
100 µA
0.637
=
141.4 V−1.4 V
RS +1000
R S = 𝟖𝟗𝟎. 𝟕 𝐤Ω
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
78. TOPIC: INSTRUMENTATION AND MEASUREMENTS
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.
A.
B.
C.
D.
40 V
50 V
60 V
70 V
ANSWER: C
75 µA
Solution: Peak f.s.d. meter current=
0.637
Now Peak f.s.d. meter current=
75 µA
Or
0.637
=
Peak applied voltage−Rectifier drop
Total circuit resistance
√2 Vr.m.s−2(0.7)
Rs+Rm
Vr.m.s = 𝟔𝟎 𝐕
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
79. TOPIC: INSTRUMENTATION AND MEASUREMENTS
A multimeter has full scale deflection current of 1 mA. Determine its sensitivity.
A.
B.
C.
10 Ω/V
100 Ω/V
1000 Ω/V
D.
10, 000 Ω/V
ANSWER: A
Solution:
Full scale deflection current, Ig = 1 mA
Multimeter sensitivity =
1
Ig
=
1
1 mA
= 1000 Ω/V
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
80. TOPIC: INSTRUMENTATION AND MEASUREMENTS
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 sensitivity x applied voltage
= 0.01 x 400= 4 mm
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
81. TOPIC: DIGITAL 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 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 Negative-AND gate is not the same as a
NAND gate. Its truth table, actually, is identical to a NOR gate:
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
82. TOPIC: DIGITAL ELECTRONICS
Suppose an AND gate is followed by an inverter. Under what conditions is the output of the resulting black box low?
A.
B.
C.
D.
If and only if both inputs are high
If and only if both inputs are low
If and only if one input is high and the other is low
Under no conditions (the output is always high)
ANSWER: A
EXPLANATION: A variation on the idea of 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).
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
83. TOPIC: DIGITAL ELECTRONICS
In Boolean algebra, addition represents
A.
B.
C.
D.
The logical NOT Operation
The logical AND operation
The logical OR operation
The logical NAND operation
ANSWER: C
EXPLANATION: Boolean algebra is a system of mathematical logic using the numbers 0 and 1 with the operations AND
(multiplication), OR (addition), and 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.
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
84. TOPIC: DIGITAL ELECTRONICS
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
ANSWER: C
EXPLANATION: An inverter or NOT gate has one input and one output. It reverses the state of the input.
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
85. TOPIC: DIGITAL ELECTRONICS
DeMorgan’s Theorem states that, for all logical statements X and Y,
A.
B.
C.
D.
–(X*Y) is equivalent to X+Y
X*Y is equivalent to –(X+Y)
(-X)+(-Y) is equivalent to X*Y
(-X)+(-Y) is equivalent to –(X*Y)
ANSWER: D
EXPLANATION: Statements on either 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: TEACH YOURSELF ELECTRICITY AND ELECTRONICS BY STAN GIBILISCO 4 TH ED.
86. TOPIC: DIGITAL ELECTRONICS
Convert the decimal number 37 to its equivalent binary number.
A.
B.
C.
D.
(101001)2
(100101)2
(100010)2
(101010)2
ANSWER: B
Solution:
Division
37/2=18
18/2=9
9/2=4
Remainder
1
0
1
4/2=2
2/2=1
1/2=0
0
0
1
Therefore, (37)10 = (𝟏𝟎𝟎𝟏𝟎𝟏)𝟐
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
87. TOPIC: DIGITAL ELECTRONICS
Convert the decimal number 76 to octal equivalent.
A.
B.
C.
D.
(114)8
(141)8
(411)8
(414)8
ANSWER: A
Solution:
Division
Remainder
76/8=9
4
9/8=1
1
1/8=0
1
Therefore,
(76)10 = (𝟏𝟏𝟒)𝟖
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
88. TOPIC: DIGITAL ELECTRONICS
Convert decimal number 423 to hexadecimal number
A.
B.
C.
D.
(1B7)16
(1C7)16
(1D7)16
(1A7)16
ANSWER: D
Solution:
Division
423/16=26
26/16=1
1/16=0
Remainder
7
10
1
Therefore, (423)10 = (𝟏𝐀𝟕)𝟏𝟔
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
89. TOPIC: DIGITAL ELECTRONICS
Using Boolean techniques, simplify the following expression
Y= AB + A (B+C) +B (B+C)
A.
B.
C.
Y= A+BC
Y=C+AB
Y=B+AC
D.
Y= A+B+C
ANSWER: C
Solution:
Y= AB + A (B+C) +B (B+C)
Y= AB + AB + AC + BB + BC
Y= AB + AB + AC + B + BC
Y= AB + AC + B + BC
Y= AB + AC + B (1+C)
Y= AB + AC + B * 1
Y= AB + AC + B
Y= B + (A+1) + AC
Y= B * 1 + AC
Y= B + AC
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
90. TOPIC: DIGITAL ELECTRONICS
Simplify the given Boolean expression:
Y= (A + B + C) * (A + B)
A.
B.
C.
D.
Y= A + B
Y= AB
Y= A + B + C
Y= ABC
ANSWER: B
Solution:
Y= (A + B + C) * (A + B)
= A*A + A * B + B * A + B * B + C * A + C * B
Y = A + AB + AB + B + AC + BC
= A + AB + B + AC + BC
= A+ B + AC + BC
= A (1 + C) + B (1 + C)
= A * 1+ B * 1
Y=A+B
REFERENCE: PRINCIPLES OF ELECTRONICS BY V.K MEHTA
91. TOPIC: OTHERS
In the case of line regulation,
A.
B.
C.
When the temperature varies, the output voltage stays constant
When the output voltage changes, the load current stays constant
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 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: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
92. TOPIC: OTHERS
The basic difference between a series regulator and a shunt regulator is
A.
B.
C.
D.
The amount of current that can be handled
The position of the control element
The type of sample circuit
The type of error detector
ANSWER: B
EXPLANATION: A basic voltage consists 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
parallel with the load.
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
93. TOPIC: OTHERS
In a switching regulator, the control transistor is conducting
A.
B.
C.
D.
Part of the time
All of the time
Only when the input voltage exceeds a set limit
Only when there is an overload
ANSWER: A
EXPLANATION: A much greater efficiency can be realized with a 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 DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
94. TOPIC: OTHERS
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
line regulation =
Solution:
(
∆VOUT
⁄V
OUT)100%
∆VIN
=
(0.25 V⁄15 V)
5V
= 𝟎. 𝟑𝟑𝟑% 𝐕
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
95. TOPIC: OTHERS
Determine the output voltage for the regulator in the given figure.
A.
B.
C.
D.
6.2 V
7.2 V
9.2 V
10.2 V
ANSWER: D
Solution:
VREF = 5.1 ⇨ the zener voltage
VOUT = (1 +
R2
R1
) VREF = (1 +
10 kΩ
10 kΩ
) (5.1 V) = 𝟏𝟎. 𝟐 𝐕
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
96. TOPIC: OTHERS
Determine the maximum current that the regulator in the figure below can provide to a load.
A.
B.
C.
D.
0.7 A
0.8 A
0.9 A
0.10 A
ANSWER: A
Solution:
IL(max) =
0.7 V
R4
=
0.7 V
1.0 Ω
= 𝟎. 𝟕 𝐀
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
97. TOPIC: OTHERS
In figure below, what power rating must R1 have if the maximum input voltage is 12.5 V?
A.
B.
C.
D.
5.10 W
6.10 W
7.10 W
8.10 W
ANSWER: C
Solution:
VR1 = VIN − VOUT = 12.5 V
PR1 =
VR1 2
R1
=
(12.5 V)2
22Ω
= 𝟕. 𝟏𝟎 𝐖
REFERENCE: ELECTRONIC DEVICES Conventional Current Version by THOMAS L. FLOYD 8th ED.
98. TOPIC: OTHERS
The primary and secondary windings of a 500 kVA transformer have resistances of 0.42 Ω and 0.0019 Ω respectively.
The primary and 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.
B.
C.
D.
98.3%
93.8%
89.3%
83.9%
ANSWER: A
Solution: Full-load secondary current is,
500×1000
≃
Full-load primary current is,
= 1250A
400
500×1000
11000
= 45.5 A
Therefore secondaryI2 R loss on full load is,
(1250)2 × 0.0019 = 2969 W
2
and primary I2R loss on full load is, (45.5) × 0.42 = 870W
Total I2R loss on full load = 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−
(1 −
losses
input power
6.74
) = 0.983 unit
406.74
= 98.3%
REFERENCE: ELECTRICAL AND ELECTRONIC TECHNOLOGY 10TH EDITION BY HUGHES
99. TOPIC: OTHERS
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.
A.
2.50 Ω
B.
C.
D.
