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英语学习电工学概论习题答案_第四章

How does 4-1. distinguish the 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.

In the modern control system is widely used in programmable control, it adopts programmable controller to complicated relay control logic conversion on the grounds of the central processor, input and output converter and user program for the amount of the switch control logic, the logic of the hardware software is realized. It not only overcomes the disadvantages of the traditional relay contactor control, but also can realize the function of numerical calculation, computer network communication, analog input and output.

Why there is no starting torque of single-phase asynchronous motor, how to solve the problem of 4-21.?

Answer: single-phase asynchronous motor of two to the opposite magnetic field with the rotor produce positive electromagnetic torque and reverse electromagnetic torque, they are similar to turn slip relationship with ordinary three-phase asynchronous motor, available figure 4.4.3 shows and curve said. Under the same rotating speed, the synthetic torque is calculated and the curve in the graph is obtained. Visible from the figure, the speed of the rotating speed in the 0 range, the synthetic torque >0, so that the motor is transferred. In the same way, the speed of the rotating speed in the range of 0, synthetic torque <0, so that the motor reverse. When the rotor is at rest, n=0, s=1, synthetic torque is zero, so the motor does not have the starting torque, and can not start on its own.

4-22. single phase induction motor only single phase pulsating magnetic field, there is no rotating magnetic field, why can make the rotor rotation?

A single phase induction motor of the two opposite magnetic field respectively with the rotor to generate forward electromagnetic torque and reverse electromagnetic torque. At the speed of 0 in the range, the synthetic torque >0, so that the motor is rotating. In the same way, the speed of the rotating speed in the range of 0, synthetic torque <0, so that the motor reverse.

4-23. capacitor split phase starting, capacitor split phase operation and shaded pole single-phase induction motors are used in what place? Which type of single phase induction motor performance is best?

A: capacitor split phase starting refers to the stator of a single-phase asynchronous motor, in addition to the original winding A (referred to as the working winding or the main winding), plus a starting winding B (secondary winding), both in the space difference. After the motor is switched on, the starting winding can be left in the circuit, and the centrifugal switch K or voltage and current type relay can be used to disconnect the starting winding and the power supply. Designed and manufactured by the former is called the capacitor running motor, it has a good operating characteristics, the power factor of about 0.9, the maximum power of about 0.75kW. According to the latter is made of a capacitor starting motor, it has a larger starting torque, the use of small pumps, fans, etc., the maximum power of more than 0.75kW. The cover pole type motor has the advantages of small starting torque, low loss, low efficiency, simple structure, reliable operation and low manufacturing cost, and is used for occasions with low starting torque

requirement, such as wind and fan.

The performance of single phase induction motor with capacitor phase splitting operation is the best.

Method for changing the direction of a single-phase asynchronous motor with capacitor split phase operation in 4-24.

Answer: as shown below, the structure of motor two winding ax, by exactly the same, can cross the main winding and the secondary winding, capacitor C connected to the winding between a and B, when the switch S is placed 1, series capacitor C and winding by, by starting winding, the current ahead of the motor is rotating, when the switch S is placed in position 2, series capacitor C and winding ax, the current ahead, and the motor rotates in the reverse direction.

Method for regulating speed of electric fan by 4-25. What kind of fan is the special fan speed governor?

Answer: by changing the stator winding tap to change the stator magnetic field strength, such as common table fan, stand fan, Hongyun fan. The external series is used for adjusting the adjustable inductance coil in order to reduce the terminal voltage of the motor, and the motor is directly connected with the power supply after the induction coil is removed, and the motor is running at full speed. The common ceiling and wall fan, to remote control the start stop and speed fan. With the special fan governor is a fan capacitor phase asynchronous motor (such as: ceiling and wall fan) matching.

4-26. DC motor which are several excitation methods? What kind of excitation is commonly used?

Answer: DC motor according to the classification of excitation: he excited motor excitation winding current by the other power supplied to the motor. The current self excitation motor excitation winding is composed of motor armature winding of the power supply. Self excitation motorCan be divided into three categories, shunt motor excitation winding and the armature parallel. On excited motor excitation winding and the armature series, compound excited motor with two excitation winding, a parallel with the armature, another with the armature series.

In practical work, the application of more is the shunt motor and the motor.

4-27. DC series motor has what characteristics? Why is it dedicated to the transmission of electric locomotive?

