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FERapid Preparation for theFundamentals of Engineering Exam

www.ppi2pass.com/extraFERM

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 by

the NCEES effective October 2005. This may indicate that the subjects are no longer

required to successfully prepare for the examination. Alternatively, the subjects may be

implicitly 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 andElectronics

1. Three-Phase Systems . . . . . . . . . . . . . . . . . . . . 1-12. E lectronics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 A

P 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 circuit

in 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 , V p, orcoil voltage . (Three-phase voltages are almost alwaysstated as effective values.) Because of the location of the windings, the three sinusoids are 120 degrees apartin phase as shown in Fig. 1.1(b). If V a 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.

V a ¼ V pff01:1

V b ¼ V pffÀ1201:2

V c ¼ V pffÀ2401:3

Equation 1.1 through Eq. 1.3 define an ABC or positive sequence . That is, V a reaches its peak before V b, and V b

peaks beforeV

c . 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

–

–

–

+

+

+

V a

V b V c

winding

magnet

(a) alternator

V V a V b V c

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,

V CA ¼ V pff01:4

V AB ¼ V pffÀ1201:5

V BC ¼ V pffÀ240 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 greater — ffiffiffi

3p

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

V AB ¼ ffiffiffi

3p

V pff301:7

V BC ¼ ffiffiffi

3p

V pffÀ901:8

V C A ¼ ffiffiffi3

p V

pffÀ210

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 calculated

from the line voltage (same as the phase voltage). For apositive sequence,

IAB ¼ VAB

ZAB

1:10

IBC ¼ VBC

ZBC

1:11

IC A ¼ VCA

ZCA

1:12

The line currents are not the same as the phase currents

but are ffiffiffi3p times the phase current and out of phase

À30 degrees from the phase current.

jIAj ¼ jIAB À ICA j ¼ ffiffiffi

3p

I AB 1:13

jIB j ¼ jIBC À IAB j ¼ ffiffiffi

3p

I BC 1:14

jIC j ¼ jIC A À IBC j ¼ ffiffiffi

3p

I CA 1:15

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

Figure 1.2 Delta Source

∼

∼ ∼

V

A

V

B

C

V

+a–a

+b –c

+c –b

Figure 1.3 Wye Source

A

V

V

B

N

C

V

+a

+b

+c

–a

–c –b

∼

∼

∼

Figure 1.4 Delta-Connected Loads

A

B

V

V

V

C

Z AB

Z C A Z B C

I A

I B

I AB

I BC

I C I CA

P P I*


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.

P t ¼ 3P p ¼ 3V pI 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.

P t ¼ ffiffiffi

3p

V l I l cos 1:17

Wye-Connected LoadsFigure 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 ¼ IAN ¼ VAN

ZAN

¼ V ffiffiffi3

p ZAN

1:18

IB ¼ IBN ¼ VBN

ZBN

¼ V ffiffiffi3

p ZBN

1:19

IC

¼IC N

¼VCN

ZCN ¼V

ffiffiffi3p ZCN

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 but may have a phase difference that couldbe as much as 180 degrees. For instance, an invertingamplifier is one for which v out ¼ ÀAV v in. 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.

v o ¼ AV ðv inþ À v inÀÞ 1:21

Figure 1.5 Positive ABC Balanced Delta Load Vector Diagram

30°30°

θ

θ

θ

V CA

I CA

I C

I CB

V AB

I AB

I A

I AC

V BC

I BC

I B

I BA

Figure 1.6 Wye-Connected Loads

A

B

V

V

V

C

Z CN

Z A N Z B

N

I A

I B

I C

I N

N

Figure 1.7 Equivalent Circuit for an Ideal Operational Amplifier

i i

i i

R i

R o

AV (v in+ − v in−)