2.05 Ω
5.02 Ω
5.20 Ω
ANSWER: B
Solution:
𝑉
𝑅𝑒 = 𝑅1 + 𝑅2′ = 𝑅1 + 𝑅2 ( 1 )2 ⇨ single resistance in the primary circuit equivalent to the
𝑉2
primary and secondary resistances of the actual transformer
400 2
)
80
𝑅𝑒 = 0.3 + 0.01(
= 0.55Ω
𝑉
𝑋𝑒 = 𝑋1 + 𝑋2′ = 𝑋1 + 𝑋2 ( 1 )2 ⇨ single reactance in the primary circuit
𝑉2
400 2
)
80
𝑋𝑒 = 1.1 + 0.035(
= 1.975Ω
𝑍𝑒 = √𝑅𝑒 2 + 𝑋𝑒 2 ⇨ equivalent impedance of the primary and secondary windings
referred to the primary circuit
𝑍𝑒 = √0.552 + 1.9752 = 𝟐. 𝟎𝟓Ω
REFERENCE: ELECTRICAL AND ELECTRONIC TECHNOLOGY 10 TH EDITION BY HUGHES
100. TOPIC: OTHERS
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
ANSWER: A
Solution:
I2′ × 1000 = 280 × 200
⇨ I2′ = 56 A
Cos ∅2 = 0.8 ⇨ sin ∅2 = 0.6
Cos ∅0 = 0.2 ⇨ sin ∅0 = 0.98
I1 cos ∅1 = I2′ cos ∅2 + I0 cos ∅0
= (56 × 0.8) + (3 × 0.2) = 45.4 A
I1 sin ∅1 = I2′ sin ∅2 + I0 sin ∅0
= (56 × 0.6) + (3 × 0.98) = 36.54 A
36.54
Tan ∅1 =
= 0.805
45.4
∅1 = 38°50′
cos ∅1 = cos 38°50′ = 𝟎. 𝟕𝟖 𝐥𝐚𝐠𝐠𝐢𝐧𝐠
REFERENCE: ELECTRICAL AND ELECTRONIC TECHNOLOGY 10 TH EDITION BY HUGHES
AC/DC CIRCUITS
101. A 95-V battery is connected in series with three resistors: 20 𝛺, 50 𝛺 and 120 𝛺. Find the voltage across 120-ohms
resistor.
A.
B.
C.
D.
60 V
58 V
70 V
80 V
ANSWER: A
Solution:
RT = R1 + R2 + R3
RT = 20 + 50 + 120
RT = 190 𝛺
VT = IRT
I = VT / RT
I = 95 V / 190 𝛺
I = 0.5 A
In series circuit, the current is the same in each part; that is, I=0.5 A through each resistor.
V1 = (0.5A)(20𝛺) = 10 V
V2 = (0.5A)(50 𝛺) = 25 V
V3 = (0.5A)(120 𝜴) = 60 V
REFERENCE: SCHAUMS OUTLINES, BASIC CIRCUIT ANALYSIS
102. In the circuit shown, find the total power dissipated by R1 and R2.
A.
B.
C.
D.
140 W
250 W
240 W
150 W
ANSWER: C
Solution:
I = VT / RT
I=4A
;
I = 60 V/ (5 𝛺+ 10 𝛺)
PT = IVT
PT = (4 A) (60 V
PT = 240 W
REFERENCE: SCHAUMS OUTLINES, BASIC CIRCUIT ANALYSIS
103. Resistors, R1, R2 and R3 are in series with a 100-V source. The total voltage drop across R1 and R2 is 50 V, and that
across R2 and R3 is 80 V. Find the three resistances if the total resistances is 50 𝛺.
A.
B.
C.
D.
R1=10 𝛺; R2=15 𝛺;R3=25 𝛺
R1=5 𝛺; R2=10 𝛺;R3=35 𝛺
R1=10 𝛺; R2=18 𝛺;R3=22 𝛺
R1=10 𝛺; R2=20 𝛺;R3=20 𝛺
ANSWER: A
Solution:
I = VT / RT
I = 100 V / 50𝛺
I=2A

Since the voltage drop across R1 and R2 is 50 V; then the voltage drop across R3 is 100V-50V = 50 V.
By ohms’ law: R3 = 50/2 = 25 𝛺

Resistors R2 and R3 have 80 V across them, leaving 100 V -80 V = 20 V across R1
By ohms’ law: R1 = 20/2 = 10 𝛺

R2 = RT – R1 – R3
R2 = 50 𝛺 – 25 𝛺 – 10 𝛺
R2 = 15 𝛺
REFERENCE: SCHAUMS OUTLINES, BASIC CIRCUIT ANALYSIS
104. What is the total resistance of thirty 6- 𝛺 resistor connected in series?
A.
B.
C.
D.
160 𝛺
170 𝛺
180 𝛺
190 𝛺
ANSWER: C
Solution:
The total resistance is the number of resistor times the common resistance of 6- 𝛺.
Total Resistance = 30 x 6 𝛺 = 180 𝛺
REFERENCE: SCHAUMS OUTLINES, BASIC CIRCUIT ANALYSIS
105. Find the mesh current in the circuit shown below:
A.
B.
C.
D.
I1 = 5A; I2 = 7A;I3 = -6 A
I1 = 6A; I2 = -8A;I3 = 5 A
I1 = 4A; I2 = 7A;I3 = 5 A
I1 = -4A; I2 = 7A;I3 = -6 A
ANSWER: A
Solution:
At Mesh1
9I1 – 5I2= 10 (Equation 1)
At Mesh2
-5I1+ 11I2 = 52 (Equation 2)
By Elimination:
5(9I1 – 5I2 = 10)
9(-5I1 + 11I2 = 52)
9
Then: I2 = 518/74 = 7 AI
45I1 -25I2 =50
-45I1 + 99I2 = 468
1
By substitution:
–
9I1 – 5I2 = 10
Then: I1 = 5A
9I1 – 5 (7A) = 10
9
I
5
I
From the original circuit, I2 –1 I3 = 13 A
2
I3 = I2 -13
I3 = 7 -13
I3 = -6 A
–
=
5
I 1 BASIC CIRCUIT ANALYSIS
REFERENCE: SCHAUMS OUTLINES,
0
2
106. Find the power dissipated =in the 20- 𝛺 resistor of the circuit by using nodal analysis.
A.
B.
C.
D.
8.466 W
6.544 W
7.466 W
8.544 W
1
0
Answer: P = 7.466 Wlxdz
Solution:
To use the nodal analysis, we convert to the 10-V
source to an equivalent current source to obtain
The figure (b).
For node 1:
1=
𝑉1
10
+
𝑉1
4
For node 2:
+
𝑉1−𝑉2
20
𝑉1−𝑉2
20
=2+
𝑉2
8
Solving for V1 and V2 yields V1=1.11 V and V2 = -11.11 V
I20 ohms =
𝑉1−𝑉2
20
=
1.11−(−11.11)
20
= 0.611 𝐴
P20 ohms = (0.611 A)2 (20) = 7.466 W
REFERENCE: SCHAUMS OUTLINES, BASIC CIRCUIT ANALYSIS
107. It is a circuit analysis method that allows us to convert any linear circuit, or more often a portion of a circuit, into a
simple equivalent circuit.
A.
B.
C.
D.
Kirchhoff’s Law
Millman’s Theorem
Thevenin’s Theorem
Norton’s Theorem
ANSWER: C
EXPLANATION: Thevenin’s Theorem is a linear circuit analysis technique that reduces a circuit (or a portion of a
circuit) to an equivalent circuit. The equivalent circuit consists of a single voltage source (VTH) and a single series
resistance (RTH).
REFERENCE: FUNDAMENTAL OF ELECTRONICS DC/AC CIRCUITS
108. Obtain the Norton equivalent circuit for the network and determine the current in the 50-ohm resistor.
A.
B.
C.
D.
0.70 A
0.75 A
0.80 A
1.5 A
ANSWER: B
Solution:
With the 50-ohms resistor short-circuited, the total current will flow through the short circuit.
Thus, IN = 0.75 + 0.75 = 1.5 A
Next, with the sources and the 50-ohms resistor removed, the two 100-ohms become in parallel resulting in a net
resistance of 50 ohms.
Therefore,
GN = 1/50 = 0.02 S
I50-ohms = IN (
𝐺50
𝐺50 + 𝐺𝑁
)= (1.5 A)(
0.02
0.02 + 0.02
) = 0.75 A
REFERENCE: SCHAUMS OUTLINES, BASIC CIRCUIT ANALYSIS
109. This theorem produces an equivalent circuit that consists of a single current source and a single parallel resistor.
A.
B.
C.
D.
Kirchhoff’s Law
Millman’s Theorem
Thevenin’s Theorem
Norton’s Theorem
ANSWER: B
EXPLANATION: Millman’s Theorem is another circuit analysis tool that is helpful to a technician for certain types of
problems. Millman’s Theorem is particularly well suited to simplifying circuits that have several parallel branches and
multiple voltage and/or current sources. Figure 7-56 shows such a circuit. Applying Millman’s Theorem results in an
equivalent circuit that consists of a single current source (Im) and a single parallel resistance (Rm).
REFERENCE: FUNDAMENTAL OF ELECTRONICS DC/AC CIRCUITS, MILLER
110. Through _______, it is often possible to convert a complex circuit into a standard series-parallel circuit.
A.
