McqMate
| 201. |
The armature shaft of a DC motor must be able to withstand |
| A. | bending moment due to weight of the armature. |
| B. | any unbalanced magnetic pull on the armature core. |
| C. | twisting stains due to transmission of torque. |
| D. | bending moment, unbalanced magnetic pull and twisting stains |
| Answer» D. bending moment, unbalanced magnetic pull and twisting stains | |
| 202. |
In DC machines the residual magnetism is present. The order of residual magnetism is |
| A. | 2 to 3 per cent |
| B. | 10 to 15 per cent |
| C. | 20 to 25 per cent |
| D. | 50 to 75 per cent |
| Answer» A. 2 to 3 per cent | |
| 203. |
How many steps are involved in the construction of single phase induction motor? |
| A. | 3 |
| B. | 4 |
| C. | 5 |
| D. | 6 |
| Answer» C. 5 | |
| 204. |
What kind of motor employs the skein winding made use of? |
| A. | maximum horse power single phase induction motor |
| B. | fractional horse power single phase induction motor |
| C. | minimum horse power single phase induction motor |
| D. | zero horse power single phase induction motor |
| Answer» B. fractional horse power single phase induction motor | |
| 205. |
A three-point starter is used for |
| A. | shunt motors |
| B. | shunt as well as compound motors |
| C. | shunt, compound and series motors |
| D. | not for dc motors |
| Answer» B. shunt as well as compound motors | |
| 206. |
The starting resistance of a DC shunt motor is generally |
| A. | low |
| B. | around 0.5 kΩ |
| C. | around 5 kΩ |
| D. | infinitely large |
| Answer» A. low | |
| 207. |
What will happen if DC motor is used without starter? |
| A. | heavy sparking at brushes |
| B. | it’ll start smoothly |
| C. | will not start at all |
| D. | depends on load |
| Answer» A. heavy sparking at brushes | |
| 208. |
In shunt and compound motor starting the shunt field should be made on with full starting resistance in |
| A. | series with field |
| B. | parallel with field |
| C. | series with armature |
| D. | parallel with armature |
| Answer» C. series with armature | |
| 209. |
A starter is required for a 220-V shunt motor. The maximum allowable current is 55 A and the minimum current is about 35 A. The armature resistance of the motor is 0.4 Ω. What will be the number of sections of starter resistance required? |
| A. | 5 |
| B. | 4 |
| C. | 6 |
| D. | 8 |
| Answer» C. 6 | |
| 210. |
A starting resistance is inserted at the starting in an induction motor as well as dc motor. |
| A. | induction motor has to control starting torque whereas in dc motor, it is done to avoid large current |
| B. | to limit starting current in both the machines |
| C. | to limit starting speed |
| D. | all of the mentioned |
| Answer» A. induction motor has to control starting torque whereas in dc motor, it is done to avoid large current | |
| 211. |
2 TYPICAL CONTROL CIRCUITS FOR SHUNT AND SERIES MOTORS |
| A. | starting resistance would burn |
| B. | field winding would burn |
| C. | speed will rise steeply |
| D. | any of the mentioned |
| Answer» A. starting resistance would burn | |
| 212. |
The shunt motor starters that can be used is/are |
| A. | 3-point and 4-point starter |
| B. | 5-point starter |
| C. | 4-point starter |
| D. | 5-point and 3-point starter |
| Answer» A. 3-point and 4-point starter | |
| 213. |
Thyristor controlled starter is preferred over DOL starter due to |
| A. | lesser losses |
| B. | controlled direction |
| C. | least resistance offered |
| D. | all of the mentioned |
| Answer» A. lesser losses | |
| 214. |
For a 7.46 kW, 200 V dc shunt motor with full load efficiency of 85% has armature resistance of 0.25 ohms. Calculate the value of starting resistance in ohms for a current 1.5 times of the full load current. |
| A. | 2.788 |
| B. | 3.038 |
| C. | 2.688 |
| D. | 2.588 |
| Answer» A. 2.788 | |
| 215. |
The post effects of the armature reaction is |
| A. | main field distortion |
| B. | shift in mna |
| C. | reduction in main field |
| D. | none of the mentioned |
| Answer» D. none of the mentioned | |
| 216. |
If the students give a forward shift of 10° to the dc generator, then it |
| A. | reduces flux per pole |
| B. | improves flux per pole |
