McqMate
| 101. |
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 | |
| Explanation: it is analogous to transformer under the shorted terminals of the secondary. | |
| 102. |
No load current in induction motor is 10- 20% of full load current and the no load current of transformer is 2-6%. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: the air gap in induction motor is more prominent than transformer and so the flux requirement will also be more. | |
| 103. |
no load estimated current for the machine? |
| A. | 12 a |
| B. | 20 a |
| C. | 30 a |
| D. | 5 a |
| Answer» A. 12 a | |
| Explanation: no load current in induction motor is 10-20% of full load current. | |
| 104. |
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 | |
| Explanation: it is due to the low reluctance path, the flux requirement is low in the transformer. | |
| 105. |
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 | |
| Explanation: it is due to the higher | |
| 106. |
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 | |
| Explanation: at no load the lower power factor is low and lagging in nature. | |
| 107. |
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 | |
| Explanation: air gap is kept lower to avoid the low value power factor, lagging behaviour and to reduce the magnetizing current. | |
| 108. |
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 | |
| Explanation: the low no-load power factor reduces full load operating pf. | |
| 109. |
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 | |
| Explanation: running stably at low speed around 1/7th of normal motor speed is known as crawling. | |
| 110. |
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 | |
| Explanation: the outer cage has higher resistance and low reactance while vice versa for the inner cage. so at starting the current is confined to outer cage with the reduction in starting current improving the torque. | |
| 111. |
The rotor of 3-phase slip ring induction motor is fed from a 3-phase supply with its stator winding short circuited having rotor rotating clockwise at a speed of Nr, then the |
| A. | speed of air gap field w.r.t. stator is ns-nr anticlockwise |
| B. | speed of air gap field w.r.t. stator is ns-nr clockwise |
| C. | speed of airgap field w.r.t rotor is ns clockwise |
| D. | speed of airgap field w.r.t. stator is ns-nr clockwise |
| Answer» A. speed of air gap field w.r.t. stator is ns-nr anticlockwise | |
| Explanation: the air gap field of stator = - (ns-nr). | |
| 112. |
(II) If a large 3–phase squirrel-cage induction motor with no load is started at full voltage, it will be damaged. |
| A. | i is true, ii is false |
| B. | i is true and ii is also true |
| C. | i is false, ii is true |
| D. | i and ii are false |
| Answer» A. i is true, ii is false | |
| Explanation: large current of short duration are not harmful to induction motor but it may cause voltage drop in the power supply. | |
| 113. |
The PWM control of DC motor varies |
| A. | linearly with speed |
| B. | inversely with speed |
| C. | parabolically with speed |
| D. | exponentially with speed |
| Answer» A. linearly with speed | |
| Explanation: in pwm technique duty ratio is a linear function with respect to speed. | |
| 114. |
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 | |
| Explanation: ward-leonard system of system of speed control is not recommended for constant speed operation. | |
| 115. |
Mark the wrong option. Which of the following cause and effect behaviour in speed control is correct when field resistance is increased. |
| A. | decrease in flux |
| B. | increase in armature current |
| C. | increase in emf |
| D. | decrease in speed |
| Answer» C. increase in emf | |
| Explanation: when field resistance is increased, emf actually decreases, not increases due to fall in flux. | |
| 116. |
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 | |
| Explanation: at low speed increase in armature current caused by decrease in emf, will not be enough to compensate for decrease in flux in induced torque equation. | |
| 117. |
by |
| A. | armature voltage control |
| B. | field resistance control |
| C. | any of the methods |
| D. | none of the mentioned |
| Answer» A. armature voltage control | |
| Explanation: for the small dc motors it is difficult to vary the speed using their field control. | |
| 118. |
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 | |
| Explanation: it must be separately excited mahcine to implement the armature voltage control. | |
| 119. |
works well for speed |
| A. | below, above |
| B. | above, below |
| C. | base speed. |
| Answer» A. below, above | |