v in –

v in +

+

–

v o

+–

P P I*


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

v o v i

v o v i

H

A+

–

=

(a) feedback system

v o v i

R i

R f

+–

(b) inverting amplifier

R i

R f

+

–

(c) non-inverting amplifier

v o v i

R

C

+

–

v o = v i dt

(e) integrator

v o

v 2

v 3

v 1

R f

v o = −R f

+

–

(d) summing amplifier

R 1

R 2

R 3

+ +

v o v i

R i

C

+

–

(g) low-pass filter

v o v i

C

+

–

(f) differentiator

R

R f

[sinusoidal input]

feedback element

( )v 1R 1

v 2R 2

v 3R 3

v o

v i

A

1 + AH

=v o

v i

= −v o

v i

R f R i

R f + R i R i

−1

RC v o = −RC

dv i dt

=v o

v i

−R f

R i (1 + jωR i C )

P P I*





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The characteristics of an ideal op amp are infinite posi-tive gain, AV , infinite input impedance, R i , zero outputimpedance, R o, 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 terminals — an inverting terminal marked “À”

and a noninverting terminal marked “+”. From

Eq. 1.21,

v oAV

¼ ðv inþ À v inÀÞ 1:22

As the gain is considered infinite in an op amp,

v oAV

¼ 0 1:23

Combining Eq. 1.22 and Eq. 1.23,

v inþ À v inÀ ¼ 0 1:24

v inþ ¼ v inÀ 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 i i 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 = 105 to 108). If v inþ and v inÀ

are forced to be different, then by Eq. 1.21 the outputwill tend to be very large, saturating the op amp ataround ±10À15 V.

The input impedance of an op amp circuit is the ratio of the applied voltage to current drawn (v in/i in). 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. Kirchhoff’s 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 À j50 .

A

B

Z

Z

Z

C

I A

I C

I AB

I B

I BC

I CA

Problem 1

What is the phase current I AB ?

(A) 1:17

ffÀ211:03 A

(B) 1:17ffÀ90:91 A

(C) 1:17ffÀ60:9 A

(D) 1:17ff29:05 A

Solution

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

IAB ¼ VAB

ZAB

VAB

¼120

ff0 V

½given

Z ¼ 90 À j50 ½given¼ 102:96ffÀ29:05

IAB ¼ 120ff0 V102:96ffÀ29:05

¼ 1:166ff29:05 A ð1:17ff29:05 AÞ

Similarly,

IBC ¼ 1:166ffÀ90:95 A

ICA ¼ 1:166ffÀ210:95 A

Answer is D.

Problem 2

What is the line current I A?

(A) 1:17ff0:99 A

(B) 2:02ff89:01 A

(C) 2:02ffÀ0:95 A

(D) 2:40ffÀ150:01 A

P P I*





Solution

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

not equal. The line currents are ffiffiffi

3p

times the phasecurrent and are out of phase by À30 degrees.

IA ¼ IAB À ICA

¼ ð ffiffiffi3p

Þð1:166 A

Þff29:05

À30

¼ 2:02ffÀ0:95 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.

P p ¼ I 2pR ¼ ð1:166 AÞ2ð90 Þ¼ 122:36 W

P t ¼ 3P p ¼ ð3Þð122:36 WÞ¼ 367:08 W

Alternatively, the power can be calculated using the linevalues.

P t ¼ ffiffiffi3p V l I l cos

¼ ð ffiffiffi3p Þð120 VÞð2:02 AÞcos29:05

¼ 367:03 W ð367:0 WÞ

Answer is A.

Problem 4

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

20 Ω

15 Ω

3 Ω

5 Ω25 V30 V

A–+

ideal

(A) À18.13 V

(B) À6.07 V

(C) 6.07 V

(D) 15.45 V

Solution

By voltage division,

v inþ ¼ ð

25 VÞ

3

5 þ 3 ¼

9:375 V

By the virtual short circuit between the input terminals,

v inÀ ¼ 9:375 V

Using Ohm’s law, the current through the 15 resistor is

I 15 ¼ 30 V À 9:375 V15

¼ 1:375 A

The input impedance is infinite; therefore,

I inÀ ¼ 0I 15 ¼ I 20

Use Kirchhoff’s voltage law to find the output voltageat A.

v A ¼ v inÀ À 20I 20

¼ 9:375 V À ð20 Þð1:375 AÞ¼ À18:125 V ð18:13 VÞ

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 of the 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.

