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Electrical Machines – Questions and Answers

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    Information courtesy of ALSTOM. INDUCTION MOTOR CONTROL GEAR 1. What are the main functions of control gear? 1. To provide a means of starting and stopping the motor and, at the same time, of limiting the starting current if required. 2. To give adequate protection to the motor under all conditions. 3. To allow speed changing when required. 4. To provide means of braking the motor when required. 5. To reverse the direction of rotation when required.

    Protection of the motor must be automatic, but the other operations may be arranged to be under the control of an operator, or may be partly or fully-automatic. 2. What devices are required to give adequate protection to the motor? 1. Under-voltage release to prevent automatic restarting after a stoppage due to a drop in voltage or failure of the supply, where unexpected restarting of the motor might cause injury to an operator. 2. Overload relays for protection against excessive current in the motor windings – e. g. in the event of overload or failure of the motor. 3. Earth fault. 4.

    Single phase protection. 3. What provision must be made for short-circuit conditions in motor circuits? Since overload relays are not designed to operate and clear the circuit in the event of a short-circuit. Circuit-breaker or fuse protection of sufficient breaking capacity to deal with any possible short-circuit that may occur must be provided. 4. What are the usual forms of overload relay in motor-control gear? In small contactor starters, generally thermal relays, either of the ‘solder pot’ or bimetal type. With large contactors or oil switches, magnetic relays of the solenoid type with dashpots.

    Either type of overload relay may be used within intermediate sizes. 5. How do thermal relays work? The bimetallic thermal relay consists of a small bimetallic strip that is heated by an element connected in series with the supply. When the current rises above a preset value, the movement of the strip releases a catch which opens the trip contacts. In recent years more modern electronic relays are used which simulate the thermal overload. Many of these relays also incorporate a memory, i. e. simulates the temperature rise / cooling curve of the winding. 6.

    How does the magnetic overload relay operate? A solenoid connected in series with the supply contains a plunger whose movement is damped by a dashpot. When the safe current is exceeded, the solenoid pulls the plunger up – disconnecting the supply. The damping provided by the dashpot prevents unwarranted tripping on short-time overloads. 7. How many overload relays are required in the control gear? On three-phase supplies where the neutral point of the system is connected to earth, as is usually the case, three overload relays (one in each line) are necessary for complete protection.

    For 2-phase 3-wire and 4-wire supplies, two overload relays are required, one in each phase line, none being connected in any neutral or earth conductor. With single-phase motors one overload relay in any conductor except an earthed conductor or neutral. 8. What happens when one of the three lines supplying a three-phase induction motor becomes open-circuited? The motor, if already running, will continue to run as a single-phase motor on the remaining single-phase supply. The condition is called single-phasing.

    If the motor is loaded to more than about 30 per cent of full load, the currents in the motor windings tend to become excessive and overheating occurs. With one line broken, the motor will not start up and, due to the heavy standstill current, burn-out is likely unless the motor is quickly disconnected. 9. What currents flow in a single-phasing delta-connected motor? Assuming that supply line L1 is open circuited as shown, typical line and phase currents, given as percentages of normal full-load three-phase current, at various loads will be:- [pic]

    Fig. 22-Open-circuited supply line of delta-connected motor. |  |Percentage of full load current at | | |1/2 Load |3/4 Load |Full Load | |Lines L2 and L3 |98 |155 |250 | |Phase W |118 |187 |285 | |Phases U and V |55 |90 |147 |

    Thus, phase W connected across the two operative lines carries nearly three times normal current under single-phasing conditions at full load, while phases U and V, which are in series, carry more than full-load current. 10. What currents flow in a single-phasing star-connected motor? Assuming that line L1 is open-circuited as shown, the current flowing at full load in lines L2 and L3 and through the two phases in series will be of the order of 250 percent of normal full-load current, 155 per cent on 3/4-load and 98 per cent on 1/2-load. [pic] Fig. 3- Open-circuited supply line of star-connected motor. 11. Will normal overload relays trip on single-phasing? If correctly set, the normal overloads will trip when the motor is fully loaded due to the rise in current passing through the closed supply lines. With a delta-connected motor partially loaded, the rise in line current may not be sufficient to operate the overload trip and one phase may became excessively overheated. 12. What special protection can be provided against single-phasing? One method is to incorporate a combined overload and single-phase relay in the control gear.

    A typical relay of this type includes three overload relays with trip contacts so arranged that it will trip if the displacement of one overload element differs from that of the others. This type of relay will operate if single-phasing occurs at or near full load with the same time delay as on overload, but at light loads, the time delay for single-phase protection is longer. Another device is a phase-failure relay in the control gear. Its principle is based on the fact that the currents in the supply lines or the voltages between them at the motor terminals are unbalanced when the motor is single-phasing.

    The phase-failure relay may be of the current or voltage-operated type which trips out the line switch when one of the supply lines becomes open circuited. 13. What are the alternatives to the use of overload releases? Direct protection against overheating or burning-out of motor windings may be built into the motor. Built-in protectors may take the form of thermostats or thermistors embedded in the end windings of the stator while the motor is under construction. These devices are sensitive to the winding temperatures and are arranged in a suitable circuit so as to cause the motor to be switched off if the windings heat up excessively. 4. How are built in thermal overload protector arranged? On smaller motors LV mush winding motors, these detectors are embedded in the overhang of the winding. On the medium voltage motors these are placed in between bottom and top coils in the slot portion of the core. 15. How do built-in thermal overload protectors work? Thermistors are very small semiconductor devices whose resistance changes rapidly with temperature. Three thermistors are inserted in the end-windings of the stator, one in each phase, and are connected in series.

    The two thermistor terminals at the motor are connected to an electronic-amplifier-control unit in the starter, through which the tripping circuit of the starter is operated. The response of the thermistors to temperature change is extremely rapid, allowing this type of protection to be effective under all motor overload conditions. 16. Resistance temperature devices (RTD) This is a resistance which increases linearly with temperature rise. The most commonly used in motors is embedded in an epoxy glass type wedge which can be inserted between the upper and lower coils.

    The resistance is measured with an electronic amplifier control unit which is converted to temperature. This unit has adjustable settings to allow for alarm and trip with contacts which are then used in the motor starter circuit. 17. Thermocouples A thermocouple is two dissimilar metals which are joined together and with a change in temperature, creates a voltaic action. This gives out a milli-volt signal which is then measured with an electronic amplifier control unit converting the measurement to temperature. 18.

    When is direct-on-line starting used for three-phase squirrel-cage motors? It is usual for small LV machines; for larger motors it is often necessary to use other methods of starting in order to ovoid excessive starting currents. HV motors are usually DOL started. (since amps are low) 19. What are the connections for direct-on-line starters? The scheme of connections is merely three line leads in and three motor leads out. Direct-on-line contactor starters are designed round the basic circuit shown. An isolating switch may be incorporated in the starter.

