Các bài tập cho sinh viên điện tử

APPENDIX A
Exercises
EA.1

Given Z 1 = 2 − j 3 and Z 2 = 8 + j 6, we have:

Z 1 + Z 2 = 10 + j 3
Z 1 − Z 2 = −6 − j 9

Z 1 Z 2 = 16 − j 24 + j 12 − j 2 18 = 34 − j 12
2 − j 3 8 − j 6 16 − j 12 − j 24 + j 2 18
Z1 / Z2 =
×
=
= −0.02 − j 0.36
8+ j6 8− j6
100
EA.2

Z 1 = 15∠45 o = 15 cos( 45 o ) + j 15 sin( 45 o ) = 10.6 + j 10.6
Z 2 = 10∠ − 150 o = 10 cos( −150 o ) + j 10 sin( −150 o ) = −8.66 − j 5
Z 3 = 5∠90 o = 5 cos(90 o ) + j 5 sin(90 o ) = j 5

EA.3

Notice that Z1 lies in the first quadrant of the complex plane.

Z 1 = 3 + j 4 = 32 + 4 2 ∠ arctan( 4 / 3) = 5∠53.13o
Notice that Z2 lies on the negative imaginary axis.
Z 2 = − j 10 = 10∠ − 90 o
Notice that Z3 lies in the third quadrant of the complex plane.

Z 3 = −5 − j 5 = 52 + 52 ∠(180 o + arctan( −5 / − 5)) = 7.07 ∠225 o = 7.07 ∠ − 135 o
EA.4

Notice that Z1 lies in the first quadrant of the complex plane.

Z 1 = 10 + j 10 = 10 2 + 10 2 ∠ arctan(10 / 10) = 14.14∠45 o = 14.14 exp( j 45 o )
Notice that Z2 lies in the second quadrant of the complex plane.

Z 2 = −10 + j 10 = 10 2 + 10 2 ∠(180 o + arctan( −10 / 10))
= 14.14∠135 o = 14.14 exp( j 135 o )

685


EA.5

Z 1Z 2 = (10∠30 o )(20∠135 o ) = (10 × 20)∠(30 o + 135 o ) = 200∠(165 o )
Z 1 / Z 2 = (10∠30 o ) /(20∠135 o ) = (10 / 20)∠(30 o − 135 o ) = 0.5∠( −105 o )
Z 1 − Z 2 = (10∠30 o ) − (20∠135 o ) = (8.66 + j 5) − ( −14.14 + j 14.14)
= 22.8 − j 9.14 = 24.6∠ − 21.8o

Z 1 + Z 2 = (10∠30 o ) + (20∠135 o ) = (8.66 + j 5) + ( −14.14 + j 14.14)
= −5.48 + j 19.14 = 19.9∠106o

Problems
PA.1

Given Z 1 = 2 + j 3 and Z 2 = 4 − j 3, we have:

Z1 + Z2 = 6 + j 0
Z 1 − Z 2 = −2 + j 6
Z 1 Z 2 = 8 − j 6 + j 12 − j 2 9 = 17 + j 6
Z1 / Z2 =

PA.2

2 + j 3 4 + j 3 − 1 + j 18
=
= 0.04 + j 0.72
×
4 − j3 4 + j3
25

Given that Z 1 = 1 − j 2 and Z 2 = 2 + j 3, we have:

Z1 + Z2 = 3 + j 1
Z 1 − Z 2 = −1 − j 5
Z1 Z2 = 2 + j 3 − j 4 − j 2 6 = 8 − j 1
Z1 / Z2 =

1 − j2 2 − j3 − 4 − j 7
=
= −0.3077 − j 0.5385
×
2 + j3 2 − j3
13

686

PA.3

Given that Z 1 = 10 + j 5 and Z 2 = 20 − j 20, we have:

Z 1 + Z 2 = 30 − j 15
Z 1 − Z 2 = −10 + j 25
Z 1 Z 2 = 200 − j 200 + j 100 − j 2 100 = 300 − j 100
Z1 / Z2 =