| Explanation: after the base speed is attained, the power consumed becomes constant as the speed remains constant in a shunt machine. | |
| 120. |
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 | |
| Explanation: the torque limit of the shunt motor remains constant till the base speed. | |
| 121. |
A laboratory group was working with a set of 3-hp shunt DC motor. But there was a mistake that it was fused with a 0.3A fuse instead 3A fuse. Then it was started |
| A. | a flash occurred instantly |
| B. | it ran for 3s and fuse was blown |
| C. | it ran normal |
| D. | none of the mentioned |
| Answer» B. it ran for 3s and fuse was blown | |
| Explanation: due to the wrong usage of the fuse wire, it could not sustain the current and in few seconds it was blown off. | |
| 122. |
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 | |
| Explanation: the runaway in large machine is avoided by using compensating winding. | |
| 123. |
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 | |
| Explanation: all of the reasons cause the dc motor to increase in speed when under loaded. | |
| 124. |
Concentrated winding differ from distributed winding with the concern of |
| A. | identical magnetic axis |
| B. | two magnetic axis |
| C. | no magnetic axis |
| D. | physical spacing |
| Answer» A. identical magnetic axis | |
| Explanation: concentrated winding have all turns on one magnetic axis. | |
| 125. |
We can place closed windings in |
| A. | ac commutator machine |
| B. | stepper motor |
| C. | ac machine |
| D. | dc machine |
| Answer» A. ac commutator machine | |
| Explanation: closed windings are placed only in commutator as it does not need any reconnections later. | |
| 126. |
Which of the following machines can be used to place open slot winding? |
| A. | ac machine |
| B. | ac commutator machine, ac machine |
| C. | dc machines |
| D. | all of the mentioned |
| Answer» A. ac machine | |
| Explanation: only ac machines are used for open type of winding inside the machine. | |
| 127. |
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) | |
| Explanation: for a progressive winding, forward pitch – backward pitch = winding pitch | |
| 128. |
A 200V dc machine has 4 poles and 40 coils, having simplex lap winding. The number of commutator segments which required in the given machine will be? |
| A. | 40 |
| B. | 20 |
| C. | 80 |
| D. | 26 |
| Answer» A. 40 | |
| Explanation: it is a simplex winding and so one segment of commutator will be needed for each coil. | |
| 129. |
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 | |
| Explanation: commutator segments are placed as per the coil sides in the machine. | |
| 130. |
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 | |
| Explanation: dummy coils are placed inside so that to provide mechanical support to dc machine and they do not take part in the emf development. | |
| 131. |
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 | |
| Explanation: double layer winding enables more usage of the space and lesser air gaps so the leakage flux also reduces and efficiency improves. | |
| 132. |
For an electrical machine with C number of coils and P poles, the distance between the coils connected by an equalizer ring is? |
| A. | c/p |
| B. | c/2 |
| C. | 2c/p |
| D. | c/2p |
| Answer» C. 2c/p | |
| Explanation: backward pitch for an equalizer ring = number of coils/pole pairs = c/p/2 = 2c/p. | |
| 133. |
With a P-pole DC machine with armature current Ia, the current per brush arm for a lap connected windings is? |
| A. | ia/p |
| B. | ia/2p |
| C. | 2ia/p |
| D. | ia/4p |
| Answer» C. 2ia/p | |
| Explanation: current per brush (lap winding) | |
| 134. |
unused component in the machine, then we must |
| A. | s/p/2 should not be an integer |
| B. | s/p/2 should be integer |
| C. | s/p should be integer |
| D. | s/p should not be an integer |
| Answer» A. s/p/2 should not be an integer | |
| Explanation: if slots per pole pair is not integer than all the winding will be completed without adding any dummy coil. | |
| 135. |
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 | |
| Explanation: in a lap winding configuration, if the machine as air gap variations due to the design aspects, it will give circulating current in the windings. | |
| 136. |
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 | |
| Explanation: taking kirchoff’s law into study here, the voltage across the winding will remain same. | |
| 137. |
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 | |
| Explanation: ward leonard method is the combination of armature control method and field control method, which can also be called as voltage control method. this is the most efficient method of speed control over wide range. | |
