Solution Manual For Electrical Engineering: Concepts and Applications, 1st Edition

Name: Solution Manual For Electrical Engineering: Concepts and Applications, 1st Edition
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Solution Manual For Electrical Engineering: Concepts and Applications, 1st Edition provides you with expert textbook solutions that ensure you understand every concept thoroughly.

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Chapter 2 Solutions
Section 2.1 Introduction
2.1 Current source
2.2 Voltage source
2.3 Resistor
2.4 Capacitor
2.5 Inductor
Section 2.2 Charge and Current
2.6 b)
The current direction is designated as the direction of the movement of positive
charges.
2.7 The relationship of charge and current is
( ) ( ) ( )0
0
tqdttitq
t
t
+= ∫
so
( ) ( ) ( )0
0
10sin2 tqdtttq
t
t
+= ∫
π
( ) ( ) ( )010cos
10
2
0
tqttq
t
t
+⎥⎦
⎤
⎢⎣
⎡ −
=
π
π
2.8 The coulomb of one electron is denoted by e and
( ) ( ) ( )0
0
tqdttitq
t
t
+= ∫
So
( ) ( )0
0
12
1
/)( tqdtt
e
etqtn
t
t
+== ∫
If t 0 = 0 and ( ) 00 =tq ,
( ) 26 t
e
tn =
2.9

( ) ∫= idttq
( ) tdtq
t
∫=
0
5
( ) ttq 5=
2.10
( ) [ ] ( ) ( ) 2505555
5
0 =−== ttq Coulombs
2.11 Using the definition of current-charge relationship, the equation can be rewritten
as:
e
t
n
dt
dq
i Δ
Δ
==
Thus, the current flow within t1 and t2 time interval is,
Ai 3)106.1(
2
10)275.5( 19
19
−=×−
×−
= −
The negative sign shows the current flow in the opposite direction with respect to the
electric charge.
2.12 Assuming the area of the metal surface is S, The mass of the nickel with depth d
= 0.15mm is
m = ρ × d × S
Meanwhile, using the electro-chemical equivalent, the mass of the nickel can be
expressed as
m = k × I × t
where I = σ × S.
Equating the two expressions of the mass, the coating time is found:
t = ρ × d / σ = 1.24 × 105 s ≈ 34.4 hour
Section 2.3 Voltage
2.13 By the definition of voltage, when a positive charge moves from high voltage to
low voltage, its potential energy decreases.
So a is “+”, b is “-”. In other words, uab =1V.
2.14 The current i(t) is defined as:
( ) ⎭
⎬
⎫
⎩
⎨
⎧ ≤<
= elsewhere
t
ti 0
103
Therefore, the charge

33
1
0
∫ == dtg C
The energy in Joules is given by:
JCVJ 1535 =×=×=
2.15
1 electron = 19
106.1 −
×− Coulombs.
Therefore, there are 18
1025.6 × electrons in a coulomb.
Coulombs of 16
105× electrons = 3
18
16
108
1025.6
105 −
×=
×
×
Therefore, the voltage is
1875
108
15 3 =
×
== −
C
J
V V
2.16
( ) dtttq ⎟
⎠
⎞
⎜
⎝
⎛
= ∫
π
2
3
sin2
1
0
C4244.0
2
3
cos
3
4
1
0
=⎥
⎦
⎤
⎢
⎣
⎡ ⎟
⎠
⎞
⎜
⎝
⎛−
= t
π
π
J122.24244.05 =×=×= CVJ
2.17
C
V
J
q 10
2
20 ==
A
s
C
dt
dq
i 5.2
4
10 ===
Section 2.4 Respective Direction of Voltage and Current
2.18 True
2.19 False
2.20 True
Section 2.5 Kirchoff’s Current Law
2.21 According to KCL