PA.4

PA.5

PA.6

10 + j 5 20 + j 20 100 + j 300
=
= 0.125 + j 0.375
×
20 − j 20 20 + j 20
800

(a)

Z a = 5 − j 5 = 7.071∠ − 45o = 7.071 exp(− j 45o )

(b)

Z b = −10 + j 5 = 11.18∠153.43o = 11.18 exp(j 153.43o )

(c)

Z c = −3 − j 4 = 5∠ − 126.87 o = 5 exp(− j 126.87 o )

(d)

Z d = − j 12 = 12∠ − 90 o = 12 exp(− j 90 o )

(a)

Z a = 5∠45o = 5 exp(j 45o ) = 3.536 + j 3.536

(b)

Z b = 10∠120 o = 10 exp(j 120 o ) = −5 + j 8.660

(c)

Z c = 15∠ − 90 o = 15 exp(− j 90 o ) = − j 15

(d)

Z d = −10∠60 o = 10 exp(− j 120 o ) = −5 − j 8.660

(a)

Z a = 5e j 30 = 5∠30 o = 4.330 + j 2.5

(b)

Z b = 10e − j 45 = 10∠ − 45o = 7.071 − j 7.071

(c)

Z c = 100e j 135 = 100∠135o = −70.71 + j 70.71

o

o

o

687


PA.7

(d)

Z d = 6e j 90 = 6∠90 o = j 6

(a)

Z a = 5 + j 5 + 10∠30 o = 13.66 + j 10

(b)

Z b = 5∠45o − j 10 = 3.536 − j 6.464

(c)

Zc =

10∠45 o
10∠45 o
2∠ − 8.13o = 1.980 − j 0.283
=
3 + j4
5∠53.13o

(d)

Zd =

15
= 3∠ − 90 o = − j 3
o
5∠90

o

688


APPENDIX C
PC.1

Because the capacitor voltage is zero at t = 0, the charge on the
capacitor is zero at t = 0. Then using Equation 3.5 in the text, we have

t

q(t ) = ∫ i (t )dx + 0
0
t

= ∫ 3dx = 3t
0

For t = 2 µs, we have
q(3) = 3 × 2 × 10 −6 = 6 µC
PC.2

Refer to Figure PC.2 in the book. Combining the 10-Ω resistance and the
20-Ω resistance we obtain a resistance of 6.667 Ω, which is in series
with the 5-Ω resistance. Thus, the total resistance seen by the 15-V
source is 5 + 6.667 = 11.667 Ω. The source current is 15/11.667 = 1.286
A. The current divides between the 10-Ω resistance and the 20-Ω
resistance. Using Equation 2.27, the current through the 10-Ω
resistance is

i 10 =

20
× 1.286 = 0.8572 A
20 + 10

Finally, the power dissipated in the 10-Ω resistance is
2
P10 = 10i 10

= 7.346 W

PC.3

The equivalent capacitance of the two capacitors in series is given by

Ceq =

1
= 4 µF
1 / C1 + 1 / C 2

The charge supplied by the source is
q = CeqV = 200 × 4 × 10 −6 = 800 µC

689


PC.4

The input power to the motor is the output power divided by efficiency

2 × 746
= 1865 W
0.80
η
However the input power is also given by
Pin =

Pout

=

Pin = Vrms I rms cos(θ )
in which cos(θ ) is the power factor. Solving for the current, we have

I rms =

Pin
1865
=
= 11.30 A
Vrms cos(θ ) 220 × 0.75

PC.5

j
= 30 + j 40 − j 80 = 30 − j 40 = 50∠ − 53.1o
ωC
Thus the impedance magnitude is 50 Ω.