| 138. |
Reducing the armature voltage will give us |
| A. | variable torque speed control |
| B. | constant torque speed control |
| C. | variable and constant both can be achieved |
| D. | cannot comment on torque |
| Answer» B. constant torque speed control | |
| Explanation: as seen from speed torque characteristics, reducing armature voltage | |
| 139. |
Reducing the field current will give us |
| A. | constant torque and variable power speed control |
| B. | constant torque speed control |
| C. | variable power speed control |
| D. | constant power speed control |
| Answer» B. constant torque speed control | |
| Explanation: as seen from speed torque characteristics, reducing field current will increase the speed of the motor above base value but power will remain same. thus, reducing armature voltage will give constant power speed control, with variable torque. | |
| 140. |
Speed-power characteristic for Ward Leonard speed control method |
| A. | will start from origin |
| B. | will start from some positive value on power axis |
| C. | will start from some positive value on speed axis |
| D. | depends on other parameters |
| Answer» A. will start from origin | |
| Explanation: speed power characteristic of dc motor is plotted when, ward leonard speed control method is employed. for speed equal to zero, which is less than base speed, we get constant torque but variable power operation. thus, power will start increasing from origin. | |
| 141. |
Efficiency of Ward Leonard method is |
| A. | higher than rheostatic control method but lower than shunted field control method |
| B. | lower than rheostatic control method |
| C. | higher than rheostatic control method and shunted field control method |
| D. | depends on load |
| Answer» C. higher than rheostatic control method and shunted field control method | |
| Explanation: unlike all other methods, external resistance is not added in the circuit of control system. thus, efficiency of ward leonard control method is always highest at various different speeds. | |
| 142. |
Ward Leonard method is an ideal choice for motor which undergoes frequent starting, stopping, speed reversal. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: absence of external resistance increases efficiency. also, when the generator emf becomes less than the back emf of the motor, electrical power flows back from motor to generator, is converted to | |
| 143. |
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 | |
| Explanation: no special starting gear is required in ward leonard method of speed control. as the induced voltage by generator is gradually raised from zero, the motor starts up smoothly. speed reversal is smoothly carried out. | |
| 144. |
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 | |
| Explanation: ward leonard method of speed control is most efficient method of speed control in all aspects. we can get constant torque operation and constant power operations as well, with this method. | |
| 145. |
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 | |
| Explanation: speed control by ward leonard method, gives uniform speed variation in both the directions and above and below of normal speed as well. speed reversal is carried out smoothly by this control method. | |
| 146. |
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 | |
| Explanation: ward leonard speed control method requires large number of building blocks like generators, motors. thus, installing cost and maintenance cost of the whole unit is very high. apart from cost, it gives low efficiency at very high loads. | |
| 147. |
TOPIC 5.1 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 | |
| Explanation: the relative speed between rotor magnetic field and stator conductors is sip speed and hence the frequency of induced | |
| 148. |
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 | |
| Explanation: given a number of poles = 8. supply frequency is 50 hz. rotor speed is 720 rpm. ns = 120×f÷p=120×50÷8 = 750 rpm. s=ns-nr÷ns = 750 – 720÷750 = .04. | |
| 149. |
Calculate the moment of inertia of the thin spherical shell having a mass of 73 kg and diameter of 36 cm. |
| A. | 1.56 kgm2 |
| B. | 1.47 kgm2 |
| C. | 1.38 kgm2 |
| D. | 1.48 kgm2 |
| Answer» A. 1.56 kgm2 | |
| Explanation: the moment of inertia of the thin spherical shell can be calculated using the formula i=mr2×.66. the mass of the thin spherical shell and diameter is given. i= | |
| 150. |
Calculate the phase angle of the sinusoidal waveform z(t)=78sin(456πt+2π÷78). |
| A. | π÷39 |
| B. | 2π÷5 |
| C. | π÷74 |
| D. | 2π÷4 |
| Answer» A. π÷39 | |
