This set of DC Machines Multiple Choice Questions & Answers (MCQs) focuses on “Speed Control Using Field Control of Shunt Motor”.

1. The speed of a DC shunt motor can be increased by ______

a) Increasing the resistance in armature circuit

b) Increasing the resistance in field circuit

c) Reducing the resistance in the field circuit

d) Reducing the resistance in the armature circuit

View Answer

Explanation: Speed of the DC motor is directly proportional to the back emf and inversely proportional to the flux produced by field. Where, flux produced is directly proportional to the current passing through the field winding (linear magnetization).

2. What will happen if excitation of DC shunt motor is changed?

a) Torque will remain constant

b) Torque and power both will change

c) Torque will change but power will remain constant

d) Torque, power and speed, all will change

View Answer

Explanation: The motor will accelerate the mechanical load connected during this period but no increase in the mechanical load as P

_{load}= T

_{1}W

_{1}= T

_{2}W

_{2}where W

_{2}>W

_{1}. So, at the higher speed there is less electrical torque for the same mechanical load / power.

3. If the speed of a DC shunt motor is increased, the back emf of the motor will ___________

a) Increase

b) Decrease

c) Remain same

d) Become zero

View Answer

Explanation: From, the speed-current characteristics of DC shunt motor we know that speed of the motor is directly proportional to the back emf and inversely proportional to the flux. So, for more speed there will be more back emf generated.

4. The speed of a DC shunt motor can be made more than full load speed by __________

a) Reducing the field current

b) Decreasing the armature current

c) Increasing the armature current

d) Increasing the excitation current

View Answer

Explanation: Speed of the DC motor obtained from speed equation is inversely proportional to flux produced by the field. So, reducing the field current flux produced by armature will decrease, and speed will increase.

5. Speed regulation of DC shunt motor is calculated by ratio of difference of full load speed and no-load speed with full load speed.

a) True

b) False

View Answer

Explanation: Speed regulation is defined as a ratio of difference of no-load speed and full load speed with full load speed. Remember that no-load speed is more than the full load speed.

6. Which speeds can be obtained from field control of DC shunt motor?

a) Lower than rated speeds

b) Greater than rated speeds

c) Lower and greater than rated speeds

d) Neither lower nor greater than rated speeds

View Answer

Explanation: Speeds greater than rated speeds can be obtained by lowering the flux of shunt field motor. Field cannot be made any stronger, it can only be weakened by this method. Thus, speed lower than the rated speed can’t be obtained.

7. No load speed of the DC shunt motor is 1322 rpm while full load speed is 1182 rpm. What will be the speed regulation?

a) 12.82 %

b) 11.8 %

c) 16.6 %

d) 14.2 %

View Answer

Explanation: Speed regulation is equal to (No-load speed – Full load speed) / (Full load speed). By substituting all the values, speed regulation= (1322-1182)/ 1182. Speed regulation is given by 0.118. In percentage notation SR= 11.8 %.

8. Speed regulation of a DC shunt motor is equal to 10%, at no load speed of 1400 rpm. What is the full load speed?

a) 1233 rpm

b) 1273 rpm

c) 1173 rpm

d) 1123 rpm

View Answer

Explanation: Speed regulation is equal to 0.1 which is also equal to (no-load – full load speed) divided by full load speed. Thus, by substituting all known quantities we get full load speed = 1400/1.1 = 1272.7 rpm so, speed equal to 1273 rpm.

9. Where will speed-torque characteristics will lie when armature reaction is considered?

a) Below the speed-torque characteristics when armature reaction is not considered

b) Above the speed-torque characteristics when armature reaction is not considered

c) On the speed-torque characteristics when armature reaction is not considered

d) Can be anywhere with the speed-torque characteristics when armature reaction is not considered

View Answer

Explanation: The speed-torque characteristic which has a small linear drop due to the second term (Ra effect) and translates upwards as the field is weakened due to the armature reaction. The demagnetizing effect of the armature reaction causes the characteristics to somewhat bend upwards with increasing torque (increasing load current).

10. Working range of the speed-torque characteristic, with increasing speed will ___________

a) Reduce

b) Increase

c) Remain same

d) Cannot comment

View Answer

Explanation: The working range of the speed-torque characteristic reduces with increasing speed in order for the armature current not to exceed the full-load value with a weakening field. Thus, armature current gives the bound limit for curve.

11. For speed x rpm, we get field current I_{f1} and for speed y rpm, we get the field current I_{f2}. If y is greater than x then, ________________

a) I_{f1} *f2
b) I _{f1} >I_{f2}
c) I_{f1} =I_{f2}
d) Cannot comment on I_{f1}, I_{f2}
View Answer*

Explanation: When speed-torque characteristic for different speeds is plotted on the same graph, we get the curve limited by armature currents also. For any value of field current flux through the field is directly proportional current, while flux is inversely proportional to speed.

12. 400-V dc shunt motor takes a current of 5.6 A on no-load and 68.3 A on full-load. Armature reaction weakens the field by 3%. What is the ratio of full-load speed to no-load speed? Given: Ra = 0.18 Ω, brush voltage drop= 2 V, R_{f} = 200 Ω.

a) 1.2

b) 0.8

c) 1.4

d) 1

View Answer

Explanation: If = 400/200= 2 A

No-load:

I

_{a0}= 5.6 – 2 = 3.6 A

E

_{a0}= 400 – 0.18 \ 3.6 – 2 = 397.4 V

Full-load:

Ib>

_{a}(fl) = 68.3 – 2 = 66.3 A

E

_{a}(fl) = 400 – 0.18 / 66.3 – 2 = 386.1 V

n (fl)/n (nl) = [386.1/397.4] [1/0.97] = 1.

13. In which of the following method, effect of armature reaction is more?

a) Field weakening method

b) Armature resistance control

c) Same in both methods

d) Cannot be determined

View Answer

Explanation: In field weakening method we are reducing the working flux to increase the speed, by reducing the field current. Therefore, effect of armature flux on main field flux will increase in case of field weakening method.

**Sanfoundry Global Education & Learning Series – DC Machines.**

To practice all areas of DC Machines, __here is complete set of 1000+ Multiple Choice Questions and Answers__.

**If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]**

**Related Posts:**

- Practice Electrical Engineering MCQs
- Check DC Machines Books
- Check Electrical Engineering Books
- Apply for Electrical Engineering Internship