PC.6

We have

Z = R + jωL −

Apparent power = Vrms I rms
Also, the power factor is cos(θ ) = 0.6 from which we find that θ = 53.13 o.
(We selected the positive angle because the power factor is stated to be

lagging.) Then we have

Q = Vrms I rms sin(θ ) = ( Apparent power) × sin(θ ) = 2000 × 0.8 = 1600 VAR
PC.7

For practical purposes, the capacitor is totally discharged after twenty
time constants and all of the initial energy stored in the capacitor has
been delivered to the resistor. The initial stored energy is

W = 21 CV 2 = 21 × 150 × 10 −6 × 100 2 = 0.75 J
PC.8

ω = 2πf = 120π
Z = R + jωL −

j
= 50 + j 56.55 − j106.10 = 50 − j 49.55 = 70.39∠ − 44.74 o
ωC

690


I rms =
PC.9

Vrms
110
=
= 1.563 A
70.39

Z

See Example 4.2 in the book. In this case, we have K 2 = K 1 = VS R = 1 A
and τ = L R = 0.5 s. Then the current is given by

i (t ) = 1 − exp( −t / τ ) = 1 − exp(−2t )
PC.10

We have VBC = −VCB = −50 V and VAB = VAC −VBC = 200 − ( −50) = 250. The

energy needed to move the charge from point B to point A is
W = QVAB = 0.2(250) = 50 J.

691


CONTENTS
Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221
Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329
Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286
Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347
Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359
Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407
Chapter 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 458
Chapter 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487

Chapter 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 520
Chapter 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 572
Chapter 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608
Chapter 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 646
Appendix A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 685
Appendix C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 689
Complete solutions to the in-chapter exercises, answers to the end-ofchapter problems marked by an asterisk *, and complete solutions to the
Practice Tests are available to students at www.pearsonhighered.com/hambley


CHAPTER 1
Exercises
E1.1

Charge = Current × Time = (2 A) × (10 s) = 20 C

E1.2

i (t ) =

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.

dq (t ) d
=
(0.01sin(200t) = 0.01 × 200cos(200t ) = 2cos(200t ) A
dt

dt

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
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

E1.6

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.
(a) pa (t ) = v a (t )ia (t ) = 20t 2
10

10

20t 3
w a = ∫ pa (t )dt = ∫ 20t dt =
3
0
0

10

2

=
0

20t 3
= 6667 J
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

0

0

w b = ∫ pb (t )dt = ∫ (20t − 200)dt = 10t 2 − 200t

1

10
0

= −1000 J


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.

E1.11

The resistance of a wire is given by R =

substituting values, we have:
9. 6 =

1.12 × 10 −6 × L
π (1.6 × 10 − 3 )2 / 4

ρL

A

. Using A = πd 2 / 4 and

⇒ L = 17.2 m

E1.12

P =V 2 R

⇒

R =V 2 / P = 144 Ω

E1.13

P =V 2 R

⇒

V = PR = 0.25 × 1000 = 15.8 V

⇒

I = V / R = 120 / 144 = 0.833 A

I = V / R = 15.8 / 1000 = 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.
2

Problems
P1.1

Four reasons that non-electrical engineering majors need to learn the
fundamentals of EE are:
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.

P1.2

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.

P1.3

Eight subdivisions of EE are:
1.
2.
3.
4.
5.
6.
7.
8.

Communication systems.
Computer systems.
Control systems.
Electromagnetics.
Electronics.
Photonics.
Power systems.
Signal Processing.

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.

3


(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) An open switch is analogous to a closed valve.
(c) A resistance is analogous to a constriction in a pipe or to a pipe with
friction.
(d) A battery is analogous to a pump.

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.
2 coulomb/s
= 12.5 × 1018
1.60 × 10 − 19 coulomb/electron

P1.8

Electrons per second =

P1.9*

i (t ) =

P1.10

The positive reference for v is at the head of the arrow, which is
terminal b. The positive reference for vba is terminal b. Thus, we have
v ba = v = −10 V. Also, i is the current entering terminal a, and iba is the

dq (t ) d
(2t + t 2 ) = 2 + 2t A
=
dt
dt

current leaving terminal a. Thus, we have i = −iba = −3 A. The true

polarity is positive at terminal a, and the true current direction is
entering terminal a. Thus, current enters the positive reference and
energy is being delivered to the device.

P1.11

To cause current to flow, we make contact between the conducting parts
of the switch, and we say that the switch is closed. The corresponding
fluid analogy is a valve that allows fluid to pass through. This
corresponds to an open valve. Thus, an open valve is analogous to a closed

4


switch.

