CHAPTER 1
Exercises
E1.1 Charge = Current Time = (2 A) (10 s) = 20 C
dq (t ) d
E1.2 i (t ) (0.01sin(20 0t) 0.01 200cos(200t ) 2cos(200t ) A
dt dt
E1.3 Because i2 has a positive value, positive charge moves in the same
direction as the reference. Thus, positive charge moves downward in
element C.
Because i3 has a negative value, positive charge moves in the opposite
direction to the reference. Thus positive charge moves upward in
element E.
E1.4 Energy = Charge Voltage = (2 C) (20 V) = 40 J
studygrid
Because vab is positive, the positive terminal is a and the negative
terminal is b. Thus the charge moves from the negative terminal to the
positive terminal, and energy is removed from the circuit element.
E1.5 iab enters terminal a. Furthermore, vab is positive at terminal a. Thus
the current enters the positive reference, and we have the passive
reference configuration.
E1.6 (a) pa (t ) v a (t )ia (t ) 20t 2
10 10 10
20t 3 20t 3
w a pa (t )dt 20t dt 2
6667 J
0 0
3 0
3
(b) Notice that the references are opposite to the passive sign
convention. Thus we have:
pb (t ) v b (t )ib (t ) 20t 200
10 10
10
w b pb (t )dt (20t 200)dt 10t 2 200t 0
1000 J
0 0
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1
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, E1.7 (a) Sum of currents leaving = Sum of currents entering
ia = 1 + 3 = 4 A
(b) 2 = 1 + 3 + ib ib = -2 A
(c) 0 = 1 + ic + 4 + 3 ic = -8 A
E1.8 Elements A and B are in series. Also, elements E, F, and G are in series.
E1.9 Go clockwise around the loop consisting of elements A, B, and C:
-3 - 5 +vc = 0 vc = 8 V
Then go clockwise around the loop composed of elements C, D and E:
- vc - (-10) + ve = 0 ve = -2 V
E1.10 Elements E and F are in parallel; elements A and B are in series.
ρL
E1.11 The resistance of a wire is given by R . Using A d and
A
substituting values, we have: studygrid
1.12 10 6 L
9.6 L = 17.2 m
(1.6 10 3 )
E1.12 P V 2 R R V 2 / P 144 I V / R 0.833 A
E1.13 P V 2 R V PR 0.25 1000 15.8 V
I V / R 15. 15.8 mA
E1.14 Using KCL at the top node of the circuit, we have i1 = i2. Then, using KVL
going clockwise, we have -v1 - v2 = 0; but v1 = 25 V, so we have v2 = -25 V.
Next we have i1 = i2 = v2/R = -1 A. Finally, we have
PR v 2i2 (25) (1) 25 W and Ps v 1i1 (25) (1) 25 W.
E1.15 At the top node we have iR = is = 2A. By Ohm’s law we have vR = RiR = 80
V. By KVL we have vs = vR = 80 V. Then ps = -vsis = -160 W (the minus sign
is due to the fact that the references for vs and is are opposite to the
passive sign configuration). Also we have PR v R iR 160 W.
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is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:
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, Problems
P1.1 Broadly, the two objectives of electrical systems are:
1. To gather, store, process, transport, and display information.
2. To distribute, store, and convert energy between various forms.
P1.2 Eight subdivisions of EE are:
1. Communication systems.
2. Computer systems.
3. Control systems.
4. Electromagnetics.
5. Electronics.
6. Photonics.
7. Power systems.
8. Signal Processing.
P1.3 Four important reasons that non-electrical engineering majors need to
learn the fundamentals of EE are:
studygrid
1. To pass the Fundamentals of Engineering Exam.
2. To be able to lead in the design of systems that contain
electrical/electronic elements.
3. To be able to operate and maintain systems that contain
electrical/electronic functional blocks.
4. To be able to communicate effectively with electrical engineers.
P1.4 Responses to this question are varied.
