G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Chapter 18: Principles of Electromechanics – Instructor Notes
The last part of the book presents an introduction to electro-magneto-mechanical systems. Some of the
foundations needed for this material (AC power) were discussed in Chapter 7; the polyphase AC power
material in Chapter 7 may be introduced prior to covering Chapter 18, or together with the AC machine
material of Chapter 18.
The emphasis in this chapter (and the next two) is on preparing the student for the use of electro-magnetomechanical systems as practical actuators for industrial applications. Thus, more emphasis is placed on describing
the performance characteristics of linear motion actuators and of rotating machines than on a description of their
construction details. The material in Chapters 18-20 has been used by several instructors over the last many years in
a second (quarter-length) course in system dynamics (System Dynamics and Electromechanics) designed for
mechanical engineering juniors.
Section 18.1 reviews basic laws of electricity and magnetism, which should already be familiar to the
student from an earlier Physics course. The box Focus on Measurements: Linear Variable Differential Transformer
(pp. 929-930) presents an example related to sensors with a discussion of the LVDT as a position transducer.
Section 18.2 discusses approximate linear magnetic circuits and the idea of reluctance, and introduces magnetic
structures with air gaps and simple electro-magnets. The box Focus on Problem Solving: Magnetic Structures and
Magnetic Equivalent Circuits (p. 938) summarizes the analysis methods used in this section. A magnetic reluctance
position sensor is presented in Focus on Measurements: Magnetic Reluctance Position Sensor (pp. 945-946) and
Focus on Measurements: Voltage Calculation in Magnetic Reluctance Position Sensor (pp. 946-947). The nonideal properties of magnetic materials are presented in Section 18.3, where hysteresis, saturation, and eddy currents
are discussed qualitatively. Section 18.4 introduces simple models for transformers; more advanced topics are
presented in the homework problems.
Section 18.5 is devoted to the analysis of forces and motion in electro-magneto-mechanical structures
characterized by linear motion. The boxes Focus on Problem Solving: Analysis of Moving-Iron Electromechanical
Transducers (pp. 957-958) and Focus on Problem Solving: Analysis of Moving-Coil Electromechanical Transducers
(p. 970) summarize the analysis methods used in this section. The author has found that it is pedagogically
advantageous to introduce the Bli and Blu laws for linear motion devices before covering these concepts for rotating
machines: the student can often visualize these ideas more clearly in the context of a loudspeaker or of a vibration
shaker. Example 18.9 (pp. 958-959) analyzes the forces in a simple electromagnet, and Examples 18.10 (pp. 959961) and 18.12 (p. 965) extend this concept to a solenoid and a relay. Example 18.11 (p. 961-962) ties the material
presented in this chapter to the transient analysis topics of Chapter 5. Example 18.13 (pp. 970-971) performs a
dynamic analysis of a loudspeaker, showing how the frequency response of a loudspeaker can be computed from an
electromechanical analysis of its dynamics. Finally, The box Focus on Measurements: Seismic Transducer (pp.
973-974) presents the dynamic analysis of an electromechanical seismic transducer.
The homework problems are divided into four sections. The first reviews basic concepts in electricity and
magnetism; the second presents basic and more advanced problems related to the concept of magnetic reluctance;
the third offers some problems related to transformers. Section 5 contains a variety of applied problems related to
electromechanical transducers, and is divided into two separate sections on moving-iron and moving-coil
transducers; some of the problems in this last section emphasize dynamic analysis (18.32, 18.41-18.44, and 18.4853) and are aimed at a somewhat more advanced audience.
Learning Objectives
1.
2.
3.
4.
5.
Review the basic principles of electricity and magnetism. Section 18.1.
Use the concepts of reluctance and magnetic circuit equivalents to compute magnetic flux and
currents in simple magnetic structures. Section 18.2.
Understand the properties of magnetic materials and their effects on magnetic circuit models.
Section 18.3.
Use magnetic circuit models to analyze transformers. Section 18.4.
Model and analyze force generation in electro-magneto-mechanical systems. Analyze moving iron
transducers (electromagnets, solenoids, relays), and moving-coil transducers (electro-dynamic
shakers, loudspeakers, seismic transducers). Section 18.5.
18.1
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Section 18.1: Electricity and Magnetism
Problem 18. 1
Solution:
Known quantities:
The system parameters are shown in Figure P18.1.
Find:
a)
The flux density in the core.
b)
Sketch the magnetic flux lines and indicate their
direction.
c)
The north and south poles of the magnet.
Assumptions:
The flux is uniform over the cross-section area A.
Analysis:
a)
B
=
Φ 4 × 10−4
=
= 0.04 T
A
0.01
b)
The magnetic flux lines are depicted in the figure on the side:
c)
See above.
Problem 18.2
Solution:
Known quantities:
As shown in Figure P18.2.
Find:
With the current directed as shown in the figure, is there a
resultant force? If so, in what direction? Why?
Assumptions:
None.
Analysis:
Yes, the resultant force on the single coil is in the downward direction. If the coils are thought of as electromagnets,
there is a north pole from the lower coil attracting a south pole from the upper coil.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.3
Solution:
Known quantities:
A LVDT is connected to a resistive load RL . Rs and Ls represent secondary coil parameters.
Find:
The LVDT equations.
Assumptions:
Assume both secondary windings have resistance Rs and inductance Ls . The current at the primary side is indicated
as i .
Analysis:
The secondary coils are descripted by the following set of equations:
di
di
− ( RsiL + Ls L )
dt
dt
di
di
v2 = M 2 + ( RsiL + Ls L )
dt
dt
∴vout =v1 − v2
v1 = M 1
=
( M1 − M 2 )
di
di
− 2( RsiL + Ls L )
dt
dt
The load current iL can be calculated as following:
iL =
vout
RL
Therefore, the transfer function is:
∴( RL + 2( Rs + sLs ))v=
sRL ( M 1 − M 2 )i
out
vout
sRL ( M 1 − M 2 )
=
i
RL + 2( Rs + sLs )
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.4
Solution:
Known quantities:
Equations of "Focus on Measurements: Linear Variable Differential Transformer", and the results of Problem 18.3.
Find:
The frequency response of the LVDT and the range of frequencies for which the device will have maximum
sensitivity for a given excitation.
Assumptions:
Assume that the primary winding has resistance Rp and inductance Lp . The voltage at the primary side is indicated
as v ex .
Analysis:
The voltage equation at the primary side is:
di
+ R pi
dt
∴vex ( s ) =( R p + Lp s )i( s )
=
vex Lp
( RL + 2( Rs + sLs ))vout= sRL ( M 1 − M 2 )i
where M
= M 1 − M 2 Therefore,
vout ( s )
sRL ( M 1 − M 2 )
=
vex ( s ) ( R p + Lp s )( RL + 2( Rs + sLs ))
To determine the maximum sensitivity of the excitation output voltage, it is possible to compute the derivative of
H( s ) =
∂H( s )
vout ( s )
with respect to s . Thus setting
= 0 , the equation is solved for s = jω and this procedure
∂s
vex ( s )
will yield the excitation frequency for which the sensitivity of the output is maximum. However, it may be more
useful to numerically compute the frequency response H( jω ) , to visualize the range of frequencies over which the
sensitivity is acceptable.
18.4
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.5
Solution:
Known quantities:
i=
λ
0.5 + λ
a.
=
λ 1V ⋅ s
b.
R = 1Ω
i( t ) 0.625 + 0.01 sin 400t A
=
Find:
a)
b)
The energy, coenergy and incremental inductance.
The voltage across the terminals on the inductor.
Assumptions:
None.
Analysis:
λ 1V ⋅ s , the current is:
a) For =
λ
=
i = 0.667 A
0.5 + λ
The energy is:
1.0
0.5 
 λ 

=
−
=
−
+
W m=
dλ
dλ
λ
.
ln
.
λ
1
0
5
0
5
(
)




∫0  0.5 + λ  ∫0  0.5 + λ 
0
1.0
1.0
W m= 1.0 − 0.5 ln 0.5 + 1.0 − 0 + 0.5 ln 0.5 = 0.4507 J
The coenergy is given by:
The incremental inductance is yield by:
−1
 di 
LΔ =   |λ =0.625
 dλ 
di 0.5 + λ − λ
dλ
=
=

→
2
dλ ( 0.5 + λ )
di
( 0.5 + λ )
2
0.5
( 0.5 + λ )=
dλ
=
LΔ =
|λ 0.=
|λ 0.625 4.5H
=
625
di
0.5
2
b) The voltage across the inductor is computed considering a standard RL series model:
V=
Ri( t ) + LΔ
L( t )
di
dt
di
= 4 cos ( 400t )
dt
VL ( t ) =
( 1Ω )( 0.625 + 0.01 sin( 400t )) + ( 4.5H )( 4 cos( 400t ))
VL ( t=
) 0.625 + 18 sin( 400t + 90° )
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
It is important to observe that this is the inductor terminal voltage only for values of flux linkage in the
.
neighborhood of
Problem 18.6
Solution:
Known quantities:
λ2
0.5 + λ2
a) =
λ 1V ⋅ s
R = 1Ω
b)
=
i( t ) 0.625 + 0.01 sin 400t A
i=
Find:
a) The energy, coenergy and incremental inductance.
b) The voltage across the terminals on the inductor.
Assumptions:
None.
Analysis:
2
λ
λ 1V ⋅ s : The current is:
a) For =
=
i =
0.667 A
2
=
The energy is given by: W
m
du
∫a +u
Integral of form: =
2
2
0.5 + λ
1.0
 λ2 
dλ
∫0  0.5 + λ2 =
u
1
tan −1 + c
a
a
1.0
0.5