P1.12*
P1.13

∞

∞

0

0

Q = ∫ i (t )dt = ∫ 2e −t dt = −2e −t | ∞0 = 2 coulombs
(a) The sine function completes one cycle for each 2π radian increase in
the angle. Because the angle is 200πt , one cycle is completed for each
time interval of 0.01 s. The sketch is:

(b) Q =

0.01

0.01

0

0

∫ i (t )dt = ∫ 10 sin(200πt )dt = (10 / 200π ) cos(200πt )

=0 C
(b) Q =

0.015

0.015

0

0

0

∫ i (t )dt = ∫ 10 sin(200πt )dt = (10 / 200π ) cos(200πt )

= 0.0318 C

P1.14*

0.01

0.015
0

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

5


v =

2 × Energy

m

= 587.9 m/s

(c) The energy density of the battery is
5.184 × 10 6
= 172.8 × 10 3 J/kg
30
which is about 0.384% of the energy density of gasoline.

dq (t ) d
(2t + e −2t ) = 2 − 2e −2t A
=
dt
dt

P1.15

i (t ) =

P1.16

The number of electrons passing through a cross section of the wire per
second is
5
N =
= 3.125 × 1019 electrons/second
− 19
1.6 × 10
The volume of copper containing this number of electrons is
volume =

3.125 × 1019
= 3.125 × 10 −10 m3
1029

The cross sectional area of the wire is

A=

πd 2
4

= 3.301 × 10 − 6 m2

Finally, the average velocity of the electrons is
volume
u =
= 94.67 µm/s

A

P1.17*

Q = current × time = (10 amperes) × (36,000 seconds) = 3.6 × 10 5 coulombs
Energy = QV = (3.6 × 10 5 ) × (12.6) = 4.536 × 10 6 joules

P1.18

Q = current × time = (2 amperes) × (10 seconds) = 20 coulombs
Energy = QV = (20) × (5) = 100 joules
Notice that iab is positive. If the current were carried by positive charge,
it would be entering terminal a. Thus, electrons enter terminal b. The
energy is delivered to the element.

6


P1.19

The electron gains 1.6 × 10 −19 × 120 = 19.2 × 10 −18 joules

P1.20*

(a) P = -vaia = 30 W

Energy is being absorbed by the element.

(b) P = vbib = 30 W

Energy is being absorbed by the element.

(c) P = -vDEiED = -60 W

Energy is being supplied by the element.

P1.21

If the current is referenced to flow into the positive reference for the
voltage, we say that we have the passive reference configuration. Using
double subscript notation, if the order of the subscripts are the same
for the current and voltage, either ab or ba, we have a passive reference
configuration.

P1.22*

Q = w V = (600 J) (12 V) = 50 C .
To increase the chemical energy stored in the battery, positive charge
should move through the battery from the positive terminal to the
negative terminal, in other words from a to b. Electrons move from b to
a.

P1.23

The amount of energy is W = QV = (4 C) × (15 V) = 60 J. Because the

reference polarity for vab is positive at terminal a and the voltage value is
negative, terminal b is actually the positive terminal. Because the charge
moves from the negative terminal to the positive terminal, energy is
removed from the device.

P1.24*

Energy =

P =

Cost
$60
=
= 500 kWh
Rate 0.12 $/kWh

Energy 500 kWh
=
= 694.4 W
Time
30 × 24 h

Reduction =

I =

60
× 100% = 8.64%
694.4

7

P 694.4
=
= 5.787 A
120
V


P1.25

Notice that the references are opposite to the passive configuration.
p (t ) = −v (t )i (t ) = −30e −t W
∞

Energy = ∫ p (t )dt = 30e −t | 0∞ = − 30 joules
0

Because the energy is negative, the element delivers the energy.
P1.26

( a) p (t ) = v ab iab = 50 sin(200πt ) W

(b) w =

0.0025

∫ p (t )dt

=

0.0025

∫ 50 sin(200πt )dt

0

0

= 79.58 mJ
(c) w =

0.01

∫ p (t )dt
0

=0 J

=

0.01

∫ 50 sin(200πt )dt
0

0.0025

= − (50 / 200π ) cos(200πt ) 0

0.01

= −(50 / 200π ) cos(200πt ) 0

*P1.27

(a) P = 50 W taken from element A.
(b) P = 50 W taken from element A.
(c) P = 50 W delivered to element A.

