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FERapid Preparation for theFundamentals of Engineering Exam

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This publication accumulates and presents a portion of the material that appeared in the Fundamentals of Engineering Review Manual, first edition, by Michael R. Lindeburg.The subjects covered do not appear explicitly in the outline of topics published bythe NCEES effective October 2005. This may indicate that the subjects are no longer required to successfully prepare for the examination. Alternatively, the subjects may beimplicitly included in other named examination subjects. Only time will tell.

Michael R. Lindeburg, PE

Copyright 2011 by Professional Publications, Inc. (PPI). All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher.

Professional Publications, Inc. Belmont, California

ReviewManualExtra

Table of Contents

P P I w w w . p p i 2 p a s s . c o m

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Three-Phase Systems and Electronics . . . . . . . . 1-1

Base Number Conversions . . . . . . . . . . . . . . . . . . 2-1

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1 Three-Phase Systems andElectronics1. Three-Phase Systems . . . . . . . . . . . . . . . . . . . . 1-12. Electronics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-3

Sample Problems . . . . . . . . . . . . . . . . . . . . . . . . 1-5FE-Style Exam Problems . . . . . . . . . . . . . . . . 1-7Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-9

Nomenclature

A gain C capacitance Fi(t) time-varying current AI rms value of current AP power WR resistance v(t) time-varying voltage VV rms value of voltage VZ impedance

Symbols

efficiency phase angle deg! angular frequency rad/s

Subscripts

f feedbacki input to the circuitin input to the op ampl lineN neutralo output to the circuitout output to the op ampp phaset totalV voltage

1. THREE-PHASE SYSTEMS

Three-phase energy distribution systems use fewer andsmaller conductors and, therefore, are more economicaland efficient than multiple single-phase systems provid-ing the same power. When rectified, three-phase voltagehas a smoother waveform and less ripple to be fil-tered out.

The principle of operation of an AC generator thatproduces three sinusoidal voltages of equal amplitudebut different phase angles is shown in Fig. 1.1(a). Eachgenerated voltage is known as the phase voltage, Vp, orcoil voltage. (Three-phase voltages are almost alwaysstated as effective values.) Because of the location ofthe windings, the three sinusoids are 120 degrees apartin phase as shown in Fig. 1.1(b). If Va is chosen as the

reference voltage, then Eq. 1.1 through Eq. 1.3 representthe phasor forms of the three sinusoids. At any moment,the vector sum of these three voltages is zero.

Va Vp0 1:1Vb Vp120 1:2Vc Vp240 1:3

Equation 1.1 through Eq. 1.3 define an ABC or positivesequence. That is, Va reaches its peak before Vb, and Vbpeaks before Vc. With a CBA (also written as ACB) ornegative sequence, obtained by rotating the field magnetin the opposite direction, the phase of the generatedsinusoids is reversed.

Although a six-conductor transmission line could beused to transmit the power generated by the three volt-ages, it is more efficient to interconnect the windings.The two methods are commonly referred to as delta(mesh) and wye (star) connections.

Figure 1.2 illustrates a delta-connected source. The volt-age across any two of the lines is known as the line

Figure 1.1 Generation of Three-Phase Voltage

S

N

+

+

+

Va

Vb Vc

winding

magnet

(a) alternator

V Va Vb Vc

120 240

(b) ABC (positive) sequence

t

P P I * w w w . p p i 2 p a s s . c o m

voltage (system voltage) and is equal to the phasevoltage. Any of the coils can be selected as the refer-ence (=0) as long as the sequence is maintained.For a positive (ABC) sequence,

VCA Vp0 1:4VAB Vp120 1:5VBC Vp240 1:6

A wye-connected source is illustrated in Fig. 1.3. Whilethe line-to-neutral voltages are equal to the phase volt-

age, the line voltages are greater3

ptimes the phase

voltages. The ground wire (neutral) is needed to carrycurrent only if the system is unbalanced. For an ABCsequence, the line voltages are

VAB 3

pVp30 1:7

VBC 3

pVp90 1:8

VCA 3

pVp210 1:9

Although the magnitude of the line voltage depends onwhether the generator coils are delta- or wye-connected,each connection results in three equal sinusoidal voltages,120 degrees out of phase with one another.