A: the wiring diagram of DC series motor is shown in figure 4.7.1. The mechanical properties are shown in figure 4.7.2. The excitation current and the armature winding are connected in series with the armature winding.

The armature current is equal, that is, the

I=== (4.7.1)

Phi obviously changes with magnetic flux motor armature current, the

When will the electromagnetic torque of T= phi and phi ~.

Relevant department. When is small, the magnetic circuit wiring diagram of unsaturated and 4.7.1 series motor

In direct proportion to T, T= is proportional to the square.

When a large, magnetic saturation is basically unchanged,

Then T=. This in general, figure 4.7.2 on mechanical characteristics of torque motor

The multiple of the increase is greater than the current increase. This

The nature is suitable for electric traction need to be on the occasion of heavy load starting.

4-28. DC motor in the operation of the armature current size and what factors? What is the speed of the motor and what factors?

A: the armature current is related to the magnetic flux and the load torque. When the flux is constant, the armature current is proportional to the load torque, which is a characteristic of the DC motor.

From knowledge, under the load torque is certain, change armature voltage u, magnetic flux phi, the armature circuit resistance, can change the rotational speed of the motor n.

How does 4-29. change the steering of a DC motor?

A: the direction of rotation of the DC motor depends on the force direction of the armature conductor in the magnetic field. To reverse the motor, as long as the change in the direction of the armature current direction or the direction of the current field, the two can only change the.

4-30. why does the excitation circuit can not be disconnected when the motor is in no load and load?

A: in operation, the excitation circuit of the shunt motor can not be disconnected. Otherwise, = 0, there is only a very small magnetic flux on the pole, when the motor is in no-load or light load operation, and will increase sharply and exceed the allowable value, so that the motor is damaged and dangerous.

4-31. DC motor should be used in what way? What will be the consequences if the shunt motor is directly connected to the power supply?

Answer: DC motor starting, the voltage is rated, the armature resistance is very small, then the starting current is very large, the value of up to 10 to 20 times the current rated current. Such a large starting current will not only have a great impact on the power supply, and will burn out the motor commutator and armature winding. Therefore, the DC motor is absolutely not allowed to start, the starting time must be in the armature circuit in series starting resistance, to limit the armature starting current in the range of 1.5 ~ 2.5 times the rated current.

4-32. DC motor which have several speed control methods? Which method is the most ideal speed performance?

Answer: DC motor with which several timing methods are (1) change the armature circuit resistance, (2). The change of magnetic flux, and (3). To change the power supply voltage (U). Due to reducing the power supply voltage, the mechanical properties of the same hardness, and can be in a wide range of uniform speed, so it is an ideal speed control method.

4-33. a shunt motor, rated power of known = 2.2kW, rated voltage = 220V, rated current = 12.5A, rated speed = 3000r / min, excitation power = 77W, sought to: (1) excitation current; (2) rated armature current; (3) rated torque; (4) E Dingyun efficiency.

Solution: (1) excitation current =/=0.35A

(2) the rated armature current ==12.5-0.35=12.15A

(3) rated torque =9550=7.0

(4) rated operating efficiency ==80%

4-34. a 15kW, 220V and shunt motor = 85.3%, = 0.2, = 44. How many times the current of the rated current is the direct starting current? If the starting current is limited to 1.5 times the rated current, how much resistance starter?

Solution: by =, =8A

=5+1100=1105A

Is about 138.1 times the rated current.

If the starting current is limited to 1.5 times the rated current, the

= 12A

=125=7A

Therefore, the resistance for starting rheostat.

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2424 34311515155b b b b V V R R V V R R --I = I =-+I = I = 1234 1243:b b b b b KCL V V V V R R R R V I =I +I +I +515-6- 5- = + + 求方程中2121+9+9 ==50k 100k 9 :=150k 100k b b b b b b b V V V V R R V V KCL V V 6-6-I = I = 6-+ =b :650KI+100KI 9=0 1 100KI=15 I=A 10k 1 =650k =1V 10k KVL V -+--? 习题 1.1 （a ） 5427x A I =+-= （b ） 10.40.70.3x A I =-=- 20.30.20.20.1x A I =-++= (C) 40.230 x ?I = =0.1A 60 30.20.10.3x A I =+= 2x 10?0.3+0.2?30 I = =0.6A 1.5 10.30.60.9x A I =+= 1.2 30.010.30.31A I =+= 4 9.610.319.3A I =-=

