3.1 The rotor of Fig.3.25 is similar to that of Fig.3.2 (EX3.1) except that it has two
coils instead of one. The rotor is nonmagnetic and is placed in a uniform magnetic field of magnitude B0 . The coil sides are of radius R and are uniformly spaced around the rotor surface. The first coil is carrying a current I 1 and the second coil is carrying a current I 2 . Assuming that the rotor is 0.3m long , R=0.13m , and B0 =0.85 T , find the -directed torque as a function of rotor position
for (a)
I 1 = 0 A and I 2 = 5 A , (b) I 1 = 5 A and I 2 = 0 A , (c) I 1 = 8 A and I 2 = 8 A .
Fig.3 two-coil rotor for problem 3.1 Sol:
(a) = 0 A and = 5 A , (b) = 5 A and = 0 A , (c) = 8 A and = 8 A
T 2 B0 Rl [ I 1 sin I 2 cos ]
6.63 102 [ I 1 sin I 2 cos ] N .m (a ) T 0.53 cos (b) T 0.53 sin
8 cos ] (c) T 0.53[8 sin
.
3.2 The The winding current of the rotor of
rotor angle
Problem 3.1 are controlled as a function of
such that I 1 8 sin A and I 2 8 sin A .Write an expression
for the rotor torque as a function of the rotor position
.
Sol
I 1 8 sin and I 2 8 cos T 2 B0 Rl [ I 1 sin I 2 cos ]
6.63 102 [ I 1 sin I 2 cos ] N .m 6.63 10 2 [8 sin 2 8 cos 2 ] N .m 0.5204 N .m
3.3 Calculate the magnetic stored energy in the magnetic circuit of Example 1.2. Sol:
由例題 1.2 知 =0.13(wb),I=10(A) 求出電感值 L
N I
1000 0.13 10
13( H )
依 1.47 式得 1
W LI 2 2
13 10 2 2
650 650 ( J )
3.4 An inductor has an inductance which is found experimentally to be of form
L
2 L0 1 x x0
where L0 =30mH, x 0 =0.87mm,and x is the displacement of moveable
element. Its winding resistance is measured and found to equal 110mΩ. a. The displacement x is held constant at 0.9mm, and the current is increased from 0
to 6.0A. Find the resultant magnetic stored energy in the inductor. b. The current is then held constant at 6.0A, and the displacement is increased to 1.80mm.Find the corresponding change in magnetic stored energy. Sol:
For x 0.9mm
帶入上述公式求得電感值 L
2 30 0.9
1
29.5( mH )
0.87
在依 1.47 式得 1 1 W LI 2 29.5 10 3 6 2 0.531( J ) 2 2
For x 1.8mm
電感值為 L
2 30 1. 8
1
19.6( mH )
0.87
其儲能為 W
1 2
19.6 10 3 6 2 0.352( J )
兩者之差 W 0.352 0.531 0.179( J )
3.5 Repeat Problem 3.4, assuming that the inductor is connected to a voltage source
which increases from 0 to 0.4V (part a ) and then is held constant at 0.4 V (part b ). For both calculations, assume that all electric transients can be ignored.
Sol:
For a coil voltage of 0.4V, so coil current will equal I
0.4 0.11
3.7 A
(a.) If L 29.5mH :
1 1 W fld LI 2 29.5 10 3 3.7 2 0.202 Joule 2 2 If L 19.6mH :
1 1 W fld LI 2 19.6 10 3 3.7 2 0.134 Joules 2 2 (b.)
W fld 0.134 0.202 0.068 ( Joules)
3.6 The inductor of Problem 3.4 is driven by a sinusoidal current source of the form
i t I 0 sin t Where I 0 5.5 A and 100 50 Hz . With the displacement held fixed at x x0 , calculate (a) the time-averaged magnetic stored energy ( W fld ) in the inductor and (b) the time – averaged power dissipated in the winding resistance. Sol:
For x x0 , L L0 30mH . The rms. current is equal I rms (a.)