P1.28

(a) P = 50 W delivered to element A.
(b) P = 50 W delivered to element A.
(c) P = 50 W taken from element A.

P1.29

The current supplied to the electronics is i = p /v = 25 / 12.6 = 1.984 A.
The ampere-hour rating of the battery is the operating time to discharge
the battery multiplied by the current. Thus, the operating time is
T = 80 / 1.984 = 40.3 h. The energy delivered by the battery is
W = pT = 25(40.3) = 1008 Wh = 1.008 kWh. Neglecting the cost of
recharging, the cost of energy for 250 discharge cycles is
Cost = 85 /(250 × 1.008) = 0.337 $/kWh.

8

P1.30

The power that can be delivered by the cell is p = vi = 0.45 W. In 10
hours, the energy delivered is W = pT = 4.5 Whr = 0.0045 kWhr. Thus,
the unit cost of the energy is Cost = (1.95) /(0.0045) = 433.33 $/kWhr
which is 3611 times the typical cost of energy from electric utilities.

P1.31

The sum of the currents entering a node equals the sum of the currents
leaving. It is true because charge cannot collect at a node in an electrical
circuit.

P1.32

A node is a point that joins two or more circuit elements. All points
joined by ideal conductors are electrically equivalent. Thus, there are
five nodes in the circuit at hand:

P1.33

The currents in series-connected elements are equal.

P1.34*

Elements E and F are in series.

P1.35

For a proper fluid analogy to electric circuits, the fluid must be

incompressible. Otherwise the fluid flow rate out of an element could be
more or less than the inward flow. Similarly the pipes must be inelastic
so the flow rate is the same at all points along each pipe.

P1.36*

At the node joining elements A and B, we have ia + ib = 0. Thus, ia = −2 A.

For the node at the top end of element C, we have ib + ic = 3 . Thus,
9


ic = 1 A .

Finally, at the top right-hand corner node, we have

3 + ie = id . Thus, id = 4 A . Elements A and B are in series.
P1.37* We are given ia = 2 A, ib = 3 A, id = −5 A, and ih = 4 A. Applying KCL, we find
ic = ib − ia = 1 A
ie = ic + ih = 5 A
if = ia + id = −3 A
i g = if − ih = −7 A
P1.38

(a) Elements C and D are in series.
(b) Because elements C and D are in series, the currents are equal in
magnitude. However, because the reference directions are opposite, the
algebraic signs of the current values are opposite. Thus, we have ic = −id .
(c) At the node joining elements A, B, and C, we can write the KCL
equation i b = i a + ic = 4 − 1 = 3 A . Also, we found earlier that

i d = −ic = 1 A.

P1.39

We are given ia = 1 A, ic = −2 A, i g = 7 A, and ih = 2 A. Applying KCL, we find

ib = ic + ia = −1 A
id = if − ia = 8 A

ie = ic + ih = 0 A
if = i g + ih = 9 A

P1.40

If one travels around a closed path adding the voltages for which one
enters the positive reference and subtracting the voltages for which one
enters the negative reference, the total is zero. KCL must be true for
the law of conservation of energy to hold.

