A. open circuit
B. decrease in speed
C. short circuit
A. Using non-magnetic material for frame
C. Providing laminations in stator
D. Providing laminations in armature core
A. torque will remain constant
C. torque and power both will change
D. torque, power and speed, all will change
A. the voltage across the terminals
C. the speed of the motor
D. none of the above
B. this will make armature to take heavy current, possibly burning it
C. nothing will happen to the motor
A. up to 20 H.P
B. up to 15 H.P
D. up to 10 H.P
A. the brushes
C. the field
D. the commutator
A. Cumulatively compounded motor
C. Shunt motor
D. Differentially compounded motor
A. varies as (Ia)
B. proportional to la
D. varies as la
B. Shunt motor
C. Cumulatively compounded motor
B. over load relay is connected in parallel and no volt relay in series with the load
C. over load relay and no volt relay are both connected in series with the load
D. over load relay and no volt relay are both connected in parallel with the load
A. eddy current losses
C. field copper losses
D. windage losses
A. four point starter is used
B. three point starter is used
D. all above can be used
A. decreases.
B. remain same
A. Shunt motors
C. Compound motors
D. Series motors
A. Coloumbs law
B. Lenzs law
C. Faradays law
A. a resistance is connected parallel to the armature
B. a high value resistor is connected across the field winding
D. armature is temporarily open circuited
A. Loss of efficiency
B. Rise in temperature of ventilating air
D. Excessive heating of core
A. increasing the excitation current
B. none of the above methods
D. decreasing the armature current
E. increasing the armature current
A. Field control
B. Shunt armature control
C. Mechanical loading system
B. series
C. shunt
D. cumulativelycompounded
A. Shunt motor
B. Series motor
A. frequency control method
C. field control method
D. field diverter
A. the gross power
B. the power drawn in kW
C. the power drawn in kVA
B. machine shops
C. pumping sets
D. air compressors
B. directly proportional to the armature current
C. inversely proportional to the armature current
D. proportional to the square of the current
A. become zero
C. decrease
A. as separately wound unit
B. in parallel with armature winding
C. in parallel with field winding
B. to increase the generated e.m.f.
C. to reduce copper losses
D. to improve cooling
A. supply voltage = back e.m.f
B. back e.m.f. = 2 x supply voltage
D. supply voltage = | x back e.m.f
B. Differential compound motor
C. Cumulative compound motor
A. motor is run at reduced speed
B. motor is reversed in direction
A. Starting torque
C. Operating speed
D. All of the above
A. shunt motor
B. compound motor
C. any of the above
B. Copper losses
C. Windage losses
D. Hysteresis losses
A. Windage loss
B. None of the above
D. Field copper loss
B. rheostat control method
C. flux control method
D. voltage control method
B. field copper loss
C. hysteresis loss
D. eddy current loss
A. the motor will work as series motor and run at slow speed in the same direction
B. the motor will not work and come to stop
C. the motor will work as series motor and run at high speed in the same direction
A. dc generator
B. all of the above
D. transformer
A. any of the above
C. dynamic braking
E. regenerative
A. None of the above
B. Induction motor
C. Synchronous motor
D. Single phase capacitor start
B. Torque is proportional to armature current
C. Torque and speed are inversely proportional to armature current
D. Torque is proportional to square root of armature current
B. constant voltage drive
C. none of the above
D. constant current drive
B. The motor will stop
C. The motor will run noisy
D. The motor will continue to run
A. sparking
B. corona losses
D. iron losses
A. reducing the resistance in the armature circuit
B. reducing the resistance in the field circuit
D. increasing the resistance in armature circuit
C. shunt motor
D. series motor
A. Frame
C. Stator
D. Shaft
A. copper losses = 0
C. hysteresis losses = eddy current losses
D. eddy current losses = stray losses
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