电工学概论答案
【篇一：电工学概论习题答案_第二章(下)】
?)与u2?173sin(314t?120?)之和u?u1?u2，并写出它们的相量表示式。
?
解：u1?
?
30?
?
2
?
12
i)?
?
u2?
?
120?
?
?
(?
12
?
2
)??
u?u1?u2??90
?
因此，u?u1?u2?200sin(314t?90?)
2-32. 已知某负载的电流的有效值及初相为2a、45?，电压的有效值及初相为100v、?45?，频率为50hz，写出它们的向量表达式，并判断该负载是什么元件，元件参数为多少？
解：i?2?45??
??
45?
?
2??50i
u?100??
z?
因此，该负载是电容，元件阻抗大小为50。
??220?900v i??10?900a，(b) 2-33. 在50hz的单相交流电路中，
若(a) u
??220?900v i??10?450a，(c) u??220?900v i??10?1200a求三
种情况下电u
路中的r及x，并写出电路阻抗z的复数式。解：(a) z?
220i10i
?22
因此，电路中的r?22?，x?0
(b) z?
?
因此，电路中的r?
，x?
(c) z?
?11i
因此，电路中的r?，x?11?
2-34. 一个电感线圈接到20v的直流电源时，通过电流为0.5a。接到50hz、100v的交流电源时通过电流为1.25a，求线圈的r和l。解：r?
200.5
?40?
?1.25
?xl?4
2 又?xl?2?f?l ?l?0.2h
z?50?88.6i
设u?220?0?，则
?
?
i?
220?0
?
50?88.6i
?2.2(
12
?
2
)?2.2??60
2-36. 把上述的rl串联电路与一个电容器并联，若并联后的电路总
电流为原来rl串联电路的电流，求电容电流及电容量，作出相量图。解：如图所示
?
ic?2.2?
90?90
??
xc?
1
?
而xc?
2?f?c
，故c?
12?f?xc
?55.2uf
?
1
?c
?
12?f?c
?
?50?
设u?220?0，则
?
ir?
220?050220?0?50i
?
?
?
225225
?
??0
?
?
ic?
?
?
?
225
?90
?
i?ir?ic=
225
i45
?
2-38. 把上述的r与c并联电路与一电感元件串联，接在u?220v，
50hz的交流电源上，测得rc并联电路的端电压正好也是220v，求
电感元件两端的电压及电感量，作出向量图。解：z?
220225?225i
?25?25i
由向量图可知，xl?50?，又xl??l?2?f?l，故l?
2-39. 有两台单相交流电动机（电感性负载）并联在u?220v的电源
上所消耗的功率及功率因数为p1?1kw，cos?1?0.8，p2?0.75kw，cos?2?0.6，求每台电动机的电流，无功功率、视在功率，以及总有功功率，总无功功率，总视在功率，总电流、总功率因数。 ?2解：
由于p1?ui1cos?1p2?ui2cos
8 因此电流 i1?5.68a i2?5.6a
xl2?f
?0.159h
?2?1000 var无功功率 q1?ui1sin?1?750varq2?ui2sin
视在功率 s1?ui1?1250v?as2?ui2?1250v?a 总有功功率
p?p1?p2?1750w 总无功功率 q?q1?q2?1750var 总视在功率
s?
?2474.87v?a
a总电流i?s/u?11.25
总功率因素 cos??0.707
2-40. 在40w日光灯电路中已知电源为220v、50hz灯管端电压为100v，功率消耗为40w，
镇流器功率消耗为8w，灯管电流为0.4a，求(1)电路的总有功功率，视在功率和无功功率，功率因数。(2)镇流器的无功功率，视在功率
及端电压。(3)作出相量图。解：
l
电路如上图所示，其中l、r为镇流器电阻电感，r1为日光灯电阻
(1) 电路的总有功功率为 p = 40 + 8 = 48w
总视在功率为s?ui?220?0.4?88v?a
总无功功率为q?
?73.76var
功率因素为cos??p/s?0.54 (2) 镇流器的无功功率qz?q?73.76var
视在功率为sz?
?
?76.86v?a
端电压uz?sz/i?192.15v
(3) 向量图
ul
i
2-41. 上述电路中欲使功率因数提高到0.9，应并联多大的电容？
解：并联电容后无功变化，有功不变
变化后的无功功率为q?48?0.9?23.25var 变化量为q
变?q?q?50.51var xc?u/q变?958.2
c?