B.
C.
D.
Pi-to-tee conversion
Norton’s Theorem
Thevenin’s Theorem
Millman’s Theorem
ANSWER: A
EXPLANATION:
(a) Tee Configuration
It is called a tee because it resembles the letter “T” when drawn on a schematic.
(b) Wye Configuration
It is called a wye because it resembles the letter “Y” when drawn on a schematic.
In general terms, tee and wye are interchangeable.
(c) Pi Configuration
It is named for its shape, which in this case resembles the Greek letter pi (π)
(d) Delta Configuration
It resembles the Greek letter delta (∆).
REFERENCE: FUNDAMENTAL OF ELECTRONICS DC/AC CIRCUITS
111. A circuit with a positive impedance angle is sometimes called an _____.
A.
B.
Input impedance
Inductive circuit
C.
D.
Capacitive circuit
Conductance circuit
ANSWER: B
EXPLANATION: The impedance angle is the angle by which the input voltage leads the input current, provided that
this angle is positive. If it is negative, then the current leads the voltage. A circuit with a positive impedance angle is
sometimes called an inductive circuit because the inductive reactancedominates the capacitive reactance to cause
the input current to lag the input voltage. Similarly, a circuit that has a negative impedance angle is sometimes called
a capacitive circuit.
REFERENCE: SCHAUMS OUTLINES, BASIC CIRCUIT ANALYSIS
112. It refers to the angular displacement between different waveforms of the same frequency.
A. Phasors
B. Phase difference
C. Distance
ANSWER: B
EXPLANATION:
If the angular displacement is 0° as in (a), the waveforms are said to be in phase; otherwise, they are out of phase. By
definition, the waveform generated by the leading phasor leads the waveform generated by the lagging phasor and
vice versa.
REFERENCE: FUNDAMENTAL OF ELECTRONICS DC/AC CIRCUITS, MILLER
113. By representing voltages and currents as phasors, _____was able to define a quantity called impedance.
A.
B.
C.
D.
Charles Proteus Steinmetz
Henry Cavendish
James Clerk Maxwell
Thomas Alva Edison
ANSWER: A
EXPLANATION: By 1893, however, Steinmetz had reduced the very complex alternatingcurrent theory to, in his
words, “a simple problem in algebra.” The key concept in this simplification was the phasor—a representation based
on complex numbers. By representing voltages and currents as phasors, Steinmetz was able to define a quantity
called impedance and then use it to determine voltage and current magnitude and phase relationships in one
algebraic operation. Steinmetz wrote the seminal textbook on ac analysis based on his method, but at the time he
introduced it he was practically the only person who understood it.
REFERENCE: FUNDAMENTAL OF ELECTRONICS DC/AC CIRCUITS, MILLER
114. In a phasor-domain circuit, 220
30 V is applied across two series components, one of which is a 20-ohms resistor
and the other of which is a coil with an impedance of 40 20 ohms. Use current to find the voltage drop across 20ohms resistor.
ANSWER: 74
16.6 V
Solution:
𝑉
220
𝑍
20+ 40
I= =
30
20
= 3.72
16.6 𝐴
Each component voltage drop is the product of the current and the component impedance.
Vr = (3.72
1.66 A) (20 ohms) = 74
16.6 V
REFERENCE: SCHAUMS OUTLINES, BASIC CIRCUIT ANALYSIS
115. The power that flows into and out of a pure inductance is __________.
A.
B.
C.
D.
Reactive power only
Real power only
Active power
Both reactive and real power
ANSWER: A
EXPLANATION: For a purely inductive load,currentlagsvoltageby90°. The average power to an inductance over a full
cycle is zero, i.e., there are no power losses associated with a pure inductance. Consequently, P = 0 W and the only
power flowing in the circuit is reactive power. This is true in general, that is, the power that flows into and out of a
pure inductance is reactive power only.
REFERENCE: FUNDAMENTAL OF ELECTRONICS DC/AC CIRCUITS, MILLER
116. An inductive circuit has a _____ power factor, while a capacitive circuit has a ______ power factor.
A.
B.
C.
D.
Leading, lagging
Leading, leading
Lagging, lagging
Lagging, leading
ANSWER: LAGGING, LEADING
Explanation: For a load containing only resistance and inductance, the load current lags voltage. The power factor in
this case is described as lagging. On the other hand, for a load containing only resistance and capacitance, current
leads voltage and the power factor is described as leading. Thus, an inductive circuit has a lagging power factor, while
a capacitive circuit has a leading power factor.
REFERENCE: FUNDAMENTAL OF ELECTRONICS DC/AC CIRCUITS, MILLER
117. Losses that arise from the additional power needed to reverse the magnetic field in magnetic materials with an
alternating current are called ______.
A.
B.
C.
D.
Eddy-current
Hysteresis
Skin effect
Radiation resistance
ANSWER: B
EXPLANATION: Losses in the magnetic core are due to eddy-current losses and hysteresis losses. Eddy currents flow
in a circular path within the core material itself and dissipate as heat in the core.Hysteresis losses arise from the
additional power needed to reverse the magnetic field in magnetic materials with an alternating current. Hysteresis
losses generally are less than eddy- current losses.
REFERENCE: SCHAUMS OUTLINES, BASIC CIRCUIT ANALYSIS
118. A choke coil of negligible resistance is to limit the current through it to 50 mA when 25 V is applied across it at 400
kHz. Find its inductance.
A.
B.
C.
D.
0.25 mH
0.20 mH
0.23 mH
0.22 mH
ANSWER: B
Solution:
𝑋𝐿 =
𝐿=
𝑉𝐿
25 𝑉
=
= 500 𝛺
𝐼𝐿 50 𝑚𝐴
𝑋𝐿
500
=
= 𝟎. 𝟐𝟎 𝒎𝑯
6.28 𝑓 (6.28)(400 𝑘𝐻𝑧)
REFERENCE: SCHAUMS OUTLINES, BASIC CIRCUIT ANALYSIS
119. What is the impedance ZT of a 200-ohms R in parallel with a 400-ohms XL? Assume 400 V for the applied voltage VT.
A.
B.
C.
D.
178.6 𝛺
187.6 𝛺
168.7 𝛺
167.8 𝛺
ANSWER: A
Solution:
𝐼𝑅 =
𝑉𝑇 400
=
=2𝐴
𝑅
200
𝐼𝐿 =
𝑉𝑇 400
=
=1𝐴
𝑋𝐿 400
𝐼𝑇 = √𝐼2 + 𝐼2 = √22 + 12 = 2.24 𝐴
𝑍𝑇 =
𝑉𝑇 400
=
= 𝟏𝟕𝟖. 𝟔 𝜴
𝐼𝑇 2.24
REFERENCE: SCHAUMS OUTLINES, BASIC CIRCUIT ANALYSIS
120. An induction motor operating at a power factor of 0.8 draws 1056 W from a 1104 ac line. What is the current?
A.
B.
C.
D.
12 A
11 A
21 A
15 A
ANSWER: A
Solution:
P = VI cos 𝜃 ; I =
𝑃
𝑉 cos 𝜃
=
1056
110 ( 0.8)
= 12 𝐴
REFERENCE: SCHAUMS OUTLINES, BASIC CIRCUIT ANALYSIS
Semiconductors (Diodes/Transistors)
121. The current that exists under reverse-bias conditions is called the ________.
A.
B.
C.
D.
Forward current
Reverse saturation current
Dark current
Zener current
ANSWER: B
EXPLANATION: The reverse saturation current is seldom more than a few microamperes except for high-power
devices. In fact, in recent years its level is typically in the nanoampere range for silicon devices and in the lowmicroampere range for germanium. The term saturation comes from the fact that it reaches its maximum level
quickly and does not change significantly with increase in the reverse-bias potential.
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
122. Circuits that separate an input signal at a particular dc level and pass to the output, without distortion, the desired
upper or lower portion of the original waveform.
A.
B.
C.
D.
Clipper
Clamper
Rectifier
Filter
ANSWER: A
EXPLANATION: Diode clipping circuits separate an input signal at a particular dc level and pass to the output, without
distortion, the desired upper or lower portion of the original waveform. They are used to eliminate amplitude noise
or to fabricate new waveforms from an existing signal.
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
123. A Zener diode has the specifications VZ = 5.2V and PD Max = 260mW. Assume RZ = 0. Find the maximum allowable
current IZwhen the Zener diode is acting as a regulator.
A.
B.
C.
D.