1 1 W fld LI 2 30 10 3 3.892 0.227 Joules 2 2 (b.) 2 P diss I rms R 3.89 2 0.11 1.63W
5.5 2
3.89 A
3.11 The inductance of a phase winding of a three-phase salient-pole motor is
measured to be of the form L( m ) L0 L2 c o s2 m where m is the angular position of the rotor. a. How many poles are on the rotor of this motor? b. Assuming that all other winding currents zero and that this phase is excited by a constant current I 0, find the torque T fld(θ) acting on the rotor. Sol:
(a):
2 個極點
(b):
1 1 L11i12 L12 i1i2 L22 i22 W fld 2 2 T fld
W f ld m
d Ia 2
d 2
( L0 L2 cos 2 m ) I 02 L2 sin 2 m
3.13 Consider the plunger actuator of Fig. 3.29. Assume that the plunger is
initially fully opened( g = 2.25 cm) and that a battery is used to supply a current of 2.5 A to the winding. a. If the plunger is constrained to move very slowly (i.e., slowly compared to the electrical time constant of the actuator), reducing the gap g from 2.25 to 0.20 cm, how much mechanical work in joules will be supplied to the plunger. b. For the conditions of part(a), how much energy will be supplied by the battery(in excess of the power dissipated in the coil)?
Figure 3.29 Plunger actuator for Problem 3.12. Sol:
(a):
(b):
Work = W fld( g = 0.2 cm) −W fld( g = 2.25 cm) = 46.7 µJoules
電源只將提供在這種線圈內熱損失的能量。
3.14
N-turn winding
Electromagnet cross-sectional Area Ac g
g
Iron slab mass M
As shown in Fig.3.30 an N-turn electromagnet is to be used to lift a slab of iron of mass M. The surface roughness of the iron is such that when the iron and the electromagnet are in contact there is a minimum air gap of
gmin=0.18mm in each
leg .the electromagnet cross-sectional area Ac=32cm and coil resistance is 2.8Ω .Calculate the minimum coil voltage which
must be used to lift a slab of
mass 95kg against the force of gravity. Neglect the reluctance of the iron. Sol:
鐵心的下降力為 9.8×95Kg 線圈的感應方程式為: L 0 N 2 Ac / 2 g 但題目中並無提供條件 N 所以經由答案反推 N 求出為 450 再來計算所需要的力 Ffid=
i 2 dL 2 dg
(
0 N 2 Ac 2
4 g
)i 2 其中 g=9.8m/ sec 2
答案為 931N 求出 Ffid 之後 計算電流 imin 公式: imin (
2 g min N
)
Ff id
0 Ac
385mA
最後由歐姆定律 求出最小電壓 Vmin =385mA×2.8=1.08V
2
3.16 An inductor id made up of a 525-turn coil on a core of 14cm cross - sectional
area and gap length 0.16mm the coil is connected directly to a 120-V 60-Hz voltage
source. Neglect the coil resistance and leakage inductance .Assuming
the coil reluctance to
be negligible ,calculate the time-averaged force acting on
the core tending to close the air gap .How would this force very if the air-gap length were doubled? Sol:
先求出電感 L
0 N 2 Ac g
i 2 dL i 2 L 在 Ff id 2 dg 2 g
假設電流為均方根值 Irms=Vrms/wL 最後所需的力為 Ff ld
I rms 2
2 g L
V rms 2 2 0 N 2 Ac
115N
3.17 Fig. 3.31 shows the general nature of the slot-leakage flux produced by current I
in a rectangular conductor embedded in a rectangular slot in iron. Assume that the iron reluctance is negligible and that the slot leakage flux goes straight across the slot in the region between the top of the conductor and the top of the slot.
(a)Derive an expression for the flux density Bs in the region between the top of the conductor and the top of the slot.(b) Derive an expression for the slot-leakage sits crossing the slot above the conductor, in terms of the height x of the slot above the conductor, the slot width s, and the embedded length l perpendicular to the paper.(c) Derive an expression for the force f created by this magnetic field on a conductor of length l. In what directi on does this force act on the conductor?(d) When the conductor current is 850 A, compute the force per meter on a conductor in a slot 2.5cm wide. Sol:
part(a)
B
A NA
B S
part(b)
s B s A B s xl
0 xl s
0 i s
partc(c)
f
dW ' dx
0 li 2
0 li 2
2 s
part(d)
f
dW ' dx
f 18.1 N
2 s
m
3.18 A long, thin solenoid of radius r and height h is shown in Fig. 3.32. The magnetic
field inside such a solenoid is axially directed, essentially uniform and equal to H=Ni/h. The magnetic field outside the solenoid can be shown to be negligible. Calculate the radial pressure in newtons per square meter acting on the sides of the solenoid for constant coil current i=Io.