Balanced Loads

Three impedances are required to fully load a three-phase voltage source. The impedances in a three-phase

system are balanced when they are identical in magni-tude and angle. The magnitude of the voltages and linecurrents and real, complex, and reactive powers are allidentical in a balanced system. Also, the power factor isthe same for each phase. Therefore, balanced systemscan be analyzed on a per-phase basis. Such calculationsare known as one-line analyses.

Delta-Connected Loads

Figure 1.4 illustrates delta-connected loads. Figure 1.5illustrates the vector diagram for a balanced delta three-phase system. The phase voltages, V, are separated by120-degree phase angles, as are the phase currents, I.The phase angle, , between a phase voltage and itsrespective phase current depends on the phase imped-ance. With delta-connected resistive loads, the phaseand line currents differ in phase by 30 degrees.

The phase currents for a balanced system are calculatedfrom the line voltage (same as the phase voltage). For apositive sequence,

IAB VABZAB

1:10

IBC VBCZBC

1:11

ICA VCAZCA

1:12

The line currents are not the same as the phase currents

but are3

ptimes the phase current and out of phase

30 degrees from the phase current.

jIAj jIAB ICAj 3

pI AB 1:13

jIBj jIBC IABj 3

pI BC 1:14

jIC j jICA IBC j 3

pI CA 1:15

Each impedance in a balanced system dissipates thesame real phase power, Pp. The total power dissipated

Figure 1.2 Delta Source

V

A

V

B

C

V

+aa

+bc

+c b

Figure 1.3 Wye Source

A

V

V

B

NC

V

+a+b

+c

a

c b

Figure 1.4 Delta-Connected Loads

A

B

V

V

V

C

ZAB

ZCAZ BC

IA

IBIAB

IBCIC

ICA


1-2 E X T R A M A T E R I A L F O R T H E F E R E V I E W M A N U A L

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is 3 times the phase power. This is the same for bothdelta- and wye-connected loads.

Pt 3Pp 3VpI p cos 1:16

If line values are used instead of phase values, the powerequation can be written as Eq. 1.17, where is the phaseangle between phase values.

Pt 3

pVlI l cos 1:17

Wye-Connected Loads

Figure 1.6 illustrates three equal impedances connectedin wye configuration. The line and phase currents areequal. However, the phase voltage is less than the linevoltage. The line and phase currents are

IA IAN VANZAN

V3

pZAN

1:18

IB IBN VBNZBN

V3

pZBN

1:19

IC ICN VCNZCN

V3

pZCN

1:20

2. ELECTRONICS

Electronic circuits may contain combinations of passiveand active elements, and these may be linear or non-linear devices. Many electronic elements are semicon-ductors, which are inherently nonlinear. While thecurves are nonlinear, performance within a limited rangemay be assumed to be linear if the variations in incom-ing small signals are much less than the average (steady,

DC, etc.) values. Amplifier operation is normally in thelinear active region, but operation in other regions ispossible for some applications.

Operational Amplifiers

An amplifier produces an output signal from the inputsignal. The input and output signals can be either voltageor current. The output can be either smaller or larger(usually larger) than the input in magnitude. In a linearamplifier, the input and output signals usually have thesame waveform butmay have a phase difference that couldbe as much as 180 degrees. For instance, an invertingamplifier is one for which vout AVvin. For a sinusoidalinput, this is equivalent to a phase shift of 180 degrees.

The ratio of the amplitude of the output signal to theamplitude of the input is known as the gain or amplifica-tion factor, A: AV if the input and output are voltages,and AI if they are currents.