53 mA
55 mA
45 mA
50 mA
ANSWER: D
Solution:
𝑃𝑑
Iz max = Iz = 𝑉𝑧 =
260 𝑚𝑊
5.2 𝑉
= 50 𝑚𝐴
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
124. In reference with problem no. 3, Ifasingle-loop circuit consists of an ideal 15-V dc source Vs, a variable resistor R, and
the described Zener diode, find the range of values of R for which the Zener diode remains in constant reverse
breakdown with no danger of failure.
a.1.96 kΩ≤ R ≤ 196 kΩ
b.1.69 kΩ≤ R ≤ 169 kΩ
c.1.66 kΩ≤ R ≤ 176 kΩ
d.1.99 kΩ≤ R ≤ 166 kΩ
ANSWER: A
Solution:
Vs = Riz + Vz
𝑅 ≤
𝑅 ≥
𝑉𝑠 − 𝑉𝑍
0.1 𝑖𝑧
𝑉𝑠 − 𝑉𝑍
𝑖𝑧
𝑅=
so that
=
15 − 5.2
(0.1)( 50 𝑚𝐴)
=
15 − 5.2
( 50 𝑚𝐴)
𝑉𝑠 − 𝑉𝑍
𝑖𝑧
= 1.96 𝑘Ω
= 196 𝑘Ω
Thus, we need 1.96 kΩ≤ R ≤ 196 kΩ
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
125. A semiconductor diode is __________ when the association p-type and positive and n-type and negative has been
established.
A.
B.
C.
D.
reverse-biased
forward-biased
reverse-forward condition
forward-reversed condition
ANSWER: B
EXPLANATION: A forward-bias or “on” condition is established by applying the positive potential to the p-type
material and the negative potential to the n-type material. The application of a forward-bias potential VD will
“pressure” electrons in the n-type material and holes in the p-type material to recombine with the ions near the
boundary and reduce the width of the depletion region.
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
126. The maximum reverse-bias potential that can be applied before entering the Zener region is called ____.
A.
B.
C.
D.
junction voltage
peak inverse voltage
Forward voltage
knee voltage
ANSWER: B
EXPLANATION: The maximum reverse-bias potential that can be applied before entering the Zener region is called the
peak inverse voltage (referred to simply as the PIV rating) or the peak reverse voltage (denoted by PRV rating).
If an application requires a PIV rating greater than that of a single unit, a number of diodes of the same characteristics
can be connected in series. Diodes are also connected in parallel to increase the current-carrying capacity.
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
127. As with the dc and ac resistance levels, the ____ the level of currents used to determine the average resistance the
______ resistance level.
A.
B.
Lower, higher
Lower, lower
C.
D.
Higher, lower
Higher, higher
ANSWER: A
EXPLANATION: The average ac resistance is, by definition, the resistance determined by a straight line drawn
between the two intersections established by the maximum and minimum values of input voltage.
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
128. A light-emitting diode (LED) has a greater forward voltage drop than does a common signal diode. A typical LED can
be modeled as a constant forward voltage drop VD = 1.6V. Its luminous intensity Iv varies directly with forward
current and is described by Iv = 40id = millicandela (mcd). A series circuit consists of such an LED, a current-limiting
resistor R, and a 5-V dc source VS. Find the value of R such that the luminous intensity is 1mcd.
A.
B.
C.
D.
138 Ω
136 Ω
163 Ω
183 Ω
ANSWER: B
Solution:
𝑖𝐷 =
𝐼𝑣
40
=
1
40
= 25 𝑚𝐴
𝑉𝑠 = 𝑅𝑖𝐷 + 1.6 𝑠𝑜 𝑡ℎ𝑎𝑡 𝑅 =
𝑉𝑠 −1.6
𝑖𝐷
=
5−1.6
25 𝑥 10−3
= 136 Ω
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
129. For the Zener diode network shown below, determine Pz.
A.
B.
C.
D.
0W
1W
2W
3W
ANSWER: A
Solution:
𝑉=
𝑅𝐿 𝑉𝑖
1.2 𝑘Ω (16 𝑉)
=
= 8.73 𝑉
𝑅 + 𝑅𝐿
1𝑘Ω + 1.2𝑘Ω
Since V = 8.73 V is less than VZ = 10 V, the diode is in the “off” state.
VL = V = 8.73 V
VR = Vi – VL = 16 V – 8.73 = 7.27 V
Iz = 0 A
P z = V Z Iz = 0 W
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
130. Determine the range of values of VI that will maintain the Zener diode of figure below in the on state.
a. Vmin = 18.5 V; Vmax = 27.5 V
b.Vmin = 17.5 V; Vmax = 28.5 V
c. Vmin = 17.5 V; Vmax = 27.5 V
d.Vmin = 18.5 V; Vmax = 28.5 V
ANSWER: C
Solution:
𝑉𝑚𝑖𝑛 =
(𝑅𝐿 + 𝑅)𝑉𝑍
(1200 Ω + 200 Ω)(15 𝑉)
=
= 17.5 𝑉
𝑅𝐿
1200 Ω
𝐼𝐿 =
𝑉𝐿
15 𝑉
=
= 12.5 𝑚𝐴
𝑅𝐿
1.2 𝑘Ω
𝐼𝑅𝑀𝑎𝑥 = 𝐼𝑍𝑀 + 𝐼𝐿 = 50 𝑚𝐴 + 12.5 𝑚𝐴 = 62.5 𝑚𝐴
𝑉𝑖𝑚𝑎𝑥 = 𝐼𝑟𝑚𝑎𝑥 𝑅 + 𝑉 = (62.5 𝑚𝐴)(0.2𝑘Ω) + 15 𝑉 = 27.5 𝑉
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
131. Determine the I for the series dc configuration of figure below.
A.
B.
C.
D.
2.50 mA
2.30 m
2.04 mA
2.25 mA
ANSWER: C
Solution:
𝐼=
𝐸1 + 𝐸2 − 𝑉𝑑
10 𝑉 + 5 𝑉 − 0.7 𝑉
=
= 𝟐. 𝟎𝟒 𝒎𝑨
𝑅1 + 𝑅2
5𝐾 + 2𝐾
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
132. In reference with problem no. 11, find the Vout.
A.
B.
C.
D.
-1 V
0.92 V
1V
-0.92 V
ANSWER: D
Solution:
V1 = IR1 = (2.04 mA) (5kΩ) = 10.2 V
V2 = IR2 = (2.04 mA) (2kΩ) = 4.08 V
Vo = V2 – E2 = 4.08 V – (5 V) = -0.92 V
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
133. Find the resistor R to ensure a current of 20 mA through “on” diode for the configuration below. Both diodes have a
reverse breakdown voltage of 3 V and an average turn-on voltage of 2 V.
A.
B.
C.
D.
200 Ω
300 Ω
400 Ω
250 Ω
ANSWER: B
Solution:
𝐼 = 20 𝑚𝐴 =
𝑅=
𝐸 − 𝑉𝐿𝐸𝐷
8𝑉−2𝑉
=
𝑅
𝑅
6𝑉
= 𝟑𝟎𝟎 Ω
20 𝑚𝐴
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
134. These are circuits employed to maintain a relatively low transformer peak voltage while stepping up the peak output
voltage to two, three, four, or more times the peak rectified voltage.
A.
B.
C.
D.
Voltage-multiplier
Rectifier
Filter
Transformer
ANSWER: A
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
135. The process of giving off light by applying an electrical source of energy is called _______.
A.
B.
C.
D.
luminous intensity
electroluminescence
efficacy
light intensity
ANSWER: B
EXPLANATION: The light-emitting diode (LED) is a diode that will give off visible light when it is energized. In any
forward-biased p-n junction there is, within the structure and primarily close to the junction, a recombination of holes
and electrons. This recombination requires that the energy possessed by the unbound free electron be transferred to
another state. In silicon and germanium, the greater percentage is given up in the form of heat and the emitted light
is insignificant. In other materials, such as gallium arsenide phosphide (GaAs) or gallium phosphide (GA), the number
of photons of light energy emitted is sufficient to create a very visible light source.
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
136. In the _______, the collector-base and base-emitter junctions are forward-biased.
A.
B.
C.
D.
active region
saturation region
cut-off region
inverse
ANSWER: B
EXPLANATION: In the active region the collector-base junction is reverse-biased, while the base-emitter junction is
forward-biased.
In the cutoff region the collector-base and base-emitter junctions of a transistor are both reverse-biased.
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
137. A BJT has α = 0.99, IB = 25µA, and ICBO = 200nA. Find (a)the dc collector current.
A.
B.
C.
D.
2.354 mA
3.02 mA
2.495 mA
2.823 mA
ANSWER: C
Solution:
α = 0.99
β=
𝛼
1− 𝛼
=
0.99
1− 0.99
= 99
𝐼𝑐 = 𝛽𝐼𝐵 + (𝛽 + 1)𝐼𝐶𝐵𝑂 = (99)(25𝜇𝐴) + (99 + 1)(200 𝑛𝐴) = 𝟐. 𝟒𝟗𝟓 𝒎𝑨
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
138. It is a dc network in which the current through a load is controlled by a current at another point in the network.
ANSWER: CURRENT MIRROR
EXPLANATION: Current mirror is a dc network in which the current through a load is controlled by a current at
another point in the network. That is, if the controlling current is raised or lowered the current through the load will
change to the same level.
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
139. For a certain BJT, β = 50;ICEO = 3µA, and Ic = 1.2mA. Find IE.
A.
B.
C.
D.