12?f?xc
?3.32uf
2
解：该负载需要的无功功率及视在功率分别为：
p?uicos??90kw
q?uisin??90000?0.6?0.8?
120kw s?
?150kv?a
若要求该电网的功率因数提高到0.9，则需要并联电容，其大小为
c?
p
?u
2
(tan?1?tan?2)?
900002??50?220
2
(1.333?0.484)?5.03mf
2-43. 解释rlc串联电路中的串联谐振现象，串联谐振有哪几个特征？通过哪些参数调节可以出现串联谐振？
解：在rlc串联电路中，当xl?xc时，电路两端的电压与电流同相，即为串联谐振
串联谐振有以下特征： (1)
电路的阻抗|z|?
|z|为最小值。
(2) 由于电源电压与电路中的电流同相，因此电路对电源呈电阻性。
(3) 由于xl?xc，有ul?uc。
(4) 串联谐振电路对频率有选择性，可在无线电选择接收信号频率时
加以利用。由谐振条件xl?xc有2?fl?
12?fc
，因此通过调节l、c、f可出现串联谐振。
2-44. 解释电感线圈与电容并联电路中出现的并联谐振现象？此种
谐振现象的最主要特征是什么？其谐振频率与哪些参数有关？
解：在电感线圈与电容并联电路中，当电路端电压与电流相位相同，则电路出现并联谐振现象。并联谐振的主要特征有：
(1) 并联谐振时，电路的阻抗z?流i接近最小。
(2) 并联谐振时，由于电源电压与电流相同时(??0)，因此，电路对
电源呈电阻性，即|z|相当于一个电阻。
(3) 在谐振时电感线圈和电容器的电流近似相等，而比总电流大许多倍。
(4) 如并联谐振电流改由恒流源供电，当电源为某一频率时电路发生
谐振，电路的阻抗为最大，这样起到了选频的作用。
lrc
接近最大，在电源电压u一定时，电路中的总电
【篇二：英语学习电工学概论习题答案_第四章】
cage form and the winding pattern from the structural characteristics of the three-phase asynchronous motor?
answer: rotor winding induced electromotive force is generated, currents and electromagnetic torque, the structure type with cage and wound rotor two, squirrel cage rotor of
each rotor slot insert a piece of copper conducting bar, in an extended at both ends of the core at the notch, with two short circuit copper rings respectively, the all guide bar ends are welded together. if the core is removed, the overall shape of
the winding is like a cage, so it is called a cage rotor. the rotor winding and the stator winding is similar to that of the
insulated conductor embedded in the rotor slot and connected into a star three-phase symmetric winding, winding three
outlet terminals are respectively connected to the rotor shaft three slip ring (ring and ring, the ring and the shaft are
insulated from each other), the carbon brush to draw out the current.
4-2. how to change the three-phase induction motor to change?
answer: any of the two will be connected with three-phase power supply connected three wire switch, three-phase asynchronous motor direction change.
4-3. known a three-phase cage induction motor rated power = 3kw, rated speed = 2880r/min. find the number of pole pairs (1);
(2) when the rated slip ratio; (3) rated torque.
solution: (1) synchronous speed, so the logarithm of the motor pole p is 1;
(2)
(3) =9.95
technical data of 4-4. known y112m-4 type asynchronous motor = 4kw, delta connection method, rated voltage = 380v, = 1440r / min, rated current = 8.8 a, power factor = 0.82,
efficiency = 84.5%. find the number of pole pairs (1); (2) input power during normal operation; (3) when the rated slip ratio; (4) rated torque.
solution: (1) synchronous speed, so the logarithm of the motor pole p is 2;
(2)
(3)
(4) =26.5
4-5. known y132m-4 type asynchronous motor rated power
7.5kw, rated current = 15.4a, rated speed = 1440r / min, rated voltage = 380v, rated power factor of 0.85 and rated efficiency = 0.87, starting torque and rated torque = 2.2, the starting current and rated current = 7.0, the maximum torque / rated torque = 2.2. tries (1) rated power input; (2) rated torque; (3) rated slip; (4) the
starting current; (5) starting torque; (6) the maximum torque.
solution: (1) rated input power
(2) =49.74
(3)
(4) =7=107.8a
(5) =2.2=109.43
(6) =2.2=109.43
4-6. in the three-phase asynchronous motor start instant (s = 1), why the rotor current is large, and the power factor of the rotor circuit is small?
a: the electric motor is connected to the power supply, the induction electromotive force and the induced current of the rotor circuit are the largest, which is called the starting current or the blocking current. in general, the starting current of the small and medium sized three phase induction motor is 5 to 7 times of the rated current. although the motor starting current is very large, but the power factor of the rotor is very low, in fact, the starting torque of the motor is not large.