P1.41*

Summing voltages for the lower left-hand loop, we have − 5 + v a + 10 = 0,
which yields v a = −5 V. Then for the top-most loop, we have

v c − 15 − v a = 0, which yields v c = 10 V. Finally, writing KCL around the
outside loop, we have − 5 + v c + v b = 0, which yields v b = −5 V.
P1.42*

Applying KCL and KVL, we have
ic = ia − id = 1 A

v b = v d − v a = −6 V

ib = −ia = −2 A
vc = vd = 4 V

The power for each element is
PA = −v a ia = −20 W
PC = v c ic = 4 W
Thus, PA + PB + PC + PD = 0

PB = v b ib = 12 W
PD = v d id = 4 W

10


P1.43

(a) Elements A and B are in parallel.
(b) Because elements A and B are in parallel, the voltages are equal in
magnitude. However because the reference polarities are opposite, the
algebraic signs of the voltage values are opposite. Thus, we have
v a = −v b .
(c) Writing a KVL equation while going clockwise around the loop
composed of elements A, C and D, we obtain v a − v d − v c = 0. Solving for

v d and substituting values, we find v d = 5 V. Also, we have
v b = −v a = −12 V.
P1.44

There are two nodes, one at the center of the diagram and the outer
periphery of the circuit comprises the other. Elements A, B, C, and D are
in parallel. No elements are in series.

P1.45

We are given v a = 15 V, v b = −7 V, vf = 10 V, and v h = 4 V. Applying KVL, we
find

P1.46

vd = v a + v b = 8 V
ve = −v a − vc + vd = 22 V

vc = −v a − vf − v h = −29 V
v g = ve − v h = 18 V

The points and the voltages specified in the problem statement are:

Applying KVL to the loop abca, substituting values and solving, we obtain:
v ab − v cb − v ac = 0
15 + 7 − v ac = 0
v ac = 22 V
Similiarly, applying KVL to the loop abcda, substituting values and solving,
we obtain:
v ab − v cb + v cd + v da = 0
15 + 7 + vcd + 10 = 0
vcd = −32 V

11

P1.47

(a) In Figure P1.36, elements C, D, and E are in parallel.
(b) In Figure P1.42, elements C and D are in parallel.
(c) In Figure P1.45, no element is in parallel with another element.

P1.48

Six batteries are needed and they need to be connected in series. A
typical configuration looking down on the tops of the batteries is shown:

P1.49

(a) The voltage between any two points of an ideal conductor is zero
regardless of the current flowing.
(b) An ideal voltage source maintains a specified voltage across its
terminals.
(c) An ideal current source maintains a specified current through itself.
(d) The voltage across a short circuit is zero regardless of the current
flowing. When an ideal conductor is connected between two points, we say
that the points are shorted together.
(e) The current through an open circuit is zero regardless of the voltage.

P1.50

Provided that the current reference points into the positive voltage
reference, the voltage across a resistance equals the current through
the resistance times the resistance. On the other hand, if the current

reference points into the negative voltage reference, the voltage equals
the negative of the product of the current and the resistance.

P1.51

Four types of controlled sources and the units for their gain constants
are:
1. Voltage-controlled voltage sources. V/V or unitless.
2. Voltage-controlled current sources. A/V or siemens.
3. Current-controlled voltage sources. V/A or ohms.
4. Current-controlled current sources. A/A or unitless.
12


P1.52*

P1.53

P1.54

The resistance of the copper wire is given by RCu = ρCu L A , and the

resistance of the tungsten wire is RW = ρW L A . Taking the ratios of the

respective sides of these equations yields RW RCu = ρW ρCu . Solving for

RW and substituting values, we have
RW = RCu ρW ρ Cu

= (1.5) × (5.44 × 10 -8 ) (1.72 × 10 −8 )

= 4.74 Ω
P1.55

Equation 1.10 gives the resistance as

R =

ρL

A

(a) Thus, if the length of the wire is doubled, the resistance doubles to
20 Ω .
(b) If the diameter of the wire is doubled, the cross sectional area A is
increased by a factor of four. Thus, the resistance is decreased by
a factor of four to 2.5 Ω .