    If reversing is required, two contactors one for each rotation, are required and are interlocked so that only one can close at a time. [pic] Fig. 24 – Basic circuit of direct-on-line contactor starter for squirrel-cage motor. 1 indicates the contactor coil and the contacts operated by it when it is energized. [pic] Fig. 25 – Connection diagram for direct-on-line contactor starter. A hand-operated oil switch with under-voltage trip coil may be used with larger motors. 20. What methods are employed to reduce the starting current of squirrel-cage motors?

    Where the starting conditions are light, the starting current can be lessened by some method of reducing the stator voltage when switching on. There are four ways of starting on reduced voltage: 1. Primary-resistance starting – introducing resistance between the supply and the stator windings. 2. Primary-reactor starting – introducing a reactor in series with the stator windings, usually connected in the star point. 3. Star-delta starting – connecting the stator windings in star for starting and in delta for running. 4. Auto-transformer starting – supplying the stator windings through tappings on an auto-transformer. pic] Fig. 26 – Primary-resistance and primary-reactor methods of reduced voltage starting for squirrel-cage motors 21. When is primary-resistance starting employed? Generally only for small motors on light-starting duty. The method is easily adjustable to suit the load and gives a smooth breakaway against low torque. If the resistance is adjustable, as in a faceplate starter, starting can be very smooth and this is useful for motors that must be started without any shock that might cause injury to the material being handled by the driven machine. 22.

    When is the primary reactor method of starting employed? Mainly for high-tension motors on very light-starting load where a fairly heavy starting current can be permitted e. g. boiler-feed pumps in a large power station. [pic] Fig. 27- Star-delta methods of reduced-voltage starting for squirrel-cage motors. Showing the switching sequence for plain and Wauchope methods. [pic] Fig. 28- Basic circuit of star-delta starter employing line contactor and hand-operated start-run switch. The start button must be depressed until the switch is moved into the ‘run’ position.

    I indicates contactor coil, line switch and maintaining contacts operated by the contactor coil. 23. When is the star-delta starter used? When the starting current has to be reduced and starting current and torque values one-third of those obtained with direct-on-line starting are suitable. It is necessary that the motor be designed to operate with the primary winding connected in delta, but with six terminals brought out to allow for connection in star during starting. The plain star-delta method is used for small and medium-sized motors on light-starting loads, e. . centrifugal pumps, fans having low inertia, line shafting and motor-generator sets. The Wauchope-type has the same uses but prevents the drop in speed when the stator is disconnected from the supply in changing from star to delta. Switching is done through resistances to maintain continuous line contact. This also obviates the momentary high current when switching from star to delta. [pic] Fig. 29- Connection diagram of air-brake hand-operated star-delta starter with line contactor. 24. What are the connections for a star-delta starter?

    Motors arranged for star-delta starting have six terminals – the two ends of each phase winding being brought out to terminals marked U1, V1, W1 and U2, V2, W2. These terminals are connected to similarly-marked terminals in the starter. The basic circuit of a typical hand-operated air-break or oil-immersed starter is shown in the diagram, the incoming supply being controlled by a line contactor. With the change-over switch in the start position, the motor windings are connected in star (U1, V1 and W1 together) and in the running position in delta (U2 to W1, V2 to U1 and W2 to V1).

    In starting the motor, the handle of the change-over switch is put into the start position, as indicated, and the ‘start’ button is pressed. This energizes the contactor coil which closes the triple-pole main switch and auxiliary switch (1). Note that the contactor coil cannot be energized unless the changeover switch has been placed in the ‘start’ position. When the motor has reached full speed, which is noticeable by sound, the handle of the change-over switch is moved to the ‘run’ position and the ‘start’ button is released.

    The motor is now directly connected to the line. In some star-delta starters, the overload units are by-passed in the ‘start’ position. A complete connection diagram of a hand-operated star-delta starter with this feature is also shown. Apart from the fact that the over-load units are brought into circuit only in the ~run~ position, the circuit is the same at the basic circuit. A fully-automatic star-delta starter has two contactors and a triple-pole line contactor with time-delay relay between ‘start’ and ‘run’ connections. 25. When is an auto-transformer starter used?

    When more flexibility is required for starting a squirrel-cage motor than is provided by the star-delta method, which is limited as far as starting torque is concerned. Auto-transformer starting permits the stator to be wound for running in star. The starting torque can be adjusted to suit the load by changing the voltage tapping on the auto-transformer. Both starting torque and current are reduced in the same proportion. It is used for motors of medium and large size on light starting loads (e. g. centrifugal pumps, fans, compressors and mills).

    Up to about 75kW the simple auto-transformer starter is employed; above this, the Korndorfer connection is recommended. 26. What does the simple auto-transformer starter consist of? The basic diagram is shown. The motor is started by connecting its primary to tappings on the starting transformer; then after a time delay, re-connecting direct to the supply. The winding on each limb of the auto-transformer usually has three taps, 60, 75 and 85 per cent of line voltage, but taps to give other percentages may be arranged as required.

    The auto-transformer may be used in conjunction with a contactor panel, or alternatively a hand-operated switch. [pic] Fig. 30 – Basic diagram of auto-transformer starter for squirrel-cage motor. [pic] Fig. 31 – Connection diagram for air-break hand-operated auto-transformer starter with line contactor. The accompanying illustration shows the wiring diagram of an auto-transformer starter consisting of a line contactor interlocked with a hand-operated change-over switch, three thermal or magnetic overload relays and an auto-transformer. 27. What are the connections for the Korndorfer system?

    The simple auto-transformer starter has the disadvantage that at the instant of transition from ‘start’ to ‘run’ the supply to the motor is interrupted. This means that the insulation may be stressed by high transient voltages. The Korndorfer method keeps the motor connected to the supply continuously by means of the connections shown in the diagram. On the first step (a), switches 1 and 2 close and the motor accelerates at a reduced voltage determined by the transformer tapping. On the second step (b), the star point of the transformer (switch 2) is opened so that the motor continues to run with part of the transformer winding in circuit.

    Next, this part is short-circuited by the ‘run’ contactor or switch (switch 3 closes) and finally the ‘start’ contactor or switch (1) is opened, as shown at (c). A fully automatic starter would comprise a triple-pole line contactor, start contactor, running contactor, three single-pole overload relays, auto-transformer with a set of links for tap-changing, a suitable timer, and ‘start’ and ‘stop’ pushbuttons. Fig. 32 – Switching sequence for auto-transformer starting by the Korndorfer method. [pic] a. Motor at reduced voltage from transformer. [pic] b. Motor with part of transformer winding in series. pic] c. Motor at full voltage. 28. What precautions should be observed when applying reduced voltage starting to a load with rising characteristic such as fans? If the specified starting current is too low, the motor may start correctly but not run fully up to speed. The result is that on changing over to the running or full-voltage condition a very high current may be taken, thus negating the low initial current. For this reason, even with fan drives, it is not desirable to pin the starting current lower than about 200 percent of full-load current. 29.