4-7. if there is a large number of asynchronous motors in the three-phase power grid at the same time, will be the impact of the grid voltage? why?
a large number of asynchronous motors start at the same time, the power grid will have a larger voltage drop. because asynchronous motor starting current is 5 ~ 7 times the rated current, if in the three-phase power grid has a large number of asynchronous motor start at the same time, will produce a large starting current, large starting current will in power line produced larger voltage landing, transformer power supply
with other work load.
4-8. three-phase induction motor torque characteristics of the shape of the curve by what factors?
answer: electromagnetic torque can be expressed as
when the s is very small, can be neglected, t is proportional to s; when the s is closer to 1, can be omitted, t is inversely proportional to s. so on the torque characteristic curve (t ~ s) as shown in figure.
t ~ s curve
at that time, said the motor at the rated operating state, at this time the rotor speed for the rated speed, torque for the rated torque, the output of the mechanical power for the rated power. at that time, the motor is in critical condition. at the same time, the electromagnetic torque is the largest value can by calculating t ~ s curve extremum method, to obtain the critical slip. visible and directly proportional to, but has nothing to do with; regardless of is proportional to. only in this way can by changing (wound rotor circuit external rheostat) and reduced
to change t ~ s curve shape, as shown in 4.2.3 and 4.2.4.
figure t increases the 4.2.3 ~ s curves of 4.2.4 to reduce the t ~ s curve
4-9. why should adopt the measure of reducing blood pressure? what is the effect of reducing the starting torque of the motor? under what circumstances can the buck start?
answer: reduced voltage starting of the purpose is to reduce the motor starting when starting the effects of electric current on the grid, the method is in starting lower voltage in the stator winding of the motor, to be motor speed is close to the stable, then the voltage returns to normal value. due to the motor torque and voltage is proportional to the square, so reduced voltage starting torque will also decrease accordingly, current step-down start the main method for star - delta (y - delta) change to connect to start and its applicable condition is normal operation of the stator winding is delta connected squirrel cage asynchronous motor.
what are some of the main methods of 4-10. reduction?
answer: the start method (1) direct starting, (2) y - delta change is connected with a step-down starting, (3) soft starting method. (4) rotor series resistance starting. starting mainly y - delta switching starting and soft start method.
why does the series resistance in the rotor circuit of 4-11. wound rotor type induction motor improve the starting performance and speed performance of the induction motor?
answer: winding type motor can be used in the rotor circuit in series resistance starting method. this can not only limit the starting current, but also increase the starting torque. so we need to start production of mechanical torque is large, such as cranes, forging machines and other commonly used linear motor drive around. after the motor is started, the starting resistance is removed by the step with the increase of the rotating speed.
4-12. three-phase induction motor in the normal operation, if the rotor suddenly stuck, what will be the consequences of the motor?
answer: three-phase asynchronous motor in normal operation, if the rotor suddenly stuck (stall), motor current increased immediately for several times of the rated current, if no protective measures promptly cut off the power supply, the motor will be burnt.
4-13. three-phase asynchronous motor in normal operation, if suddenly there is a phase power, the motor can run? if the motor is under heavy load, what is the change of the motor current?
answer: three-phase asynchronous motor in normal operation, if suddenly there is a phase power, the motor can run. if the motor is overloaded in case of slip increases, the induced electromotive force increases, the motor current increases. long run may burn the motor.
4-14. three phase induction motor which are several speed control method? which method is the best performance of the speed control?
answer: (1) the number of magnetic poles of the speed change, (2) change of the rotating slip speed, (3) variable frequency speed regulation are the main speed control method of three-phase asynchronous motor. in many speed regulation methods of ac induction motor, the performance of variable frequency speed regulation is the best, which is characterized by high speed range, good stability and high efficiency.