∫  1 − 0.5 + λ d λ
0
2
1.0
1.0

0.5 
0.5
 λ 

W m=
tan −1 
λ−

∫0  1 − 0.5 + λ2 d λ =
0.5
 0.5   0

0.5
0.5
 1 
 0 
tan −1 
tan −1 
−0+
 = 0.3245 J
0.5
0.5
 0.5 
 0.5 
The coenergy is yield by: Wm' =i λ − Wm =0.3425 J
W m = 1.0 −
The incremental inductance is:
LΔ =
di 2 λ ( 0.5 + λ ) − 2 λ ( λ )
=
2
dλ
( 0.5 + λ2 )
2
dλ
|λ=0.625
di
2
d λ ( 0.5 + λ )
=
λ
di
2 2
0.5 + λ2 )
(
dλ
=
LΔ =
|λ 0=
=
|λ 0.625 2.25 H
=
.625
di
λ
2
b) The voltage across the inductor is computed considering a standard RL series model:
di
di
= 4 cos ( 400t )
dt
dt
VL ( t ) =
( 1Ω )( 0.625 + 0.01 sin( 400t )) + ( 2.25 H )( 4 cos( 400t ))
V=
Ri( t ) + LΔ
L( t )
18.6
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
VL ( t ) = 0.625 + 9 sin( 400t + 89.9° )
It is important to observe that this is the inductor terminal voltage only for values of flux linkage in the
neighborhood of 1V ⋅ s .
Problem 18.7
Solution:
Known quantities:
Characteristic plot shown in Figure P18.7.
Find:
a) The energy and the incremental inductance for i = 1.0 A.
b) The voltage across the terminals of the inductor when
=
R 2Ω
=
,i( t ) 0.5 sin 2πt.
Assumptions:
None.
Analysis:
a)
The flux leakage diagram is shown in the figure below.
λ (V⋅ S)
is the area at the left of the curve as shown.
1
1
Wm = × 0.5 × 2 + 0.5 × 1 + × 0.5 × 1 =1.25 J
2
2
The incremental inductance is:
4
Δλ
2
|i ==1 = 2 H
Δi
1
3
LΔ=
b)
On the basis of
inductance can be obtained:
and
2
wm
, the voltage across the
0
dλ
=
v Ri +
dt
For i < 0.5, λ =
4i :
=
vL ( t ) sin( 2πt ) + 4 × 0.5 × 2π cos( 2πt )
= sin( 2πt ) + 4 π cos( 2πt )
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.8
Solution:
Known quantities:
Structure of Figure 18.12
A = 01
. m2
Each leg is 0.1 m in length
Find:
The reluctance of the structure.
Assumptions:
Mean magnetic path runs through the exact center of the structure
.08m
Analysis:
Using the assumption that the mean magnetic path runs through the exact
center of the structure, and since the structure is square, the mean path is
determined using the following figure:
.09m
.1m
lc =
4 × 0.09m =
0.36m
Calculation of Reluctance:
=
ℜ
lc
lc
0.36
=
=
= 1432 A − turns / Wb
.
μ A μr μo A 2000 × 4 π × 10 −7 × 01
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Section 18.2: Magnetic Circuits
Problem 18.9
Solution:
Known quantities:
a)
b)
φ=
4.2 × 10−4 Wb,mmf =
400 A ⋅ turns .
Find:
a)
b)
The reluctance of the magnetic circuit.
The magnetizing force in SI units.
Assumptions:
None.
Analysis:
a)
b)
18.9
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.10
Solution:
Known quantities:
As shown in Figure P18.10.
Find:
a) The reluctance values and show the magnetic circuit when
.
b) The inductance of the device.
is cut in the arm of length
c) The new value of inductance when an air gap of
.
d) The limiting value of inductance when the gap is increased in size (length).
Assumptions:
Neglect leakage flux and fringing effects.
Analysis:
a)
Considering circuit shown in the right, the total reluctance is yield:
b)
=
L
N2
1002
=
= 0.8 H
ℜT 12.51 × 103
c)The air gap reluctance is:
is in series with
c)
and thus:
As the air gap get longer,
will get larger and as an extreme case the circuit is made of
and
in series,
therefore:
ℜT= 18.57 × 103 H −1
L=
N2
= 0.54 H
ℜT
18.10
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.11
Solution:
Known quantities:
=
N 1000=
turns,i 0=
.2 A,lg 1 0.02=
cm,lg 2 0.04cm.
Find:
The flux and flux density in each of the legs of the magnetic
circuit.
Assumptions:
Neglect fringing at the air gaps and any leakage fields. Assume the reluctance of the magnetic core to be negligible.
Analysis:
The reluctance of each air gap is obtained as following:
ℜ g1=
0.0002
= 1.59 × 106 H −1
4π × 10−7 × ( 0.01 )2
ℜ g 2 = 2ℜ g1 =3.18 × 106 H −1
Assume the reluctance of the material can be neglected when compared to the reluctance of the air gaps, the
equivalent circuit is shown below and the flux and flux density can be calculated:
φ1
=
Ni
= 1.26 × 10 −4 Wb
ℜg1
φ1
= 1.26Wb m2
A
1
φ2
φ1 0.63 × 10 −4 Wb
=
=
2
1
=
B2 =
B1 0.63Wb m2
2
=
B
1
18.11
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.12
Solution:
Known quantities:
φ=
lsteel ==
3 × 10−4 Wb,liron =
0.3 m,A 5 × 10−4 m 2 ,N =
100 turns
Cast Iron:
Cast Steel:
Find:
The current needed to establish the flux.
Assumptions:
None
Analysis:
Reluctance for each material is calculated as follows:
The total reluctance of the structure is the sum of the reluctances for each material.
Cast Iron:
ℜ=
CI
lCI
=
μr μo A
0.3m
= 91909
.
× 10 4 A − turns / Wb
−7
−4 2
( 5195 ) ( 4 π × 10 )( 5 × 10 m )
Cast Steel:
ℜ=
CS
lCS
=
μr μo A
0.3m
= 4.7746 × 105 A − turns / Wb
(1000 ) ( 4 π × 10 −7 )( 5 × 10 −4 m2 )
Note: Cast Steel is less permeable than cast iron
Total Reluctance:
ℜT = ℜCI + ℜCS = 91909
.
× 10 4 A − turns / Wb + 4.7746 × 105 A − turns / Wb =
=
ℜT 5.6937 × 105 A − turns / Wb
From φ =
=
i
Ni
, we can compute the current.
ℜT
φRT ( 3 × 10 −4 Wb )( 5.6937 × 105 A − turns / Wb )
=
= 1.71 A
N
100turns
18.12
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.13
Solution:
Known quantities:
The magnetic circuit is shown in Figure P18.13.
2
=
I 2=
A,r 0.08 m,N
= 100,A=
0.009 m=
,μr 1000
cross
Find:
The magnetic flux
.
Assumptions:
None.
Analysis:
=
l 2=
πr 0.50265 m,μ
= μr μ0
=
R
1
A
= 4.44444 × 104
μAcross
Wb
mmf = I ⋅ N
mmf
∴
=
= 0.0045Wb
φ
R
Problem 18.14
Solution:
Known quantities:
φ=
2.4 × 10−4 Wb,lab =
lef =
0.05 m,laf =
lbe =
0.02 m,lbc =
ldc ,A =
2 × 10−4 m 2 .
The magnetic circuit is reported below and its material is sheet steel.
Find:
a) The current required to establish the flux.
b) Compare the mmf drop across the air gap to that across the rest of the magnetic circuit and discuss your results
for each material.
using the value of
Assumptions:
Assume the material is cast steel.
Analysis:
a)
18.13
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
2.4 × 10−4
12 T
=
2 × 10−4
A⋅t
H CS = 1400
m
A⋅t
9.55 × 105
H=
AG
m
=
I 1400( 0.1 + 0.02 + 0.017 ) + ( 9.55 × 105 )( 0.003 )
100
⇒I =
30.6 A
B
=
b) The mmf drop for the magnetic circuit and the air gap are here reported:
=
FCS 191.8 A ⋅ t
=
FAG 2865 A ⋅ t
Considering the μ of the materials the following consideration can be done:
μCS = 0.0009
μ AG = 0.00000126
μCS
≈ 700
μ AG
Problem 18.15
Solution:
Known quantities:
Magnet of Figure P18.15, φ =
2 × 10−4 Wb,lab =
0.2 m,lbc l fg ==
0.1 m,lcd lef =
0.099 m.
lbg =
lgh =
lha ==
The magnetic circuit is reported below and its material is sheet steel.
Find:
The value of
required to establish the flux.
Assumptions:
None.
Analysis:
By means the magnetic circuit figure it is possible to
calculate:
A1 =
2 × 10−4 m 2 ,A2 =
5 × 10−4 m 2
μr = 4000
lgap =lha − 2lcd =0.002 m
18.14
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
lgap
A
=
3.1831 × 106
μ0 A2
Wb
=
Rgap
=
636.61977 A
mmf
φ=
gap
1 Rgap
R=
R=
ef
cd
lcd
A
= 3.93908 × 104
μA2
Wb
=
7.87817 A
mmf
φ=
cd
1 Rcd
lbc
A
= 3.97887 × 104
R=
R=
fg
bc
μA2
Wb
=
7.95775 A
mmf
φ=
bc
1 Rbc
To find the mmf in the rightmost leg of the magnetic circuit,
R=
R=
ef
cd
lcd
A
= 3.93908 × 104
μA2
Wb
lab
A
7.95775 × 104
=
μA2
Wb
Rab
=
A
Wb
848.69641 A
φ=
=
mmf
T Rseries
series
Rseries
Rab 2.38732 × 105
= 3=
1.51799 × 103 A
mmf total =mmf series + mmf parallel =
=
∴i
mmf total
= 7.59 A
N
Thus, the total mmf can be yield:
mmf parallel = mmf gap + 2mmf cd + 2mmf bc = 66829161 A
=
R
bg
lbg
A
=
1.98944 × 105
μA1
Wb
=
φ2
mmf parallel
= 0.00336Wb
Rbg
φT = φ1 + φ2 = 0.00356Wb
Problem 18.16
Solution:
Known quantities:
With reference to fig. P18.16 and considering a fixed air gap,
=
N 2000 turns,g
= 10=
mm,B 1=
.2 T ,μr 4000 .
The entire device is made of sheet steel.
Find:
a) The coil current.
b) The energy stored in the air gaps.
c) The energy stored in the steel.
18.15
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Assumptions:
None.
Analysis:
a)
The equivalent circuit is reported in here reported.
Taking into account the relative permeability of sheet steel ( μr = 4000 ).
80 × 10−3
ℜ L =ℜR =
=25,984 A ⋅ t Wb
4000( 4π × 10−7 )( 35 × 17.5 ) × 10−6
=
ℜC
80 × 10−3
= 12 ,992 A ⋅ t Wb
4000( 4π × 10−7 )( 35 × 35 ) × 10−6
61.25 × 10−3
ℜ B1 =
ℜB 2 =
=
19 ,894 A ⋅ t Wb
4000( 4π × 10−7 )( 35 × 17.5 ) × 10−6
=
ℜg1
10 × 10−3