    What are the initial-starting line current and motor torque when star-delta starting? Both line current and torque are approximately one-third of the motor standstill values on full volts. 30. What are the initial starting line current and motor torque when starting with primary resistance or primary reactance? The initial starting line current is approximately equal to:- |Starting Current =  |applied voltage |x |standstill current with full volts | | |full voltage | | | The initial starting torque is approximately equal to:- Starting Torque =  |( |applied voltage |)2 |x |standstill torque with full volts | | | |full voltage | | | | 31. What are the initial-starting line current and motor torque when starting by auto-transformer? The initial-starting line current is approximately equal to:- |Starting Current =  |1. 1 ( |applied voltage |)2 |x |standstill current with full volts | | | |full voltage | | | | The factor of 1. in the above allows for the magnetizing current of the auto-transformer. The initial starting torque is approximately equal to:- |Starting Torque =  |( |applied voltage |)2 |x |standstill torque with full volts | | | |full voltage | | | | 32. Why are the above values of initial-starting current and torque approximate? Because the formulae given assume for simplicity that the standstill/reactance of a motor is constant at all voltages -that the short-circuit current varies in direct proportion to the applied voltage.

    Owing to magnetic saturation, particularly of the slot lips, the standstill reactance tends to be less on full volts than on reduced volts so the current and torque values tend to be rather less than those obtained by the formulae given. 33. How do the various methods of starting on reduced voltage compare as regards torque per ampere? Star-delta and auto-transformer methods have the advantage over primary resistance and primary reactor methods. 34. What mechanical methods of reducing starting current can be adopted?

    The starting duty can be reduced by fitting a centrifugal or other type of clutch which only picks up the load when the motor is well up to speed. 35. What is sequence starting? A system of starting by which several motors of similar rating are started in sequence off one starter in conjunction with interlocked switching. 36. How are slip-ring motors started? By first switching the supply on the stator winding with all the external rotor resistance in circuit across the slip rings and then cutting out the rotor resistance progressively as the motor speeds up until finally the rotor winding is short-circuited. 7. What is the usual arrangement of connections for a hand-operated slip-ring starter? Small slip-ring starters usually consist of a contactor for the stator circuit and a face plate-type starting resistance for the rotor circuit. The basic essentials are shown, the three wires from the stator going to slip-ring terminals R, S and T on the motor. An actual wiring diagram is also shown. The starter must be fitted with interlocks to ensure that the resistance is all-in when starting.

    With a contactor controlling the stator supply, interlocking is simply effected, as shown, through electrical contacts on the arm of the rotor starter, no current reaching the contactor coil ‘I’ unless the arm is in the starting position. The start button must be kept depressed until all resistance has been taken out; this ensures that the motor is not accidentally left running with some of the rotor resistance still in circuit. When the operating arm of the face- plate is in the ‘run’ position the start button is short-circuited. [pic] Fig. 33 – Bask diagram of contactor starter for slip-ring motor with faceplate-type secondary resistance. I’ indicates contactor coil and contacts operated by the contactor coil. If the motor is fitted with a device designed to lift the brushes and short-circuit the slip rings when the motor is up to speed, an interlock must be arranged in the control circuit to ensure that the brush-gear is in the starting position before the stator contactor can close. For larger motors, a stator oil switch is usual and may be used in conjunction with a liquid resistance or an oil-immersed grid resistance in the rotor circuit. 38. What are the essentials of a full-automatic stator-rotor starter?

    An automatic starter would include a triple pole contactor to control the stator circuit, together with rotor-resistance grids short-circuited by the necessary number of accelerating contactors, the last of which must be continuously rated to carry the full-load rotor current. Also required are the necessary number of overload relays and timers controlling the duration of the starting period. The number of timers and accelerating contactors correspond to the number of steps of rotor resistance that are provided. A wiring diagram -of an automatic slip ring motor starter with two steps of rotor resistance is shown.

    Control terminals are provided for pushbutton control from one or two positions, or alternatively, for automatic control (for use with thermostat, float-switch or similar switching) with or without a try-out switch. [pic] Fig. 34 – Connections of air-brake stator-rotor starter. [pic] Fig. 35 – Connections of automatic slip-ring motor starter. When the ‘start’ button is pressed (or the automatic switch closes), the control circuit is made through the coil of the stator contactor M. The stator contactor closes, connecting the stator to the line. At the same time the first timing relay is TR1 is energized.

    At this stage, the rotor is complete through the whole resistance since the accelerating contactors 2R and 3R are open. After an adjustable delay, the contacts of TR1 close, thus energizing the accelerating contactor 2R which short-circuits a portion of the rotor resistance and energizing the second timing relay TR2. When in turn the contacts of TR2 close the second and, in this case, final contactor 3R is energized and closes, short-circuiting the whole of the rotor resistance. The overload relays are in circuit during starting and running. For automatic (2-wire) remote control, hand-resetting overloads are essential. 39.

    How is speed control of a slip-ring motor effected? By introducing resistance into the rotor circuit similar to a starting resistance except that the heat losses in the resistance must be dissipated continuously. Unless the duty is intermittent, all except small sizes require some means of cooling the resistors. Grid resistances with a motor-driven fan may be used in conjunction with a drum controller. Alternative methods are oil-immersed resistances or a liquid resistance cooled by circulating water through cooling tubes. 40. What is liquid resistance? Insulated pots filled with a resistance solution of electrolyte, e. . caustic soda or washing soda. Plates connected to the slip rings dip into the pots and are shorted out in the full-speed position. Liquid starters and controllers are used for large motors. 41. What is the advantage of a liquid resistance for starting purposes? Resistance may be reduced continuously so that, with close control over the current as indicated on an ammetre, a very smooth start can be obtained. 42. What is a slip resistance? A fixed step of rotor resistance used to limit the current taken from the supply at the instant when peak load is applied to the motor.

    It is often desirable to do this on press drives, guillotines, etc. As the resistance value is small, it is usual to have a conventional starter so arranged that the last step of resistance is not cut out when the starting handle is right home. This last step of resistance is continuously rated. 43. What is meant by motors in synchronous tie? When the two slip-ring motors are required to run at the same speed, it is possible to do this by connecting their rotors together through the slip rings in conjunction with a single slip resistance.

    The starter for such a scheme includes a single rotor resistance, the last step of which is the continuously-rated slip resistance, and two-stator contactors, one for each motor. In order to limit the circulating current in the event of the motors being out of phase when started, a reactance is usually inserted in the interconnecting tie. The reactor is wound in two sections, and connected so that it is non-inductive to currents flowing through each half into the slip resistance but inductive to circulating currents between rotors.

    This reactance also assists load sharing when the two motors are driving a common load, as for example travel motors at opposite ends of an overhead crane. [pic] Fig. 36 – Connections of reversing-drum controller and three-phase slip-ring motor. The controller gives speed control by varying resistance in series with the rotor windings and also breaks the three-stator phase in the ‘off’ position. The moving-copper-contact rings are shown as thick horizontal lines, while the forward and reverse steps are indicated by the numbered vertical lines. The diagram below shows the connections of series limit switches when used.