4-15. constant supply voltage in the case, if the rating for the error connected y type connection delta connection, what are the consequences? and as the amount of time for a false delta connection type y connection, and what are the consequences? answer: y type equivalent connecting resistance of the same size three resistor delta after connection for direct y connection type size 1/3. therefore, at a supply voltage of constant, if rated delta connected mistakenly connected type y connection, actual larger resistance, smaller current in the circuit, resulting in lack of power. if the amount?a delta connection error word type fast connection, the actual resistance is small, the current in the circuit becomes larger, it is possible to burn out resistance.
4-16. thermal relay in the three-phase asynchronous motor control circuit to play what role? can it play the role of short circuit protection?
answer: thermal relay is used for ac asynchronous motor overload protection, it uses the thermal effect of current and action. now the production of heat relay has three groups of heating elements and has a phase failure protection function, namely in the current serious imbalance or even disconnection
of a phase, the mechanical structure can also make a trip of
the gusset plate will move off contact disconnect and protects the motor because of the phase interruption caused by phase current overload and off the ring.
4-17. test to be able to directly control the control circuit of a three-phase cage asynchronous motor at two locations.
answer:
small capacity cage type motor direct starting control circuit, which used the air switch q, ac
contactor km, button sb, thermal relay fr and fuse fu and other control electrical appliances. work first air q switch closed,
into the power, when the press the start button (dynamic close contact points), the coil of the ac contactor km energized, moving core is attracted, the three main contacts (contact closed. the speed of the motor through electric starting. when the release button contact restored to original broken position, but due to and parallel ac contactor km of auxiliary contacts (contact) and main contact is closed at the same time, so contactor coil circuit is switched on, and make contacts of the contactor is maintained in a closed position. the auxiliary contact is called the self locking contact, which has the function of self locking for the motor to run for a long time. if the stop button is stopped, the circuit of the contact coil is cut off, the moving iron core and the contact point are restored to the open position, and the motor is stopped.
the control circuit has the functions of short circuit protection, overload protection, pressure loss and under voltage protection, besides the starting and stopping control functions of the motor.
4-18. test to be able to make a point of control, but also can be used as a direct starting control (two switch conversion) control circuit.
answer:
4-19. test of the positive inversion control circuit in the contact of the contacts of the contactor what role?
answer: if the two ac contactor at the same time, their main contacts are closed, will cause short circuit. in order to ensure the two contacts by not working at the same time, in the
control circuit, connected in series with the other contactor move off of auxiliary contact points and namely positive
rotation contactor of a dynamic fault of auxiliary contact points on connected in reverse rotation contactor coil circuit, and the overturning contactor of a dynamic breaking auxiliary point connected in positive rotation contactor coil circuit. these two are called interlocking contacts. in this way, when you press a positive starting button, positive rotation contactor coils, and the main contact, motor forward. at the same time, the interlocking contacts are disconnected from the coil circuit of the reverse contactor. therefore, even if the reverse starting button is mistaken, the contactor can not be moved. at the
same time the two contactor only allows one to work in a controlled role called interlocking or interlocking.
4-20. programmable controller has what function? what is the advantage of the traditional relay contactor control method when it is used in the motor control?
answer: the relay - contactor control is the traditional
industrial control mode, it relay contacts and coils according
to the control logic relation of certain connected control circuit, control contactor on-off, followed by the motor or solenoid device drive mechanism motion. however, because the relay itself accounts for a certain size and power consumption, often a failure, coupled with the fixed wiring, so that change control logic is difficult. so the application of complex control system reliability and flexibility are relatively poor.