| Explanation: sinusoidal waveform is generally expressed in the form of v=vmsin(ωt+α) where vm represents peak value, ω represents angular frequency, α represents a phase difference. | |
| 151. |
A 50 Hz, 4poles, a single phase induction motor is rotating in the clockwise direction at a speed of 1425 rpm. The slip of motor in the direction of rotation & opposite direction of the motor will be respectively. |
| A. | 0.05, 0.95 |
| B. | 0.04, 1.96 |
| C. | 0.05, 1.95 |
| D. | 0.05, 0.02 |
| Answer» C. 0.05, 1.95 | |
| Explanation: synchronous speed, ns=120×50÷4=1500 rpm. given a number of poles = 4. supply frequency is 50 hz. rotor speed is 1425 rpm. s=ns-nr÷ns=1500- 1425÷1500=.05. sb=2-s=1.95. | |
| 152. |
The frame of an induction motor is made of |
| A. | aluminum |
| B. | silicon steel |
| C. | cast iron |
| D. | stainless steel |
| Answer» C. cast iron | |
| Explanation: the frame of an induction motor is made of cast iron. the power factor of an induction motor depends upon the air gap between stator and rotor. | |
| 153. |
The slope of the V-I curve is 5°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line. |
| A. | .3254 Ω |
| B. | .3608 Ω |
| C. | .3543 Ω |
| D. | .3443 Ω |
| Answer» D. .3443 Ω | |
| Explanation: the slope of the v-i curve is resistance. the slope given is 5° so r=tan(5°)=.3443 ω. the slope of the i-v curve is reciprocal of resistance. | |
| 154. |
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 | |
| Explanation: when the number of stator slots is an integral multiple of a number of rotor slots the machine fails to start and this phenomenon is called cogging. | |
| 155. |
rotor? |
| A. | 30 revolution per minute |
| B. | 40 revolution per minute |
| C. | 60 revolution per minute |
| D. | 50 revolution per minute |
| Answer» D. 50 revolution per minute | |
| Explanation: supply frequency=50 hz. no- load speed of motor = 1000 rpm. the full load speed of motor=950 rpm. since the no- load speed of the motor is almost 1000 rpm, hence synchronous speed near to 1000 rpm. | |
| 156. |
Calculate the active power in a 487 H inductor. |
| A. | 2482 w |
| B. | 1545 w |
| C. | 4565 w |
| D. | 0 w |
| Answer» D. 0 w | |
| Explanation: the inductor is a linear element. it only absorbs reactive power and stores it in the form of oscillating energy. the | |
| 157. |
Calculate the active power in a 788 ω resistor with 178 A current flowing through it. |
| A. | 24.96 mw |
| B. | 24.44 mw |
| C. | 24.12 mw |
| D. | 26.18 mw |
| Answer» A. 24.96 mw | |
| Explanation: the resistor is a linear element. it only absorbs real power and dissipates it in the form of heat. the voltage and current are in the same phase in case of the resistor so the angle between v & i is 90°. | |
| 158. |
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 | |
| Explanation: in the rotor injection method, when an external voltage is in phase with the main voltage net voltage increases and the value of slip decreases and the value of rotor speed increases. | |
| 159. |
R.M.S value of the periodic square waveform of amplitude 72 V. |
| A. | 72 v |
| B. | 56 v |
| C. | 12 v |
| D. | 33 v |
| Answer» A. 72 v | |
| Explanation: r.m.s value of the periodic square waveform is vm and r.m.s value of the trapezoidal waveform is vm÷3½. the peak value of the periodic square waveform is vm. | |
| 160. |
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 | |
| Explanation: given a number of poles = 2. rotor speed is 550 rpm. | |
| 161. |
phase with the main voltage then speed will |
| A. | increase |
| B. | decrease |
| C. | remain unchanged |
| D. | first increases then decrease |
| Answer» B. decrease | |
| Explanation: in the rotor injection method, when an external voltage is in opposite phase with the main voltage net voltage decreases and the value of slip increases and the value of rotor speed decreases. | |
| 162. |
The rotor injection method is a part of the slip changing technique. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: rotor injection method comes under slip changing technique. it uses an external voltage source to change the slip value. the load torque remains constant here. | |
| 163. |
synchronous speed changing technique. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: slip recovery scheme comes under the synchronous speed changing technique. it uses an external induction machine to change the frequency value. the load torque remains constant here. | |
| 164. |
relationship between voltage and current is a straight line. |
| A. | .16 Ω |
| B. | .26 Ω |
| C. | .25 Ω |
| D. | .44 Ω |
| Answer» A. .16 Ω | |
| Explanation: the slope of the v-i curve is resistance. the slope given is 9.1° so r=tan(9.1°)=.16 Ω. the slope of the i-v curve is reciprocal of resistance. | |