An operational amplifier (op amp) is a high-gain DCamplifier that multiplies the difference in input voltages.The equivalent circuit of an op amp is shown in Fig. 1.7.

vo AV vin vin 1:21

Figure 1.5 Positive ABC Balanced Delta Load Vector Diagram

3030

VCA

ICA

IC

ICB

VAB

IAB

IA

IAC

VBC

IBC

IB

IBA

Figure 1.6 Wye-Connected Loads

A

B

V

V

V

C

ZCN

ZANZ BN

IA

IB

IC

IN

N

Figure 1.7 Equivalent Circuit for an Ideal Operational Amplifier

ii

ii

Ri

Ro

AV(vin+ vin)

vin

vin +

+

vo

+


T H R E E - P H A S E S Y S T E M S A N D E L E C T R O N I C S 1-3

Figure 1.8 Operational Amplifier Circuits

vovi

vovi

H

A+

=

(a) feedback system

vovi

Ri

Rf

+

(b) inverting amplifier

Ri

Rf

+

(c) non-inverting amplifier

vovi

R

C

+

vo = vidt

(e) integrator

vov2

v3

v1Rf

vo = Rf

+

(d) summing amplifier

R1

R2

R3

+ +

vovi

Ri

C

+

(g) low-pass filter

vovi

C

+

(f) differentiator

R

Rf

[sinusoidal input]

feedback element

( )v1R1 v2R2 v3R3

vovi

A1 + AH

=vovi

= vovi

RfRi

Rf + RiRi

1RC

vo = RCdvidt

=vovi

Rf Ri(1 + jRiC)



. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .The characteristics of an ideal op amp are infinite posi-tive gain, AV, infinite input impedance, Ri, zero outputimpedance, Ro, and infinite bandwidth. (Infinite band-width means that the gain is constant for all frequenciesdown to 0 Hz.) Since the input impedance is infinite,ideal op amps draw no current. An op amp has twoinput terminalsan inverting terminal marked and a noninverting terminal marked +. FromEq. 1.21,

voAV

vin vin 1:22

As the gain is considered infinite in an op amp,

voAV

0 1:23

Combining Eq. 1.22 and Eq. 1.23,

vin vin 0 1:24vin vin 1:25

This is called a virtual short circuit, which means that, inan ideal op amp, the inverting and non-inverting termi-nals are at the same voltage. The virtual short circuit,and the fact that with infinite input impedance the inputcurrent ii is zero, simplify the analysis of op amp circuits.

With real op amps, the gain is not infinite but is never-theless very large (i.e., AV=10

5 to 108). If vin and vinare forced to be different, then by Eq. 1.21 the outputwill tend to be very large, saturating the op amp ataround 1015 V.The input impedance of an op amp circuit is the ratio ofthe applied voltage to current drawn (vin/iin). In prac-tical circuits, the input impedance is determined byassuming that the op amp itself draws no current; anycurrent drawn is assumed to be drawn by the remainderof the biasing and feedback circuits. Kirchhoffs voltagelaw is written for the signal-to-ground circuit.

Depending on the method of feedback, the op amp canbe made to perform a number of different operations,some of which are illustrated in Fig. 1.8. The gain of anop amp by itself is positive. An op amp with a negativegain is assumed to be connected in such a manner as toachieve negative feedback.

SAMPLE PROBLEMS

Problem 1 through Prob. 3 refer to the following figure.All phase voltages are 120 V rms, and Z=90 j50 .

A

B

Z

Z

Z

C

IA

IC

IAB

IB

IBC

ICA

Problem 1

What is the phase current IAB?

(A) 1:17211:03 A(B) 1:1790:91 A(C) 1:1760:9 A(D) 1:1729:05 A

Solution

This is a balanced delta-connected load. The phasecurrent is calculated from the line voltage.

IAB VABZAB

VAB 1200 V givenZ 90 j50 given 102:9629:05

IAB 1200 V

102:9629:05 1:16629:05 A 1:1729:05 A

Similarly,

IBC 1:16690:95 AICA 1:166210:95 A

Answer is D.

Problem 2

What is the line current IA?