| C. | increases the flux density in core |
| D. | none of the mentioned |
| Answer» A. reduces flux per pole | |
| 217. |
A dc machine is run at rated speed in forward direction and then in backward direction. It is observed that, speeds of the rotation are different, then it leads to the conclusion of |
| A. | incorrect brush placement |
| B. | incorrect pole and core alignment |
| C. | incorrect field supply |
| D. | all of the mentioned |
| Answer» A. incorrect brush placement | |
| 218. |
For a dc machine, its commutator has a diameter of 50 c rotating at 1000 rpm. For a brush width of 1 cm, the time commutation taken by the machine will be |
| A. | 0.382 ms |
| B. | 0.456 ms |
| C. | 0.573 ms |
| D. | 0.312 ms |
| Answer» A. 0.382 ms | |
| 219. |
An induction motor can be said analogous to |
| A. | transformer |
| B. | synchronous motor |
| C. | universal motor |
| D. | stepper motor |
| Answer» A. transformer | |
| 220. |
A 3-phase induction motor with its rotor blocked behaves similar to a |
| A. | transformer under short circuit of secondary terminals |
| B. | transformer under open circuit of secondary |
| C. | synchronous motor under slip test |
| D. | synchronous motor under open circuit |
| Answer» A. transformer under short circuit of secondary terminals | |
| 221. |
The no load current of the transformer is very less due to |
| A. | mutual flux having low reluctance iron core |
| B. | mutual flux having high reluctance iron core |
| C. | leakage flux having low reluctance iron core |
| D. | leakage flux having high reluctance iron core |
| Answer» A. mutual flux having low reluctance iron core | |
| 222. |
The no load current of the induction motor is high due to |
| A. | long and high reluctance path between stator and rotor |
| B. | mutual flux having moderate reluctance path between stator and rotor |
| C. | leakage flux having low reluctance iron core |
| D. | leakage flux having high reluctance iron core |
| Answer» A. long and high reluctance path between stator and rotor | |
| 223. |
At no load induction motor has possible power factor as |
| A. | 0.2 |
| B. | 0.5 |
| C. | 0.65 |
| D. | 0 |
| Answer» A. 0.2 | |
| 224. |
Mechanically air gaps in induction motor are kept very low to avoid |
| A. | lower power factor |
| B. | lagging nature |
| C. | magnetizing current |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| 225. |
The low no load power factor |
| A. | reduces full load operating pf |
| B. | increases full load operating pf |
| C. | reduces full load excitation voltage |
| D. | increases full load excitation voltage |
| Answer» A. reduces full load operating pf | |
| 226. |
An induction motor when started on load, it does not accelerate up to full speed but runs at 1/7th of the rated speed. The motor is said to be |
| A. | locking |
| B. | plumming |
| C. | crawling |
| D. | cogging |
| Answer» C. crawling | |
| 227. |
The great advantage of the double squirrel-cage induction motor over single cage rotor is that its |
| A. | efficiency is higher |
| B. | power factor is higher |
| C. | slip is larger |
| D. | starting current is lower |
| Answer» D. starting current is lower | |
| 228. |
Ward-Leonard system of system of speed control is not recommended for |
| A. | constant speed operation |
| B. | wide speed |
| C. | frequent-motor reversed |
| D. | very slow speed |
| Answer» A. constant speed operation | |
| 229. |
At a very low speed, increase in field resistance will |
| A. | decrease the speed of motor |
| B. | increase the speed of motor |
| C. | not have signicant effect on speed |
| D. | no effect |
| Answer» A. decrease the speed of motor | |
| 230. |
Small DC motors have best speed control by |
| A. | armature voltage control |
| B. | field resistance control |
| C. | any of the methods |
| D. | none of the mentioned |
| Answer» A. armature voltage control | |
| 231. |
To implement armature voltage control, it must be ensured that |
| A. | it is used on separately excited machine |
| B. | it is used on shunt machine |
| C. | it is used on series machine |
| D. | any of the mentioned machine |
| Answer» A. it is used on separately excited machine | |
| 232. |