    Electrical Machine Design Multiple Choice Questions of Electric Machine Design (196-210): 196. The materials in order of decreasing eddy current loss will be (A) iron, wood, aluminium (B) iron, aluminium, wood (C) aluminium, iron, wood (D) wood, aluminium, iron. Get Answer: (Show) 197. For a 5 kW DC motor the number of slots per pole should be (A) 4 (B) 8 (C) 12 (D) 16. Get Answer: (Show) 198. In a synchronous generator in order to eliminate the fifth harmonic the chording angle should be (A) 9° (B) 18° (C) 27° (D) 36°. Get Answer: (Show) 199. Inter poles in DC machines are provided to reduce (A) sparking (B) armature reaction C) iron loss (D) efficiency. Get Answer: (Show) 200. Skewing in the slots of an induction motor is provided to reduce (A) iron loss (B) noise (C) harmonics (D) temperature rise. Get Answer: (Show) 201. Sometimes a reactor is connected in series with a transformer to (A) improve regulation (B) control fault current (C) improve efficiency (D) improve power factor. Get Answer: (Show) 202. DC motor yoke is generally made of (A) wood (B) copper (C) aluminium (D) steel. Get Answer: (Show) 203. For the same rating, the cost of an induction motor as compared to that of a DC motor is (A) more (B) less (C) same (D) nearly the same.

    Get Answer: (Show) 204. In case of electrical machines, the intermittent rating as compared to its continuous rating is (A) more (B) less (C) same. Get Answer: (Show) 205. For the use of mush windings in 3? induction motors, the slot should be (A) semi-closed (B) open (C) closed (D) either (a) or (b). Get Answer: (Show) 206. A hunting sound is produced in a synchronous motor when (A) load fluctuates (B) supply frequency varies (C) both (A) and (B) above (D) none of the above. Get Answer: (Show) 207. The transformer noise is mainly because of (A) cooling oil (B) sinusoidal current (C) magnetic flux (D) all of the above. Get Answer: (Show) 08. Stampings in transformers are provided to reduce (A) hysteresis loss (B) eddy current loss (C) copper loss (D) all of the above. Get Answer: (Show) 209. Which loss occurs in the yoke of a DC machine ? (A) iron loss (B) copper loss (C) heat loss (D) no loss. Get Answer: (Show) 210. The losses occurring in the rotor of an induction motor are less than those in the stator because of (A) small diameter of rotor (B) less rotor frequency (C) slot skewing (D) none of the above. Get Answer: (Show) Get all answers at once: (Hide) 196. C —– 197. B —– 198. D —– 199. A —– 200. C —– 201. B —– 202. D —– 203. B —– 204.

    A —– 205. A —–206. C—– 207. C —– 208. B —– 209. D —– 210. B Electrical Machine Design State whether the following statements related to Electric Machine Design are true (T) or false (F): 1. Hard rubber is obtained by increased sulphur content and extended vulcanization treatment. Get Answer: (Show) 2. Ceramics are unaffected by chemical action. Get Answer: (Show) 3. B-type insulation is largely used in commutators, wedges of slots and capacitors. Get Answer: (Show) 4. H-type insulation is used in traction motors. Get Answer: (Show) 5. C-type insulation includes press board and vulcanized natural rubber.

    Get Answer: (Show) 6. Florentine, foulard and excelsior are silky fabrics. Get Answer: (Show) 7. PVC is obtained by combination of acctylene and hydrogen chloride in the presence of catalyst like oxides at a temperature of 323 K. Get Answer: (Show) 8. SF6 has high dielectric strength and is non inflammable. Get Answer: (Show) 9. Square section of the transformer core requires less copper wire than equivalent rectangular section. Get Answer: (Show) 10. Gum, starch, asphalt are useful for binding laminates of wood. Get Answer: (Show) 11. For liquid insulating materials the dielectric strength increases with rise in temperature.

    Get Answer: (Show) 12. Mineral insulating oils are obtained from the fractional distillation of crude petroleum. Get Answer: (Show) 13. Transformer oil is flammable 2nd sludging type oil. Get Answer: (Show) 14. Sulphur hexafluoride is used as insulating material in turbo-generators. Get Answer: (Show) 15. Hydrogen is used in large turbo-generators primary as a dielectric. Get Answer: (Show) 16. Oil varnish is obtained by dissolving shellac in methylated spirit Get Answer: (Show) 17. Lacquer is obtained by dissolving nitrocellulose in ketone. Get Answer: (Show) 18. The base material of black varnish is bitumen. Get Answer: (Show) 19.

    Smaller is the number of slots, the more is the distortion in the field flux. Get Answer: (Show) 20. The dielectric strength of PVC is around 30 kV/mm. Get Answer: (Show) 21. PVC can be safely used up to temperatures of 250 0 C. Get Answer: (Show) Get all answers at once: (Hide) 1. T —– 2. T —– 3. T —– 4. T —– 5. F —– 6. T —– 7. T —– 8. T —–9. T —–10. T —–11. F —– 12. T —–13. F —–14. F —– 15. F —– 16. F —– 17. T —– 18. T —– 19. T —– 20. T —– 21. F Electrical Machine Design State whether the following statements related to Electric Machine Design are true (T) or false (F): 22.

    Dielectric constant of P. T. F. E. is 5-6 at 60 Hz. Get Answer: (Show) 23. Phlogopite mica has greater thermal stability as compared to muscovite mica. Get Answer: (Show) 24. Nickel brass is used in arcing electrodes. Get Answer: (Show) 25. Electric machines up to 6. 6 kV are called low voltage machines. Get Answer: (Show) 26. Heat conducted by radiation is proportional to temperature difference. Get Answer: (Show) 27. The density of hydrogen is half that of air. Get Answer: (Show) 28. Hydrogen cooling of generators reduces windage loss. Get Answer: (Show) 29. Short time rating is more than its continuous rating in case of induction motors.

    Get Answer: (Show) 30. For petrochemical plants TEFC motors are used. Get Answer: (Show) 31. Chrysotile asbestos has low dielectric strength. Get Answer: (Show) 32. In order to transmit torque, the motor shaft should have good shear strength. Get Answer: (Show) 33. Critical speed of shaft should be far away from normal speed. Get Answer: (Show) 34. Ball and roller type bearings are also known as antifriction bearings. Get Answer: (Show) 35. Corona can possibly occur in machines designed for 440 V operation. Get Answer: (Show) 36. In simplex windings, there are as many number of parallel paths as there are number of poles.

    Get Answer: (Show) 37. In dc machines, the choice of number of poles directly depends upon the speed and supply frequency. Get Answer: (Show) 38. The lowest is the regulation of a transformer, the best it is. Get Answer: (Show) 39. Glass is an inorganic material prepared by fusion of different metallic oxides. Get Answer: (Show) 40. In cold rolling uni axial anisotropy is induced in the direction of rolling. Get Answer: (Show) 41. The converse of magnetostriction is called Villari effect. Get Answer: (Show) Get all answers at once: (Hide) 22. F —– 23. T —– 24. F —– 25. T —– 26. F —– 27. F —– 28. T —–29. T —– 30.