| 165. |
If induction motor air gap power is 1.8 KW and gross developed power is .1 KW, then |
| A. | s |
| B. | s2 |
| C. | s3 |
| D. | s4 |
| Answer» A. s | |
| Explanation: at low values of slip, the electromagnetic torque is directly proportional to slip value. due to heavy loading slip value decreases which increases the ratio of r2÷s. | |
| 166. |
KW. |
| A. | 8 sec |
| B. | 4 sec |
| C. | 7 sec |
| D. | 3 sec |
| Answer» B. 4 sec | |
| Explanation: the fundamental time period of | |
| 167. |
Calculate the total heat dissipated in a resistor of 44 Ω when 0 A current flows through it. |
| A. | 0 w |
| B. | 2 w |
| C. | 1.5 w |
| D. | .3 w |
| Answer» A. 0 w | |
| Explanation: the resistor is a linear element. it only absorbs real power and dissipates it in the form of heat. the voltage and current are in the same phase in case of the resistor so the angle between v & i is 90°. | |
| 168. |
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 | |
| Explanation: the maximum torque occurs when the slip value is equal to r2÷x2. | |
| 169. |
AC voltage controllers convert |
| A. | fixed ac to fixed dc |
| B. | variable ac to variable dc |
| C. | fixed ac to variable ac |
| D. | variable ac to fixed ac |
| Answer» C. fixed ac to variable ac | |
| Explanation: voltage controllers convert the fixed ac voltage to variable ac by changing the values of the firing angle. | |
| 170. |
devices were used for voltage control applications. |
| A. | cycloconverters |
| B. | vacuum tubes |
| C. | tap changing transformer |
| D. | induction machine |
| Answer» C. tap changing transformer | |
| Explanation: a tap changing transformer can give variable ac from fixed ac without a change in frequency. | |
| 171. |
applications. |
| A. | power generation |
| B. | electric heating |
| C. | conveyor belt motion |
| D. | power transmission |
| Answer» B. electric heating | |
| Explanation: in electric heating, variable ac supply is needed. the devices are fired appropriately to apply enough temperature. | |
| 172. |
In the principle of phase control |
| A. | the load is on for some cycles and off for some cycles |
| B. | control is achieved by adjusting the firing angle of the devices |
| C. | control is achieved by adjusting the number of on off cycles |
| D. | control cannot be achieved |
| Answer» B. control is achieved by adjusting the firing angle of the devices | |
| Explanation: switching devices is so operated that the load gets connected to ac source for a part of each half cycle. | |
| 173. |
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 | |
| Explanation: as it is half wave, it consists of one scr ( the control element) in anti parallel with one diode. | |
| 174. |
The below given 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» B. full wave controller | |
| Explanation: as it consists of two scrs, it is a full wave controller. | |
| 175. |
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 | |
| Explanation: as it consists of two scrs, it is a full wave controller and both the half cycles can be controlled by varying their respective firing angles. | |
| 176. |
Calculate the resonant frequency if the values of the capacitor and inductor are 4 F and 4 H. |
| A. | .25 rad/sec |
| B. | .26 rad/sec |
| C. | .28 rad/sec |
| D. | .29 rad/sec |
| Answer» A. .25 rad/sec | |
| 177. |
Which test is used to determine the conditions where electric motors are being used? |
| A. | thermography |
| B. | geography |
| C. | seismography |
| D. | anthropology |
| Answer» A. thermography | |
| 178. |
1 Btu/h is equal to |
| A. | .641 |
| B. | .334 |
| C. | .293 |
| D. | .417 |
| Answer» C. .293 | |
| 179. |
Full form of IDMT is |
| A. | inverse definite minimum time |
| B. | inverter definite minimum time |
| C. | inverter definite maximum time |
| D. | insert definite minimum time |
| Answer» A. inverse definite minimum time | |
| 180. |
Full form of THD is |
| A. | total harmonic diverter |
| B. | total harmonic digital |
| C. | total harmonic distortion |
| D. | total harmonic discrete |
| Answer» C. total harmonic distortion | |
| 181. |
Calculate the value of THD if the distortion factor is .8. |
| A. | 76 % |
| B. | 75 % |
| C. | 74 % |
| D. | 73 % |
| Answer» B. 75 % | |
| 182. |
Second lowest order harmonic present in 3-Φ fully controlled rectifier is |
| A. | 5th |
| B. | 7th |
| C. | 3rd |
| D. | 2nd |
| Answer» B. 7th | |
| 183. |
The flux and the internally generated voltage of a dc machine is a function of its magneto-motive force. |
| A. | non-linear |
| B. | linear |
| C. | constant |
| D. | inverse |
| Answer» A. non-linear | |