【篇三：电工学概论习题答案第四章】
txt>4-1. 怎样从三相异步电动机的结构特征来区别笼型和绕线型？答：转子绕组的作用是产生感应电动势、流过电流和产生电磁转矩，其结构型式有笼型和绕线型两种，笼型转子的每个转子槽中插入一
根铜导条，在伸出铁心两端的槽口处，用两个短路铜环分别把所有
导条的两端都焊接起来。如果去掉铁心，整个绕组的外形就像一个
笼子，所以称为笼型转子。绕线型转子的绕组和定子相似，是用绝
缘导线嵌放在转子槽内，联结成星形的三相对称绕组，绕组的三个
出线端分别接到转子轴上的三个滑环（环与环，环与转轴都互相绝缘），在通过碳质电刷把电流引出来。
4-2. 怎样使三相异步电动机改变转向？
答：将同三相电源相联接的三个导线中的任意两根的对调一下，三
相异步电动机改变转向。
4-3. 已知一台三相笼型异步电动机的额定功率＝3kw，额定转速＝2880r/min。试求(1)磁极对数；
(2)额定时的转差率；(3)额定转矩。
解：(1) 同步转速，因此电动机磁极对数p为1；
(2)
(3) =9.95
4-4. 已知y112m-4型异步电动机的技术数据为＝4kw，△接法，额定电压＝380v，＝1440r/min，额定电流＝8.8a，功率因数=0.82，
效率＝84.5%。试求(1)磁极对数; (2)额定运行时的输入功率; (3)额定
时的转差率; (4)额定转矩。
解：(1) 同步转速，因此电动机磁极对数p为2；
(2)
(3)
(4) =26.5
4-5. 已知y132m-4型异步电动机的额定功率为7.5kw，额定电流
＝15.4a，额定转速＝1440r/min，额定电压＝380v，额定时的功率
因数=0.85，额定时的效率＝0.87，起动转矩/额定转矩=2.2，起动电
流/额定电流＝7.0，最大转矩/额定转矩=2.2。试求(1)额定输入功率；
(2)额定转矩；(3)额定时的转差率；(4)起动电流；(5) 起动转矩; (6)
最大转矩。
解：(1) 额定输入功率
(2) =49.74
(3)
(4) =7=107.8a
(5) =2.2=109.43
(6) =2.2=109.43
4-6. 在三相异步电动机起动瞬间（s＝1），为什么转子电流大，而
转子电路的功率因数小？答：电动机在接通电源瞬间，转子电路的
感应电动势和感应电流为最大，这称为起动电流或堵转电流。一般
中小型三相异步电动机的起动电流为额定电流的5～7倍。尽管电动
机起动电流很大，
但由于转子的功率因数很低，实际上电动机的起动转矩是不大的。
4-7. 假如在三相电网中有一大批异步电动机同时起动，将对电网电
压造成什么影响？为什么？答：在三相电网中有一大批异步电动机
同时起动时，电网会产生较大的电压降落。因为异步电动机起动电
流为额定电流的5～7倍，假如在三相电网中有一大批异步电动机同
时起动，会产生较大的启动电流，过大的起动电流
会在电源线路上产生较大的电压降落，影响同一变压器供电的其它
负载的正常工作。
4-8. 三相异步电动机转矩特性曲线的形状受哪些因素影响？
答：电磁转矩又可以表示为
当s很小时,可略去，t正比于s；当s接近于1时，，可略去，则t
与s成反比。所以上式作出的转矩特性曲线（t～s曲线）如图所示。 t～s曲线
当时，表示电动机处于额定运行状态，此时转子转速为额定转速，
转矩为额定转矩，所输出的机械功率为额定功率。
当时，电动机处于临界运行状态，此时电磁转矩为最大值，可以通
过求取t～s曲线极值的方法，求得临界转差率，。可见与成正比，
但与无关；而与成正比，与无关，这样可以通过改变（绕线型转子
电路外接变阻器）及降低来改变t～s曲线形状，如图4.2.3及图
4.2.4所示。
图4.2.3增大的t～s曲线图4.2.4降低的t～s曲线
4-9. 为什么要采用降压起动措施？降压起动对电动机的起动转矩有
何影响?在什么情况下才可以采用降压起动？
4-10. 降压起动有哪几种主要的方法？
4-11. 绕线型异步电动机转子电路中串联电阻为什么能改善异步电
动机的起动性能和调速性能？