4.122 mA
1.782 mA
2.421 mA
1.224 mA
ANSWER: D
Solution:
𝐼𝐵 =
𝐼𝑐 − 𝐼𝐶𝐸𝑂
1.2 𝑚𝐴 − 3𝜇𝐴
=
= 23.94 𝜇𝐴
𝛽
50
IE = Ic + IB = 1.2 mA + 23.94 uA = 1.2239 mA
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
140. A transistor circuit is to be operated with a base current of 40uA and VBB = 6V. The Si transistor (VBE = 0.7V) has
negligible leakage current. Find the required value of RB.
A.
B.
C.
D.
130 kΩ
140 kΩ
132.5 kΩ
d.122.5 kΩ
ANSWER: C
Solution:
𝑅𝐵 =
𝑉𝐵𝐵 − 𝑉𝐵𝐸
𝐼𝐵
=
6−0.7
40 𝑢𝐴
= 𝟏𝟑𝟐. 𝟓 𝒌Ω
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
141. The transistor of figure below has α = 0.96 and a base current of 35 µA. Find Ie. Assume negligible leakage current.
A.
B.
C.
0.578 mA
0.785 mA
0.875 mA
D.
0.965 mA
ANSWER: C
Solution:
β=
𝛼
1− 𝛼
=
0.96
1−0.96
= 24
Ic = βIB = (24) (35 uA) = 0.84 mA
IE =
𝐼𝑐
𝛼
=
0.84 𝑚𝐴
0.96
= 𝟎. 𝟖𝟕𝟓 𝒎𝑨
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
142. In reference with problem no. 21, the transistor circuit of figure 1-21 is to be operated with a base current of 40µA
and VBB = 6 V. The Si transistor (VBE = 0.7 V) has negligible leakage current. Find the required value of RB.
a.132.5 MΩ
b. 132.5 kΩ
c. 153.2 kΩ
d. 153.2 MΩ
ANSWER: B
Solution:
VBB= IBRB + VBE
𝑅𝐵 =
𝑉𝐵𝐵 − 𝑉𝐵𝐸
6 − 0.7
=
= 𝟏𝟑𝟐. 𝟓 𝒌Ω
𝐼𝐵
40 µ𝐴
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
143. Networks that are quite stable and relatively insensitive to temperature variations have________ stability factors.
a.
b.
c.
d.
Low
High
Normal
Zero
ANSWER: A
EXPLANATION: The higher the stability factor, the more sensitive the network to variations in that parameter.
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
144. In this region the JFET can actually be employed as a variable resistor whose resistance is controlled by the applied
gate-to-source voltage.
a. Voltage-controlled resistance region
b. Current-controlled resistance region
c. Depletion region
d. linear amplification region
ANSWER: A
EXPLANATION: Ohmic or voltage-controlled resistance regionis the region to the left of the pinch-off locus.
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
145. As gate potential _______, the pinchoff voltage _________.
A.
B.
C.
D.
Decreases, increases
Increases, decreases
Decreases, decreases
Increases, increases
ANSWER: C
EXPLANATION: Above pinchoff but below avalanches breakdown, drain current ID remains nearly constant as VDS is
increased.As gate potential decreases, the pinchoff voltage, that is, the source-to-drain voltage Vp at which pinchoff
occurs, also decreases, approximately obeying the equation
VP = VPO+ VGS
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
146. A p-channel MOSFET operating in the enhancement mode is characterized by V T = -3 V and ID = -8 mA when VGS = 4.5 V. Find the VGS if ID = -16 mA.
a. -0.88 V
b. -0.78 V
c. -0.98 V
d. -0.77 V
ANSWER: A
Solution:
𝐼𝐷𝑜𝑛 =
𝐼𝐷𝑄
−8 𝑚𝐴
=
= −32 𝑚𝐴
( 1 − 𝑉𝐺𝑆 /𝑉𝑇 )2
(1 − (−4.5/−3)2
1
𝑉𝐺𝑆
𝐼𝐷𝑄 1/2
−16 2
= 𝑉𝑇 [1 − ( ) ] = (−3) [1 − (
) ] = −𝟎. 𝟖𝟖 𝑽
𝐼𝑂𝑁
−32
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
147. In reference with problem no. 25, find ID if VGS = -5 V.
a. -14.5 mA
b. -12.44 mA
c. -13.22 mA
d. -14.22 mA
ANSWER: D
Solution:
𝐼𝐷 = 𝐼𝐷𝑜𝑛 (1 −
𝑉𝐺𝑆 2
−5 2
𝑉𝑇
−3
) = (−32 𝑚𝐴) (1 −
) = −𝟏𝟒. 𝟐𝟐 𝒎𝑨
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
148. The transistor circuit below has a base current of 40 µA and ICB = 0. If VBB = 6 V, RE = 1kΩ, and β = 80. Find IE.
A.
B.
C.
D.
2.366 mA
4.226 mA
3.226 mA
2.465 mA
ANSWER: C
Solution:
𝛼=
𝐼𝐸=
𝛽
80
=
= 0.9876
𝛽+1
81
𝐼𝐵
40 𝜇𝐴
=
= 𝟑. 𝟐𝟐𝟔 𝒎𝑨
1− 𝛼
1 − 0.9876
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
149. The JFET can be an ______ device (conduction by electrons) or a ________ device (conduction by holes).
a.
b.
c.
d.
N-channel; p-channel
P-channel; n-channel
Npn; pnp
Pnp; Npn
ANSWER: A
EXPLANATION: The operation of the field-effect transistor (FET) can be explained in terms of only majority-carrier
(one-polarity) charge flow; the transistor is therefore called unipolar.The transistor can be an n-channel device
(conduction by electrons) or a p-channel device (conduction by holes); a discussion of n-channel devices applies
equally to p-channel devices if complementary (opposite in sign) voltages and currents are used.
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
150. The _______ is a two-port transistor arrangement in which the base shares a common point with the input and
output terminals.
A. common-collector (CC) connection
B. common-emitter (CE) connection
C. common-base (CB) connection
D. Emitter-follower connection
ANSWER: C
EXPLANATION: The common-base (CB) connection is a two-port transistor arrangement in which the base shares a
common point with the input and output terminals. The independent input variables are emitter current andbase-toemitter voltage. The corresponding independent output variables are collector current and base-to-collector voltage.
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
INDUSTRIAL/POWER ELECTRONICS
151. It is an arrangement of semiconductor devices all operating in the switching mode.
A.
B.
C.
D.
Control circuit
Power conditioner
Power source
Filter
ANSWER: B
EXPLANATION: The power conditioner is an arrangement of semiconductor devices all operating in the switching
mode. This means that the device is switched from cut-off to saturation ('off' to 'on') by the application of gate, or
base, drive pulses. The ideal switch would have full voltage across it when 'off', and zero voltage across it when 'on'.
REFERENCE: INTRODUCTION TO POWER ELECTRONICS, DENIS FEWSON
152. A thyristor half-wave controlled converter has a supply voltage of 240V at 60Hz and a load resistance of 120 Ω. What
are the average values of load voltage and current when the firing delay angle is 30o?
A.
B.
C.
D.
Vav = 100.8 V;Iav = 1.05 A
Vav = 101.8 V;Iav = 1 A
Vav = 100.8 V;Iav = 0.84 A
Vav = 120.8 V;Iav = 0.75 A
ANSWER: C
Solution:
Vav = (Em/2π)(1+ cos α)
Vav = (√2 𝑥 240/2𝜋) ( 1 + cos 30)
Vav = 100.8 V
Iav = Vav/R = 100.8 V/120 Ω
Iav = 0.84 A
REFERENCE: INTRODUCTION TO POWER ELECTRONICS, DENIS FEWSON
153. It is a two-terminal semiconductor device whose terminal resistance will vary (linearly) with the intensity of the
incident light.
A.
B.
C.
D.
Photodiode
photoconductive cell
LED
photons
ANSWER: B
EXPLANATION: It is frequently called a photoresistive device.The photoconductive materials most frequently used
include cadmium sulfide (CdS) and cadmium selenide (CdSe).The photoconductive cell does not have a junction like
the photodiode. A thin layer of the material connected between terminals is simply exposed to the incident light
energy.
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
154. It is the “forcing” of current through the SCR in the direction opposite to forward conduction.
A.
B.
C.
D.
Forced commutation
Anode-current interruption
Forced breakover
Holding current
ANSWER: A
EXPLANATION: The two general methods for turning off an SCR are categorized as the anode current interruption and
the forced-commutation technique.Forced commutation is the “forcing” of current through the SCR in the direction
opposite to forward conduction.During SCR conduction, the transistor is in the “off” state, that is, IB = 0 and the
collector-to-emitter impedance is very high (for all practical purposes an open circuit). This high impedance will
isolate the turn-off circuitry from affecting the operation of the SCR. For turn-off conditions, a positive pulse is applied
to the base of the transistor, turning it heavily on, resulting in a very low impedance from collector to emitter (shortcircuit representation).
