Category: DC Machines Objective Questions

DC Machines Multiple Choice Questions on “Methods of Excitation”.

1. Which winding has large number of turns?
a) Shunt field
b) Series field
c) Both have same number of turns
d) Depends on requirement
Answer: a
Clarification: The shunt field winding is provided with a large number (hundreds or even thousands) of turns of thin wire and is excited from a voltage source. The series field winding has a few turns of thick wire and is excited from armature current by placing it in series with armature.

2. Which winding contains wire with higher thickness?
a) Shunt field
b) Series field
c) Both have same number of turns
d) Depends on requirement
Answer: b
Clarification: Series field winding is used when wire is thick. Thus, by making minimum number of turns it can be used to excite a DC machine. For a given field current, control of this field is achieved by means of a diverter, a low resistance connected in parallel to series winding.

3. Which winding have higher resistance?
a) Shunt field
b) Series field
c) Both have same number of turns
d) Depends on requirement
Answer: a
Clarification: The shunt field winding is provided with a large number (hundreds or even thousands) of turns of thin wire and is excited from a voltage source. The field winding, therefore, has a high resistance and carries a small amount of current. It is usually excited in parallel with armature circuit.

4. Which is more practical method used in control of series field?
a) Diverter
b) Tappings
c) Switch relay
d) Cannot be determined
Answer: b
Clarification: For a given field current, control of series field is achieved by means of a diverter, a low resistance connected in parallel to series winding. A more practical way of a series field control is changing the number of turns of the winding by suitable tappings which are brought out for control purpose.

5. In compound excitation, which winding/s is/are excited?
a) Shunt
b) Series
c) Both
d) Cannot be determined
Answer: c
Clarification: In compound excitation, both series and shunt windings are connected to the armature. For compound excitation both of these windings are excited. Two types are cumulative excitation and differential excitation.

6. In which type of excitation air gap flux increases with armature current?
a) Differential compound
b) Cumulative compound
c) Differential and Cumulative
d) Cannot be determined
Answer: b
Clarification: In compound excitation both shunt and series field are excited. If the two field aid each other (their ampere-turn is additive), the excited is called cumulative compound. The shunt field is much stronger than the series field. The air gap flux increases with armature current.

7. In differential compound excited machine, what is the variation in air gap flux per pole with respect to armature current?
a) Increases
b) Decreases
c) Remains constant
d) Always varies
Answer: b
Clarification: If the two fields oppose each other, the excitation is called differential compound. The air gap flux/pole decreases with armature current. The series field is so designed that the increase or decrease in flux/pole is to a limited extent.

8. Which winding in compound excitation is responsible for change in air gap flux per pole?
a) Series
b) Parallel
c) Interconnected
d) No coil is responsible
Answer: a
Clarification: In a compound excited machine, both series and parallel windings are connected with armature circuit. The series winding is specially designed for increasing or decreasing flux per pole in DC machine up to certain extent.

9. How short shunt and long shunt compound winding is selected?
a) Mechanical considerations
b) Switch reversal
c) Performance
d) Mechanical considerations and switch reversal
Answer: d
Clarification: In long shunt compound, the shunt field is connected across terminals. In short shunt compound, the shunt field is connected directly across the armature. There is no significant difference in machine performance for the two types of connections. The choice between them depends upon mechanical consideration or the reversing switches.

10. If a DC compound machine connected as a motor is about to use as a generator, we reverse the series field connections.
a) True
b) False
Answer: a
Clarification: If a dc compound machine connected as a generator is run as a motor, the series field connections must be reversed as the armature current reverses. The motoring action as cumulative/ differential would then be preserved (same as in the generator). This equally applies vice versa – motor to generator.

11. For a long-shunt compound motoring, which of the following equation is correct?
a) V_t= E_a+ I_a(R_a+ R_se)
b) V_t= E_a– I_a(R_a+ R_se)
c) V_t= E_a+ I_a(R_a– R_se)
d) V_t= -E_a+ I_a(R_a+ R_se)
Answer: a
Clarification: For a long shunt compound motor Rse is connected in series with armature, while in short shunt compound motors Rse is connected in series with terminal voltage. I_L is supplied through the terminals which split into I_f and I_a.

12. V_t= E_a– I_a(R_a+ R_se) is the equation for _______________
a) Short shunt compound motoring mode
b) Short shunt compound generating mode
c) long shunt compound motoring mode
d) Cannot be determined
Answer: b
Clarification: In a DC machine, for a long shunt compound motor Rse is connected in series with armature, while in short shunt compound motors Rse is connected in series with terminal voltage. IL is supplied to the terminals which is equal to I_a – I_f.