321 iii +=
Also, at the other end of the node, we have,
324 iii +=
Then, combine these two equations we obtain,
41 ii =
2.25 The algebraic sum of currents flowing into the node in Figure 2.11 is
i1 + i2 - i3 - i4
where i1 and i2 flow in, and i3 and i4 flow out
As current is defined as the rate of variation of the charge,
i1 + i2 - i3 - i4 = dq/dt
By law of the conservation of charge, charges cannot be stored in the node and charges
can neither be destroyed nor created. Therefore the variation rate of the charge is zero and
further the KCL law holds,
i1 + i2 - i3 - i4 = 0
2.26 The resistance across the voltage source is
Ω=
+
×
+=+= 50
12040
12040
20120//4020R
So the current flowing through the voltage source is
A
V
R
V
I 1.0
50
5 =
Ω
==
Therefore,
mAi 251.0
160
40 =×=
2.27 By KCL,
xI
VV 2
35
3
0 −+
−
=
Note that,
xx IV
V
I 3
3 =→=
Therefore,
xx
x II
I 2
5
33
0 −+
−
=
xx II 5330 −−=

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A5.123 −=→−= xx II
2.28 The voltage across node C and D (node C is positive) is U2 - U 1, the voltage
across node C and B is UAB - U 1, and the voltage across node D and B is UAB - U 2
The KCL equation for node C is
U 1 /R = (U2 - U 1 ) /R + (UAB - U 1 ) /2R
which is
5U1 - 2U 2 = UAB
The KCL equation for node D is
U2 /2R + (U 2 - U 1 ) /R = (UAB - U 2 ) /R
which is
5U2 - 2U 1 = 2UAB
Solving U2 and U 1 , U 1 = 9UAB /21, U2 = 12UAB /21
Thus, by the KCL equation at node A
IAB = U 1 /R + U2 /2R
R AB = UAB /IAB = 7R /5
R
2R
R
2R
R
A
+
-
B
C D
IAB
U AB
+
-
+
-
U 1 U 2
Figure S2.28: Circuit for Problem 2.28.
Section 2.6 Kirchoff’s Voltage Law
2.29 By KVL,
DCBA vvvv +−−−=0
6620 ++−−= Av
Then
Vv A 10=
2.30 Using KVL,
146120 +−+−= Rv
Then
4=Rv V

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2−=Dv V
Using KVL
2130 −+= Av
Then
11−=Av V
2.35 Using KVL,
VVV HH 8210 =→+=
VVVV IIH 552 =→+=+
VVV CC 264 =→=+
VVVV DCD 145 =→+=+
2.36
V = 6V +
-
R1
R2
4Ω
6Ω
3Ω
i
i1
i2
i3
i4
Figure S2.36: Circuit for Problem 2.36.
a) R1 =6Ω, R2 =3Ω
By KCL,
i1 + i = i2
i4 + i = i3
Applying KVL on the loop made by 6V source, R1 and R2 ,
6i1 + 3i2 = 6
Applying KVL on the loop made by 6V source, 3Ω resistor and 6Ω resistor,
3i3 + 6i4 = 6
Applying KVL on the loop made by 3Ω resistor, 4Ω resistor and R1 ,
3i3 + 4i = 6i1
Now we have 5 equations for 5 variables and i can be solved using Cramer’s rule
or variable elimination. Then we get
i = 1/4A, i1 = 7/12A, i2 = 5/6A, i3 = 5/6A, i4 = 7/12A
b) i = 0 => i1 = i2 , i3 = i4.

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Moreover, the voltage across the 4Ω resistor is zero. This also leads to the voltage
across R1 equals to the voltage across the 3Ω resistor and the voltage across R2
equals to the voltage across the 6Ω resistor,
R1i1 =3i3
R2i2 =6i4
Therefore,
R1 / R2 = 3/6 = 1/2
2.37 The current from node A to node C is IAB - I1, the current from node D to node B
is I1 - I 2, and the current from node C to node B is IAB - I 1 + I2.
Then the KVL equation for loop ADC is
I 1 2R + I2 R = (IAB - I 1 )R
which is
3I1 + I2 = IAB
The KVL equation for loop DBC is
(I 1 - I 2 )R = I2 R + (IAB - I 1 + I2 )2R
which is
3I 1 - 4I 2 = 2IAB
Solving I1 and I2 , I 1 = 2IAB /5, I2 = -I AB /5
Thus,
U AB = I1 2R + (I1 - I 2 )R=7IAB R /5
RAB = UAB /IAB = 7R /5
R
2R
R
2R
R
A
+
-
B
C D
IAB
U AB
I1
I2
Figure S2.37: Circuit for Problem 2.37.
Section 2.7 Ohm’s Law
2.38 By KCL, the current through the 4Ω resistor is 3A. Using Ohm’s law,
1234 =×=×= RIV V
2.39 By KVL , the voltage across the resistor is 5V.
Using Ohm’s law:

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Ω=== 5.2
2
5
I
V
R
2.40 Using Ohm’s law:
A
R
V
I 3
5
15 ===
2.41 The resistivity of copper is,
m−Ω×= −8
1072.1
ρ
The resistance of the copper wire is,
Ω=
×
××
== −
−
70
102286.1
5.01072.1 10
8
A
L
R
ρ
Using the definition of Ohm’s Law, we can obtain the maximum allowable voltage,
which is:
VRIV 70701maxmax =⋅==
2.42 Using Ohm’s law, the resistance of the tissue is
Ω=
×
= − kR 5.804
1043.12
10 6
Because,
A
ρL
=R
Therefore,
23
10468.3
751
5.048 m
LmΩ
=kΩ −
×
×⋅
94.15=L m
2.43 The resistance across the voltage source is
32
32
1321 // RR
RR
RRRRR +
⋅
+=+=
Using Ohm’s law R
V
I = , and
32321
32
32
2
32
2
)(
)(
RRRRR
RRV
RR
R
I
RR
R
i s
++
+⋅
⋅
+
=⋅
+
=
32321
2
)( RRRRR
RVs
++
⋅
=

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2.44 By KCL:
A520 =I
Therefore, by Ohm’s Law:
V100520 =⋅=AV
V30103 =⋅=BV
V30152 =⋅=CV
By KVL:
V130=SV
2.45 Using Ohm’s law:
2.1
10200
240 3 =
×
=I mA
2.46 For short circuit,
v(t) = 0
Using ohm’s law
Rshort = v(t)/i(t) = 0
For open circuit
i(t) = 0
Using ohm’s law
R open = v(t)/i(t) = infinity
2.47
V out = Vs(Rs /(Rs + Rc)) - Vs(Rb /(Ra+Rb))
By product rule of derivation
dVout /dRs = VsRc/(Rs+Rc)2
When Rs = 0, the above expression is minimum. However, you cannot control over
sensor resistance, but you can choose Rc to achieve the maximum slope.
We differentiate dVout /dRs with respect to Rc,
d(dVout /dRs) /dRc =Vs(Rs-Rc) /(Rs+Rc)3
The maximum slope is achieved when Rc = Rs.
Rc can be chosen to have the same value as the nominal resistance of the sensor.
2.48 Using data pairs in the table, we get three equations in A, B, C
(1/273) = A + B ln(16330) + C (ln(16330))3
(1/298) = A + B ln(5000) + C (ln(5000))3
(1/323) = A + B ln(1801) + C (ln(1801))3
Solving those equations, we get
A = 0.001284
B = 2.364 × 10-4
C = 9.304 × 10-8

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Section 2.8 Power and Energy
2.49
a) The voltage and the current are in associated direction,
P = u×i = -3×1 = -3W => active element
b) The voltage and the current are not in associated direction,
P = -u×i = 2×2 = 4W => passive element
c) The voltage and the current are not in associated direction,
P = -u×i = -2×3 = -6W => active element
2.50
a) The voltage and the current are in associated direction,
P = u×i = 20×2.5e-2t = 50e-2t W => active element
b) The voltage and the current are not in associated direction,
P = -u×i = -20×2sint = -40sint W => t > 0, active element; t < 0, passive
element
2.51 The current can be found using the power formula:
5.0
10
5 === V
P
I A
And the resistance can then be found using Ohm’s law:
Ω=== 20
5.0
10
I
V
R
2.52 2
IRIVP ×=×=
500520 2 =×= W
2.53 Applying KCL, the current through each resistor is 4mA.
Given 2
IRP ×= ,
( ) mWP 80104105 233
1 =×××= −
( ) mWP 2241041014 233
2 =×××= −
( ) mWP 5121041032 233
3 =×××= −
2.54 According to KVL:
21 35.1 kiki =
By KCL:
213 iimA +=
mAi 12 =