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第一章习题答案 A 选择题 1.4.1（A ） 1.4.2（C ） 1.4.3（C ） 1.4.4（B ） 1.5.1（B ） 1.5.2（B ） 1.6.1（B ） 1.6.2（B ） 1.8.1（B ） 1.9.1（B ） 1.9.2（B ）1.9.3 （B ） 1.11.1(A) 1.1 2.1(B) 1.12.3 (B) 1.12.4 (B) 1.12.5 (B) B 基本题 1.4.5 （1）略（2）元件1和2为电源，元件3，4和5为负载（3）（-560-540+600+320+180）*w=0 平衡 1.4.6 380/(2110/8+R)=8/110，所以R ≈3.7K Ω，W R =（8/110）2×3.7K ≈20W 1.4.7 电阻R=U/I=6/50*310-=120Ω， a ）图. 1.4.8 解：220/（R1+315）=0.35A ，得R1≈314Ω. 220/（R2+315）=0.7A ，得R2≈0Ω. 1.4.9(1)并联R2前，I1=E/( 0R +2R e +1R )=220/（0.2+0.2+10）≈21.2A. 并联R2后，I2=E/( 0R +2R e +1R ∥2R )≈50A. (2)并联R2前，U2=R1*I1=212V,U1=(2R e +1R )*I1=216V. 并联R2后，U2=(1R ∥2R )*I1=200V,U1=2R e +1R ∥2R =210V. (3)并联R2前，P=212*21.2=4.5KW. 并联R2后，P=200*50=10KW. 1.5.3 I3=I1+I2=0.31ｕA ，I4=I5-I3=9.61-0.31=9.3ｕA ，I6=I2+I4=9.6ｕA. 1.6.3 因为电桥平衡，所以不管S 断开还是闭合 ab R =5R ∥（1R +3R ）∥（2R +4R ）=200Ω. 1.6.4 解： a U =1U =16V,b U =＜[(45+5) ≈5.5]+45＞×16/＜[(45+5) ∥5.5] ∥5.5+45＞≈1.6. c U =（45+5）∥5.5×b U /总R ≈b U /10=0.16V ，同理d R ≈ c U /10=0.016V. 1.6.5 解：当滑动端位于上端时，2U =（R1+RP ）1U /（R1+RP+R2）≈8.41V. 当滑动端位于下端时，2U =R2*1U /（R1+RP+R2）≈5.64V. 所以输出范围为5.64-8.14. 1.6.6

电工学概论习题答案_第二章

2-1. 列出以下节点的电流方程式，并求出x I 。解：(a) 由KCL 得： 1234()()()0x I I I I I +-+-+-+= ∴2341x I I I I I =++-; (b) 由KCL 得： 39(5)()0x I ++-+-= ∴7x I A = (c) 由KCL 之推广得: 69()0x I ++-= ∴15x I A = (d) 由KCL 得： 1(2)(3)40x I +-++-+= ∴0x I = 2-2. 列出以下回路的电压方程式，并标出回路循环方向，求出U 。解：(a) 回路循环方向与I 方向相同，由KVL 得： U+(2S U -)+I R ?=0 ∴U = 2S U -I R ? (b) 回路循环方向与3I 、4I 方向相同，由KVL 得： 44I R ?+33I R ?+(U -)=0 ∴U = 44I R ?+33I R ? (c) 回路循环方向为逆时针方向，由KVL 得： U+(9-)+7=0 ∴U = 2V (d) 设回路电流大小为I ，方向为逆时针方向，由KVL 得： 2S U +I ?4+I ?6+(1S U -)+I ?3+I ?7=0 ∴I = 0.3A 因此，U=2S U +I ?4 = 2-3. 求出下列电路中的电流I 。解：(a) 由KCL ，流过2Ω电阻的电流大小为(9+I)，对2Ω电阻与10V 电压源构成的回路，由KVL ，得：(9)I +?2+(10-) =0, 因此，I ＝－4A ； (b) (2)I -?3+6＝0； ∴I = 4A; (c) 流过3Ω电阻的电流 1I ＝93÷＝3A 流过6Ω电阻的电流 2I ＝(6+9) ÷6＝2.5 A (d) 由KCL ，I + 3-3＝0 ∴I =0;