P1.5 (a) Electrical current is the time rate of flow of net charge through a
conductor or circuit element. Its units are amperes, which are equivalent
to coulombs per second.
(b) The voltage between two points in a circuit is the amount of energy
transferred per unit of charge moving between the points. Voltage has
units of volts, which are equivalent to joules per coulomb.
(c) The current through an open switch is zero. The voltage across the
switch can be any value depending on the circuit.
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is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:
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, (d) The voltage across a closed switch is zero. The current through the
switch can be any value depending of the circuit.
(e) Direct current is constant in magnitude and direction with respect to
time.
(f) Alternating current varies either in magnitude or direction with time.
P1.6 (a) A conductor is analogous to a frictionless pipe.
(b) A resistance is analogous to a constriction in a pipe or to a pipe
with friction.
(c) A battery is analogous to a pump.
(d) A fluid flow is analogous to current flow.
P1.7* The reference direction for iab points from a to b. Because iab has a
negative value, the current is equivalent to positive charge moving
opposite to the reference direction. Finally, since electrons have
negative charge, they are moving in the reference direction (i.e., from a
to b).
For a constant (dc) current, charge equals current times the time
interval. Thus, Q (3 A) (3 s) 9 C.
studygrid
dq t d
P1.8* i t
dt
dt
4 + 2t 5t 2 2 10t A
P1.9* Q i (t )dt 2e t dt 2e t | 0 2 coulombs
0 0
P1.10* The charge flowing through the battery is
Q (5 amperes ) (24 3600 seconds ) 432 10 3 coulombs
The stored energy is
Energy QV (432 10 3 ) (12) 5.184 10 6 joules
(a) Equating gravitational potential energy, which is mass times height
times the acceleration due to gravity, to the energy stored in the battery
and solving for the height, we have
Energy 5.184 10 6
h 17.6 km
mg 30 9.8
(b) Equating kinetic energy to stored energy and solving for velocity, we
have
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is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
4
or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:
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Exercises
E1.1 Charge = Current Time = (2 A) (10 s) = 20 C
dq (t ) d
E1.2 i (t ) (0.01sin(20 0t) 0.01 200cos(200t ) 2cos(200t ) A
dt dt
E1.3 Because i2 has a positive value, positive charge moves in the same
direction as the reference. Thus, positive charge moves downward in
element C.
Because i3 has a negative value, positive charge moves in the opposite
direction to the reference. Thus positive charge moves upward in
element E.
E1.4 Energy = Charge Voltage = (2 C) (20 V) = 40 J
studygrid
Because vab is positive, the positive terminal is a and the negative
terminal is b. Thus the charge moves from the negative terminal to the
positive terminal, and energy is removed from the circuit element.
E1.5 iab enters terminal a. Furthermore, vab is positive at terminal a. Thus
the current enters the positive reference, and we have the passive
reference configuration.
E1.6 (a) pa (t ) v a (t )ia (t ) 20t 2
10 10 10
20t 3 20t 3
w a pa (t )dt 20t dt 2
6667 J
0 0
3 0
3
(b) Notice that the references are opposite to the passive sign
convention. Thus we have:
pb (t ) v b (t )ib (t ) 20t 200
10 10
10
w b pb (t )dt (20t 200)dt 10t 2 200t 0
1000 J
0 0
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication
is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
1
or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Mafiadocs
, E1.7 (a) Sum of currents leaving = Sum of currents entering
ia = 1 + 3 = 4 A
(b) 2 = 1 + 3 + ib ib = -2 A
(c) 0 = 1 + ic + 4 + 3 ic = -8 A
E1.8 Elements A and B are in series. Also, elements E, F, and G are in series.
E1.9 Go clockwise around the loop consisting of elements A, B, and C:
-3 - 5 +vc = 0 vc = 8 V
Then go clockwise around the loop composed of elements C, D and E:
- vc - (-10) + ve = 0 ve = -2 V
E1.10 Elements E and F are in parallel; elements A and B are in series.