P P I*





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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 Ω

4 Ω

8 Ω

v 1 = 2 V

v 2 = 3 V

v o –+

i

i f

ideal

2. What is the current, i ?

(A) À0.88 A

(B) À0.25 A

(C) 0 A

(D) 0.25 A

3. What is the output voltage, v o?

(A)

À7 V

(B) À6 V

(C) À1 V

(D) 6 V

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

1 kΩ

2 kΩ

3 kΩ

R f

v o

v i

–

+

ideal

i

i f

(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 R 4 in order to obtain a voltage gain(v o/v i ) of À120.

1 MΩ

500 kΩ R 4

R 3

R 2

R 1

100 Ω

v ov i –

+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 of inductive 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

P P I*





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, v o, of the op-ampcircuit shown?

+

–

1 kΩ

10 kΩ

v o

2 V

(A) À22 V

(B) À20 V

(C) À2 V

(D) 22 V

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

ðA:AND:BÞ:AND:ðNOT:ðC:OR:AÞÞ

(A) true

(B) false

(C) either true or false

(D) neither true nor false

16. Minimize the Boolean expression shown.

AB þ A þ BC

(A) AB þ BC

(B) AC

(C) A þ B þ BC

(D) AC

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

OUTPUT

x

y

1

2

P P I*





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(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 01 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 ¼ P o ¼ 1 kW

power in ¼ P i ¼ P o

¼ 1000 W0:80

¼ 1250 W

Find the line current.

P t ¼ ffiffiffi

3p

V l I l cos

1250 W ¼ ð ffiffiffi3p Þð110 VÞðI l Þð0:9ÞI l ¼ 1250 W

ð ffiffiffi3

p Þð110 VÞð0:9Þ

¼7:29 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,

i f ¼v

1R 1 þv

2R 2 ¼3 V8 þ

2 V4

¼ 0:875 A

v o ¼ Ài f R f

¼ Àð0:875 AÞð8 Þ¼ À7 V

Answer is A.

4. By voltage division,

v inþ ¼ v i 2 k3 k

¼ 2

3v i

By the virtual short circuit,

v inÀ ¼ v inþ ¼ 23

v i

i ¼ v inÀ

3 k¼

23

v i

3 k

P P I*





Since the op amp draws no current,

i f ¼ i

v o À v inÀ

R f ¼

2

3v i

3 k

But, v o = 5v i .

5v i À 2

3v i

R f ¼

2

3v i

3 k

13

3R f

¼2

33 k

R f ¼ 19:5 k

Answer is B.

5.R 4

R 3

R 2

R 1

v o

v i

v in−

v in+

–

+

i 4i 3

i 2

i 1

V x

v inþ is grounded, so v inÀ is also a virtual ground.

v inÀ ¼ 0

Since v inÀ ¼ 0; v i ¼ i 1R 1 and i 1 ¼ v i R 1

.

Since v inÀ ¼ 0; v x ¼ Ài 2R 2 and i 2 ¼ Àv x R 2

.

Similarly,

v x ¼ Ài 3R 3v x

Àv o

¼i 4R 4

From Kirchhoff’s current law,

i 4 ¼ i 2 þ i 3

v x À v oR 4

¼ Àv x R 2

þ Àv x R 3

Now, v o ¼ À120v i .