(A) 1:170:99 A(B) 2:0289:01 A(C) 2:020:95 A(D) 2:40150:01 A



Solution

In a delta-connected system, line and phase currents are

not equal. The line currents are3

ptimes the phase

current and are out of phase by 30 degrees.IA IAB ICA

3p 1:166 A29:05 30 2:020:95 A

Answer is C.

Problem 3

What is the real power dissipated by the entire system?

(A) 367.0 W

(B) 388.8 W

(C) 389.5 W

(D) 480.0 W

Solution

The total power dissipated is 3 times the phase power.

Pp I 2pR 1:166 A290 122:36 W

Pt 3Pp 3122:36 W 367:08 W

Alternatively, the power can be calculated using the linevalues.

Pt 3

pV lI l cos

3p 120 V2:02 Acos 29:05 367:03 W 367:0 W

Answer is A.

Problem 4

For the difference amplifier circuit shown, determine theoutput voltage at terminal A.

20

15

3

5 25 V30 V

A+

ideal

(A) 18.13 V(B) 6.07 V

(C) 6.07 V

(D) 15.45 V

Solution

By voltage division,

vin 25 V 3 5 3

9:375 V

By the virtual short circuit between the input terminals,

vin 9:375 V

Using Ohms law, the current through the 15 resistor is

I 15 30 V 9:375 V15

1:375 A

The input impedance is infinite; therefore,

I in 0I 15 I 20

Use Kirchhoffs voltage law to find the output voltageat A.

vA vin 20I 20 9:375 V 20 1:375 A 18:125 V 18:13 V

Answer is A.

Problem 5

A set of standard logic gates receive signals A, B, and Cwith values of 1, 0, and 0, respectively. What is the logiclevel of output signal D?

AB

C

D

(A) A

(B) B

(C) C

(D) C

Solution

The output of the upper AND gate is 0. The output ofthe lower NOT gate is 1. The output of the final ORgate is 1, which is the same as A, B, and C.

Answer is D.



. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .FE-STYLE EXAM PROBLEMS

1. What is the line current for a three-phase delta-connected motor with a line voltage of 110 V rms, apower factor of 90%, a power output of 1 kW, and 80%efficiency?

(A) 4.67 A

(B) 6.56 A

(C) 7.29 A

(D) 8.08 A

Problem 2 and Prob. 3 refer to the following figure.

8

4

8

v1 = 2 V

v2 = 3 V

vo+

i

if

ideal

2. What is the current, i ?

(A) 0.88 A(B) 0.25 A(C) 0 A

(D) 0.25 A

3. What is the output voltage, vo?

(A) 7 V(B) 6 V(C) 1 V(D) 6 V

4. For the ideal op amp shown, what should be thevalue of resistor Rf to obtain a gain of 5?

1 k

2 k

3 k

Rf

vovi

+

ideal

i

if

(A) 12.0 k

(B) 19.5 k

(C) 22.5 k

(D) 27.0 k

5. Evaluate the following amplifier circuit to determinethe value of resistor R4 in order to obtain a voltage gain(vo/vi) of 120.

1 M

500 k R4

R3

R2

R1

100

vovi

+ideal

(A) 25

(B) 23 k

(C) 24 k

(D) 25 k

For the following problems, use the NCEESHandbook as your only reference.

6. The windings of a three-phase transformer are con-nected in a wye configuration. The line voltage is 4160 V.What is the line-to-neutral voltage?

(A) 1390 V

(B) 2080 V

(C) 2400 V

(D) 4160 V

7. A three-phase voltage source with line voltage of 220V is connected to three wye-connected loads. The loadin each phase is 6 of resistance in series with 8 ofinductive reactance. What is the line current?

(A) 9 A

(B) 13 A

(C) 16 A

(D) 21 A

Problem 8 through Prob. 10 pertain to the followingsituation.

A balanced three-phase, wye-connected circuit with agrounded neutral has a line voltage of 12.8 kV. Thecircuit impedance is 300+ j400 per leg.