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
155. A thyristor has a linearized gate-cathode characteristic of slope 25 V/A. A gate current of 200mA turns the thyristor
on in 16 µs. The gate source voltage is 10V. The manufacturer's average maximum power for the gate is 400mW.
Pulse firing is used. Calculate the value of the gate series resistance.
A.
B.
C.
D.
30 Ω
50 Ω
20 Ω
25 Ω
ANSWER: D
Solution:
The gate resistance is rg = VGC/IG = 25 Ω
RG (total) = VG/IG = 10/0.2 = 50 Ω
Gate series resistance = 50 – 25 = 25 Ω
REFERENCE: INTRODUCTION TO POWER ELECTRONICS, DENIS FEWSON
156. In reference with problem no.5, determine the Gate power dissipation.
A. 1.5 W
B. 1 W
C. 2 W
D. 2.3 W
ANSWER: B
Solution:
Gate Power Dissipation = IG2 rg
Gate Power Dissipation = (0.2)2 (25)
Gate Power Dissipation = 1 W
REFERENCE: INTRODUCTION TO POWER ELECTRONICS, DENIS FEWSON
157. This device originally called a duo (double) base diode due to the presence of two base contacts.
A.
B.
C.
D.
Unijunction Transistor
Silicon Control Rectifier
Silicon Control Switch
Diac
ANSWER: A
EXPLANATION:The UJT is a three-terminal device having the basic construction ofslab of lightly doped (increased
resistance characteristic) n-type silicon material has two base contacts attached to both ends of one surface and an
aluminum rod alloyed to the opposite surface. The p-n junction of the device is formed at the boundary of the
aluminum rod and the n-type silicon slab. The single p-n junction accounts for the terminology unijunction. It was
originally called a duo (double) base diode due to the presence of two base contacts.
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
158. Given the relaxation oscillator. Determine the RB2at IE = 0 A.
RBB=5kΩ, ɳ=0.6
VV=1 V, Iv=10 mA, Ip=10 µA
(Rb1=100Ω during discharge phase)
A.
B.
C.
D.
4𝑘Ω
3𝑘Ω
2𝑘Ω
1.5𝑘Ω
ANSWER: C
Solution:
ɳ=
𝑅𝐵1
𝑅𝐵1 +𝑅𝐵2
0.6 =
𝑅𝐵1
𝑅𝐵2
𝑅𝐵1 = 0.6𝑅𝐵𝐵 = (0.6)(5𝑘Ω) = 3 𝑘Ω
𝑅𝐵2 = 𝑅𝐵𝐵 − 𝑅𝐵1 = 5𝑘Ω − 3 𝑘Ω = 𝟐𝒌Ω
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
159. In reference with problem no. 8, calculate Vp, the voltage necessary to turn on the UJT.
a. 9 V
b. 7 V
c. 8 V
d. 6 V
ANSWER: C
Solution:
𝑉𝑝 = 0.7 𝑉 +
𝑉𝑝 = 0.7 𝑉 +
(𝑅𝐵1 + 𝑅2 )12 𝑉
𝑅𝑏1 + 𝑅𝐵2 + 𝑅2
(3 𝑘Ω + 0.1 𝑘Ω)12𝑉
5 𝑘Ω + 0.1 𝑘Ω
𝑽𝒑 = 𝟖 𝑽
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
160. The ____ is most sensitive to light when the gate terminal is open.
a. Photodiode
b. LED
c. LASCR
d. SCS
ANSWER: C
EXPLANATION: LASCR is an SCR whose state is controlled by the light falling upon a silicon semiconductor layer of the
device.The LASCR is most sensitive to light when the gate terminal is open. Its sensitivity can be reduced and
controlled somewhat by the insertion of a gate resistor.
REFERENCE: ELECTRONIC DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
INSTRUMENTATION
161. It is the desired value of the output parameter or variable being monitored by a sensor.
A. Manipulated variable
B. Set point
C. Measured variable
D. Controlled variable
ANSWER: B
EXPLANATION: process control is the automatic control of an output variable by sensing the amplitude of the output
parameter from the process and comparing it to the desired or set level and feeding an error signal back to control an
input variable – in the case steam. Set Point is the desired value of the output parameter or variable being monitored
by a sensor. Any deviation from this value will generate an error signal.
REFERENCE: FUNDAMENTALS OF INDUSTRIAL INSTRUMENTATION AND PROCESS CONTROL, WILLIAM C. DUNN
162. A pressure gauge ranges from 0 to 50 psi, the worst case spread in readings is +4.35 psi. What is the % FSD accuracy?
A. ±8.7
B. ±9.8
C. ±7.8
D. ±6.9
ANSWER:A
Solution:
% FSD = ± (4.35
𝑝𝑠𝑖
50
𝑝𝑠𝑖) 𝑥 100 = ±𝟖. 𝟕
REFERENCE: FUNDAMENTALS OF INDUSTRIAL INSTRUMENTATION AND PROCESS CONTROL, WILLIAM C. DUNN
163. What is the capacitance between two parallel plates whose areas are 1 m2separated by a 1-mm thick piece of
dielectric with a dielectric constant of 5.5 x 10-9 F/m?
A. 6.5 µF
B. 5.5 µF
C. 4.5 µF
D. 5.4 µF
ANSWER: B
Solution:
𝐶=
𝜀𝐴
5.5𝑛𝐹/𝑚 (1 𝑚2 )
=
= 𝟓. 𝟓𝝁𝑭
𝑑
1 𝑥 10−3 𝑚
REFERENCE: FUNDAMENTALS OF INDUSTRIAL INSTRUMENTATION AND PROCESS CONTROL, WILLIAM C. DUNN
164. It is defined as the density of a material divided by the density of water.
A. Specific Weight
B. Specific Gravity
C. Static Pressure
D. Impact Pressure
ANSWER: B
EXPLANATION: Specific gravity of a liquid or solid is a dimensionless value since it is a ratio of two measurements in
the same unit. It is defined as the density of a material divided by the density of water or it can be defined as the
specific weight of the material divided by the specific weight of water at a specified temperature.
REFERENCE: FUNDAMENTALS OF INDUSTRIAL INSTRUMENTATION AND PROCESS CONTROL, WILLIAM C. DUNN
165. What is the pressure at the base of a water tower which has 50 ft. of head?
A. 21.67 psig
B. 22.67 psig
C. 26.17 psig
D. 27.16 psig
ANSWER: A
Solution:
P = 62.4 lb./ft3 x 50 ft. = 3120 psfg = 3120 psf/144 ft2/in2 = 21.67 psig
REFERENCE: FUNDAMENTALS OF INDUSTRIAL INSTRUMENTATION AND PROCESS CONTROL, WILLIAM C. DUNN
166. A displacer with a diameter of 8 in is used to measure changes in water level. If the water level changes by 1 ft. what
is the change in force sensed by the force sensor?
A. 22.9 lb.
B. 21.8 lb.
C. 23.7 lb.
D. 25.6 lb.
ANSWER: B
Solution:
F2 – F1 =
62.4 𝑙𝑏/𝑓𝑡 3 𝑥 𝜋(8 𝑓𝑡)2 𝑥 122
4
= 𝟐𝟏. 𝟖 𝒍𝒃
REFERENCE: FUNDAMENTALS OF INDUSTRIAL INSTRUMENTATION AND PROCESS CONTROL, WILLIAM C. DUNN
167. It is used in liquids that are nonconductive and have high µ.
A. Capacitive probes
B. Conductive probes
C. Bubbler devices
D. Resistive tapes
ANSWER: A
EXPLANATION: Capacitive probes are used for continuous level monitoring. It consists of an inner rod with an outer
shell. The capacitance is measured between the two using a capacitance bridge.
REFERENCE: FUNDAMENTALS OF INDUSTRIAL INSTRUMENTATION AND PROCESS CONTROL, WILLIAM C. DUNN
168. How far below the surface of the water is the end of a bubbler tube, if the bubbles start to emerge from the end of
the tube when the air pressure in the bubbler is 148 kPa?
A. 15 cm
B. 16.5 cm
C. 13.9 cm
D. 14.8 cm
ANSWER: D
Solution:
ℎ=
𝑝
148 𝑘𝑃𝑎 𝑥 10−4
=
= 𝟏𝟒. 𝟖 𝒄𝒎
𝛾
1 𝑔𝑚/𝑐𝑚3
REFERENCE: FUNDAMENTALS OF INDUSTRIAL INSTRUMENTATION AND PROCESS CONTROL, WILLIAM C. DUnn
169. It is an instrument used to measure gas flow rates.
A. Elbow
B. Dall tube
C. Pilot static tube
D. Anemometer
ANSWER: D
EXPLANATION: Mass flow instruments include constant speed impeller turbine wheel-spring combinations that relate
the spring force to mass flow and devices that relate heat transfer to mass flow. Anemometer is an instrument used
to measure gas flow rates. One method is to keep the temperature of heating element in a gas flow constant and
measure the power required. The higher the flow rate, the higher the amount of heat required.