= 6,496,120 A ⋅ t Wb
( 4π × 10 )( 35 × 35 ) × 10−6
=
ℜg 2
10 × 10−3
= 12 ,955,225 A ⋅ t Wb
( 4π × 10−7 )( 35 × 17.5 ) × 10−6
−7
Thus, the total reluctance of the circuit is the series-parallel combination of the calculated ones.
ℜT = ℜC + ℜ g1 +
ℜR + ℜ g 2 + ℜB 2
2
= 13.01 × 106 A ⋅ t Wb
The coil current is obtained as following:
φT = BA = 1.2( 35 × 35 ) × 10−6 = 1.47 × 10−3 Wb
2000 I =ℜT φT ⇒ I =9.56 A
c)
The energy stored in the air gap is yield as:
1.2
( 10 × 10−3=
) 9549.3 A ⋅ t
4π × 10−7
1
7.02 J
wg1 =
( 1.47 × 10−3 )( 9549.3 ) =
2
1 1.47 × 10−3
wg 2 =
(
)( 9549.3 ) 3.51 J
=
2
2
wg =wg1 + 2 wg 2 =14.04 j
H=
lg
c) while the energy stored in the steel:
1
wT =
( 1.47 × 10−3 )( 2000 )( 9.56 ) =
14.05 J
2
wST = wT − wg = 0.01 J
18.16
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.17
Solution:
Known quantities:
With reference to the core depicted in the figure below,
where
=
μr 2000
=
,N 100
Find:
a)
The current needed to produce φ = 0.4Wb m 2 in
the center leg.
b)
The current needed to produce φ = 0.8Wb m 2 in
the center leg.
Assumptions:
None.
Analysis:
a)
With l1= 34 cm,l2= l3= 90 cm and A
= ( 8 × 10−2 )2 cm 2 , it is possible to compute the following
quantities:
=
ℜ1
0.34
= 2.114 × 104 H −1
2000 × 4π × 10−7 × ( 8 × 10−2 )2
ℜ2 =5.595 × 104 H −1 =ℜ3
ℜT =ℜ1 +
ℜ 2 ℜ3
=4.91 × 104 H −1
ℜ 2 + ℜ3
φT =×
0.4 ( 0.08 )2 =
2.56 × 10−3 Wb
BART
Ni
From φ
, we have i =
.
=
BA
=
T
N
ℜT
The current needed to produce the desired flux density is:
i
2.56 × 10−3 × 4.91 × 104 125.7
=
N
100
= 1.257 A
b) Since the current is directly proportional to B, to produce a double flux density, the current will have to be
doubled.
18.17
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Section 18.4: Transformers
Problem 18.18
Solution:
Known quantities:
The figure P18.18 shows the transformer structure having
2
=
=
=
N 1000 turns,l
cm ,A1 4 =
cm 2 ,l2 22=
cm,A2 4 =
cm 2 ,l3 5=
cm ,A3 2 cm
,μr 1500.
16=
1
Find:
a) Construct the equivalent magnetic circuit and find the reluctance associated with each part of the circuit.
= L=
L21 ) for the pair of coils.
b) The self-inductance ( L11 , L22 ) and mutual-inductance ( M
12
Assumptions:
None.
Analysis:
a) The analogous circuit of the transformer is shown below:
The individual reluctances associate to each part are:
b) Thus, the mutual inductance can be yield as following:
L=
m1
N2
= 4.72 H
ℜ1
L=
m2
N2
= 3.43 H
ℜ2
L=
m3
N2
= 7.54 H
ℜ3
LT = Lm1 + Lm 2 + Lm 3 = 15.68 H
M
=
Lm1 Lm 2
= 1.03 =
H L=
L21
12
LT
The self-inductance can be computed as follows:
=
L1
Lm1 Lm 3
= 2.27 H
LT
=
L2
Lm 2 Lm 3
= 1.65 H
LT
L11 = L1 + Lm =3.3 H
L22 = L2 + Lm = 2.68 H
18.18
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.19
Solution:
Known quantities:
A 300Ω -resistive load is connected to a transformer and the resistance referred to the primary is 7500Ω .
The system parameters are here reported:
=
r1 20
=
Ω ,L1 1.0 mH
=
,Lm 25=
mH ,r2 20Ω
=
,L2 1.0 mH .
Find:
a) The turns ratio.
b) The input voltage, current, and power and the efficiency when this transformer is delivering 12W to the 300Ω
load at a frequency f = 10 ,000 2π Hz .
Assumptions:
Core losses are negligible.
Analysis:
The equivalent circuit of a transformer is reported on the side.
a)
The turn ratio can be obtained us following:
7500
= N 2 × 300
N =5
b)
X
=
2πfL
=
10Ω
= X L2
L1
1
X Lm = 250Ω
From
=
I L2 RL 12
=
W ,I L2 0.04 , we have:
I=
0.2∠0 A
L
V=
60∠0 V
L
18.19
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.20
Solution:
Known quantities:
A 220 20V transformer has
turns on its low-voltage side.
Find:
a)
b)
The number of turns on its high side.
The turns ratio α when it is used as a step-down transformer.
The turns ratio α when it is used as a step-up transformer.
c)
Assumptions:
None.
Analysis:
α = 220 20 = 11
a) The primary has N P = 50 × 11 = 550 turns
b) α = 11 is a step-down transformer.
c) α = 1 11 is a step-up transformer.
Problem 18.21
Solution:
Known quantities:
The high-voltage side of the transformer has
turns, and the low-voltage side
turns. The high side is
connected to a rated voltage of 120V . A rated load of 40A is connected to the low side.
Find:
a)
b)
c)
The turns ratio.
The secondary voltage.
The resistance of the load.
Assumptions:
No internal transformer impedance voltage drops
Analysis:
a)
b)
c)
R=
L
8
= 0.2 Ω
40
18.20
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.22
Solution:
Known quantities:
A transformer is used to match an
loudspeaker to a
audio line.
Find:
a)
b)
The turns ratio of the transformer.
The voltages at the primary and secondary terminals when
of audio power is delivered to the speaker.
Assumptions:
The speaker is a resistive load and the transformer is ideal.
Analysis:
a) From α RL = 500 , we have
2
V22
, we have
RL
V1 αV
=
70.7V
c)=
2
b)From 10 =
Problem 18.23
Solution:
Known quantities:
It is a step-down transformer. The high-voltage and low-voltage sides have
turns and
A voltage equal to 240V is applied to the high side. The impedance of the low side is
.
turns respectively.
Find:
a)
b)
c)
d)
The secondary voltage and current.
The primary current.
The primary input impedance from the ratio of primary voltage and current.
The primary input impedance.
Assumptions:
None.
Analysis:
The transformer ratio is obtained as following
a)
c)
.
1
V2
=
V1 30V =
I2 =
10 A
α
RL
240
=
Z in = 192
1.25
V
=
2
1
=
I 2 1.25 A
α
b)
=
I1
d)
2
=
Z in α=
RL 192
18.21
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.24
Solution:
Known quantities:
It is a step-up transformer. All the others parameters are the same as Problem 18.23.
Find:
The transformer ration of the transformer.
Assumptions:
None.
Analysis:
Problem 18.25
Solution:
Known quantities:
A 2 ,300 240 − V ,60 − Hz,4.6 − kVA transformer is considered, having an induced emf of 2.5V turn .
Find:
a)
The number of high-side turns N h and low-side turns N l .
b)
c)
The rated current of the high-voltage side I h .
The transformer ratio when the device is used as a step-up transformer.
Assumptions:
It is an ideal transformer.
Analysis:
2300
240
= 920 turns =
N l = 96 turns
2.5
2.5
3
4.6 × 10
b) I h
=
= 2A
2300
Nl
c)=
α = 0.1044
Nh
Nh
a) =
18.22
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Section 18.5: Electromechanical Transducers (a) Moving-Iron Transducers
Problem 18.26
Solution:
Known quantities:
Considering the electromagnet of Example 18-9 (Figure 18.38) with the following
electromagnet parameters:
−7
=
=
=
,μr 104 ,airgap
N 700
,μ0 4πx10=
0.5m
0
=
=
=
=
l1 0.80
m,l2 0.40m,sec
tion 5 x10−4=
m 2 ,m 10
kg ,g 9.8m / s 2
Find:
The current required to keep the bar in place
Assumptions:
Air gap becomes zero and the iron reluctance cannot be neglected
Analysis:
To compute the current we need to derive an expression for the force in the air gap.
Without neglecting the iron reluctance, we can write the expression for the reluctance as follows:
ℜ( x=
)
L
2x
+
μr μ0 A μ0 A
where L is the total length of the iron magnetic path (excluding the air gap).
Knowing the reluctance we can calculate the magnetic flux in the structure as a function of the coil current:
Φ
=
Ni
μμA
= Ni r 0
ℜ( x )
L + 2 μr x
Then, the magnitude of the force in the air gap is given by the expression
( Niμr μ0 A)=
( Niμr ) μ0 A
Φ2 d ℜ( x ) 1
2
=
f
=
2
2 2
2
2 dx
2 L + 4 Lμr x + 4 μr x μ0 A L + 4 Lμr x + 4 μr2 x 2
2
2
As x approaches zero, we can calculate the force to be:
( Niμr ) μ0 A
f (=
x 0=
)
2
L2
and the current required to maintain the gravity force is:
A
( Niμr ) μ0=
f (=
x 0=
)
mg
2
L2
L2 mg
L
mg
l + l mg
0.8 + 0.4
i=
±
=
±
=
± 1 2
=
±
2
Nμ
μ
A
Nμ
μ
A
700
× 104
( Nμr ) μ0 A
r
0
r
0
10 × 9.8
=
±0.068 A
4π × 10−7 × 5 × 10−4
18.23
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.27
Solution:
Known quantities:
Electromagnet of Example 18-9 (Figure 18.38)
2
=
N 700turns; =
f restore 8,900 N =
; Agap 0.01m
=
;L 1=
m; μr 1000
Find:
a)
b)
The current required to keep the bar in place
Initial current to lift the magnet if the bar is initially 0.1 m away from the electromagnet
Assumptions:
a)
b)
Air gap becomes zero and the iron reluctance cannot be neglected
Neglect the iron reluctance
Analysis:
a) To compute the current we need to derive an expression for the force in the air gap.
