1. In a progressive simplex, lap winding for a 4-pole, 14-slot, 2 coil-sides per slot d.c. armature, the back pitch yb, and front pitch yf will be respectively.
- 5 and 5
- 7 and 5
- 5 and 7
- 7 and 7
2. A dc shunt motor has the following characteristics, Ra = 0.5 Ω, Rf = 200 Ω, base speed = 1000 rpm, rated voltage = 250 V. On no load, it draws a current 5 A. At what speed will this run while delivering a torque of 150 Nm?
- 881 rpm
- 920 rpm
- 950 rpm
- 990 rpm
3. A speed of a dc motor is
- directly proportional to back emf and inversely proportional to flux.
- inversely proportional to back emf and directly proportional to flux.
- directly proportional to back emf as well as to flux.
- inversely proportional to back emf as well as to flux.
4. A cumulative compound dc motor runs at 1500 rpm on full load. If its series field is short circuited, its speed
- remains same
- becomes zero
5. A dc motor develops an electromagnetic torque of 150 N-m in a certain operation condition. From this operating condition, a 10% reduction in field flux and 50% increase in armature current is made. What will be new value of electromagnetic torque?
- 202.5 N-m
- 225 N-m
- 22.5 N-m
- 20.25 N-m
6. A 10 hp, 240V dc shunt motor, having armature-circuit resistance of 0.5 Ω and a full-load current of 40 A, is started by a starter, such that sections of required resistances in series with the armature-circuit should limit the starting current to 150% of the full-load current. The steady-state emf developed by the machine at full-load when the arm of the starter is moved to the next step is
- 100 V
- 120 V
- 60 V
7. The speed control of dc shunt motor in both directions can be obtained by
- Armature resistance control method
- Ward Leonard method
- Field diverter method
- Armature voltage control method
8. A dc motor running at 2000 rpm has a hysteresis loss of 500 W and eddy-current loss of 200 W. The flux is maintained constant but the speed is reduced to 1000 rpm. At the reduced speed the total iron-loss would be
- 500 W
- 400 W
- 300 W
- 200 W
9. The drawbacks of ‘Armature Resistance Control’ method of speed control a dc motor are
- A large amount of power is lost in the external resistance R.
- This method gives the speed below normal values.
- For a specified value of R, speed reduction is not constant. It varies with motor load.
Which of the above statements are correct?
- i, ii and iii
- i and ii only
- i and iii only
- ii and iii only
10. A shunt motor supplied at 250 V, runs at 900 rpm and the armature current drawn is 30 A. The resistance of the armature circuit is 0.4 Ω. The resistance required in series with the armature to reduce the speed to 600 rpm when the armature current is 20 A will be
- 2.17 Ω
- 3.17 Ω
- 4.17 Ω
- 5.17 Ω
11. A dc series motor of resistance 1 Ω across terminals runs at 1000 rpm at 250 V taking a current of 20 A. When an additional resistance of 6 Ω is inserted in series and taking the same current, the new speed would be
- 166.7 rpm
- 142.8 rpm
- 956.6 rpm
- 478.3 rpm
12. A dc series motor with a resistance between terminals of 1 Ω, runs at 800 rpm from a 200 V supply taking 15 A. If the speed is to be reduced to 475 rpm for the same supply voltage and current the additional series resistance to be inserted would be approximately
- 2.5 Ω
- 3 Ω
- 4.5 Ω
- 5 Ω
13. ‘A time-varying flux causes an induced electromotive force’. What law does this statement represent?
- Ampere’s law
- Faraday’s law
- Lenz’s law
- Field form of Ohm’s law
14. The dc series motor is best suited for traction work, because
- torque is proportional to the square of armature current and speed is inversely proportional to torque.
- torque is proportional to the square of armature current and speed is directly proportional to torque.
- both torque and speed are proportional to the square of armature current.
- torque is proportional to armature current and speed is inversely proportional to torque.
15. A dc shunt motor is required to drive a constant power load at rated speed while drawing rated armature current. Neglecting saturation and all machine losses, if both the terminal voltage and the field current of the machine are halved then
- the speed becomes 2 p.u. but armature current remains at 1 p.u.
- the speed remains at 1 p.u. but armature current becomes 2 p.u.
- both speed and armature current become 2 p.u.
- both speed and armature current remain at 1 p.u.
16. The speed of a separately excited dc motor is varied by varying the armature voltage in the range zero the base speed and by varying the field current above the base speed. It is suitable for constant
- torque drive at all speeds
- power drive at all speeds
- torque drive till base speed and constant power drive beyond base speed
- power drive till base speed and constant torque drive beyond base speed
17. In case of dc motor, maximum mechanical power is developed when back emf equals
- the applied voltage
- half the applied voltage
- one third of the applied voltage
- double the applied voltage
18. A dc series motor is running at rated speed and rated voltage, feeding a constant power load. If the speed has to be reduced to 0.25 p.u., the supply voltage should be reduced to
- 0.5 p.u.
- 0.75 p.u.
- 0.075 p.u.
- 0.25 p.u.
19. A separately excited dc generator is feeding a dc shunt motor. If the load torque on the motor is
reduced to half, then
- the armature current of both motor and generator are reduced to half.
- the armature current of motor is halved and that of generator remains unchanged
- the armature current of generator is halved and that of motor remains unchanged
- the armature current of both machines remains unchanged.
20. The speed of a dc motor is related to the back emf and flux in the following ways:
- Directly proportional to flux and inversely proportional to back emf.
- Directly proportional to back emf and inversely proportional to flux.
- Inversely proportional to flux and inversely proportional to back emf.
- Directly proportional to flux and directly proportional to back emf.