电工学2(试卷及答案)-第1套j

---○---○--- ---○---○--- 学院专业班级学号姓名 … ……… 评卷密封线 ……………… 密封线内不要答题，密封线外不准填写考生信息，违者考试成绩按0分处理 ……………… 评卷密封线 ………… 中南大学考试模拟试卷1 20 ~20 学年学期电工学II 课程时间100分钟 32学时， 2 学分，闭卷，总分100分，占总评成绩60 % 题号一二三四五六七八九十合计得分评卷人复查人一、填空题（每空1分，共20分） 1. 常见的无源电路元件有、和；常见的有源电路元件是和。 2. 正弦交流电的三要素是、和。 3. 普通二极管具有导电性。 4. 集成运算放大器具有和两个输入端，相应的输入方式有输入、输入和输入三种。 5. 在时间上和数值上均作连续变化的电信号称为信号；在时间上和数值上离散的信号叫做信号。 6. B A C B A F ++=)(＝ 7. 时序逻辑电路的输出不仅取决于的状态，还与电路的现态有关。 8. 下图所示是u A 、u B 两输入端的输入波形，输出波形F 为门的输出波形。得分评卷人 F u A t u B t

二、试求图所示电路中A 点的电位。（8分）三、用戴维宁定理计算图中的电流I 。 (本题10分) 四、试将下图所示电路化简成最简形式。 (本题8分) 得分评卷人得分评卷人得分评卷人 Ω1Ω5Ω3Ω22A 10V +-I

---○---○--- ---○---○--- 学院专业班级学号姓名 … ……… 评卷密封线 ……………… 密封线内不要答题，密封线外不准填写考生信息，违者考试成绩按0分处理 ……………… 评卷密封线 ………… 五、电路如图所示，已知u i =5sin ωt (V)，二极管导通电压U D ＝0.7V 。试画出u i 与u 0波形，并标出幅值(本题10分) 六、单管放大电路如图所示，Ω=k R B 400，Ω=k 5R C ，cc V =12V ，且电容21C ,C 足够大。晶体管的60=β，Ω=k 1r be 。试求：⑴计算该电路的静态工作点Q ；⑵画出微变等效电路，求电压放大倍数u A ?)/(i o U U ??=；⑶求解输入电阻i r 和输出电阻o r 。（14分）得分评卷人得分评卷人 C R B R CC V +1C 2C +-i U ?+-o U ?

大学电工学1试卷及答案(A)

2010-2011学年第二学期《电工学1》试卷（A ）一、单项选择题：在下列各题中，只有唯一正确的答案。将选择题答案填入下表中。(本大题分12小题, 每小题2分, 共24分) 1、在RLC 串联电路中，已知R =3Ω，X L = 4Ω，X C = 5Ω，则电路呈（）。 (a) 电阻性 (b) 电感性 (c) 电容性 2、用叠加原理分析电路，当某一电源单独作用时，其它电源处理的方法是 ( ) (a) 理想电压源短路，理想电流源开路 (b) 理想电压源开路，理想电流源短路 (c) 理想电压源短路，理想电流源短路 3、理想电压源的外接负载越大，则流过理想电压源的电流（） (a) 越大 (b) 越小（c ）不能确定 4、某RC 电路的全响应为t c e t u 2536)(--=V ，则该电路的零状态响应为（） (a) t e 2536--V (b) )1(625t e --V (c) t e 253-V 第5题图第9题图第10题图（a ）第10题图（b ） 5、电路如图所示，在开关S 断开时A 点的电位为（） (a) 6V (b)0 V (c) 3 V 6、两个交流铁心线圈除了匝数不同（N 1 = 2 N 2）外，其它参数都相同。若将这两个线圈接在同一交流电源上，它们的电流关系为 ( )。 (a) I I 12> (b) I I 12< (c) I I 12= 7、对称三相电路的有功功率?cos 3L L I U P = ，其中?角为（）： (a) 线电压与线电流之间的相位差 (b) 相电压与相电流之间的夹角 (c) 线电压与相电流的夹角 8、有一台星形连接的三相交流发电机，额定相电压为660V ，若测得其线电压为U A B =1143V ，U B C =660V ，U C A =660V (a) A 相绕组接反 (b) B 相绕组接反 (c) C 相绕组接反 9、在图示电路中，R = X L = 10Ω, 欲使电路的功率因数 = 1，则X C 的值为( )。 (a) 20Ω (b) 10Ω (c) 5Ω 10、R ,L 串联电路与电压为8V 的恒压源接通，如图(a)所示。在t =0瞬间将开关S 闭合，换路前电感没有储能。当电阻分别为10Ω，30 Ω，20Ω时得到的3条u R (t)曲线如图(b)，其中30Ω电阻所对应的u R (t)曲线是（）。 11、某三相电源的电动势分别为)16314sin(20 +=t e A V ，)104314sin(20 -=t e B V ，)136314sin(20 +=t e C V ，当t = 13s ，该三相电动势之和为 ( )。 (a) 20V (b) 2 20V (c) 0V 12、某三相异步电动机在额定运行时的转速为1440r/min ，电源频率为50Hz ，此时转子电流频率为（）。 (a) 50 Hz (b) 48 Hz (c) 2 Hz 二非客观题( 本大题12分 ) 电路如图，已知 E =10V 、I S =1A ，R 1 = 10Ω ，R 2 = R 3= 5 Ω，（1）用戴维宁定理求流过 R 2的电流 I 2；（2）计算理想电流源 I S 两端的电压 U S ；（3）判断理想题目一二三四五六七八总计得分题目 1 2 3 4 5 6 7 8 9 10 11 12 答案