8. What are the line currents?

(A) 0 A

(B) 15 A

(C) 26 A

(D) 77 A



9. What are the phase currents?

(A) 0 A

(B) 15 A

(C) 26 A

(D) 77 A

10. What is the current in the neutral?

(A) 0 A

(B) 15 A

(C) 26 A

(D) 77 A

11. What would be the phase-to-neutral voltage from athree-phase, wye-connected generator if the line-to-linevoltage was 15 kV?

(A) 5 kV

(B) 6.7 kV

(C) 8.7 kV

(D) 19 kV

12. The real power dissipated in one leg of a balanced,wye-connected three-phase circuit is 8.7 kW. What totalpower is supplied by the generator?

(A) 5.0 kW

(B) 12 kW

(C) 15 kW

(D) 26 kW

13. What of the following could serve as a polyphaseinductor?

(A) an induction motor

(B) heating coils

(C) a synchronous capacitor

(D) a plating tank

14. What is the output voltage, vo, of the op-ampcircuit shown?

+

1 k

10 k

vo

2 V

(A) 22 V(B) 20 V(C) 2 V(D) 22 V

15. Given A= true, B= true, and C= false, what is thevalue of the following logical expression?

A:AND:B:AND:NOT:C:OR:A

(A) true

(B) false

(C) either true or false

(D) neither true nor false

16. Minimize the Boolean expression shown.

ABA BC

(A) AB BC(B) AC

(C) A B BC(D) AC

17. Which truth table represents the outcome of thefollowing sets of logic gates?

OUTPUT

x

y

1

2



. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(A) X Y OUTPUT

0 0 0

0 1 1

1 0 0

1 1 1

(B) X Y OUTPUT

0 0 0

0 1 1

1 0 1

1 1 0

(C) X Y OUTPUT

0 0 0

0 1 0

1 0 1

1 1 1

(D) X Y OUTPUT

0 0 1

0 1 0

1 0 0

1 1 1

SOLUTIONS

1. Determine the input power.

power out Po 1 kWpower in Pi Po

1000 W0:80

1250 WFind the line current.

Pt 3

pVlI l cos

1250 W 3p 110 VI l0:9I l 1250 W 3p 110 V0:9 7:29 A

Answer is C.

2. The input current in an op amp is so small that it isassumed to be zero.

Answer is C.

3. This op amp circuit is a summing amplifier. Sincei=0,

if v1R1

v2R2

3 V8

2 V4

0:875 A

vo if Rf 0:875 A8 7 V

Answer is A.

4. By voltage division,

vin vi 2 k3 k

23vi

By the virtual short circuit,

vin vin 23vi

i vin3 k

23vi

3 k



Since the op amp draws no current,

if i

vo vinRf

23vi

3 k

But, vo=5vi.

5vi 23vi

Rf

23vi

3 k

133Rf

23

3 k

Rf 19:5 k

Answer is B.

5.R4

R3

R2

R1vo

vivin

vin+

+

i4i3

i2

i1

Vx

vin is grounded, so vin is also a virtual ground.

vin 0

Since vin 0; vi i1R1 and i1 viR1

.

Since vin 0; vx i2R2 and i2 vxR2

.

Similarly,

vx i3R3vx vo i4R4

From Kirchhoffs current law,

i4 i2 i3vx voR4

vxR2

vxR3

Now, vo 120vi.

Also, i1 i2, so

viR1 vx

R2

vx R2R1

vi

R2R1

vi 120viR4

R2R1

vi

R2

R2R1

vi

R3

120 R1R2

1

R4 1R2

1R3

R2 R3R2R3

R4 120 R1

R2

1

R2 R3R2R3

120 1 10

6

5 105

1

5 105 100 5 105 100

2:39 104 24 k

Answer is C.

6. A wye-connected source has a line voltage of3

ptimes the phase voltage. The phase voltage is equal tothe line-to-neutral voltage.