答：绕线型电动机可以采用在转子回路中串电阻的起动方法。这样
既可以限制起动电流，同时又增大了起动转矩。因此对要求启动转
矩较大的生产机械，例如起重机、锻压机等常采用绕线型电动机拖动。电动机起动结束后，随着转速的上升将起动电阻逐段切除。
4-12. 三相异步电动机在正常运行中，如果转子突然被卡住，将会
对电动机造成什么后果？答：三相异步电动机在正常运行中，如果
转子突然被卡住(堵转)，电动机中的电流立即升高为额定电流的数倍，如果没有保护措施及时切断电源，电动机将被烧毁。
4-13. 三相异步电动机在正常运行中，如果突然有一相断电，电动
机是否还
能运行？若电动机是在重载情况下，电动机电流有何变化？
答：三相异步电动机在正常运行中，如果突然有一相断电，电动机
可以运行。若电动机是在重载的情况下，转差率增大，感应电动势
增大，电动机电流增大。长时间运转有可能烧毁电机。
4-14. 三相异步电动机有哪几种调速方法？其中哪种方法的调速性能
最好？
答：三相异步电动机调速方法主要有(1)改变磁极对数调速，(2)改变
转差率调速，(3) 变频调速。在交流异步电动机的诸多调速方法中，
变频调速的性能最好，其特点是调速范围大、稳定性好、运行效率高。
4-15. 在电源电压不变的情况下，如果额定时为△联接的误接成y
型联接，会造成什么后果？又如额定时为y型联接的误接成△联接，又会造成什么后果？
答：相同大小三个电阻△联接后的等效y型联接电阻为直接y型联
接大小的1/3。因此，在电源电压不变的情况下，如果额定时为△联
接的误接成y型联接，实际电阻偏大，电路中的电流变小，造成功
率不足。如果额定时为y型联接的误接成△联接，则实际电阻偏小，电路中的电流变大，有可能烧坏电阻。
4-16. 热继电器在三相异步电动机控制电路中起到什么作用？它能
否起短路保护作用？
答：热继电器是用于交流异步电动机过载保护的，它利用电流的热
效应而动作。现在生产的热继电器都具有三组热元件，且具有断相
保护功能，即在三项电流严重不平衡甚至断开一相的情况下，其机
械结构亦能使扣板脱扣将动断触点断开，保护了电动机因断相造成
二相电流过载而脱环的情况。
4-17. 试拟出能在两个地点分别对一台三相笼型异步电动机进行直
接起动控制的控制电路。答：
小容量笼型电动机直接起动的控制线路，其中用了空气开关q、交
流接触器km、按钮sb、热继电器fr及熔断器fu等控制电器。工作
时先将空气开关q闭合，引入电源，当按下起动按
钮(动合触点)时，交流接触器km的线圈通电，动铁心被吸合，将三
个主触点(动合触点)闭合，使电动机通电起动。当松开时，按钮触点
恢复到原来断开位置，但是由于与并联的交流接触器km的辅助触点(动合触点)和主触点同时闭合，因此接触器线圈的电路仍然接通，而
使接触器触点保持在闭合的位置。这个辅助触点称为自锁触点，起
到使电动机能长时间运行的自锁作用。如按停止按钮(动断触点)，则
将接触线圈的电路切断，动铁心和触点恢复到断开的位置，电动机
停车。
上述控制电路除具有对电动机实行远距离的起动、停车控制功能外，还具有短路保护、过载保护、失压和欠压保护作用。
4-18. 试
拟出既能作点动控制，又能作直接起动控制（二者用开关转换）的
控制电路。
答：
4-19. 试说明正反转控制电路中接触器的互锁触点起什么作用？
答：如果两个交流接触器同时工作，它们的主触点都闭合，必将造
成电源短路。为了保证两个接触器不同时工作，在控制电路中，串
接了对方接触器的动断辅助接触点和，即把正转接触器的一个动断
辅助接触点串接在反转接触器的线圈电路中，而把翻转接触器的一
个动断辅助点串接在正转接触器的线圈电路中。这两个动断触点称
为联锁触点。这样一来，当按下正转起动按钮时，正转接触器线圈
通电，主触点逼和，电动机正转。与此同时，联锁触点断开了反转
接触器的线圈电路。因此，即使误按了反转起动按钮，反转接触器
也不能动作。在同一时间里的两个接触器只允许一个工作的控制作
用称为互锁作用或联锁作用。
4-20. 可编程控制器具有什么功能？它用于电动机控制时较传统的