Sol:
0 H 2 0 r 02 N 2 2 Coil i W 2 2h '
0 r 0 N 2 2 I 0 f dr 0 h dW '
0 N 2 2 I 0 P 2 r 0 h 2h 2 f
3.22 The two-winding magnetic circuit of Fig.3.36 has a winding on a fixed yoke and
a second winding on a moveable element .The moveable element is constrained to motion such that the length of both air gaps remain equal. a.
Find the self-inductances of winding 1 and 2 in terms of the core dimension and the number of turns
b.
find the mutual inductance between the two winding.
c.
Calculate the coenergy W’fld(i1,i2)
d.
Find an expression for the force acting on the movable element as a function of the winding currents.
_
+
Moveable element
2
i1
Winding 2, N2 turns
u
g0
g0 Cross-sectional area A
u
Yoke
i2 1
+
_
Winding 1,N1 turns
SOL: (a)
F N i N 1 i1 N 2 i 2
u A N i N 1 i1 N 2 i2 0 R 2 g 0
線圈1之合成磁通鍊可表示為 1 N 1 N 1
2
u 0 A u A i1 N 1 N 2 0 i 2 2 g 0 2 g 0
上式亦可寫成 1 L11 i1 L12 i 2
其中 L11為線圈 1 的自感表示式為下
L11 N 1
2
u 0 A 2 g 0
則 L12為線圈 1與線圈 2 互感表示式為下
L12 N 1 N 2
u 0 A 2 g 0
同理線圈2之合成磁通鍊可表示為
2 N 2 N 2
2
u 0 A 2 g 0
i 2 N 1 N 2
u 0 A 2 g 0
i1
上式亦可寫成
2 L22 i 2 L21i1 其中 L22為線圈 2 的自感表示式為下
L22 N 2
2
u 0 A 2 g 0
(b)
L12 N 1 N 2
u 0 A 2 g 0
i2
(c) W
'
1
fld
1
L i L i L12 i1i2
u 0 A
u 0 A
4 g 0 4 g 0
2
2 11 1
2
2 22 2
( N 12 i12 N 22 i 22 2 N 1 N 2 i1i 2 )
N 1i1 N 2 i2 2
u 0 N 12 A 4 g 0
i 2 1
u 0 N 22 A 4 g 0
i 22
2u 0 N 1 N 2 i1i 2 4 g 0
(d) ' fld
W
' W fld
g 0
u 0 A
i1 ,i2
(4 g 0 ) 0 4u 0 A N 1i1 N 2 i2
2 N i N i 1 1 2 2 2
4 g 0
(4 g 0 ) 2
2
3.24 Two windings, one mounted on a stator and the other on a rotor,have self-and
mutual inductances of L11=4.5H
L22=2.5H
L12=2.8cos H
where is the angle between the axes of the windings. The resistances of the windings may be neglected. Winding 2 is short-circuited, and the current in winding 1 as a function of time is i1=10sinwt A. a. Derive an expression for the numerical value in newton-met ers of the instantaneous torque on the rotor in terms of the a ngle . b. Compute the time-averaged torque in newton-meters where 45 0 c. If the rotor is allowed to move ,will it rotate continuously or will it tend to cone to rest ? If the latter, at what value of 0 sol : (a) 1 1 W ' fld L11i12 L22 i22 L12i1i2 2 2
W ' fld T fld 2.8i1i2 sin N m 又因 為線圈2短路所以
2 L12 i1 L22i2 0 L12
i2 (
L22
)i1 (
2.8 cos H 2.5 H
)i1 1.12 i1 cos
題目Ｉi1 10 sin wt , T fld 3.14i12 sin cos 314 sin 2 ( wt ) sin cos 又因sin 2 wt T fld
1 - cos2wt
, sin2 2sin cos 所以 2 78.5(1- cos(2wt))sin(2 ) N m(1)
(b) 將 45 0 帶入(1)上式 T fld 78.5 N m
(c )如果轉子停止運轉 , 並且 為正向角 Tfld 0
得 90 or 270 0
d Tfld
and
d 0
0
3.25
Fig 3.37 A loudspeaker is made of a magnetic core of infinite permeability and circular symmetr y，as shown in Fig 3.37a and b. The air-gap length g is much less than the radius r 0 of the central core. The voice coil is constrained to move only in the x direction and is attached to the speaker cone ，which is not shown in the figure. A constant radial magnetic field is produced in the air gap by a direct current in coil 1 ， i1 I 1 . An audio-frequency