V 1 3

p V 1N V 1N V 1

3p 4160 V

3p

2402 V 2400 V

Answer is C.

7. The impedance in each phase is

Zphase R2 XL XC 2

q

6 2 8 0 2

q 10

For wye-connected loads, the line voltage is3

ptimes

the phase voltage. The phase and line currents are thesame.



I line I phase V phaseZphase

220 V 3p 10 12:7 A 13 A

Answer is B.

8. The reactance is

Z 300 2 400 2

q 500

The line current is

I l V l3

pZ12:8 kV 1000 V

kV

3p 500 14:8 A 15 A

Answer is B.

9. In a wye-connected circuit, the line and phase cur-rents are the same.

Answer is B.

10. In a balanced three-phase circuit, the current iszero in the neutral.

Answer is A.

11. For a wye-connected generator,

Vp V l3

p 15 kV3

p

8:66 kV 8:7 kVAnswer is C.

12. Regardless of how the load and generator are con-nected, the energy lost must be supplied by the genera-tor. Since there are three phases, the total power is threetimes the phase power.

Pt 3Pp 38:66 kW 26 kWAnswer is D.

13. Heating coils are essentially pure resistors anddo not have a large inductive effect. A synchronouscapacitor is a synchronous motor that runs unloadedand draws a leading current, used for power factorcorrection. A plating tank is not an inductive load.

The stator of a three-phase induction motor consistsof three sets of primary windings connected in eitherwye or delta. The induction motor is essentially athree-phase inductor.

Answer is A.

14. This is a standard inverting amplifier.

vo RfRi

vi

10 k1 k

2 V

20 V

Answer is B.

15. Evaluate the terms within the parentheses first.

A:AND:B true:AND:true trueC:OR:A false:OR:true trueNOT:C:OR:A NOT:true falseA:AND:B:AND:NOT:C:OR:A true:AND:false

false

Answer is B.

16. The minimization (simplification) twice makes useof the following rule:

x xy x y

The minimization proceeds as follows.

AB A BC originalB A BC using x AA B BC commutative

Answer is C.

17. It is expedient to assign values. For example, letX=Y=0. Both X and Y enter AND gate 1. The outputof AND gate 1 is

X AND Y 0 AND 0 0

Both X and Y enter the OR gate. The output of theOR gate is

X OR Y 0 OR 0 0



The output of AND gate 1 is NOTed. The output of theNOT gate is

NOT X AND Y NOT 0 1

The output of AND gate 2 is

0 AND 1 0The remaining entries are found similarly.

X Y OUTPUT

0 0 0

0 1 1

1 0 1

1 1 0

This is seen to be the gate implementation of theBoolean expression in Problem 30, and is known as ahalf-adder.

Answer is B.



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2 Base Number ConversionsSAMPLE PROBLEMS

Problem 1

What is the binary (base-2) representation of (135)10?

(A) 101111010

(B) 010111101

(C) 010000111

(D) 101111001

Solution

13510 1 27 0 26 0 25 0 24 0 23 1 22 1 21 1 20

13510 10000111 same as 010000111Answer is C.

Problem 2

What is the base-10 equivalent of the binary number0101110?

(A) 18

(B) 30

(C) 46

(D) 47

Solution

0 26 1 25 0 24 1 23 1 22 1 21 0 20 46

Answer is C.

Problem 3

What is the octal (base-8) equivalent of the binarynumber (1101101)2?

(A) (109)8

(B) (155)8

(C) (550)8

(D) (660)8

Solution

Since 8= 23, separate the bits into groups of three bits,starting with the least significant bits. Add leading zerosif necessary for consistency.

001 101 101

Convert each group of three bits into its octalequivalent.

001 101 101

18 58 58

The octal equivalent is 155.

Answer is B.