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VkmAV 111 =Ω×=
2.55 From the problem 2.54, we have mA21 =i and mA12 =i .
Recall RIP 2
= . Therefore,
( ) mW6105.1102 323
5.1 =×⋅×= −
kP
( ) mW2102101 323
2 =×⋅×= −
kP
( ) mW1101101 323
1 =×⋅×= −
kP
2.56
+ -V
I
Figure S2.56: Resistor with current direction and voltage polarity.
Consider Figure S2.56, the power absorbed by the resistor is
P=VI
By Ohm’s law,
V=RI
Thus
P=I 2 R=V2 /R
Therefore P is always a nonnegative number when R is positive.
2.57 The voltage and the current are in associated direction, so
P = u×i
and the waveform of the power is shown below
P (W)
1 20
10
-10
t (s)
Figure S2.57: Power curve.
The consumed energy is
∫= 2
0 PW =0

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which is resulted from the negative symmetry of P around t = 1.
2.58
a) Let the current through the -5V voltage source be in associated direction with the
voltage, which is from the positive side to the negative side.
By KCL at the node below the -5V voltage source, the current through it equals
2A+4A=6A.
Therefore the power of the -5V voltage source is
P = -5V × 6A =30W
b) To zero the power above, the current through the -5V voltage has to be zero. So
the 4A current source needs to be changed to -2A by KCL.
Section 2.9 Independent and Dependent Sources
2.59 In Figure S2.59, i axis is the current through the source and u axis is the voltage
across the source. us denotes the voltage provided by the ideal voltage source and is
denotes the current provided by the ideal current source. It is shown that for ideal
voltage source the voltage does not change with the current through it; for ideal
current source the current does not change with the voltage.
i
u
us
i s
Ideal voltage source
Ideal current source
Figure S2.59: Voltage VS current curves.
2.60 Because the two current sources are in series, the currents through them are the
same,
i = 1A
2.61 Because the 3Ω resistor is in series with the current source, the current through
the resistor is 1A and, by Ohm’s law, its voltage is
u = 3Ω × 1A = 3V
By KVL, the voltage across the current source is
u + 2V = 5V
The power supplied by the current source is (the current and the voltage are not in
associated direction)

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P = -5V × 1A = -5W
2.62 Using KVL:
xx VV 3100 ++−=
Therefore,
5.2=xV
2.63 Using KVL and Ohm’s law,
( ) xx II 20105.360 3 +××−−=
xI3
1048.36 ×−=
mAI x 724.1−=
2.64 The voltage across the 2Ω resistor is -1V, where the voltage polarity is associated
with the current direction. So by Ohm’s law
I = -1V/2Ω = -0.5A
Therefore, the current of the dependent source is
2I = -1A
2.65 According to KCL, the current through R3 is xi
β
According to Ohm’s law,
30 Riv x
β=
ix is given by:
21 RR
v
i s
x +
=
Therefore,
21
30 RR
v
Rv s
+
=
β
2.66 According to Ohm’s law, the current flow (in associated direction with the
voltage) through the 5Ω resistor is
i1 = 4.9V/5Ω = 0.98A.
Because the CCCS is in series with the 5Ω resistor,
i1 = 0.98i => i = 1A
which is the current through the 6Ω.
By KCL, the current through the 0.1Ω resistor is
i-0.98i = 0.02A.
So the voltage across the 0.1Ω resistor

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Solution Manual For Electrical Engineering: Concepts and Applications, 1st Edition

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