signal i2 I 2 cos t is then applied to the voice coil. Assume the voice coil to be of negligible thickness and composed of N 2 turns uniformly distributed over its height h. Also assume that its displacement is such that it remains in the air gap（ 0 x l h ）
（a）Calculate the force on the voice coi l，using the Lorentz Force Law （Eq.3.1） （ b）calculate the self-inductance of each coil. （c）calculate the mutual inductance between the coils. (Hint：Assume that current is applied to the voice coil ，and calculate the flux linkages of coil 1. Note that these flux linkages very with the displacement x) ' . （d）calculate the force on the voice coil from the coenergy W fid
Sol:
（a） g << r 0 B r , 1
φ
A
f B l
μ 0 N 1
g1 f z 2 r 0 N 2 Br ,1i2
2 r 0 0 N 1 N 2 g
i1i2
N 2 0 Ag N 2 （ b） L g R g 0 A g N 2
L22
2 r 0 0 N 22 g
(l x
2h 3
L11
2 r 0 l 0 N 12 g
)
（c） L12
2 r 0 0 N 1 N 2 g
( x
h 2
l )
（d） w fld
r 0 0 N 22 2 2 r 0 0 N 1 N 2 d 1 1 2 2 L11i1 L22 i2 L12 i1i 2 i2 i1i 2 dx 2 2 g g
將 i1 I 1 , i2 I 2 cos t 帶回上式 w fld
r 0 0 N 22 g
I 22 cos 2 t
2 r 0 0 N 1 N 2 g
I 1 I 2 cos t
3.29
Fig 3.40
Figure 3.40 shows a circularly symmetric system in which a moveable plunger
（constrained to move only in the vertical direction ） is supported by a spring of spring constant K=5.28 N/m. The system is excited by a samarium-cobalt permanent-magnet in the shape of a washer of outer radius R3 ，inner radius R2 ，and thickness tm . The system dimension are ： R1=2.1cm
R2=4cm
R3=4.5cm
h=1cm
g=1mm
tm=3mm
The equilibrium position of the plunger is observed to be x=1.0mm
（a）Find the magnetic flux density Bg in the fixed gap and Bx in the variable gap （b）Calculate the x-directed magnetic force pulling down on the plunger （c）The spring force is of the form f spring K X 0 x . Find X 0
（a） B g 0 H g R 1.05 0
B x 0 H x
H c' 712 KA
B m R ( H m H c )
m
0 R1 ( H c t m ) 0562T B g 2 0 R12 ht m 2hx gR1 2 2 ( R R ) R 3 2
2h B x B g 0.5 3 T 5 R 1 （ b）
NAB
2 2 2 hR N 0 1 N 2 R1h B g i L i 2 2 0 R1 ht m 2hx gR1 2 2 R ( R3 R2 ) f fld
i 2 dL 2 dX
2 0 hR1 Ni 2
f K X 0 x
f K
2mm
2 0 hR1 H c t m 2
2
2 0 R ht m 2hx gR1 2 2 R ( R3 R2 ) 2 1
（c）
X 0 x
2
2
2 0 R ht m 2hx gR1 2 2 R ( R3 R2 ) 2 1
2
0.0158 N
3.30 The plunger of a solenoid is connected to a spring. The spring force is given by
f K 0 0.9a x ,where x is the air-gap length. The inductance of the solenoid is of the form L L0 1 x / a , and its winding resistance is R. The plunger is initially stationary at position x 0.9a when a dc voltage of magnitude V 0 is applied to the solenoid. a. Find an expression for the force as a function of time required to hold the plunger at position a / 2 b. If the plunger is then released and allowed to come to equilibrium, find the equilibrium position X 0. You may assume that this position falls in the range 0 X 0 a . Sol :
part (a):
假如柱塞固定在
x 0. 9a ，所以電感 L L0 1 x / a 0.1 L0
因此
i(t )
V 0 R
e t /
, L / R
力的表示式為 f fld
i 2 dL 2 dx 2
V 0 t / e R d L0 1 x / a 2
dx
2
V 0 t / e R L0 2 a
L0 V 0
2
e 2t / 2a R
part (b):
平衡位置 X 等於 0
L V X 0 0.9a 0.9a 0 0 K 0 2aK 0 R f
2