FE-STYLE EXAM PROBLEMS

1. What is the binary (base-2) representation of(143)10?

(A) 101111010

(B) 010111101

(C) 010001111

(D) 101111001

2. What is the octal (base-8) equivalent of the binarynumber (1101111)2?

(A) (109)8

(B) (157)8

(C) (550)8

(D) (660)8


3. What is the binary equivalent of the base-5 number(103144)5?

(A) 0 1101 1101 1101

(B) 1 0011 0111 1100

(C) 1 0111 0110 1101

(D) 1 1100 1000 0011

4. What is the binary (base-2) representation of thehexadecimal (base-16) number (7704)16?

(A) 0101 1011 0000 0100

(B) 0110 0111 0000 0100

(C) 0111 0011 0000 0100

(D) 0111 0111 0000 0100

5. What is the binary equivalent of the base-5 number(213144)5?

(A) 0 1101 1001 0101

(B) 1 0011 0111 1100

(C) 1 0111 0110 1101

(D) 1 1100 1000 0011

6. What is the value of (120)10 in base 5?

(A) 44

(B) 440

(C) 625

(D) 660

7. Convert the binary number 1011011101 to decimalformat.

(A) 125

(B) 733

(C) 792

(D) 847

8. Convert the decimal number 3510 to binary (base-2)format.

(A) 011101

(B) 100011

(C) 100111

(D) 110011

9. Convert the octal number 3728 to binary (base-2)format.

(A) 010111011

(B) 011001010

(C) 011110010

(D) 011111010

10. Convert the hexadecimal (base-16) number 43C16to binary (base-2) format.

(A) 010000011010

(B) 010000111100

(C) 100010101010

(D) 110000110100

11. Convert the binary (base-2) number 011101010102to octal (base-8).

(A) 1146

(B) 1352

(C) 1534

(D) 1652


2-2

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1. 14310 1 27 0 26 0 25 0 24 1 23 1 22 1 21 1 20

14310 10001111

Answer is C.

2. Since 8= 23, separate the bits into groups of threebits, starting with the least significant bits. Add leadingzeros if necessary for consistency.

001 101 111

Convert each group of three bits into its octalequivalent.

001 101 111

18 58 78The octal equivalent is 157.

Answer is B.

3. Convert the base-5 number to base-10.

1031445 1 55 0 54 3 53 1 52 4 51 4 50

3125 0 375 25 20 4 354910

Convert the base-10 number to base-2.

354910 1 211 1 210 0 29 1 28 1 27 1 26 0 25 1 24 1 23 1 22 0 21 1 20

The base-2 equivalent is 1101 1101 1101.

Answer is A.

4. Since 16= 24, each hexadecimal digit will expandinto four bits.

716 01112716 01112016 00002416 01002

Combine the bits.

0111 0111 0000 0100

Answer is D.

5. Convert the base-5 number to base-10.

2131445 2 55 1 54 3 53 1 52 4 51 4 50

6250 625 375 25 20 4 729910

Convert the base-10 number to base-2.

729910 1 212 1 211 1 210 0 29 0 28 1 27 0 26 0 25 0 24 0 23 0 22 1 21 1 20

The base-2 equivalent is 1 1100 1000 0011.

Answer is D.

6. 1205

24 remainder 0245

4 remainder 445

0 remainder 412010 4405

Answer is B.


B A S E N U M B E R C O N V E R S I O N S 2-3

7. Work from right to left.

1 20 = 10 21 = 01 22 = 41 23 = 81 24 = 160 25 = 01 26 = 641 27 = 1280 28 = 01 29 = 512TOTAL 733

Answer is B.

8.352 17 remainder 1

172 8 remainder 1

82 4 remainder 0

42 2 remainder 0

22 1 remainder 0

12 0 remainder 1

Answer is B.

9. 8=23, so each octal digit can be converted to threebits.

38 011278 111228 0102

3728 011111010Answer is D.

10. 16= 24, so each hexadecimal digit can be convertedto four bits.

416 01002316 00112C16 11002

43C16 0100001111002Answer is B.

11. 8=23, so every three bits can be converted to anoctal digit. Start at the LSD.

0102 281012 581102 680012 18

011101010102 16528Answer is D.


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