    F —– 31. T —– 32. T —– 33. T —– 34. T —– 35. F —– 36. T —– 37. F —– 38. F —– 39. T —– 40. T —– 41. T State whether the following statements related to Electric Machine Design are true (T) or false (F): 42. Lubrication of ball bearings used in electric motors is done-by grease. Get Answer: (Show) 43. In induction motors a large value of air gap flux density will increase/iron losses decrease/efficiency. Get Answer: (Show) 44. Power factor of small induction motors is around 0. 8 and that of large motors around 0. 9. Get Answer: (Show) 45. Polytetra-flouroethylene (PTFE) or Teflon is highly resistant to xidation. Get Answer: (Show) 46. In an induction motor, the noises mainly due to variation of leakage flux path of Zig-Zag leakage flux. Get Answer: (Show) 47. In an induction motors as the length of the air gap is increased, the reluctance variation due to slots will decrease. Get Answer: (Show) 48. To avoid synchronous cusps (Ss = no. of stator slots and Sr = no. of rotor slots) in an induction motor, Ss – Sr = ± p. Get Answer: (Show) 49. In induction motors, for good starting the rotor resistance should be small but for good running efficiency, the rotor resistance should be large. Get Answer: (Show) 50.

    Skewing is done to eliminate the effect of harmonics. Get Answer: (Show) 51. In induction motors, rotor slots give low reluctance and less magnetizing current. Get Answer: (Show) 52. The friction and windage loss in a properly designed induction motor should not exceed 5%. Get Answer: (Show) 53. A circle diagram is usually drawn for distribution transformers only. Get Answer: (Show) 54. The power factor of an induction motor depends upon its magnetizing current and short circuit current. Get Answer: (Show) 55. Harmonic losses, skin effect and pulsation losses in induction motor are accounted for as stray losses. Get Answer: (Show) 6. The phenomenon of skin effect occurs mostly in stator and rotor windings of a squirrel cage motor. Get Answer: (Show) 57. A high slip, high torque squirrel cage or slip ring motor is preferred for coal crushers. Get Answer: (Show) 58. In a synchronous motor when the winding is chorded by 300 it minimizes the effect of 5th and 7th harmonic. Get Answer: (Show) 59. A single conductor of large cross-section is not used because it results in excessive I2R loss due to conductor eddy currents due to pulsation of leakage flux. Get Answer: (Show) 60. Nichrome and constantan are basically insulating materials. Get Answer: (Show)

    Get all answers at once: (Hide) 42. T —– 43. T —– 44. T —– 45. T —– 46. T —– 47. T —– 48. F —– 49. F —– 50. T —–51. T —– 52. T —– 53. F —– 54. T —– 55. T —– 56. T —– 57. T —– 58. T —–59. T —– 60. F Synchronous Generator Multiple Choice Questions of Synchronous Generator (16-30): 16. As the speed of an alternator increases (A) the frequency increases (B) the frequency decreases (C) the frequency remains constant but power factor decreases (D) none of the above. Get Answer: (Show) 17. For an alternator when the power factor of the load is unity (A) the armature flux will have square waveform B) the armature flux will be demagnetising (C) the armature flux will be cross-magnetising (D) the armature flux will reduce to zero. Get Answer: (Show) 18. The driving power from the prime mover driving the alternator is lost but the alternator remains connected to the supply network and the field supply also remains on. The alternator will (A) get burnt (B) behave as an induction motor but will rotate in the opposite direction (C) behave as a synchronous motor and will rotate in the same direction (D) behave as a synchronous motor but will rotate in a reverse direction to that corresponding to generator action. Get Answer: (Show) 9. If the input of the prime mover of an alternator is kept constant but the excitation is changed, then (A) the active component of the output is changed (B) the reactive component of the output is changed (C) power factor of the load remains constant (D) power factor of the load changes from lagging to leading. Get Answer: (Show) 20. For 50 Hz system the maximum speed of an alternator can be (A) approximately 3600 rpm (B) approximately 3000 rpm (C) 3600 rpm (D) 3000 rpm. Get Answer: (Show) 21. Voltage characteristic of an alternator is shown in figure. Which curve represents the characteristics for leading power factor ? pic] (A) A (B) B (C) C (D) D. Get Answer: (Show) 22. In the above figure, the characteristic for unity power factor is represented by the curve maked (A) A (B) B (C)C (D )D. Get Answer: (Show) Questions 23 to 26 refer to the following data: In a 50 kVA, star connected 440 V, 4-phase 50 Hz alternator, the effective armature resistance is 0. 25 ohm per phase. The synchronous reactance is 3. 2 ohm per phase and leakage reactance is 0. 5 ohm per phase. 23. Full load output current at unity power factor will be (A) 65. 6 A (B) 55. 4 A (C) 45. 6 A (D) 35. 4 A. Get Answer: (Show) 24. Full load line voltage will be (A) 500 V (B) 471 V C) 450 V (D) 435 V. Get Answer: (Show) 25. No load line voltage will be (A) 600 V (B) 599 V (C)592V (D) 580 V. Get Answer: (Show) 26. Percentage regulation of the alternator is approximately (A) 55% (B) 45% (C) 35% (D) 25%. Get Answer: (Show) 27. In order that two alternators be put in parallel, which of the following factors should be identical for both ? (A) Voltage (B) Frequency (C) Phase sequence (D) All of the above. Get Answer: (Show) 28. When two alternators are running in parallel, their RKVA load share is changed by changing their ……………. while their kW load share is changed by changing their ………….. A) excitation, driving torque (B) driving torque, excitation (C) excitation, excitation (D) driving torque, driving torque. Get Answer: (Show) 29. Two-alternators are running in parallel. If the driving force of both the alternators is changed, this will result in change is (A) frequency (B) back emf (C) generated voltage (D) all of the above. Get Answer: (Show) 30. A three phase alternator has a phase sequence of RYB for its three output voltages. In case the field current is reversed, the phase sequence will become (A) RBY (B) RYB (C) YRB (D) none of the above. Get Answer: (Show) Get all answers at once: (Hide) 6. A —– 17. C —– 18. C —– 19. B —– 20. D —– 21. D —– 22. C —– 23. A —–24. B —–25. C —– 26. D —– 27. D —–28. A —–29. A —– 30. B 31. The armature reaction of an alternator influences (A) windage losses (B) operting speed (C) generated voltage per phase (D) waveform of voltage generated. Get Answer: (Show) 32. For the same power rating, a lower voltage alternator will be (A) more efficient (B) larger in size (C) operating at high rpm (D) more costly. Get Answer: (Show) 33. An alternator is supplying 10A to an inductive load at 220 V, while running at 1000 rpm.