The torque limit of speed for a shunt motor |
| A. | remains constant till base speed |
| B. | remains constant after base speed |
| C. | varies linearly after base speed |
| D. | varies inversely till base speed |
| Answer» A. remains constant till base speed | |
| 233. |
Run-away for large DC machine is avoided by using |
| A. | a turn or two of cumulative compounding |
| B. | compensating winding |
| C. | stablizer winding |
| D. | all of the mentioned |
| Answer» B. compensating winding | |
| 234. |
Why does DC motor sometimes run too fast when under-loaded? |
| A. | due to weak field |
| B. | due to high line voltage |
| C. | due to brush-shifted to neutral |
| D. | all of the mentioned |
| Answer» A. due to weak field | |
| 235. |
DC machines have windings and synchronous machines use windings. |
| A. | closed,open |
| B. | open, closed |
| C. | open, open |
| D. | closed, closed |
| Answer» A. closed,open | |
| 236. |
The simplex lap winding has the range of winding pitch of |
| A. | (-2,2) |
| B. | (-1,1) |
| C. | more than 2 |
| D. | less than 1 |
| Answer» A. (-2,2) | |
| 237. |
The commutator pitch for a simplex lap winding is equal to |
| A. | 1 and -1 |
| B. | 1 |
| C. | -1 |
| D. | 2 to -2 |
| Answer» A. 1 and -1 | |
| 238. |
Commutator segments in a DC shunt machine is equal to the number of |
| A. | coil sides |
| B. | turns |
| C. | coils |
| D. | slots |
| Answer» A. coil sides | |
| 239. |
We can employ dummy coils in a DC machine to |
| A. | compensate reactance voltage |
| B. | reduce armature reaction |
| C. | provide mechanical balance to the armature |
| D. | improve the waveforms generated inside the commutator |
| Answer» A. compensate reactance voltage | |
| 240. |
In AC machines we should prefer double layer winding over single layer windings because |
| A. | it requires identical coils |
| B. | it is economical to use |
| C. | it offers lower leakage reactance |
| D. | all of the mentioned |
| Answer» A. it requires identical coils | |
| 241. |
While doing regular checks on the dc machine with the lap connected winding, it is reported to have ammeter fluctuations, this can be due to |
| A. | different air gaps under poles |
| B. | variable reluctances in the core |
| C. | irregular design deformations |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| 242. |
A 6-pole lap wound DC generator has a developed power of P watts and brush voltage of E volts. Three adjacent brushes of the machine had been found worn out and got open circuited. If the machine is operated with the remaining brushes, voltage and power that could be obtained from the machine are |
| A. | e, p |
| B. | e, 2p/3 |
| C. | e/p, 2p/3 |
| D. | e, p/3 |
| Answer» A. e, p | |
| 243. |
1 = 0.0917. |
| A. | reduce armature voltage |
| B. | increase armature voltage |
| C. | increase field current |
| D. | decrease field current |
| Answer» A. reduce armature voltage | |
| 244. |
Ward Leonard method is |
| A. | armature control method |
| B. | field control method |
| C. | combination of armature control method and field control method |
| D. | totally different from armature and field control method |
| Answer» C. combination of armature control method and field control method | |
| 245. |
Starting gear used in Ward Leonard method |
| A. | is of small size |
| B. | is of large size |
| C. | size depends on application |
| D. | is absent |
| Answer» D. is absent | |
| 246. |
To get the speed of DC motor below the normal speed without wastage of electrical energy we use |
| A. | ward leonard control |
| B. | rheostatic control |
| C. | any of the ward leonard or rheostatic method can be used |
| D. | not possible |
| Answer» A. ward leonard control | |
| 247. |
Speed control by Ward Leonard method, can give uniform speed variation |
| A. | in both directions |
| B. | in one direction |
| C. | below normal speed only |
| D. | above normal speed only. |
| Answer» A. in both directions | |
| 248. |
The disadvantage of the Ward Leonard control method is |
| A. | high initial cost |
| B. | high maintenance cost |
| C. | low efficiency at high loads |
| D. | high cost, high maintenance and low efficiency |