3.31 Consider the solenoid system of Problem 3.30. Assume the following parameter
values: L0 4.0 mH
a 2.2 cm R 1.5 K 0 3.5 N / cm
The plunger has mess M 0.2 Kg . Assume the coil to be connected to a dc source of magnitude 4A. Neglect any effects of gravity. a. Find the equilibrium displacement X 0. b. Write the dynamic equations of motion for the system. c. Linearize these dynamic equations for incremental motion of the system around its equilibrium position. d. If the plunger is displaced by an incremental distance
from its
equilibrium position X 0 and released with zero velocity at time t 0 , find (i) The resultant motion of the plunger as a function of time, and ( ii) The corresponding time-varying component of current induced across the coil terminals. Sol :
part (a):
電流
i I 0 4 A
力等於
f
因此可求出
i 2 L0
2a
X 0 0.9a
16 4 10 3 2 2.2 10 2
f K 0
0.9 2.2
1.45 N 1.45 3 .5
1.56 cm
part (b):
移動的動態方程式為： M
d 2 x dt 2
f K 0 0.9a x
0.2
d 2 x dt 2
v I 0 R I 0
1.45 3.50.9 2.2 x 5.48 3.5 x N dL dt
I 0 R
v 6 0.182
L0 dx a dt
dx dt
part (c):
線性化
x X 0 x (t ) , v V 0 x (t )
0.2
d 2 x 2
dt
5.48 3.5 x
d 2 x dt 2
17.5 x
v 6 0.182
dx dt
v 0.182
part (d)
x (t ) c o s t m
17.5 4.18 rad / sec v (t ) 0.76 sin t V
d x dt
3.32 The solenoid of Problem 3.31 is now connected to a dc voltage source of
magnitude 6V. a. Find the equilibrium displacement X 0 . b. Write the dynamic equations of motion for the system. c. Linearize these dynamic equations for incremental motion of the system around its equilibrium position. Sol:
(a)
I 0
V 0 R
4 A
電流與問題 3.31 的電流相同，因此平衡點相同，故 X 0 (b)
V 0 iR
d dt
flux linkage
λ
L(x) i
dynamic equation:
V 0 iR
d
(Li) dt di dL iR L i dt dt di dL dx iR L i dt dx dt x di dL dx iR L0( 1 ) i a dt dx dt L x di dx iR L0( 1 ) ( 0 )i a dt a dt
or 3
6 1.5i 4 10 ( 1
x
)
di
2.2 dt
(
1.5i 4 10 3 ( 1 0.455 x)
di dt
4 10
3
2.2
0.182 i
)i
dx dt
dx dt
(解答中為 40 是錯誤)
1.56 cm .
and d 2 x
f M
M
dt 2
d 2 x
K 0( 0.9a x)
f K 0( 0.9a x)
dt 2
i L0
(解答中為正號是錯誤)
2
2a
K 0( 0.9a x)
or
0.2
d 2 x
dt 2
0.2
d 2 x
i 2 4 10 3 2 2.2
3.5( 0.9 2.2 x)
0.9091 10 3 6.93 3.5 x
dt 2
(解答中為數據是錯誤) (c) The equation can be linearized by letting x X 0 x ' t and i I 0 i ' t . The result is
X 0 di' L0 dx 0 i R L0 1 I 0 a dt a dt '
or 0 1.5i 1.5 10 '
3
di ' dt
0.728
and
M
I L 0 0 i ' K 0 x' d t a
d 2 x' 2
or 0 .2
d 2 x ' 2
dt
0.727 i ' 350 x '
dx dt
3.33 Consider the single-coil rotor of Example 3.1. Assume the rotor winding to be
carrying a constant current of I 8 A and the rotor to 2 have a moment of inertia J 0.0125 kg m
.
a. Find the equilibrium position of the rotor. Is it stable? b. Writer the dynamic equations for the system. c. find the natural frequency in hertz for incremental rotor motion around this equilibrium position. Sol:
(a) Rotor current =8A Torque T T 0 sin T 0 2 IB0 Rl 2 8 0 .02 0.05 0.3 0.0048 N m The stable equilibrium position will be at 0 . (b) J
d 2 α dt 2
T 0 sin α
(c) The incremental equation of motion is J
d 2 α
T 0 α
dt 2
and the natural frequency is
T 0 J
0.0048
0.0125
0.62r a d / s c
Corresponding to a frequency of
2 f
f
2
0.62 2
0.099 Hz