Without neglecting the iron reluctance, we can write the expression for the reluctance as follows:
ℜ( x=
)
2x
L
+
μr μ0 A μ0 A
where L is the total length of the iron magnetic path (excluding the air gap).
Knowing the reluctance we can calculate the magnetic flux in the structure as a function of the coil current:
=
φ
Ni
μμA
= Ni r 0
ℜ( x )
L + 2 μr x
Then, the magnitude of the force in the air gap is given by the expression
( Niμr μ0 A)=
( Niμr ) μ0 A
φ2 d ℜ( x ) 1
2
=
f
=
2
2 2
2
2 dx
2 L + 4 Lμr x + 4 μr x μ0 A L + 4 Lμr x + 4 μr2 x 2
2
2
( Niμr ) μ0 A
As x approaches zero, we can calculate the force to be: f (=
x 0=
)
2
L2
and the current required to maintain the given force is:
( Niμr ) μ0 A
f (=
x 0=
)
2
L2
=
L2 f ( x 0=
)
f ( x 0)
L
i=
±
=
±
2
Nμr
μ0 A
( Nμr ) μ0 A
Assuming that the total length of the magnetic path is L=1 m and that  r = 1,000, we can calculate a value for the
current to be 1.20 A.
b) Since the bar is initially 0.1 m away from the structure, the reluctance of the air dominates the reluctance of the
2x
μ0 A
structure. The reluctance is calculated as: ℜ( x ) =
Knowing the reluctance we can calculate the magnetic flux in the structure as a function of the coil current:
=
φ
Ni
μA
= Ni 0
ℜ( x )
2x
18.24
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Then, the magnitude of the force in the air gap is given by the expression
φ2 d ℜ( x ) 1 ( Niμ0 A) 2
=
=
=
f
μ0 A
2 dx
2 4 x2
2
( Ni ) μ0 A
2x
N
2
4 x2
2 ( 0.1m )
8900 N
=
±
=
±240.3 A
μ0 A
700
4π × 10−7 × 0.01m 2
f
Finally, the current required is: i =
±
Note that the holding current is significantly smaller than the current required to lift the bar from the initial distance
of 0.1 m.
Problem 18.28
Solution:
Known quantities:
The problem it is based on the electromagnet circuit depicted in
fig. P18.28, where:
7 108 ( 0.002 + x ) H −1 ,N =
980 t,R =
30Ω ,Va =
120V
ℜ( x ) =×
Find:
a)
The energy stored in the magnetic field for x = 0.005 m .
b) The magnetic force for x = 0.005 m .
Assumptions:
None.
Analysis:
a)
L(x) =
N2
.
ℜ( x )
'
W=
W=
m
m
L( x )i 2
2
The current is:
120
= 4A
30
9802 × 42
=
∴Wm
=
1.568 J
2 × 7 × 108 × 0.007
=
I dc
b)
i2
N 2 d ℜ( x )
F=
−
=
−224 N
2 ( ℜ( x ))2 dx
The minus sign indicates that the force F is in a direction opposite to that indicated in the figure.
18.25
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.29
Solution:
Known quantities:
With reference to the solenoid of Example 18.10 (fig. 18.40) and the reported parameters:
=
a 0.=
01m,lgap 0=
.001m,k 10 N / m
Find:
The best combination of current magnitude and wire diameter to reduce the volume of the solenoid coil.
Will this minimum volume result in the lowest possible resistance?
How does the power dissipation of the coil change with the wire gauge and current value?
Assumptions:
Use of Copper wire in solenoid
Analysis:
In order to access the effects of the wire diameter and current magnitude to the volume, resistance, and power
dissipated, mathematical expressions need to be developed for each variable.
Volume:
Vcoil = lcoil Acoil
The length of the coil is given by the circumference:
lcoil = πNd coil
The area of the coil:
Acoil =
π
d coil 2
4
π
π 2 Nd coil 3
2
=
Vcoil πNd
=
d
coil
coil
4
4
From the results of Example 18.10, the relationship between the current and the number of turns is given as:
=
Ni 56.4 A − turns
Hence:
Vcoil
56.4 ) π 2 d coil 3
(
=
4i
Resistance:
The resistance of the wire is given as:
R=
ρlcoil
Acoil
where  is the resistivity of the wire, which is assumed to be copper with=
a ρ 1.725 × 10−8 Ω / m
Using the previous derivations:
=
R
ρπNd coil 4 ρN 4 ( 56.4 ) ρ
= =
π
d coil
id coil
d coil 2
4
Power:
The dissipated power is given by:
2
=
P i=
R
4 ( 56.4 ) iρ
d coil
18.26
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Using a chart for AWG wire gauge and current carrying capacity, a table can be developed relating the wire gauge
and current carrying capacity to the volume, resistance, and power dissipated.
AWG
Gauge
Wire
Diameter
[in]
Wire
Diameter
[m]
Current
Capacity
[A]
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
1.019E-01
8.081E-02
6.408E-02
5.082E-02
4.030E-02
3.196E-02
2.530E-02
2.010E-02
1.590E-02
1.260E-02
1.000E-02
8.000E-03
6.300E-03
5.000E-03
4.000E-03
3.100E-03
2.500E-03
2.000E-03
1.600E-03
1.200E-03
1.000E-03
2.588E-02
2.053E-02
1.628E-02
1.291E-02
1.024E-02
8.118E-03
6.426E-03
5.105E-03
4.039E-03
3.200E-03
2.540E-03
2.032E-03
1.600E-03
1.270E-03
1.016E-03
7.874E-04
6.350E-04
5.080E-04
4.064E-04
3.048E-04
2.540E-04
30.0
20.0
15.0
10.0
5.0
3.3
2.5
1.25
0.83
0.63
0.31
0.21
0.16
7.81E-02
5.16E-02
3.91E-02
1.95E-02
1.29E-02
9.77E-03
4.88E-03
3.22E-03
Required
Number
of Turns
[turns]
2
3
4
6
11
17
23
45
68
90
180
273
361
722
1094
1444
2888
4375
5775
11551
17501
Coil
Volume
[m3]
Resistance
[ ]
Power
Dissipated
[W]
8.041E-05
6.017E-05
4.001E-05
2.993E-05
2.986E-05
2.256E-05
1.477E-05
1.481E-05
1.111E-05
7.299E-06
7.297E-06
5.661E-06
3.649E-06
3.649E-06
2.831E-06
1.739E-06
1.824E-06
1.415E-06
9.565E-07
8.070E-07
7.076E-07
5.012E-06
9.480E-06
1.594E-05
3.015E-05
7.603E-05
1.453E-04
2.422E-04
6.098E-04
1.168E-03
1.946E-03
4.903E-03
9.286E-03
1.556E-02
3.922E-02
7.428E-02
1.265E-01
3.138E-01
5.943E-01
9.806E-01
2.615E+00
4.754E+00
4.511E-03
3.792E-03
3.586E-03
3.015E-03
1.901E-03
1.582E-03
1.514E-03
9.528E-04
7.950E-04
7.600E-04
4.788E-04
3.950E-04
3.800E-04
2.394E-04
1.975E-04
1.931E-04
1.197E-04
9.875E-05
9.351E-05
6.234E-05
4.938E-05
As the wire gauge increases, the wire diameter and current carrying capacity both decrease and thus the number of
turns required increases, and the coil volume decreases. However, the resistance also increases with an increase in
wire gauge. Hence, the minimum volume will not result in the lowest possible resistance. Finally, the power
dissipation decreases with the increase of wire gauge due to decrease in current capacity.
18.27
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.30
Solution:
Known quantities:
Solenoid of Example 18.10 (Figure 18.40)
a = 0.01m , lgap = 0.001m , K = 10 N / m
Find:
L,Wm ,mmf using equation 18.46 and equation 18.30
Assumptions:
The reluctance of the iron is negligible, Neglect fringing
At x = 0 , the plunger is in the gap by an infinitesimal displacement
Analysis:
From Example 18.10, the reluctance as function of the distance x is here defined:
2l
ℜ ( x ) =gap
μo ax
Thus, it is possible to compute the inductance of the magnetic circuit as a function of reluctance (eq. 18.30):
=
L
N2
N 2 μo ax
=
2lgap
ℜ( x)
The stored magnetic energy is:
=
Wm
1 2 1 N 2i 2 μo ax
=
Li
2
2 2lgap
Finally, using equation 18.46 to write the expression for the magnetic force:
dWm
N 2i 2 μo a
fe =
−
=
−
dx
4lgap
This matches the derivation from example 18.10 using the relationship between the magnetic flux, the reluctance of
the structure, and the magnetic force.
The calculation for the required mmf is identical to example 18.10:
f gap =
kx =
ka =
0.1N
(10 N / m ) × ( 0.01m ) =
18.28
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.31
Solution:
Known quantities:
With reference to the solenoid used in Example 18.11 (Figure 18.40), the following parameters are considered:
a = 0.01m , lgap = 0.001m , K = 10 N / m , N = 1000turns , V = 12V , Rcoil = 5Ω
Find:
L,Wm ,mmf using equation 18.46 and equation 18.30.
Assumptions:
The reluctance of the iron is negligible, Neglect fringing
The inductance of the solenoid is approximately constant and is equal to the midrange value (plunger displacement
equal to a / 2 ).
Analysis:
From Example 18.11:
2lgap
ℜ gap ( x ) =
μo ax
Compute inductance if the magnetic circuit as a function of reluctance (equation 18.30):
L
=
N2
N 2 μo ax
=
2lgap
ℜ gap ( x )
Compute stored magnetic energy:
=
Wm
1
1 N 2i( t )2 μo ax
=
Li( t )2
2
2
2lgap
Finally, use equation 16.46 to write the expression for the magnetic force:
−7
dWm
N 2i( t )2 μo a (1000 ) × 4π × 10 × 0.01
f gap =
i( t )2 =
πi( t )2
−
=
−
=
4lgap
4 × 0.001
dx
With the assumption that the inductance is constant with x = a / 2 :
2
N 2 μo a 2 (1000 ) × 4π × 10 × ( 0.01 )
=
=
= 31.4mH
L≈
ℜ gap ( x )
4lgap
4 × 0.001
2
N2
−7
2
From Example 18.11:
− t
V
12
− Rt
−3
i( t ) = 1 − e L =  1 − e 6.3×10  A
R
5 