13


P1.56

P1.57

P1.58*

P1.59

R

(
V1 )2
=

P2

(V )
= 2

=

P1

R

1002
= 100 Ω
100

2

=

902
= 81 W for a 19% reduction in power
100

The power delivered to the resistor is
p (t ) = i 2 (t ) R = 10 exp( − 6t )
and the energy delivered is

∞

∞

 10



∞

10

w = ∫ p (t )dt = ∫ 10 exp( −6t )dt = 
exp( −6t ) =
= 1.667 J
−
6
6


0
0
0
P1.60

The power delivered to the resistor is
p (t ) = v 2 (t ) / R = 5 sin 2 (2πt ) = 2.5 − 2.5 cos(4πt )
and the energy delivered is
2

w = ∫ p (t )dt =
0

2

∫ [2.5 − 2.5 cos(4πt )]dt
0

2

2. 5


= 2.5t −
sin(4πt ) = 5 J
4π

0

14


P1.61

(a) Ohm's law gives iab = vab/2.
(b) The current source has iab = 2 independent of vab, which plots as a
horizontal line in the iab versus vab plane.
(c) The voltage across the voltage source is 5 V independent of the
current. Thus, we have vab = 5 which plots as a vertical line in the iab
versus vab plane.

(d) Applying KCL and Ohm's law, we obtain iab = v ab / 2 + 1 .

(e) Applying Ohm's law and KVL, we obtain v ab = iab + 2 which is equivalent
to iab = v ab − 2.

The plots for all five parts are shown.

P1.62*

(a) Not contradictory.
(b) A 2-A current source in series with a 3-A current source is
contradictory because the currents in series elements must be equal.
(c) Not contradictory.
(d) A 2-A current source in series with an open circuit is contradictory
because the current through a short circuit is zero by definition and
currents in series elements must be equal.
(e) A 5-V voltage source in parallel with a short circuit is contradictory
because the voltages across parallel elements must be equal and the
voltage across a short circuit is zero by definition.

15


P1.63*

As shown above, the 2 A current circulates clockwise through all three
elements in the circuit. Applying KVL, we have

v c = v R + 10 = 5iR + 10 = 20 V
Pcurrent − source = −v c iR = −40 W. Thus, the current source delivers power.

PR = (iR ) 2 R = 22 × 5 = 20 W. The resistor absorbs power.
Pvoltage − source = 10 × iR = 20 W. The voltage source absorbs power.
P1.64*

Applying Ohm's law, we have v 2 = (5 Ω ) × (1 A) = 5 V . However,v 2

is the voltage across all three resistors that are in parallel. Thus,

i3 =

v2

= 1 A , and i2 =

v2

= 0.5 A . Applying KCL, we have
10
i1 = i2 + i3 + 1 = 2.5 A . By Ohm's law: v 1 = 5i1 = 12.5 V . Finally using KVL,
5

we havev x = v 1 + v 2 = 17.5 V .

P1.65

The power for each element is 30 W. The voltage source delivers power
and the current source absorbs it.

16

P1.66

This is a parallel circuit, and the voltage across each element is 15 V
positive at the top end. Thus, the current flowing through the resistor is
15 V
iR =
= 3A
5Ω
Applying KCL, we find that the current through the voltage source is 6 A
flowing upward. Computing power for each element, we find

Pcurrent − source = 45 W
Thus, the current source absorbs power.

PR = (iR )2 R = 45 W
Pvoltage − source = −90 W
The voltage source delivers power.
P1.67

Ohm’s law for the right-hand 5-Ω resistor yields: v1 = 5 V. Then, we have

i1 = v1 / 5 = 1 A. Next, KCL yields i2 = i1 + 1 = 2 A. Then for the 10-Ω

resistor, we have v 2 = 10i2 = 20 V. Using KVL, we have v 3 = v1 + v 2 = 25 V.
Next, applying Ohms law, we obtain i3 = v3 / 10 = 2.5 A. Finally applying
KCL, we have I x = i2 + i3 = 4.5 A.

P1.68

(a) The 3-Ω resistance, the 2-Ω resistance, and the voltage source Vx are
in series.
(b) The 6-Ω resistance and the 12-Ω resistance are in parallel.
(c) Refer to the sketch of the circuit. Applying Ohm's law to the 6-Ω
resistance, we determine that v1 = 12 V. Then, applying Ohm's law to the
12-Ω resistance, we have i1 = 1 A. Next, KCL yields i2 = 3 A. Continuing,

17