    Now if the speed of the alternator is reduced to 750 rpm but the field current remains unchanged, the load current will become (A) 18 A (B) 13. 3 A (C) 10 A (D) 7. 5 A. Get Answer: (Show) 34. Dampers in a large generator (A) increase stability (B) reduce voltage fluctuations (C) reduce frequency fluctuations. Get Answer: (Show) 35. An alternator is rated for 75 kW at 0. 8 power factor. It means that (A) alternator has 4 poles (B) alternator can supply 75 kW at 0. 8 power factor (C) alternator can supply power only to loads having power factor 0. 8 only (D) the peak efficiency of alternator occurs only at 75 kW load having 0. lagging power factor. Get Answer: (Show) 36. The regulation of an alternator is (A) the reduction in terminal voltage when alternator is loaded (B) the variation of terminal voltage under the conditions of maximum and minimum excitation (C) the increase in terminal voltage when load is thrown off (D) the change in terminal voltage from lagging power factor to leading power factor. Get Answer: (Show) 37. A magnetisation curve represents the relationship between (A) reactive and non-reactive components of voltage (B) exciting currents and terminal voltage (C) power factor and terminal voltage (D) magnetic flux and armature current.

    Get Answer: (Show) 38. In an alternator if the armature reaction produces demagnetisation of the main field, the power factor should be (A) Zero, lagging load (B) Zero, leading load (C) Unity. Get Answer: (Show) 39. In an alternator if the armature reaction produces magnetisation of the main field the power factor should be (A) Zero, lagging load (B) Zero, leading load (C) Unity. Get Answer: (Show) 40. When an alternator is supplying unity power factor load, the armature reaction will produce (A) magnetisation of the main field (B) demagnetisation of the main field (C) distortion of the main field. Get Answer: (Show) 41.

    An alternator has full load regulation of 4% when the power factor of the load is 0. 8 lagging while alternator runs at 1500 rpm. The full load regulation of 1400 rpm for 0. 8 pf lagging load will be (A) 15/14 x 4 percent (B) 14/15 x 4 percent (C) 4 percent (D) Depends on other factors also. Get Answer: (Show) 42. The Potier’s triangle separates the (A) iron losses and copper losses (B) field mmf and armature mmf (C) stator voltage and rotor voltage (D) armature leakage reactance and armature reaction mmf. Get Answer: (Show) 43. In the Potier’s triangle, the Potier reactance drop per phase is 22 volts per phase at 88 amperes per phase.

    The Potier’s reactance per phase is (A) 0. 22 (B) 0. 25 (C) 0. 30 (D) 0. 44. Get Answer: (Show) 44. Two alternators are running in parallel. The excitation of one of the alternator is increased. The result will be (A) machine with excess excitation will burn (B) both machines will start vibrating (C) power output will decrease (D) wattless component will change. Get Answer: (Show) 45. The power output of an alternators is 100 kW. In order that the tangent of pf angle may be 0. 8 lagging, the KVAR rating must be (A) 80 cos? KVAR (B) 80 sin ? KVAR (C) 80 KVAR (D) -80 KVAR. Get Answer: (Show) Get all answers at once: (Hide) 1. C —– 32. B —– 33. C —– 34. A —– 35. B —– 36. C —– 37. B —– 38. A —– 39. A —– 40. C —–41. C —– 42. D —–43. B —–44. D —– 45. D 46. The power output of an alternator is 40 kW and KVAR component is – 25. What will be the value of tan? (? being the power factor angle) ? (A) 0. 625 lagging (B) 0. 625 leading (C) 0. 375 lagging (D) 0. 375 leading. Get Answer: (Show) 47. When short pitch coils of 160 are used in an alternator, which harmonic component will not be present in the output emf ? (A) third (B) fifth (C) seventh (D) ninth. Get Answer: (Show) 48.

    A 120 MW turbo alternator is supplying power to 80 MW load at p. f. lagging. Suddenly the steam supply to the turbine is cut off and the alternator remains connected to the supply network and the field supply also remains on. What will happen to the alternator ? (A) The stator winding of the alternator will get burnt (B) The rotor winding of the alternator will get burnt (C) The alternator will continue to run as a synchronous motor rotating in the same direction (D) The alternator will continue to run as a synchronous motor rotating in the opposite direction. Get Answer: (Show) 49. The figure shows the characteristics of an alternator.

    Which curve represents synchronous impedance ? [pic] (A) curve A (B) curve B (C) curve C (D) curve D. Get Answer: (Show) 50. In the above figure (Figure of Question 49) which curve represents short circuit ? (A) curve A (B) curve B (C) curve C (D) curve D. Get Answer: (Show) 51. In the above figure which curve represents open circuit voltage ? (A) curve A (B) curve B (C) curve C (D) curve D. Get Answer: (Show) 52. For a peripheral speed of 314 m/s, a 2 pole cylindrical machine will have maximum diameter of (A) 255 cm (B) 235 cm (C) 200 cm (D) 170 cm. Get Answer: (Show) 53. The rotor of the salient pole alternator has 24 poles.

    The number of cycles of emf in one revolution would be (A) 24 (B) 12 (C) 6 (D) 4. Get Answer: (Show) 54. Two alternators A and B are sharing an inductive load equally. If the excitation of alternator A is increased (A) alternator B will deliver more current and alternator A will deliver less current (B) alternator B will deliver less current and alternator A will deliver more current (C) both will continue to share load equally (D) both will deliver more current. Get Answer: (Show) 55. Desirable feature for the parallel operation of two alternators is (A) both should have same resistance (B) both should have same reactance C) both should have less of resistance as compared to synchronous reactance (D) both should have more of resistance as compared to synchronous reactance. Get Answer: (Show) 56. Alternators used in aircraft systems usually have frequency of (A) 25 Hz (B) 50 Hz (C) 100 Hz (D) 400 Hz. Get Answer: (Show) 57. High frequency on aircraft alternators is selected in order to (A) free the systems from external disturbance (B) compensate for high speeds (C) compensate for high altitudes (D) reduce the bulk. Get Answer: (Show) 58. A 20 pole ac generator rotates at 600 rpm. The periodic time of current in seconds per cycle is (A) 0. 09 (B) 0. 004 (C) 0. 008 (D) 0. 01. Get Answer: (Show) 59. What kind of rotor is most suitable for turbo alternators ? (A) salient pole type (B) non-salient pole type (C) both (A) and (B) above (D) none of the above. Get Answer: (Show) 60. The synchronizing power developed in one of the alternators, when two alternators are running in parallel, will load the same alternator in which it is developed and reduce its speed (A) True (B) False Get Answer: (Show) Get all answers at once: (Hide) 46. B —– 47. D —– 48. C —– 49. A —– 50. C —– 51. D —– 52. C —– 53. B —– 54. B —– 55. C —– 56.

    D —– 57. D —– 58. D —– 59. B —– 60. A Multiple Choice Questions of Synchronous Generator (61-75): 61. If the input to the prime mover of an alternator is kept constant but the excitation is changed then the (A) reactive component of the output is changed (B) active component of the output is changed (C) power factor of load remains constant. Get Answer: (Show) 62. If two machines are running in synchronism and the voltage of one machine is suddenly increased (A) the machines will burn (B) both machines will stop (C) synchronising torque will be produced to restore further synchronism. Get Answer: (Show) 63.