(
)
− t
−3
 12

=
f gap π   1 − e 6.3×10  

5
2
This matches the derivation from example 18.11 using the relationship between the magnetic flux, the reluctance of
the structure, and the magnetic force. The response curves are shown below:
18.29
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.32
Solution:
Known quantities:
Solenoid of Example 18.11 (Figure 18.40)
a = 0.01m , lgap = 0.001m , K = 10 N / m , N = 1000turns , V = 12V , Rcoil = 5Ω , m = 0.5kg , μr = 1000
Find:
Generate a simulation program taking into acocunt the fact that the solenoid inductance is not constant, but is a
function of plunger position
Compare graphically the current and force step responses of this system to the step response obtained in Example
18.11.
Assumptions:
The reluctance of the iron is negligible, Neglect fringing
Neglect damping on the plunger
Analysis:
From Example 18.11:
2lgap
ℜ gap ( x ) =
μo ax
The inductance is a function of plunger position, as shown in the following equation:
=
L
N2
N 2 μo ax
=
ℜ gap ( x )
2lgap
The differential equation of the current is based on the
equivalent circuit shown below:
v( t=
) Ri + L
di
dt
To find the equation of the force, compute stored magnetic
energy as a function of current:
Wm
=
1
1 N 2i( t )2 μo ax
Li( t )2
=
2
2
2lgap
Finally, use equation 16.46 to write the expression for the
magnetic force:
−7
dW
N 2i( t )2 μo a (1000 ) × 4π × 10 × 0.01
f gap =
i( t )2 =
πi( t )2
− m =
−
=
dx
4lgap
4 × 0.001
2
This matches the derivation from example 18.11 using the relationship between the magnetic flux, the reluctance of
the structure, and the magnetic force.
Simulink Block Diagram:
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Comparison of Step Responses for Constant Inductance and Variable Inductance Systems:
Note the quicker response of the variable inductance system due to the smaller inductance initially. The larger
constant inductance results in a delayed response.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.33
Solution:
Known quantities:
Relay of Example 18.12 (Figure 18.46)
N 10,000turns;
f restore 5=
N ; Agap
=
=
m ) ; x 0.05
m; μr 1000
=
( 0.01=
2
m= 0.05kg; x= ε= 0.001m
Find:
Required holding current to keep the relay closed.
Assumptions:
Air gap becomes zero and the iron reluctance cannot be neglected.
Analysis:
To compute the current we need to derive an expression for the force in the air gap.
Without neglecting the iron reluctance, we can write the expression for the reluctance as follows:
ℜ( x=
)
L
2x
+
μr μ0 A μ0 A
where L is the total length of the iron magnetic path (excluding the air gap).
Knowing the reluctance we can calculate the magnetic flux in the structure as a function of the coil current:
=
φ
Ni
μμA
= Ni r 0
ℜ( x )
L + 2 μr x
Then, the magnitude of the force in the air gap is given by the expression
( Niμr μ0 A)=
( Niμr ) μ0 A
φ2 d ℜ( x ) 1
2
=
2
2 2
2
2 dx
2 L + 4 Lμr x + 4 μr x μ0 A L + 4 Lμr x + 4 μr2 x 2
2
=
f
2
As x approaches zero, we can calculate the force to be:
( Niμr ) μ0 A
2
f (=
x 0=
)
L2
and the current required to maintain the given force is:
=
L2 f ( x 0=
)
f ( x 0)
L
i=
±
=
±
2
Nμr
μ0 A
( Nμr ) μ0 A
The total length of the magnetic path:
L = 0.05m + 0.05m + 0.10m + 0.10 = 0.30m
Thus, the current can be yield as following:
( 0.30 )
i=
±
(10000 )(1000 )
( 5)
( 4π × 10 ) ( 0.01)
−7
2
=
±0.0060 A =
±6.0mA
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.34
Solution:
Known quantities:
Relay Circuit shown in Figure P18.34
=
N 500turns;
=
Agap 0.001
=
m 2 ;L 0=
.02m;k 1000 N=
/ m;R 18Ω
=
μ μ=
4π ⋅ 10−7
0
Find:
Minimum DC supply voltage v for which the relay will make
contact when the electrical switch is closed
Assumptions:
Neglect the iron reluctance
Analysis:
The reluctance of the gap is:
2x
ℜ( x) =
μ0 Agap
dℜ( x )
2
=
dx
μ0 Agap
Magnetic Flux:
=
φ
Niμ0 Agap
Ni
=
2x
ℜ( x)
Magnetic force:
φ2 d ℜ ( x )  Niμ0 Agap  1 2
= 
=
2 dx
 2 x  2 μ0 Agap
2
=
f gap
( Ni ) μ0 Agap
2
4 x2
The force that must be overcome is the spring force, fk:
f gap
= f=
kx
k
Equating the two force equations and solving for the current:
i= ±
2
N
kx 3
μ0 Agap
The voltage is determined using Ohm’s law, and x = L:
2 (18 ) (1000 )( 0.02 )
2R
kx 3
±
=
±
=
v=
iR =
182.0V
N μ0 Agap
( 500 ) 4π × 10−7 ( 0.001)
3
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.35
Solution:
Known quantities:
The simplified representation of a surface roughness sensor shown in Figure P18.35
Find:
Derive an expression for the displacement x as a function of
the various parameters of the magnetic circuit and of the
measured emf e
Assumptions:
The flux =
φ β / ℜ ( x ) where β is a known constant and
ℜ ( x ) is the gap reluctance.
Frictionless contact between the moving plunger and the
magnetic structure.
The plunger is restrained to vertical motion only.
Analysis:
dφ
e=N
;
dt
β
φ=
;
ℜ( x )
The reluctance is function of the displacement x
)
ℜ( x=
1
L−x
x
+ =
 L + ( μr − 1) x 
μr μ0 A μ0 A μr μ0 A 
β
μr μ0 A
;
=
ℜ( x )  L + ( μr − 1) x 
Nμ μ Aβ ( μr − 1) dx
dφ
dφ dx
e= N
= N
= − r 0
2
dt
dx dt
 L + ( μr − 1) x  dt
φ
=
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.36
Solution:
Known quantities:
As shown in Figure P18.36. The air gap between the shell and the plunger is uniform and equal to 1mm . The
diameter is d = 25 mm . The exciting current is
.
N = 200 .
Find:
The force acting on the plunger when x = 2 mm .
Assumptions:
None.
Analysis:
Note that there are multiple air gaps: the one around the plunger, as well as the one between the plunger and the top
of the electromagnet.
We assume that the first air gap is negligible because its cross section area (which wraps the plunger) is bigger than
the plunger cross section area. The second air gap has the following cross section area (which is on top of the
plunger):
2
 25 ×10−3 
d 
−4
2
=
A π  = π 
= 4.9 ×10 m
2
2