电工学简明教程第二版答案(第一章)

第一章习题答案 A 选择题 1.4.1（A ） 1.4.2（C ） 1.4.3（C ） 1.4.4（B ） 1.5.1（B ） 1.5.2（B ） 1.6.1（B ） 1.6.2（B ） 1.8.1（B ） 1.9.1（B ） 1.9.2（B ）1.9.3 （B ） 1.11.1(A) 1.1 2.1(B) 1.12.3 (B) 1.12.4 (B) 1.12.5 (B) B 基本题 1.4.5 （1）略（2）元件1和2为电源，元件3，4和5为负载（3）（-560-540+600+320+180）*w=0 平衡 1.4.6 380/(2110/8+R)=8/110，所以R ≈3.7K Ω，W R =（8/110）2×3.7K ≈20W 1.4.7 电阻R=U/I=6/50*310-=120Ω，应选者（a ）图. 1.4.8 解：220/（R1+315）=0.35A ，得R1≈314Ω. 220/（R2+315）=0.7A ，得R2≈0Ω. 1.4.9(1)并联R2前，I1=E/( 0R +2R e +1R )=220/（0.2+0.2+10）≈21.2A. 并联R2后，I2=E/( 0R +2R e +1R ∥2R )≈50A. (2)并联R2前，U2=R1*I1=212V,U1=(2R e +1R )*I1=216V. 并联R2后，U2=(1R ∥2R )*I1=200V,U1=2R e +1R ∥2R =210V. (3)并联R2前，P=212*21.2=4.5KW. 并联R2后，P=200*50=10KW. 1.5.3 I3=I1+I2=0.31ｕA ，I4=I5-I3=9.61-0.31=9.3ｕA ，I6=I2+I4=9.6ｕA. 1.6.3 因为电桥平衡，所以不管S 断开还是闭合 ab R =5R ∥（1R +3R ）∥（2R +4R ）=200Ω. 1.6.4 解： a U =1U =16V,b U =＜[(45+5) ≈5.5]+45＞×16/＜[(45+5) ∥5.5] ∥5.5+45＞≈1.6. c U =（45+5）∥5.5×b U /总R ≈b U /10=0.16V ，同理d R ≈c U /10=0.016V.

电工学概论习题答案_第四章

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N P ＝3kW ，额定转速N n ＝2880r/min 。试求 (1)磁极对数；(2)额定时的转差率N s ；(3)额定转矩N T 。解：(1) 同步转速03000/min n r =，因此电动机磁极对数p 为1； (2) 00300028804%3000 N n n s n --=== (3) 9.55 N N N P T n ==9.95N m ? 4-4. 已知Y112M-4型异步电动机的技术数据为N P ＝4kW ，△接法，额定电压N U ＝380V ，N n ＝1440r/min ，额定电流N I ＝8.8A ，功率因数cos N ?=0.82，效率N η＝84.5%。试求(1)磁极对数; (2)额定运行时的输入功率1N P ; (3)额定时的转差率N s ; (4)额定转矩N T 。解：(1) 同步转速01500/min n r =，因此电动机磁极对数p 为2； (2) 1 4.73N N N P P kW η== (3) 00150014404%1500 N n n s n --=== (4) 9.55 N N N P T n ==26.5N m ? 4-5. 已知Y132M-4型异步电动机的额定功率N P 为7.5kW ，额定电流N I ＝15.4A ，额定转速