REFERENCE: FUNDAMENTALS OF INDUSTRIAL INSTRUMENTATION AND PROCESS CONTROL, WILLIAM C. DUNN
170. It is the transition of matter from the solid to the liquid to the gaseous states.
A. Specific heat
B. Heat change
C. Phase change
D. Thermal change
ANSWER: C
EXPLANATION: Heat is a form of energy; as energy is supplied to a system the vibration amplitude of its molecules
and its temperature increases. The temperature increase is directly proportional to the heat energy in the system.
Phase change is the transition of matter from the solid to the liquid to the gaseous states; matter can exist in any of
these three stages. However, for matter to make the transition from one step up to the next, it has supplied with
energy, or energy removed if the matter is going from gas to liquid to solid.
REFERENCE: FUNDAMENTALS OF INDUSTRIAL INSTRUMENTATION AND PROCESS CONTROL, WILLIAM C. DUNN
DIGITAL ELECTRONICS
171. Which of the following is not a BCD codes?
A. 8421 BCD code
B. 6421 BCD code
C. 4221 BCD code
D. 5421 BCD code
ANSWER: B
EXPLANATION: The binary-coded decimal (BCD) code makes conversion to decimals much easier.The three weighted
BCD codes are 8421 BCD code, 4221 BCD code and 5421 BCD code. The most significant bit, has a weight of 8, and the
least significant bit has a weight of only 1. This code is more precisely known as the 8421 BCD code. The 8421 part of
the name gives the weighting of each place in the 4-bit code.
REFERENCE: SCHAUMS OUTLINE OF THEORY AND PROBLEMS OF DIGITAL PRINCIPLES, 3 RD EDITION
172. The 5421 BCD equivalent of decimal 75 is _________.
A. 10101000
B. 11101000
C. 10101100
D. 10101010
Answer: 10101000
Solution: The 5421 BCD equivalent of decimal 75 is 10101000.
REFERENCE: SCHAUMS OUTLINE OF THEORY AND PROBLEMS OF DIGITAL PRINCIPLES, 3 RD EDITION
173. The XS3 number is always _________ than the 8421 BCD number.
A. 3 less
B. 3 more
C. Equal
D. Less
Answer: 3 more
Explanation: The excess-3 (XS3) code is related to the 8421 BCD code because of its binary-coded-decimal nature. In
other words, each 4-bit group in the XS3 code equals a specific decimal digit.
Reference: Schaums Outline of Theory and problems of Digital Principles, 3rd Edition
174. Convert the 8421 BCD number 0111 to their XS3 code equivalents.
A. 1100
B. 1010
C. 1101
D. 1001
ANSWER: B
Solution:
0111
+0011
----------------------1010
REFERENCE: SCHAUMS OUTLINE OF THEORY AND PROBLEMS OF DIGITAL PRINCIPLES, 3 RD EDITION
175. The 7-bit alphanumeric code serves as the industry standard for input/output on micro-computers.
A. EBCDIC
B. ASCII
C. BCDIC
D.
Hollerith
ANSWER: B
EXPLANATION: The ASCII code is used extensively in small computer systems to translate from the keyboard
characters to computer language.The alphanumeric ASCII code is the modern code for getting information into and
out of microcomputers. ASCII is used when interfacing computer keyboards, printers, and video displays. ASCII has
become the standard input/output code for microcomputers.
REFERENCE: SCHAUMS OUTLINE OF THEORY AND PROBLEMS OF DIGITAL PRINCIPLES, 3RD EDITION
176. It is a form of symbolic logic that shows how logic gates operate.
A. Digital Electronics
B. Boolean Algebra
C. Binary numbers
D. ASCII
ANSWER: B
EXPLANATION: The logic gate is the basic building block in digital systems. Logic gates operate with binary numbers.
Gates are therefore referred to as binary logic gates. Boolean algebra is a form of symbolic logic that shows how logic
gates operate. A Boolean expression is a “shorthand” method of showing what is happening in a logic circuit.
REFERENCE: SCHAUMS OUTLINE OF THEORY AND PROBLEMS OF DIGITAL PRINCIPLES, 3 RD EDITION
177. The exclusive-OR gate is referred to as the _______ gate.
A. “any but not all”
B. “all or nothing”
C. “inverter”
D. “any or all”
ANSWER: A
EXPLANATION: The exclusive-OR gate is referred to as the “any but not all” gate. The exclusive-OR term is often
shortened to read as XOR.The XOR gate is enabled only when an odd number of 1s appear at the inputs.The XOR gate
could be referred to as an odd-bits check circuit.
REFERENCE: SCHAUMS OUTLINE OF THEORY AND PROBLEMS OF DIGITAL PRINCIPLES, 3 RD EDITIOn
178. Write the Boolean expression for a 3-input NOR gate.
A. ̅̅̅̅̅̅̅̅̅̅̅̅̅
A+B+C = Y
B. ̅̅̅̅̅̅̅̅̅̅
A ∙B∙C=Y
C. ̅̅̅̅̅̅̅̅̅̅̅̅̅̅
A⊕B⊕C=Y
D. A ⊕ B ⊕ C = Y
ANSWER: A
EXPLANATION:
The Boolean expression at the input to the inverter is A +B + C. The inverter then
complements the ORed terms, which are shown in the Boolean expression with an
overbar. Adding the overbar produces the Boolean Expression ̅̅̅̅̅̅̅̅̅̅̅̅̅
A + B + C = Y This is a
not-OR function. The not-OR function can be drawn as a single logic symbol called a NOR
gate.
REFERENCE: SCHAUMS OUTLINE OF THEORY AND PROBLEMS OF DIGITAL PRINCIPLES, 3 RD EDITION
179. 1010.1012 = ___________10
A. 10.62510
B. 15.87510
C. 14.62510
D. 15.62510
ANSWER: A
EXPLANATION:
Binary
1
0
1
0 . 1
Decimal
8 + 0 + 2 + 0 .
0
1
0.5 + 0 + 0.125 = 10.625
REFERENCE: SCHAUMS OUTLINE OF THEORY AND PROBLEMS OF DIGITAL PRINCIPLES, 3 RD EDITION
180. 34.7510 = ______2
A. 100011.112
B. 010001.112
C. 100010.112
D. 110010.112
ANSWER: C
EXPLANATION:
34 ÷ 2 = 17 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 0
17 ÷ 2 = 8 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 1
8 ÷ 2 = 4 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 0
4 ÷ 2 = 2 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 0
2 ÷ 2 = 1 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 0
1 ÷ 2 = 0.5 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 1
0.75 x 2 = 1.5
0.5 x 2 = 1
34.7510 = 100010.112
REFERENCE: SCHAUMS OUTLINE OF THEORY AND PROBLEMS OF DIGITAL PRINCIPLES, 3 RD EDITION
181. An op amp has saturation voltage Vosat = 10V, an open-loop voltage gain of -10-5. Find thevalue of vd that will just
drive the amplifier to saturation.
A. ±0.2 𝑚𝑉
B. ±0.1 𝑚𝑉
C. ±0.3 𝑚𝑉
D. ±0.5 𝑚𝑉
ANSWER: B
Solution:
𝑣𝑑 =
± 𝑉𝑜𝑠𝑎𝑡
± 10
=
= ±0.1 𝑚𝑉
𝐴𝑂𝐿
− 105
Reference: Electronics Devices and Circuit Theory, Robert Boylestad
182. Typical values for CMRR range from _______.
A. 100 to 10,000
B. 1000 to 10,000
C. 99 to 1000
D. 1001 to 10,000
ANSWER: A
EXPLANATION: Common-mode gain sensitivity is frequently quantized via the common-mode rejection ratio (CMRR),
𝐴
defined as𝐶𝑀𝑅𝑅 = 𝑂𝐿 . Typical values for the CMRR range from 100 to 10,000, with corresponding CMRRdb values of
𝐴𝐶𝑀
from 40 to 80db.
REFERENCE: ELECTRONICS DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
183. Determine the output voltage of an op-amp for input voltages of Vi1 = 150 µV, Vi2 = 140 µV. The amplifier has a
differential gain of Ad = 4000 and the value of CMRR is 100.
A. 45.8 mV
B. 46.5 mV
C. 48.5 mV
D. 47.5 mV
ANSWER: A
Solution:
Vd = Vi1 – Vi2 = (150 – 140) µV = 10 µV
1
150 𝜇𝑉+140 𝜇𝑉
2
2
Vc = (𝑉𝑖1 + 𝑉𝑖2 ) =
𝑉𝑜 = 𝐴𝑑 𝑉𝑑 (1 +
= 145 𝜇𝑉
1 𝑉𝑐
1 145 𝜇𝑉
) = (4000)(10𝜇𝑉) (1 +
) = 𝟒𝟓. 𝟖 𝒎𝑽
𝐶𝑀𝑅𝑅 𝑉𝑑
100 10 𝜇𝑉
REFERENCE: ELECTRONICS DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
184. Calculate the output voltage of an op-amp summing amplifier for the following sets of voltages and resistors. Use Rf
= 1 MΩ. V1 = -2 V, V2= +3 V, V3 = +1 V, R1 = 200 kΩ, R2 = 500 kΩ, R3 = 1 MΩ.