2
The expression for the reluctance as a function of the plunger displacement x is the following:
lg
x
x
+
=
=
1.62 ×109 x
−7
−4
µ0 A µ0 Ag 4π ×10 × 4.9 ×10
The resulting force for a displacement of the plunger x = 0.002 m is the following:
R ( x) =
d R ( x)
1 2
1  Ni  d R ( x )
=
=
φ ( x)
f
 =

dx x = 2×10−3 2  R ( x ) 
dx
2
x = 2×10−3
2
1  200 × 7.5 
9
=
=
1.62 ×10
173.6 N

9 
2  1.62 ×10 x 
−3
2
x = 2×10
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.37
Solution:
Known quantities:
As shown in Figure P18.37, the double-excited electromechanical system moves horizontally. The cross section of
the structure is w × w .
Find:
a) The reluctance of the magnetic circuit.
b) The magnetic energy stored in the air gap.
c) The force on the movable part as a function of its
position.
Assumptions:
Resistance, magnetic leakage and fringing are negligible.
The permeability of the core is very large.
Analysis:
a)
The reluctance of the magnetic circuit is:
=
ℜ
lg
x
+
2
μ0 w
μ0 w( w − x )
b)
The magnetic energy stored in the air gap is:
Wm =
( N1 + N 2 )2 i 2
2ℜ
c)
The magnetic force is:
lg
 x

+
( N1 + N 2 )2 
2
2 
μ0 w( w − x ) 
i
 μ0 w
f =
2
2
lg
 x

+


2
μ0 w( w − x ) 
 μ0 w
2
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.38
Solution:
Known quantities:
The flux density in the cast steel pathway is 1.1T . The diameter of the plunger is 10mm . The solenoid is depicted
in the figure below.
Find:
The force F between the faces of the poles.
Assumptions:
Air gap is negligible between walls and plunger
(from Table 18.1).
Since the pathway is cast steel,
Assume that the reluctance of the steel is negligible.
Analysis:
Using Equation 18.50:
F=
φ2 d ℜ ( x )
2 dx
The flux is determined from the flux density and the area:
φ = BA
Since the plunger is cylindrical and the air gap between the plunger and the coil is negligible, the reluctance is
calculated as:
x
ℜ( x) =
μA
where x is the gap between the plunger and back wall of the solenoid. The reluctance of the cast steel pathway can
be neglected due to the low reluctance of the structure.
The derivative of the reluctance is:
dℜ( x ) 1
=
dx
μA
The area is the cross-sectional area of the plunger:
A=
π 2
d
4
Combining all of the equations into the force equation:
φ2 d ℜ ( x )
=
F
=
2 dx
BA) 1
(=
2
2
μA
πd 2 B 2 1 π ( 0.010 ) (1.1)
=
= 37.8 N
8 ( 4π × 10−7 )
( 2 )( 4 ) μo
2
2
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.39
Solution:
Known quantities:
A force of 10 ,000 N is required to support the weight. The cross-sectional area of the magnetic core is 0.01 m 2 .
The coil has 1000turns .
Find:
The minimum current that can keep the weight from falling for x = 1.0 mm .
Assumptions:
Negligible reluctance for the steel parts and negligible fringing in the air gaps.
Analysis:
2x
=
159.15 × 106 x
4π × 10−7 ( 0.01 )
The variable reluctance is given by: ℜT ( x ) =
2ℜ( x ) =
N 2i 2 d ℜT ( x )
2 RT ( x ) dx
The force is related to the reluctance by: f =
−10000 =
− 2
Therefore,
=
i
=
3.18 1.784 A
Problem 18.40
Solution:
Known quantities:
The 12 −VDC control relay is made of sheet steel. Average length of the magnetic circuit is 12 cm . The average
cross section of the magnetic circuit is 0.60 cm 2 . The coil has 250turns and carries 50mA .
Find:
The flux density B in the magnetic circuit of the relay when the coil is energized.
The force F exerted on the armature to close it when the coil is energized.
a.
b.
Assumptions:
None.
Analysis:
NI
) H( 12 × 10−2 ) ⇒ H
= Hl 250( 250 × 10−3=
= 104.2
from the curve, B = 0.75 T .
a.
b.
=
F
1 AB 2 1 ( 0.6 × 10−4 )( 0.75 )2
=
= 13.4 N
2 μ0
2
4π × 10−7
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.41
Solution:
Known quantities:
As shown in Figure P18.41.
Find:
The differential equations describing the system.
Assumptions:
None.
Analysis:
The equation for the electrical system is:
=
v iR + L( x )
di
dt
N2
N 2 μ0 A
=
ℜT ( x )
2x
where: =
L( x )
The equation for the mechanical system is:
=
Fm m
d 2x
+ kx
dt 2
where Fm is the magnetic pull force. To calculate this force we use the following equation:
Fm = −
dWm
dx
where Wm is the energy stored in the magnetic field.
Let
and
be the magnetomotive force acting on the structure and its reluctance, respectively; then:
φ2ℜ( x )
F2
N 2i 2 μA
Wm =
= =
2
2ℜ( x )
4x
dWm N 2i 2 μA
Fm =
−
=2
dx
4x
Finally, the differential equations governing the system are:
N 2 μ0 A di
2 x dt
2
d x
N 2i 2 μA
m 2 + kx = 2
dt
4x
=
v iR +
This system of equations could be solved using a numerical simulation, since it is nonlinear.
18.40
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.42
Solution:
Known quantities:
As shown in Figure P18.42. The solenoid has a cross section
of 5cm 2 and N = 100 turns .
Find:
a.
The force exerted on the plunger when the distance x
is 2 cm and the current is 5 A . Relative permeabilities of the
magnetic material and the nonmagnetic sleeve are 2 ,000 and
1.
b.
Develop a set of differential equations governing the
behavior of the solenoid.
Assumptions:
The fringing and leakage effects are negligible.
Analysis:
a.
x
ℜx =
μ0 Ax
ℜ=
g
lg
0.005
=
= 7.96 × 106 H −1
−7
−4
μ0 Ag ( 4π × 10 )( 5 × 10 )
ℜ1 =ℜ2 =
0.335
l1
=
=2.67 × 105 H −1
μr μ0 Ag 2000( 4π × 10−7 )( 5 × 10−4 )
ℜ=
3
0.055
l3
=
= 2.19 × 104 H −1
μr μ0 Ax 2000( 4π × 10−7 )( 10 × 10−4 )
=
ℜm
lx
lx
=
μr μ0 Ax 2000( 4π × 10−7 )( 10 × 10−4 )
= ( 3.98 × 105 )( 0.095 − x ) H −1
The equivalent circuit is shown on the right:
ℜeq=
ℜ g + ℜ1
lg
l1
=
+
= 4.11 × 106 H −1
2
2 μ0 Ag 2 μr μ0 A
ℜT = ℜ3 + ℜ x + ℜm + ℜeq =
l
l3
x
( 0.095 − x )
l1
+
+
+ g +
2 μ0 Ag 2 μr μ0 Ag
μr μ0 Ax μ0 Ax
μr μ0 Ax
Since Ax = 2Ag:
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
l3 + xμr + ( 0.095 − x ) + lg μr + l1
=
2 μr μ0 Ag
ℜT
0.055 + 0.095 − x + 2000 x + ( 0.005)( 2000 ) + 0.335 10.485 + 1999 x
= =
2.51 × 10−6
2 ( 2000 ) ( 4π × 10−7 )( 5 × 10−4 )
=
= 4.18 × 106 + 7.96 × 108 x
N 2i 2 ( 2 μr μ0 Ag )
( μr − 1) =
φ2 d ℜT ( x )
N 2i 2 d ℜT
−
=
−
=
−
fm =
2
2
dx
2
2 RT dx
2 ( l3 + xμr + ( 0.095 − x ) + lg μr + l1 ) 2 μr μ0 Ag
2
μr μ0 Ag N 2i 2 ( μr − 1)
=
−
=
2
( l3 + xμr + ( 0.095 − x ) + lg μr + l1 )
( 2000 ) ( 4π × 10−7 )( 5 × 10−4 ) (100 ) ( 5) ( 2000 − 1)
= −
( 0.055 + 0.095 − x + 2000 x + ( 0.005)( 2000 ) + 0.335 )2
2
2
For x = 0.02, f = −0.25N .
b.
Electrical subsystem:
=
vs ( t ) L( x )
di( t )
+ Ri( t )
dt
dx( t ) k
− x( t )
dt
l
l3 + xμr + ( 0.095 − x ) + lg μr + l1
Reluctance: ℜT =
2 μr μ0 Ag
 t ) =f m ( x ) − d
Mechanical subsystem: mx(
Flux:
=
φ
2 μr μ0 Ag Ni( t )
Ni( t )
=
ℜm ( x ) l3 + xμr + ( 0.095 − x ) + lg μr + l1
Magnetic force:
N 2i 2 ( 2 μr μ0 Ag )
( μr − 1) =
φ2 d ℜT ( x )
N 2i 2 d ℜT
fm ( x ) =
=
−
=
2
2
dx
2
2 RT dx
2 ( l3 + xμr + ( 0.095 − x ) + lg μr + l1 ) 2 μr μ0 Ag
2
=
μr μ0 Ag N 2i 2 ( μr − 1)
( l + xμ + ( 0.095 − x ) + l μ + l )
3
r
g
r
2
1
N 2 2 μr μ0 Ag
N2
Inductance:=
L( x) =
ℜT ( x ) l3 + xμr + ( 0.095 − x ) + lg μr + l1
Substituting the expressions for fm and L(x) in the two differential equations, we have the final answer.