    In an alternator, at 0. 8 lagging power factor, the generated voltage per phase is 240 V to give a rated terminated voltage ‘ V ‘. If the power factor of load increases to unity, the generated voltage per phase must be (A) 260 V (B) 250 V (C) 240 V (D) 225 V. Get Answer: (Show) 64. The advantage of salient poles in an alternator is (A) reduce noise (B) reduced windage loss (C) adoptability to low and medium speed operation (D) reduce bearing loads and noise. Get Answer: (Show) 65. Magnetisation curves for no load and full load unity power factor are shown in figure below. Which is the magnetisation curve for full load 0. power factor ? [pic] (A) curve A (B) curve B (C) curve C (D) curve D. Get Answer: (Show) 66. At a particular instant a turbo alternator is generating 80 MW at 0. 8 power factor lagging. Now if the steam supply valve to the steam turbine is further opened and the excitation is not changed (A) the speed of the alternator will increase but kW delivered will remain unchanged (B) the speed of the alternator will increase and kW delivered will also increase (C) the speed of the alternator will remain unchanged but it can meet more kW demand (D) the speed of the alternator will remain unchanged but it will deliver more kVA.

    Get Answer: (Show) 67. Two alternators A and B are sharing a resistive load (p. f. = 1 ) equally. Now if the excitation of alternator A is increased (A) alternator A will become lagging and alternator B will become leading (B) alternator A will become leading and alternator B will become lagging (C) both alternators will continue to operate on unity power factor (D) both alternators will operate on lagging power factor (E) both alternators will operate on leading power factor. Get Answer: (Show) 68. The advantage of providing damper winding in alternators is (A) elimination of harmonic effects B) provide a low resistance path for the currents due to unbalancing of voltage (C) oscillations are provided when two alternators operate in parallel (D) all of the above. Get Answer: (Show) 69. When two alternators are running in exactly synchronism, the synchronising power wil be (A) zero (B) sum of the output of two (C) unity (D) 0. 707. Get Answer: (Show) Questions 70 to 72 refer to the figure given below : [pic] 70. Load characteristic curves for an alternator are shown. The curves are drawn for 0. 8 pf lagging, 0. 8 p. f. leading, 0. 7 p. f. leading and 0. 9 p. f. lagging.

    Which curve represents the characteristics for 0. 8 p. f. leading ? (A) curve A (B) curve B (C) curve C (D) curve D. Get Answer: (Show) 71. Which curve represents the data for 0. 8 p. f. lagging ? (A) curve A (B) curve B (C) curve C (D) curve D. Get Answer: (Show) 72. Which curve represents the data for 0. 9 p. f. lagging ? (A) curve A (B) curve B (C) curve C (D) curve D. Get Answer: (Show) 73. The balanced short circuit current of a three phase alternator is 25 amperes at 1500 rpm. For the same field current, the balanced short current at 1400 rpm will be (A)18 A (B) 27 A (C) 20 A D) 20*(2)1/2A Get Answer: (Show) 74. A three phase alternator has a phase sequence of RYB for its three output voltages, for clockwise rotation. Now if the alternator is rotated anticlockwise, the phase sequence will be (A) RYB (B) RBY (C) BYR (D) none of the above. Get Answer: (Show) 75. In a synchronous machine, if the field flux axis is ahead of the armature field axis, in the direction of rotation, the machine working as (A) asynchronous motor (B) asynchronous generator (C) synchronous motor (D) synchronous generator. Get Answer: (Show) Get all answers at once: (Hide) 61. A —– 62.

    C —– 63. D —– 64. C —– 65. D —– 66. C —– 67. A —– 68. D —– 69. A —– 70. C —– 71. A —– 72. B —–73. C —– 74. B —– 75. D 76. In synchronous alternator, which of the following coils will have emf closer to sine waveform ? (A) concentrated winding in full pitch coils (B) concentrated winding in short pitch coils (C) distributed winding in full pitch coils (D) distributed winding in short pitch coils. Get Answer: (Show) 77. An alternator has rated field current of 4 A. The alternator develops 180 V while drawing a field current of 2 A at 750 rpm.

    If the field current is made 4 A at 750 rpm generated voltage could be (A) 400 V (B) 380 V (C) 60V (D) 330 V. Get Answer: (Show) 78. The armature reaction of an alternator will be completely magnetizing in case the load power factor is (A) unity (B) 0. 707 (C) zero lagging (D) zero loading. Get Answer: (Show) 79. Which of the following is not an integral part of synchronous generator system ? (A) prime mover (B) distribution transformer (C) excitation system (D) protection system. Get Answer: (Show) 80. For turbo generators the range of excitation voltage is (A) 10 to 20 V (B) 30 to 100 V C) 100 to 800 V (D) 1000 to 1800 V. Get Answer: (Show) 81. In case of low speed hydrogenerators, the short circuit ratio is usually (A) 0. 1 to 0. 5 (B) 0. 5 to 0. 6 (C) 0. 6 to 1. 0 (D) 1. 0 to 1. 5. Get Answer: (Show) 82. The permissible duration for which a generator of rated frequency 50 Hz can run at 46 Hz is (A) zero (B) one cycle (C) one second (D) one minute. Get Answer: (Show) 83. The permissible duration in supply . frequency is (A) ± 2 % (B) ± 5 % (C) ± 10 % (D) ± 25 %. Get Answer: (Show) 84. The regulation of an alternator is likely to be negative in case of (A) high speed alternators B) slow speed alternators (C) lagging power factor of the load (D) leading power factor of the load. Get Answer: (Show) Questions 85 to 88 refer to the data given below: A phase, 50 Hz, 6600 V, alternator is rated at 6600 kW at 0. 8 power factor and a full load efficiency of 90%. 85. kVA is rating of the alternator is (A) 750 kVA (B) 7500 kVA (C) 75000 kVA (D) 750000 kVA. Get Answer: (Show) 86. The current rating of the alternator is (A) 65. 63 A (B) 656. 3 A (C) 6563 A (D) 65630 A. Get Answer: (Show) 87. The input to the alternator is (A) 666. 6 kW (B) 6666 kW (C) 66660 kW D) 666,600 kW. Get Answer: (Show) 88. If the input. to an alternator remains unaltered, but excitation is changed then which of the following will not change ? (A) kVA output (B) kW output (C) power factor (D) all of the above. Get Answer: (Show) 89. Which of the following method is likely to give the voltage regulation more than the actual value ? (A) Synchronous reactance method (B) MMF method (C) Zero power factor method (D) None of the above. Get Answer: (Show) 90. The effect of cross magnetization in an alternator field is to make the output (A) true sinusoidal B) non-sinusoidal (C) harmonic free (D) none of the above. Get Answer: (Show) Get all answers at once: (Hide) 76. D —– 77. D —– 78. D —– 79. B —– 80. C —– 81. D —– 82. C —– 83. A —– 84. D —– 85. B —– 86. C —– 87. B —– 88. B —– 89. A —– 90. B 91. In order to reduce the harmonics in the emf generated in an alternator (A) slots are skewed (B) salient pole tips are chamfered (C) winding is well distributed (D) all of the above. Get Answer: (Show) 92. The maximum power in a synchronous machine is obtained when the load angle is (A) 0° (B) 85° C) 120° (D) 135°. Get Answer: (Show) 93. The emf generated due to nth harmonic component of flux in an alternator will be (A) n times the fundamental emf (B) same as fundamental emf (C) less than the value of fundamental emf. Get Answer: (Show) 94. Synchronizing torque comes into operation under all of the following cases EXCEPT (A) phase difference between two voltages (B) frequency difference between two voltages (C) voltage difference between two voltages (D) reduction in exciting current in one of the alternators. Get Answer: (Show) 95. Unbalanced 3-phase stator currents cause A) double frequency currents in the rotor (B) healing of rotor (C) vibrations (D) all of the above. Get Answer: (Show) 96. In large generators protection provided against external faults is (A) biased differential protection (B) sensitive earth fault protection (C) inter-turn fault protection (D) all of the above. Get Answer: (Show) 97. Pitch factor is the ratio of the emfs of (A) short pitch coil to full pitch coil (B) full pitch winding to concentrated winding (C) full pitch winding to short pitch winding (D) distributed winding to full pitch winding. Get Answer: (Show) 98.