继电器接触器控制方法有什么优点？
答：继电—接触器控制是传统的工业控制模式，它把继电器触点及
线圈按一定的控制逻辑关系连成控制线路，控制接触器通断，然后
由电动机或电磁装置带动机构运动。但由于继电器本身占一定体积
并消耗电能，经常会出故障，加上接线固定，使变更控制逻辑比较
困难。所以应用复杂的控制系统中可靠性、灵活性都比较差。
现代控制系统中普遍采用可编程控制，它采用可编程控制器把复杂
的继电器控制逻辑转换为由中央处理器、输入、输出变换器及用户
程序进行处理的开关量控制逻辑，实现了硬件逻辑的软件化。不仅
克服了传统继电—接触器控制的弊病，而且在控制器中还可以实现
数值运算、与计算机联网通信、模拟量输入、输出等功能。
4-21. 单相异步电动机为什么没有起动转矩，怎样才能解决它的起
动问题？
答：单相异步电动机的两个转向相反的磁场分别同转子作用产生正
向电磁转矩和反向电磁转矩，它们与转差率的关系跟普通三相异步
电动机相似，可用图4.4.3所示的及曲线表示。在同一转速条件下，
求出合成转矩并作出图中的曲线。由图可见，在即转速在～0范围内，
合成转矩0，使电动机正转。同理，在即转速在～0范围内，合成转
矩0，使电动机反转。当转子静止时，n=0，s=1，合成转矩为零，
因此电动机没有起动转矩，不能自行起动。
4-22. 单相异步电动机中只有单相脉动磁场，没有旋转磁场，为什么
能使转子转动？
答：单相异步电动机的两个转向相反的磁场分别同转子作用产生正
向电磁转矩和反向电磁转矩。在即转速在～0范围内，合成转矩0，
使
电动机正转。同理，在即转速在～0范围内，合成转矩0，使电动
机反转。
4-23. 电容分相起动、电容分相运转及罩极式单相异步电动机分别
用于什么地方？其中哪一种类型的单相异步电动机性能最好？
答：电容分相起动指在单相异步电动机的定子内，除原来绕组a
（称为工作绕组或主绕组）外，再加一个起动绕组b（副绕组），两者在空间相差。电动机转起来之后，起动绕组可以留在电路中，也
可以利用离心式开关k或电压、电流型继电器把起动绕组与电源断开。按前者设计制造的叫做电容运转电动机，它具有良好的运行特性，功率因数大约0.9左右，其功率最大约为0.75kw。按后者制造
的叫做电容起动电动机，它具有较大起动转矩，使用于小型水泵、
风机等，最大功率可超过0.75kw。罩极式电动机起动转矩小、损耗大、效率低，但结构简单，工作可靠，制造成本低，使用于对起动
转矩要求不高的场合，如风扇等设备中。
其中电容分相运转单相异步电动机性能最好。
4-24. 说明改变电容分相运转单相异步电动机转向的方法。
答：如下图所示，电动机的二个绕组ax、by的结构完全相同，可
以互为主、副绕组，电容器c接在绕组a、b之间，当开关s置于1时，电容器c与绕组by串联，则by为起动绕组，其电流超前于，
电动机正向旋转，当开关s置于位置2时，电容器c与绕组ax串联，其电流超前于，电动机将反方向旋转。
4-25. 说明电风扇调速的方法。专门的风扇调速器是与哪种风扇相
配套的？
答：通过改变其定子绕组的抽头来改变定子磁场的强弱，如常用的
台扇、落地扇、鸿运扇等。在外部串联用于调速的可调电感线圈，
以降低电动机的端电压，当调速开关把电感线圈切除后，电动机直
接与电源接通，而全速运转。常见的吊扇和壁扇，人必须远距离控
制电扇的开停和调速。专门的风扇调速器是与采用电容分相运转异步电动机风扇（如：吊扇和壁扇）相配套的。
4-26. 直流电动机有哪几种励磁方法？常用的是哪种励磁？
答：直流电动机按励磁方式的分类：⑴他励电动机励磁绕组的电流由其它电源供给的电动机。⑵自励电动机励磁绕组的电流是由电枢绕组的电源供给的电动机。自励电动机又可分为三类，并励电动机励磁绕组与电枢并联，串励电动机励磁绕组与电枢串联，复励电动机具有两个励磁绕阻，一个与电枢并联，另一个与电枢串联。
在实际工作中，应用较多的是并励电动机和他励电动机。
4-27. 直流串励电动机有什么特点？为什么它专用于电力机车的传动？
答：直流串励电动机的接线图如图4.7.1所示，机械