 di( t )
N 2 2 μr μ0 Ag
=
+ Ri( t )
Electrical subsystem: vs ( t ) 
 l + xμ + ( 0.095 − x ) + l μ + l  dt
r
g
r
3
1


μr μ0 Ag N 2i 2 ( μr − 1)
dx( t ) k
 t )
Mechanical
subsystem: mx(
=
−d
− x( t )
2
dt
l
+
+
−
+
+
l
xμ
(
0
.
095
x
)
l
μ
l
(3 r
g r
1)
Note that these equations are very nonlinear!
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Problem 18.43
Solution:
Known quantities:
The relay shown in Figure P18.43
Find:
Derive the differential equations (electrical and mechanical)
for the relay
Assumptions:
The inductance is a function of x
The iron reluctance is negligible
Analysis:
Electrical subsystem:
=
v( t ) L( x )
di( t )
+ Ri( t )
dt
Mechanical subsystem:
 t ) = f m ( x ) − b
mx(
dx( t )
− kx( t )
dt
Next, we calculate the magnetic force and inductance as functions of x.
Reluctance:
2x
ℜ( x ) =
μA
Flux:
=
φ
Ni( t ) Ni( t )μA
=
ℜ( x )
2x
Magnetic force:
=
fm ( x )
φ2 d ℜ( x ) N 2i 2 μA
=
2 dx
4 x2
Inductance:
=
L( x)
N2
N 2 μA
=
2x
ℜ( x )
Substituting the expressions for fm and L(x) in the two differential equations, we have the final answer.
Electrical subsystem:
v( t )
=
N 2 μA di( t )
+ Ri( t )
2 x( t ) dt
 t ) =
Mechanical subsystem: mx(
N 2i 2 μA
dx( t )
−b
− kx( t )
2
4 x( t )
dt
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 18
Note that these equations are very nonlinear!
Problem 18.44
Solution:
Known quantities:
The relay shown in Figure P18.44
Find:
Derive the differential equations.
Assumptions:
The inductance is a function of x
The iron reluctance is negligible
Analysis:
The equation for the electrical system is:
VS =
i (t ) R +
d ( L ( x ) i (t ))
dt
=
i (t ) R + L ( x )
di ( t )
dL ( x )
di ( t )
dL ( x ) dx ( t )
+ i (t )
=
i (t ) R + L ( x )
+ i (t )
dt
dt
dt
dx
dt
where:
N2
N 2 μ0 A
=
ℜT ( x )
2x
=
L( x )
dL( x )
N 2 μ0 A
= −
dx
2 x2
The equation for the mechanical system is:
Fm = m
d 2 x (t )
dx ( t )
+b
+ kx ( t )
2
dt
dt
where Fm is the magnetic pull force. To calculate this force we use the following equation:
Fm = −
dWm
dx
where Wm is the energy stored in the magnetic field.
Let F and
=
Wm
be the magnetomotive force acting on the structure and its reluctance, respectively, then:
φ2ℜ( x )
F2
N 2i 2 μA
= =
2
2ℜ( x )
4x
dW
N 2i 2 μA
Fm =
− m =
−
dx
4 x2
Finally, the differential equations governing the system are:
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N 2 μ0 A di ( t )
N 2 μ0 A dx ( t )
−i
2x
dt
2 x2
dt
2 2
d 2 x (t )
dx ( t )
N i μA
m
+b
+ kx ( t ) =
−
dt 2
dt
4 x2
VS ( t ) =
Ri ( t ) +
This system of equations could be solved using a numerical simulation, since it is nonlinear.
b) Moving-Coil Transducers
Problem 18.45
Solution:
Known quantities:
Length of the wire is 20cm ; Flux density is 0.1T ; The
position of the wire is x( t ) = 0.1 sin 10t m
Find:
The induced emf across the length of the wire as a function of time.
Assumptions:
None.
Analysis:
From e = Blu , we have
dx
=
e( t ) Bl
= 0.02 cos( 10t )V
dt
Problem 18.46
Solution:
Known quantities:
Emf: e1( t ) = 0.02 cos 10t ; the length of the second wire:
0.1 m ; the position of the second wire: x( t )= 1 − 0.1 sin 10t .
Find:
The induced emf e( t ) defined by the difference in emf's
e1( t ) and e2 ( t ) .
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Assumptions:
None.
Analysis:
We have
e(
=
t ) e1( t ) − e2 ( t )
=
e2 ( t ) ( 0.1 )( 0.1 )( −1 cos 10t )V
e( t ) = 0.02 cos( 10t ) + 0.01 cos( 10t ) = 0.03 cos( 10t )V
Problem 18.47
Solution:
Known quantities:
The system parameters are shown in fig. P18.48 and
=
I 4=
A,B 0.3Wb m 2 .
Find:
The magnitude and direction of the force induced on the conducting bar.
Assumptions:
None.
Analysis:
f = Bli= 0.3 × l × 4= 1.2l N
The force will be to the left if current flows upward.
Problem 18.48
Solution:
Known quantities:
Consider the electrodynamic bar of fig. P18.48, immersed in a
uniform field B = 0.3Wb m 2 .
Find:
The magnitude and direction of the induced voltage in the
wire.
Assumptions:
None.
Analysis:
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Problem 18.49
Solution:
Known quantities:
The electrodynamic shaker shown in Figure P18.49
Shaker mass, m; air gap dimension, d; number of coil turns,
N; spring parameter, k; armature resistance, R and inductance,
L
Find:
a.
Reluctance of the structure and flux density B
b.
Dynamic equations of the shaker
c.
Transfer function and frequency response of the
shaker velocity to input voltage Vs.
Assumptions:
Iron reluctance is negligible. Neglect fringing. Assume no
damping in this system.
Analysis:
2d
μ0 A
φ NI f NI f μ0
Compute flux density: B
= =
=
A Aℜ
2d
a.
Reluctance of structure: ℜ =
b.
Electrical Subsystem, using KVL: Vs =Ri + L
Mechanical Subsystem: m
c.
di
dx
+ Bl
dt
dt
d 2x
= Bli − kx
dt 2
Laplace Transform:
Vs ( s ) =
( R + Ls )I( s ) + BlsX ( s )
0=
− BlI( s ) + ( ms 2 + k )X ( s )
 ( R + Ls )( ms 2 + k )