| 184. |
Choose the best option which identifies about the below characteristics. |
| A. | (1)series motor (2) cumulative compound motor (3) shunt motor |
| B. | (1)cumulative compound motor (2) series motor (3) shunt motor |
| C. | (1)cumulative compound motor (2) shunt motor (3) series motor |
| D. | (1)shunt motor (2) series motor (3) cumulative compound motor |
| Answer» A. (1)series motor (2) cumulative compound motor (3) shunt motor | |
| 185. |
The speed torque of the differential compound dc motor is shown below. What conclusions can be made? |
| A. | this is an unstable machine |
| B. | there is regenerative increment in the speed |
| C. | this is impractical to be used |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| 186. |
A student is given a differential compound motor and he has been asked to make it start. How will he try? |
| A. | by shorting series field at start |
| B. | to run as shunt motor at start |
| C. | by making rated current at start |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| 187. |
For a 100 hp 250 V, compound dc motor with compensating winding has a field current of 5 A to produce a voltage of 250 V at 1200 rpm. What will be the shunt field current of this machine at no load? |
| A. | 5 a |
| B. | 5.6 a |
| C. | 4 a |
| D. | 0 a |
| Answer» A. 5 a | |
| 188. |
Load torques can be classified into how many types? |
| A. | three |
| B. | two |
| C. | four |
| D. | five |
| Answer» B. two | |
| 189. |
What is the relationship between torque and speed in constant type loads? |
| A. | torque is independent of speed |
| B. | torque linearly increases with increase in speed |
| C. | torque non-linearly increases with an increase in speed |
| D. | torque non-linearly decreases with an increase in speed |
| Answer» A. torque is independent of speed | |
| 190. |
Which of the following is the plugging method of braking? |
| A. | reversal of field connections |
| B. | reversal of armature connections |
| C. | addition of equal and opposite field |
| D. | removal of field circuit from current machine circuit |
| Answer» B. reversal of armature connections | |
| 191. |
Electrical braking of any variety becomes less effective as |
| A. | speed increases |
| B. | speed decreases |
| C. | independent of speed |
| D. | depends on supply voltage |
| Answer» B. speed decreases | |
| 192. |
Plugging is used in |
| A. | small motors only |
| B. | small and medium powered |
| C. | only in large heavy machines |
| D. | everywhere |
| Answer» A. small motors only | |
| 193. |
Braking time in the dynamic braking is the function of |
| A. | system inertia |
| B. | load torque |
| C. | motor rating |
| D. | all- system inertia, load torque and motor rating |
| Answer» D. all- system inertia, load torque and motor rating | |
| 194. |
In dynamic braking, when braking is applied system acts as |
| A. | freely running machine |
| B. | motor with slow speed |
| C. | generator |
| D. | motor with same speed in opposite direction |
| Answer» C. generator | |
| 195. |
What will happen if DC shunt motor is connected across AC supply? |
| A. | will run at normal speed |
| B. | will not run |
| C. | will run at lower speed |
| D. | burn due to heat produced in the field winding |
| Answer» D. burn due to heat produced in the field winding | |
| 196. |
What will happen, with the increase in speed of a DC motor? |
| A. | back emf increase but line current falls. |
| B. | back emf falls and line current increase. |
| C. | both back emf as well as line current increase. |
| D. | both back emf as well as line current fall. |
| Answer» A. back emf increase but line current falls. | |
| 197. |
In DC motor, which of the following part can sustain the maximum temperature rise? |
| A. | field winding |
| B. | commutator |
| C. | slip rings |
| D. | armature winding |
| Answer» A. field winding | |
| 198. |
Direction of rotation of motor is determined by |
| A. | faraday’s law |
| B. | lenz’s law |
| C. | coulomb’s law |
| D. | fleming’s left-hand rule |
| Answer» D. fleming’s left-hand rule | |
| 199. |
Which power is mentioned on a name plate of a motor? |
| A. | gross power |
| B. | power drawn in kva |
| C. | power drawn in kw |
| D. | output power available at the shaft |
| Answer» D. output power available at the shaft | |
| 200. |
An electric motor is having constant output power. So, motor will have a torque speed characteristic |
| A. | circle about the origin. |
| B. | straight line parallel to the speed axis. |
| C. | straight line through the origin. |
| D. | rectangular hyperbola |
| Answer» D. rectangular hyperbola | |
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