电工学试题及答案套

电工学练习题（A ）一、单项选择题：在下列各题中，有四个备选答案，请将其中唯一正确的答案填入题干的括号中。(本大题共5小题，总计10分) 1、图示电路中, 若电压源U S =10 V, 电流源I S =1 A, 则( ) A. 电压源与电流源都产生功率 B. 电压源与电流源都吸收功率 C. 电压源产生功率, 电流源不一定 D. 电流源产生功率, 电压源不一定 2、电路如图所示, U S 为独立电压源, 若外电路不变, 仅电阻R 变化时, 将会引起 ( ) A. 端电压U 的变化 B. 输出电流I 的变化 C. 电阻R 支路电流的变化 D. 上述三者同时变化 3、电路如图所示, 支路电流I AB 与支路电压U AB 分别应为( ) A. 05. A 与15. V B. 0 A 与1 V C. 0 A 与-1 V D. 1 A 与0 V 4、图示正弦交流电路中，已知R L C ==ωω1 ， i t 13245=+?cos()ωA ， i t 24245=-?cos()ωA ，则i 3为 ( )A. 5281 cos(.)ωt -? A B. 5281cos(.)ωt +? A C. 52cos ωt A D. 7245cos()ωt +? A 5、可以通过改变电容来调节RLC 串联电路的谐振频率，若要使谐振频率增大一倍，则电容应( ) A.大4倍 B.大2倍 C. 减至21 D. 减至4 1 二、填空题：（共20分）（要求写出计算过程） 1、电路如图所示，欲使电压源输出功率为零，则电阻R 为____Ω, 所吸收功率为______W 。

2、若图(a)的等效电路如图(b)所示, 则其中I S 为__________A, R S 为__________Ω。 3、 3、图示正弦交流电路中，已知&I R =∠-23 π A ，则&I L =____________A 三、非客观题 ( 本大题10分 ) 电路如图所示，应用KCL 与KVL 求电流I 、电压U 及元件X 吸收的功率。四、非客观题 ( 本大题15分 ) 写出图示电路端口的电压电流关系式，并画出其等效电路及伏安特性曲线。（15分） U 五、非客观题 ( 本大题10已知图示正弦交流电路中，R =C L ωω1==100，?∠=02R I &A 。求S U &和电路有功功率P 。六、图示对称三相电路中，已知电源线电压&U AB =∠?3800V ，线电流&.I A =∠-?173230A ，第一组星形联接负载的三相功率P =，cos 1?=0866.(滞后)，求第二组星形联接负载的三相功率P 。（15分）电工学练习题（B ）一、单项选择题：在下列各题中，有四个备选答案，请将其中唯一正确的答案填入题干的括号中。 (本大题共5小题，总计10分) 4、电路如图所示, 若电流源的电流I S >1 A, 则电路的功率情况为( )

电工学试题及答案

第 2 页共 19 页课程名称电工学A(上) 考试日期考生姓名学号专业或类别题号一二三四五六七八总分累分人签题分 30 8 10 8 12 10 10 12 100 得分考生注意事项：1、本试卷共页，请查看试卷中是否有缺页。 2、考试结束后，考生不得将试卷、答题纸和草稿纸带出考场。一、单项选择题：将下列各题中唯一正确的答案代码填入括号内(每小题2分) 得分评卷 1、( C )已知图 1所示电路中的 U S =10 V ， I S = 13 A 。电阻 R 1 和 R 2 消耗的功率由 ( )供给。 (a) 电压源 (b) 电流源 (c) 电压源和电流源 U U I R R S S S S 1 211Ω Ω . .. . +图1

2、( C )在图2中，N 是一线性无源网络。当 U 1 =1V，I 2 = 1A时，U 3 = 0V；当U 1 =10V，I 2 = 0A 时，U 3 =1V。则当U 1 = 0V，I 2 =10A时，U 3 = （）V A、 0 B、 1 C、－1 D、 3 图2 第 3 页共 19 页

福州大学 2007～2008学年第一学期考试A卷第 4 页共 19 页

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