ρL
E1.11 The resistance of a wire is given by R . Using A d and
A
substituting values, we have: studygrid
1.12 10 6 L
9.6 L = 17.2 m
(1.6 10 3 )
E1.12 P V 2 R R V 2 / P 144 I V / R 0.833 A
E1.13 P V 2 R V PR 0.25 1000 15.8 V
I V / R 15. 15.8 mA
E1.14 Using KCL at the top node of the circuit, we have i1 = i2. Then, using KVL
going clockwise, we have -v1 - v2 = 0; but v1 = 25 V, so we have v2 = -25 V.
Next we have i1 = i2 = v2/R = -1 A. Finally, we have
PR v 2i2 (25) (1) 25 W and Ps v 1i1 (25) (1) 25 W.
E1.15 At the top node we have iR = is = 2A. By Ohm’s law we have vR = RiR = 80
V. By KVL we have vs = vR = 80 V. Then ps = -vsis = -160 W (the minus sign
is due to the fact that the references for vs and is are opposite to the
passive sign configuration). Also we have PR v R iR 160 W.
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication
is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
2
or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Mafiadocs
, Problems
P1.1 Broadly, the two objectives of electrical systems are:
1. To gather, store, process, transport, and display information.
2. To distribute, store, and convert energy between various forms.
P1.2 Eight subdivisions of EE are:
1. Communication systems.
2. Computer systems.
3. Control systems.
4. Electromagnetics.
5. Electronics.
6. Photonics.
7. Power systems.
8. Signal Processing.
P1.3 Four important reasons that non-electrical engineering majors need to
learn the fundamentals of EE are:
studygrid
1. To pass the Fundamentals of Engineering Exam.
2. To be able to lead in the design of systems that contain
electrical/electronic elements.
3. To be able to operate and maintain systems that contain
electrical/electronic functional blocks.
4. To be able to communicate effectively with electrical engineers.
P1.4 Responses to this question are varied.
P1.5 (a) Electrical current is the time rate of flow of net charge through a
conductor or circuit element. Its units are amperes, which are equivalent
to coulombs per second.
(b) The voltage between two points in a circuit is the amount of energy
transferred per unit of charge moving between the points. Voltage has
units of volts, which are equivalent to joules per coulomb.
(c) The current through an open switch is zero. The voltage across the
switch can be any value depending on the circuit.
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication
is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
3
or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Mafiadocs
, (d) The voltage across a closed switch is zero. The current through the
switch can be any value depending of the circuit.
(e) Direct current is constant in magnitude and direction with respect to
time.
(f) Alternating current varies either in magnitude or direction with time.
P1.6 (a) A conductor is analogous to a frictionless pipe.
(b) A resistance is analogous to a constriction in a pipe or to a pipe
with friction.
(c) A battery is analogous to a pump.
(d) A fluid flow is analogous to current flow.
P1.7* The reference direction for iab points from a to b. Because iab has a
negative value, the current is equivalent to positive charge moving
opposite to the reference direction. Finally, since electrons have
negative charge, they are moving in the reference direction (i.e., from a
to b).
For a constant (dc) current, charge equals current times the time
interval. Thus, Q (3 A) (3 s) 9 C.
studygrid
dq t d
P1.8* i t
dt
dt
4 + 2t 5t 2 2 10t A
P1.9* Q i (t )dt 2e t dt 2e t | 0 2 coulombs
0 0
P1.10* The charge flowing through the battery is
Q (5 amperes ) (24 3600 seconds ) 432 10 3 coulombs
The stored energy is
Energy QV (432 10 3 ) (12) 5.184 10 6 joules
(a) Equating gravitational potential energy, which is mass times height
times the acceleration due to gravity, to the energy stored in the battery
and solving for the height, we have
Energy 5.184 10 6
h 17.6 km
mg 30 9.8
(b) Equating kinetic energy to stored energy and solving for velocity, we
have
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication
is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
4
or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Mafiadocs