=
+ Bls  X ( s )
Vs ( s ) 
Bl


X( s )
Bl
=
3
2
Vs ( s ) mLs + mRs + ( kL + B 2l 2 )s + Rk
U( s ) SX ( s )
Bls
= =
3
2
Vs ( s ) Vs ( s ) mLs + mRs + ( kL + B 2l 2 )s + Rk
Frequency Response: s = jω
U ( jω )
Bljω
=
3
Vs ( jω ) mL ( jω ) + mR ( jω ) 2 + ( kL + B 2l 2 ) jω + Rk
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U ( jω )
Bljω
=
2
Vs ( jω ) R( k − mω ) + j ( kL + B 2l 2 )ω − mLω 3 
Problem 18.50
Solution:
Known quantities:
The electrodynamic shaker shown in Figure P18.49 is used to perform vibration testing of an electrical connector.
2
=
=
=
B 1,000 Wb/m
; l 5=
m 1 kg;
m; k 1000 N/m;
=
=
b 5 N-s/m;
L 0=
.8 H; R 0.5 Ω;
The test consists of shaking the connector at the frequency ω
= 2π × 100rad / s
Find:
The peak amplitude of sinusoidal voltage Vs required to generate an acceleration of 5g (49m/s2) under the stated test
conditions
Assumptions:
Connector has negligible mass when compared to the platform.
Analysis:
Applying KVL around the coil circuit: L
di
dx
+ Ri + Bl
=
VS
dt
dt
Next, we apply Newton’s Second Law:
− Bli + m
d 2x
dx
+ b + kx =
0
dt 2
dt
To derive the frequency response, Laplace transform the two equations to obtain:
VS ( s )
( sL + R ) I( s ) + BlsX ( s ) =
− BlI( s ) + ( ms 2 + bs + k )X ( s ) =
0
We can write the above equations in matrix form and resort to Cramer’s rule to solve for U(s) as a function of V(s):
Bls
( sL + R )
  I( s )  V ( s )
=


2
( ms + bs + k )  X ( s )  0 
 − Bl
with solution
X( s )
( Ls + R ) V ( s )
det 

0 
 − Bl
=
Bls
( Ls + R )

det 

2
( ms + bs + k )
 − Bl
BlV ( s )
( Ls + R ) ( ms 2 + bs + k ) + ( Bl ) s
2
To obtain the acceleration response, we multiply the numerator by
s2:
s 2 X ( s ) X ( s )
= =
V( s )
V( s )
Bls 2
( Lm ) s 3 + ( bL + Rm ) s 2 + (bR + kL + ( Bl ) ) s + ( kR )
2
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− Blω2
X ( jω )
=
V ( jω )  kR − ( bL + Rm ) ω2  + j  bR + kL + ( Bl )2 ω − ( Lm ) ω3 




(
)
The magnitude of this complex number, evaluated at  =2  100 is 0.62.
Thus, to obtain the desired acceleration (peak) of 49 m/s2, we wish to have a peak voltage amplitude |VS| = 49/0.62 ≈
78 V.
Problem 18.51
Solution:
Known quantities:
As described in Example 18.13.
Find:
Derive and sketch the frequency response of the loudspeaker in the following two cases. Describe qualitatively how
the loudspeaker frequency response changes as the spring stiffness k increases and decreases. Find the limit of the
frequency response and the kind of the speaker as k approaches zero.
a.
k = 50,000 N m .
b.
c.
k = 5 × 106 N m .
k = 0N m
Assumptions:
None.
Analysis:
a.
For k = 50 ,000 N m , the transfer function is:
U
1.478 × 105 s
= 3
V s + 3075s 2 + 9 × 106 s + 4 × 109
The magnitude frequency response is plotted below:
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This response would correspond to a midrange speaker.
b.
For k = 5 × 106 N m , the transfer function is :
U
1.478 × 105 s
= 3
V s + 3075s 2 + 5.04 × 108 s + 4 × 1011
The magnitude frequency response is plotted below:
It should be apparent that this response would enhance the treble range, and is the response of a "tweeter".
c.
For k = 0 , the transfer function is :
U
1.478 × 105
= 2
V s + 3075s + 4 × 106
The magnitude frequency response is plotted below:
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In this case, the speaker acts as a "woofer", emphasizing the low frequency range. In practice,
cannot be
identically zero, so the actual response of a woofer would resemble that of a midrange speaker, shifted towards the
lower frequencies.
Problem 18.52
Solution:
Known quantities:
Loudspeaker of Example 18.13 (Figure 18.52)
=
rcoil 0.05=
m;L 10mH
=
;R 8Ω
=
;N 47
=
;B 1T ;
m=
0.01kg;b =
22.75N − s / m;k =
5 × 104 N / m
Find:
Modify the parameters of the loudspeaker (mass, damping, and spring rate), so as to obtain a loudspeaker with a
bass response centered on 400 Hz.
Demonstrate that your design accomplishes the intended task, using frequency response plots.
Assumptions:
None.
Analysis:
From Example 18.13, the system transfer function is:
U (s)
Bls
=
V ( s ) ( Lm ) s 3 + ( Rm + Lb ) s 2 +  Rb + kL + ( Bl )2  s + kR


The frequency response is:
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U ( jω )
jBlω
=
V ( jω ) kR − ( Rm + Lb ) ω2 + j  Rb + kL + ( Bl )2  ω − ( Lm ) ω3


where: l = 2πNrcoil
{
}
Using MATLAB, it is easy to adjust the mechanical parameters one-by-one in order to see each parameters’ effect
on the system frequency response.
Mass: Increasing the mass decreases the frequency response center frequency.
Damping: Increasing the damping also decreases the center frequency, but also widens the response bandwidth.
Spring Rate: Decreasing the spring rate also decreases the center frequency.
There are many possible combinations of mechanical parameters that could be substituted to generate the desired
response.
One such set of parameters is: m =
0.05kg;b =
30 N − s / m;k =
5 × 103 N / m
The frequency response for this system is given below, with the response centered over 400 Hz:
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Problem 18.53
Solution:
Known quantities:
Electrodynamic shaker as shown in Figure 18.53, with parameters. The test
consist in shaking the circuit at the frequency ω = 2π( 100 )rad / s
2
=
=
= 0.2kg;
B 200Wb / m
;l 5=
m;k 100 N / m;m
b =−
5N s / m;L =
8mH ;R =
0.5Ω;
Find:
a) Write down dynamic equations and indicate system input and output.
b) Find the frequency response function of the table acceleration in response to the applied voltage.
c) Determine the peak amplitude of the sinusoidal voltage VS required to generate an acceleration of 5g
(
).
Assumptions:
None.
Analysis:
a) Applying KVL around the coil circuit: L
di
dx
+ Ri + Bl
=
VS
dt
dt
Next, we apply Newton’s Second Law :
− Bli + m
d 2x
dx
+ b + kx =
0
2
dt
dt
Voltage applied is the input of the system, while acceleration response of the shaker are the output of the system.
b) To derive the frequency response we Laplace transform the two equations to obtain:
VS ( s )
( sL + R ) I( s ) + BlsX ( s ) =
− BlI( s ) + ( ms 2 + bs + k )X ( s ) =
0
We can write the above equations in matrix form and resort to Cramer’s rule to solve for U(s) as a function of V(s):
Bls
( sL + R )
  I( s )  V ( s )
=


2
( ms + bs + k )  X ( s )  0 
 − Bl
with solution
X( s )
( Ls + R ) V ( s )
det 

0 
 − Bl
=
Bls
( Ls + R )

det 

2
( ms + bs + k )
 − Bl
BlV ( s )
( Ls + R ) ( ms 2 + bs + k ) + ( Bl ) s
2
To obtain the acceleration response, we multiply the numerator by s2:
s 2 X ( s ) X ( s )
= =
V( s )
V( s )
Bls 2
( Lm ) s 3 + ( bL + Rm ) s 2 + (bR + kL + ( Bl ) ) s + ( kR )
2
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− Blω2
X ( jω )
=
V ( jω )  kR − ( bL + Rm ) ω2  + j  bR + kL + ( Bl )2 ω − ( Lm ) ω3 




(
)
c) The magnitude of this complex number, evaluated at  =2  100 is 0.62.
Thus, to obtain the desired acceleration (peak) of 49 m/s2, we wish to have a peak voltage amplitude
=
|VS | 49 / 0.62 ≈ 78 V .
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