Also, i 1 ¼ i 2, so

v i R 1

¼ Àv x R 2

v x ¼ À R 2R 1

v i

À R 2R 1

v i À ðÀ120v i ÞR 4

¼R 2R 1

v i

R 2þ

R 2R 1

v i

R 3

ð120Þ R 1R 2

À 1

R 4¼ 1

R 2þ 1

R 3

¼ R 2 þ R 3R 2R 3

R 4

¼

ð120Þ R 1R 2

À 1

R 2 þ

R 3

R 2R 3

¼ð120Þ 1 Â 106

5 Â 105

À 1

5 Â 105 þ 100

ð5 Â 105 Þð100 Þ¼ 2:39 Â 104 ð24 kÞ

Answer is C.

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

ffiffiffi3

p

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

V 1 ¼ ffiffiffi3

p ðV 1ÀN ÞV 1ÀN ¼ V 1 ffiffiffi

3p ¼ 4160 V ffiffiffi

3p

¼ 2402 V ð2400 VÞ

Answer is C.

7. The impedance in each phase is

Z phase ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiðR Þ2 þ ðX L ÀX C Þ2

q

¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffið6 Þ2 þ ð8 À 0 Þ2

q ¼ 10

For wye-connected loads, the line voltage is ffiffiffi

3p

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

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I line ¼ I phase ¼ V phase

Z phase

¼ 220 V

ð ffiffiffi3

p Þð10 Þ¼ 12:7 A ð13 AÞ

Answer is B.

8. The reactance is

Z ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffið300 Þ2 þ ð400 Þ2

q ¼ 500

The line current is

I l ¼ V l ffiffiffi3

p Z

¼ð12:8 kVÞ 1000

VkV

ð ffiffiffi3

p Þð500 Þ¼ 14:8 A ð15 AÞ

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,

V p ¼ V l ffiffiffi3

p ¼ 15 kV ffiffiffi3

p

¼ 8:66 kV ð8:7 kVÞAnswer 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.

P t ¼ 3P p ¼ ð3Þð8:66 kWÞ ¼ 26 kW

Answer 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.

v o ¼ À R f

R i

v i

¼ À 10 k1 k

ð2 VÞ

¼ À20 V

Answer is B.

15. Evaluate the terms within the parentheses first.

ðA:AND:BÞ true:AND:true ¼ true

ðC:OR:AÞ false:OR:true ¼ true

ðNOT:ðC:OR:AÞÞ NOT:true ¼ false

ðA:AND:BÞ:AND:ðNOT:ðC:OR:AÞÞ true:AND:false

¼ false

Answer is B.

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

x þ xy ¼ x þ y

The minimization proceeds as follows.

AB þ A þ BC ½originalB þ A þ BC ½using x ¼ AA þ B þ BC ½commutative

Answer is C.

17. It is expedient to assign values. For example, let

X = Y = 0. Both X and Y enter AND gate 1. The outputof AND gate 1 is

ðX AND YÞ ¼ ð0 AND 0Þ ¼ 0

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

ðX OR YÞ ¼ ð0 OR 0Þ ¼ 0

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

NOT ðX AND YÞ ¼ NOT ð0Þ ¼ 1

The output of AND gate 2 is

0 AND 1 ¼ 0

The 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 the

Boolean expression in Problem 30, and is known as ahalf-adder.

Answer is B.

P P I*





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2 Base Number Conversions

SAMPLE PROBLEMS

Problem 1

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

(A) 101111010

(B) 010111101

(C) 010000111

(D) 101111001

Solution

ð135Þ10 ¼ 1 Â 27

þ 0 Â 26

þ 0 Â 25

þ 0 Â 24

þ 0 Â 23

þ 1 Â 22

þ 1 Â 21

þ 1 Â 20

ð135Þ10 10000111 ½same as 010000111

Answer is C.

Problem 2

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

(A) 18

(B) 30

(C) 46

(D) 47

Solution

0 Â 26

þ 1 Â 25

þ 0 Â 24

þ 1 Â 23

þ 1 Â 22

þ 1 Â 21

þ 0 Â 20 ¼ 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

ð1Þ8 ð5Þ8 ð5Þ8

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

P P I*




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 43C16

to binary (base-2) format.