| Answer» D. high cost, high maintenance and low efficiency | |
| 249. |
1 SPEED CONTROL OF THREE PHASE INDUCTION MOTOR |
| A. | slip frequency |
| B. | supply frequency |
| C. | the frequency corresponding to rotor speed |
| D. | zero |
| Answer» A. slip frequency | |
| 250. |
An 8-pole, 3-phase, 50 Hz induction motor is operating at a speed of 720 rpm. The frequency of the rotor current of the motor in Hz is |
| A. | 2 |
| B. | 4 |
| C. | 3 |
| D. | 1 |
| Answer» A. 2 | |
| 251. |
Calculate the moment of inertia of the disc having a mass of 54 kg and diameter of 91 cm. |
| A. | 5.512 kgm2 |
| B. | 5.589 kgm2 |
| C. | 5.487 kgm2 |
| D. | 5.018 kgm2 |
| Answer» B. 5.589 kgm2 | |
| 252. |
In an induction motor, when the number of stator slots is equal to an integral number of rotor slots |
| A. | there may be a discontinuity in torque slip characteristics |
| B. | a high starting torque will be available |
| C. | the maximum torque will be high |
| D. | the machine may fail to start |
| Answer» D. the machine may fail to start | |
| 253. |
In the rotor voltage injection method, when an external voltage source is in phase with the main voltage then speed will |
| A. | increase |
| B. | decrease |
| C. | remain unchanged |
| D. | first increases then decrease |
| Answer» A. increase | |
| 254. |
A 2-pole, 3-phase, Hz induction motor is operating at a speed of 550 rpm. The frequency of the rotor current of the motor in Hz is 2. |
| A. | 9.98 |
| B. | 9.71 |
| C. | 9.12 |
| D. | 9.37 |
| Answer» D. 9.37 | |
| 255. |
Calculate the average value of the sinusoidal waveform x(t)=848sin(1.65πt+2π÷0.68). |
| A. | 0 |
| B. | 78 v |
| C. | 15 v |
| D. | 85 v |
| Answer» A. 0 | |
| 256. |
In the rotor voltage injection method, when an external voltage source is in opposite phase with the main voltage then speed will |
| A. | increase |
| B. | decrease |
| C. | remain unchanged |
| D. | first increases then decrease |
| Answer» B. decrease | |
| 257. |
The slip recovery scheme is a part of the synchronous speed changing technique. |
| A. | true |
| B. | false |
| Answer» A. true | |
| 258. |
If induction motor air gap power is 1.8 KW and gross developed power is .1 KW, then rotor ohmic loss will be KW. |
| A. | 1.7 |
| B. | 2.7 |
| C. | 3.7 |
| D. | 4.7 |
| Answer» A. 1.7 | |
| 259. |
Calculate the time period of the waveform v(t)=12sin(8πt+8π÷15)+144sin(2πt+π÷6)+ 445sin(πt+7π÷6). |
| A. | 8 sec |
| B. | 4 sec |
| C. | 7 sec |
| D. | 3 sec |
| Answer» B. 4 sec | |
| 260. |
The value of slip at which maximum torque occurs |
| A. | r2÷x2 |
| B. | 4r2÷x2 |
| C. | 2r2÷x2 |
| D. | r2÷3x2 |
| Answer» A. r2÷x2 | |
| 261. |
In AC voltage controllers the |
| A. | variable ac with fixed frequency is obtained |
| B. | variable ac with variable frequency is obtained |
| C. | variable dc with fixed frequency is obtained |
| D. | variable dc with variable frequency is obtained |
| Answer» A. variable ac with fixed frequency is obtained | |
| 262. |
A single-phase half wave voltage controller consists of |
| A. | one scr is parallel with one diode |
| B. | one scr is anti parallel with one diode |
| C. | two scrs in parallel |
| D. | two scrs in anti parallel |
| Answer» B. one scr is anti parallel with one diode | |
| 263. |
In the below given voltage controller circuit |
| A. | the positive half cycle at the load is same as the supply vs |
| B. | the negative half cycle at the load is same as the supply vs |
| C. | the positive and negative half cycles at the load are identical to the supply |
| D. | none of the mentioned |
| Answer» B. the negative half cycle at the load is same as the supply vs | |
| 264. |
The below shown controller circuit is a |
| A. | half wave controller |
| B. | full wave controller |
| C. | none of the mentioned |
| D. | will depend upon the firing angle |
| Answer» A. half wave controller | |
| 265. |
In the below given voltage controller circuit |
| A. | only the negative cycle can be controlled |
| B. | only the positive cycle can be controlled |
| C. | both the cycles can be controlled |
| D. | none of the mentioned |
| Answer» C. both the cycles can be controlled | |
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