    In an alternator if the winding is short pitched by 50 electrical degrees, its pitch factor will be (A) 1. 0 (B) 0. 866 (C) 0. 75 (D) 0. 50. Get Answer: (Show) 99. The Potier’s triangle separates (A) stator losses and rotor losses (B) fixed losses and variable losses (C) armature voltage and field voltage (D) armature leakage reactance and armature reaction mmf. Get Answer: (Show) 100. If a single phase alternator has 8 slots per pole uniformly speed, but the winding is arranged with the middle two left empty, the breadth coefficient will be (A) 0. 99 (B) 0. 88 (C) 0. 67 (D) 0. 53. Get Answer: (Show) 01. Two alternators are running in parallel. If the field of one of the alternator is adjusted, it will (A) reduce its speed (B) change its load (C) change its power factor (D) change its frequency. Get Answer: (Show) 102. A generator is operating by itself supplying the system loads. The reactive power supplied by the generator will (A) depend on prime mover rpm (B) depend on type of insulation used (C) depend on the amount demanded by the load (D) depend on inter-coil inductance. Get Answer: (Show) 103. Which of the following part plays important role in over speed protection of a generator ? A) Over current relay (B) Alarm (C) Differential protection (D) Governor. Get Answer: (Show) 104. Which type of protection is provided on a generator to protect against stator insulation failure ? (A) Differential protection (B) Thermocouple actuated alarm (C) Over current relay (D) Reverse power relay. Get Answer: (Show) 105. Which relays comes into operation in the event of the failure of prime mover connected to ihe generator ? (A) Reverse power relay (B) Differential relay (C) Buchholz relay (D) None of the above. Get Answer: (Show) Get all answers at once: (Hide) 91. D —– 92.

    B —– 93. C —– 94. D —– 95. D —– 96. D —– 97. A —– 98. B —– 99. D —–100. D —–101. C —– 102. C —–103. D —–104. A —– 105. A 106. In alternators, the distribution factor is defined as the ratio of emfs of (A) distributed winding to connected winding (B) full pitch winding to distributed winding (C) distributed winding to full pitch winding (D) concentrated winding to distributed winding. Get Answer: (Show) 107. One of the advantages of distributing the winding in alternator is to (A) reduce noise (B) save on copper (C) improve voltage waveform D) reduce harmonics. Get Answer: (Show) 108. In case of a uniformly distributed winding, the value of distribution factor is (A) 0. 995 (B) 0. 80 (C) 0. 75 (D) 0. 50. Get Answer: (Show) 109. The advantage of a short pitch winding is (A) low noise (B) increased inductance (C) suppression of harmonics (D) reduced eddy currents. Get Answer: (Show) 110. Two alternators are connected in parallel. Their kVA and kW load share can be changed by changing respectively their (A) driving torque and excitation (B) excitation and driving torque (C) excitations only (D) driving torques only.

    Get Answer: (Show) 111. In case of alternators, the dark and bright lamp method is used for (A) phase sequence (B) load balancing (C) synchronizing (D) load transfer. Get Answer: (Show) 112. The advantage of using short pitched windings in an alternator is that it (A) suppresses the harmonics in generated emf (B) reduces the total voltage around the armature coils (C) saves copper used in windings (D) improves cooling by better circulation of air. Get Answer: (Show) 113. For the same power rating, an alternator operating at lower voltage will be (A) less noisy (B) costlier (C) larger in size D) more efficient. Get Answer: (Show) 114. Which of the following is the common synchronous speed in rpm between 60 Hz and 50 Hz alternators ? (A) 900 (B) 600 (C) 375 (D) 225. Get Answer: (Show) 115. All of the following losses for a synchronous machine are fixed EXCEPT (A) Bearing friction loss (B) Copper loss (C) Windage loss (D) Core loss. Get Answer: (Show) 116. Salient pole type rotors as compared to cylindrical pole type are (A) smaller in diameter and larger in axial length (B) larger in diameter and smaller in axial length (C) larger in diameter as well as axial length D) small in diameter as well as axial length. Get Answer: (Show) 117. In a synchronous machine, the field flux axis is ahead of the armature field axis in the direction of rotation, the machine is working as (A) asynchronous alternator (B) asynchronous motor (C) synchronous motor (D) synchronous alternator. Get Answer: (Show) 118. Which of the following is not a common synchronous speed in rpm between a 50 Hz and 25 Hz alternator ? (A) 750 (B) 375 (C) 250 (D) 200. Get Answer: (Show) 119. The effective voltage in one phase of an alternator having 240 turns per phase, frequency of 60 Hz and flux per pole of 2. 8 x 106 lines will be (A) 332. 5 V (B) 665 V (C) 1330 V (D) 2660 V. Get Answer: (Show) 120. The maximum current that can be supplied by an alternator depends on (A) speed of the exciter (B) number of poles (C) exciter current (D) strength of the magnetic field. Get Answer: (Show) Get all answers at once: (Hide) 106. A —– 107. C —– 108. A —– 109. C —– 110. B —– 111. C —– 112. B —– 113. C —– 114. B —– 115. B —–116. B —– 117. D —–118. D —–119. C —– 120. D 121. The windings for an alternator are I. 36 slots, four poles, span 1 to 8 II. 2 slots, six poles, span 1 to 10 III. 96 slots, six poles, span 1 to 12. The windings having pitch factors of more than 0. 9 are (A) I and II only (B) II and III only (C) I and II only (D) I, II and III. Get Answer: (Show) Questions 122 to 124 refer to data given below: A 500 kVA ,2300 volt three phase star connected alternator has a full load armature-resistance drop per phase of 50 volts and a combined armature reactance plus armature-reaction drop of 500 volts per phase 122. The percent regulation of the alternator at unity power factor is (A) 1. 05 (B) 10. 5 (C) 21. 5 (D) 27. 5.

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