(A) 010000011010

(B) 010000111100

(C) 100010101010

(D) 110000110100

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

to octal (base-8).

(A) 1146

(B) 1352

(C) 1534

(D) 1652

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2-2



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SOLUTIONS

1. ð143Þ10 ¼ 1 Â 27

þ 0 Â 26

þ 0 Â 25

þ 0 Â 24

þ 1 Â 23

þ 1 Â 22

þ 1 Â 21

þ 1 Â 20

ð143Þ10 10001111

Answer is C.

2. Since 8 = 23, separate the bits into groups of threebits, starting with the least significant bits. Add leading

zeros if necessary for consistency.

001 101 111

Convert each group of three bits into its octalequivalent.

001 101 111

ð1Þ8 ð5Þ8 ð7Þ8

The octal equivalent is 157.

Answer is B.

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

ð103144Þ5 ð1 Â 55Þ þ ð0 Â 54Þ þ ð3 Â 53Þ

þ ð1 Â 52Þ þ ð4 Â 51Þ þ ð4 Â 50Þ

¼ 3125 þ 0 þ 375 þ 25 þ 20 þ 4

¼ ð3549Þ10

Convert the base-10 number to base-2.

ð3549Þ10 ð1 Â 211Þ þ ð1 Â 210Þ þ ð0 Â 29Þ

þ ð1 Â 28Þ þ ð1 Â 27Þ þ ð1 Â 26Þ

þ ð0 Â 25Þ þ ð1 Â 24Þ þ ð1 Â 23Þ

þ ð1 Â 22Þ þ ð0 Â 21Þ þ ð1 Â 20Þ

The base-2 equivalent is 1101 1101 1101.

Answer is A.

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

ð7Þ16 ¼ ð0111Þ2

ð7Þ16 ¼ ð0111Þ2

ð0Þ16 ¼ ð0000Þ2

ð4Þ16 ¼ ð0100Þ2

Combine the bits.

0111 0111 0000 0100

Answer is D.

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

ð213144Þ5 ð2 Â 55Þ þ ð1 Â 54Þ þ ð3 Â 53Þ

þ ð1 Â 52Þ þ ð4 Â 51Þ þ ð4 Â 50Þ

¼ 6250 þ 625 þ 375 þ 25 þ 20 þ 4

¼ ð7299Þ10

Convert the base-10 number to base-2.

ð7299Þ10 ð1 Â 212Þ þ ð1 Â 211Þ þ ð1 Â 210Þ þ ð0 Â 29Þ

þ ð0 Â 28

Þ þ ð1 Â 27

Þ þ ð0 Â 26

Þþ ð0 Â 25Þ þ ð0 Â 24Þ þ ð0 Â 23Þ

þ ð0 Â 22Þ þ ð1 Â 21Þ þ ð1 Â 20Þ

The base-2 equivalent is 1 1100 1000 0011.

Answer is D.

6. 1205

¼ 24 remainder 0

245

¼ 4 remainder 4

45

¼ 0 remainder 4

ð120Þ10 ¼ ð440Þ5

Answer is B.

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

TOTAL 733

Answer is B.

8.352

¼ 17 remainder 1

17

2

¼ 8 remainder 1

82

¼ 4 remainder 0

42

¼ 2 remainder 0

22

¼ 1 remainder 0

12

¼ 0 remainder 1

Answer is B.

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

bits.

38 ¼ 0112

78 ¼ 1112

28 ¼ 0102

3728 ¼ 011111010

Answer is D.

10. 1 6 = 24, so each hexadecimal digit can be convertedto four bits.

416 ¼ 01002

316 ¼ 00112

C16 ¼ 11002

43C16 ¼ 0100001111002

Answer is B.

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

0102 ¼ 28

1012 ¼ 58

1102 ¼ 68

0012 ¼ 18

011101010102 ¼ 16528

Answer is D.

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