A. - 3 V
B. +2 V
C. +3 V
D. -2 V
ANSWER: C
Solution:
𝑅𝑓
𝑅𝑓
𝑅𝑓
𝑉𝑜 = − ( 𝑉𝑖 +
𝑉 +
𝑉)
𝑅𝑖
𝑅2 2 𝑅3 3
1𝑀Ω
1𝑀Ω
1𝑀Ω
(+1))
𝑉𝑜 = − (
(−2) +
(+3) +
200𝑘Ω
500 kΩ
1 MΩ
𝑽𝒐 = +𝟑 𝑽
REFERENCE: ELECTRONICS DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
185. Determine the cutoff frequency of an op-amp having specified values B1 = 2MHz and AVD = 400 V/mV.
a. 3 Hz
b. 10 Hz
c. 5 Hz
d. 8 Hz
ANSWER: C
Solution:
𝑓𝑐 =
𝑓1
2 𝑀𝐻𝑧
=
= 𝟓 𝑯𝒛
𝐴𝑉𝐷
400 𝑉/𝑚𝑉
REFERENCE: ELECTRONICS DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
186. The _______ provides a parameter specifying the maximum rate of change of the output voltage when driven by a
large step-input signal.
a. slew rate
b.
c.
d.
CMRR
Roll-off
Unity follower
ANSWER: A
EXPLANATION: Another parameter reflecting the op-amp’s ability to handling varying signals is slew rate, defined as
slew rate = maximum rate at which amplifier output can change in volts per microsecond (V/ µs)
REFERENCE: ELECTRONICS DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
187. When a number of stages are connected in series, the overall gain is the _______ of the individual stage gains.
A. sum
B. difference
C. product
D. quotient
ANSWER:C
EXPLANATION: For multiple-stage gains, when a number of stages are connected in series, the overall gain is the
product of the individual stage gains.
A = A1A2A3
REFERENCE: ELECTRONICS DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
188. Determine the output voltage for the circuit below.
A.
B.
C.
D.
V1 + V2
V1 - V2
V2 – V1
V2+ V1
ANSWER: B
Solution:
20 𝑘Ω
100 𝑘Ω + 100 𝑘Ω
100 𝑘Ω
𝑉𝑜 = (
)(
) 𝑉1 −
𝑉
20 𝑘Ω + 20 𝑘Ω
100 𝑘Ω
100 𝑘Ω 2
𝑽 𝒐 = 𝑽𝟏 − 𝑽𝟐
The resulting output voltage is seen to be the difference of the two input voltages.
REFERENCE: ELECTRONICS DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
189. A _______ circuit provides a means of isolating an input signal from a load by using a stage having unity voltage gain.
A. Integrator
B. voltage buffer
C. Amplifier
D. Voltage multiplier
ANSWER: B
EXPLANATION: A voltage buffer circuit provides a means of isolating an input signal from a load by using a stage
having unity voltage gain, with no phase or polarity inversion, and acting as an ideal circuit with very high input
impedance and low output impedance.
REFERENCE: ELECTRONICS DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
190. A filter that provides a constant output from dc up to a cutoff frequency fOH and then passes no signal above that
frequency is called an ______.
A. High-pass filter
B. ideal low-pass filter
C. bandpass filter
D. High-pass active filter
ANSWER: B
EXPLANATION: A filter circuit can be constructed using passive components: resistors and capacitors. An active filter
additionally uses an amplifier to provide voltage amplification and signal isolation or buffering. A filter that provides a
constant output from dc up to a cutoff frequency fOH and then passes no signal above that frequency is called an
ideal low-pass filter.
REFERENCE: ELECTRONICS DEVICES AND CIRCUIT THEORY, ROBERT BOYLESTAD
OTHERS
191. According to Nakano, a ______ is a simple mechanical arm without AI and can work in groups if their actions are
synchronized.
A. First-generation robot
B. Second-generation robot
C. Third-Generation robot
D. Fourth-generation robot
ANSWER: A
EXPLANATION: First-Generation robots have the ability to make precise motions at high speed, may times, for a long
time. They have found widespread industrial application and have been around for more than a half a century. These
are fast-moving systems that install rivets and screws in assembly lines, that solder connections on printed circuits.
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS, 3RD EDITION BY GIBILISCO
192. It refers to the number of different ways in which a robot manipulator can move.
A. Degrees of rotation
B. Degrees of freedom
C. Articulated geometry
D. Coordinate geometry
ANSWER:B
EXPLANATION: The term degrees of freedom refers to the number of different ways in which a robot manipulator can
move. Most manipulators move in three dimensions, but often they have more than three degrees of freedom.
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS, 3RD EDITION BY GIBILISCO
193. It is the ability of a machine to see in dim light or to detect weak impulses at invisible wavelengths.
A. Resolution
B. Selectivity
C. Sensitivity
D. Vision system
ANSWER: C
EXPLANATION: Sensitivity is the ability of a machine to see in dim light or to detect weak impulses at invisible
wavelengths. In some environments, high sensitivity is necessary. In others, it is not needed and might not be wanted.
A robot that works in bright sunlight doesn’t need to be able to see well in a dark cave. A robot designed for working
in mines, pipes, or caverns must be able to see in dim light, using a system that might be blinded by ordinary daylight.
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS, 3RD EDITION BY GIBILISCO
194. It is a means by which a machine can locate objects in three-dimensional space.
A. Epipolar navigation
B. Vision systems
C. Mapping
D. Telepresence
ANSWER: A
EXPLANATION: Epipolar navigation is a form of electronic spatial perception requiring only one observation point, but
that point must be moving. It can also navigate, and figure out its own position and path. Epipolar navigation works
by evaluating the way an image changes as the viewer moves.
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS, 3RD EDITION BY GIBILISCO
195. An android takes the form of:
A. An insect
B. A human body
C. A simple robot arm
D. Binocular vision
ANSWER: C
EXPLANATION: An android is a robot, often very sophisticated, that takes a more or less human form. An android
usually propels itself by rolling on small wheels in its base. The technology for fully functional arms is under
development, but the software needed for their operation has not been made cost-effective for small robots.An
android has a rotatable head equipped with position sensors. Binocular, or stereo, vision allows the android to
perceive depth, thereby locating objects anyplace within a large room. Speech recognition and synthesis are
common.
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS, 3RD EDITION BY GIBILISCO
196. A robot that has its own computer, and can work independently of other robots or computers, is called an:
A. Android
B. Insect robot
C. Automated guided vehicle
D. Autonomous robot
ANSWER: D
EXPLANATION: A robot is autonomous if it is self-contained, housing its own computer system, and not depending on
a central computer for its commands. It gets around under its own power, usually by rolling on wheels or by moving
on two, four, or six legs.The most autonomous robots have AI. The ultimate autonomous robot will act like a living
animal or human.
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS, 3RD EDITION BY GIBILISCO
197. Robots that operate in large numbers under the control of a central AI system.
A. Android
a.
b.
c.
Insect robot
Automated guided vehicle
Autonomous robot
ANSWER: B
EXPLANATION: Insect robots operate in large numbers under the control of a central AI system. A mobile insect robot
has several legs, a set of wheels, or a track drive. The first machines of this type, developed by Brooks, looked like
beetles. They ranged in size from more than a foot long to less than a millimeter across. Most significant is the fact
that they worked collectively.
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS, 3RD EDITION BY GIBILISCO
198. A manipulator is also known as a:
A. Track drive
B. Robot arm
C. Vision system
D. Robot controller
ANSWER: B
EXPLANATION: Robot arms are technically called manipulators. Some robots, especially industrial robots, are nothing
more than sophisticated manipulators. A robot arm can be categorized according to its geometry. Some manipulators
resemble human arms. The joints in these machines can be given names like “shoulder,” “elbow,” and “wrist.”
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS, 3RD EDITION BY GIBILISCO
199. A telechir is used in conjunction with:
A. An automated guided vehicle
B. Telepresence
C. An insect robot
D. An autonomous robot
ANSWER: B
EXPLANATION: Telepresence is a refined, advanced form of robot remote control. The robot operator gets a sense of
being “on location,” even if the remotely controlled machine, or telechir, and the operator are miles apart. Control
and feedback are done by means of telemetry sent over wires, optical fibers, or radio.
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS, 3RD EDITION BY GIBILISCO
200. It is the measurement of distances to objects in a robot’s environment in a single dimension.
A. Range sensing
B. Range plotting
C. Embedded path
D. Edge detection
ANSWER: A
EXPLANATION: For one-dimensional (1-D) range sensing, a signal is sent out, and the robot measures the time it takes
for the echo to come back.Two-dimensional (2-D) range plotting involves mapping the distance to various objects, as
a function of their direction.Three-dimensional (3-D) range plotting is done in spherical coordinates: azimuth
(compass bearing), elevation (degrees above the horizontal), and range (distance).
REFERENCE: TEACH YOURSELF ELECTRICITY AND ELECTRONICS, 3RD EDITION BY GIBILISCO