13. How shunt field is controlled?
a) Diverter resistor in parallel
b) Tapped field winding
c) Series regulating resistance
d) Other methods
Answer: c
Clarification:
Control of Excitation: 1) Shunt field: by a series regulating resistance.2) Series field: For small armature by a diverter resistance connected in parallel with series field. For large armature by tapped field winding so the winding turns can be changed.

14. The generator is called flat compounded if _____________
a) The series field ampere turns are such as to produce the same voltage at rated load as at no load
b) The series field turns are such as that the no load voltage is smaller than the rated load voltage
c) The rated voltage is less than the no load voltage
d) Cannot be determined
Answer: a
Clarification: According to the operating characteristics of a DC compound generator, if series field mmf produces same voltage at rated load as that of no load then it is called as flat compounded generator.

DC Machines, .

DC Machines online quiz on “Starting of DC Motors – 2”.

1. In three-point starter, as a starting handle is rotated __________
a) The resistance is added into armature circuit
b) The resistance is removed from field circuit
c) The resistance is added into field circuit
d) Resistance is neither added nor removed
Answer: c
Clarification: The starting resistance is arranged in steps between conducting raised studs. As the starting handle is rotated about its fulcrum, it moves from one stud to the next, one resistance step is cut out, and it gets added to the field circuit.

2. How much torque is ensured by resistance adding arrangement in 3-point starter?
a) Low and non-zero
b) Infinite
c) High finite
d) Zero
Answer: c
Clarification: As a starting handle is rotated one resistance step is added into field circuit. There is a short time wait at each stud for the motor to build up speed. This arrangement ensures a high average starting torque.

3. The resistance of NVC is _______
a) Small
b) Large
c) Infinite
d) Zero
Answer: a
Clarification: The resistance of no volt coil (NVC) is small. NVC resistance forms a part of field resistance when resistor rotating handle of three-point starter is moved from min. to max. position. Starting resistance is also added to the field circuit.

4. NVC will release the handle electromagnetically when ________
a) In the case of failure of field current
b) If the resistance is very high
c) At the end of each rotation
d) At the start of each rotation
Answer: a
Clarification: In case of failure of field current (due to accidental or otherwise open circuiting) NVC coil releases the handle (held electromagnetically), which goes back to the OFF position under the spring action.

5. Over-load coil performs the function when __________
a) In the case of failure of field current
b) If the resistance is very high
c) If armature current increases beyond certain value
d) At the start of each rotation
Answer: c
Clarification: The contact of this relay at armature current above a certain value (over load/ short circuit) closes the NVC ends, again bringing the handle to OFF position. NVC and OL release are protections incorporated in 3-point starter.

6. In 4-point starters the resistance is added in series with NVC because __________________
a) To increase field current
b) To increase armature current
c) To limit the NVC current
d) To limit armature current
Answer: c
Clarification: To overcome the problem caused when the field current is low, NVC is connected across the two lines, one line connected to F terminal through the starter and other directly to the second line from another L terminal of the starter. To limit the NVC current a protective resistance R is connected in series with it.

7. What will be the γ value for starter taking 4 steps, where ratio of resistance at maximum allowable current to armature resistance is equal to 1.8?
a) 1.2164
b) 1.8
c) 2.2468
d) 0.8220
Answer: a
Clarification: γ is defined as the ratio of upper current limit to the lower current limit in starters of DC machine. γ ^n-1 = ratio of resistance at maximum allowable current to the armature resistance. Substituting values for n=4, we get γ=1.2164.

8. Maximum allowable current for a 240-V DC shunt motor is equal to 65 A. Minimum allowable current for same is equal to 40 A. What will be the value of γ?
a) 0.6153
b) 1.265
c) 1.625
d) 2.652
Answer: c
Clarification: γ is defined as the ratio of upper limit current value to the lower limit current value. So, γ will be the ratio of 65/40. From calculations, we get γ= 1.625. Note that γ is unitless quantity.

9. For a certain machine having γ = 1.8, we are using starter with 5 steps. What will be the resistance at step 3 if step 2 resistance is equal to 2 Ω?
a) 2.111
b) 1.111
c) 3.6
d) 10.8
Answer: b
Clarification: γ is defined so as to calculate step resistance from the given maximum limit of current and minimum limit of the same. This γ = r_n/r_n-1. So, for calculating resistance at step 3 we’ll substitute the corresponding values in the equation, which will give step 3 resistance as 1.1111Ω.

10. Which of the following is the correct formula for calculating step resistance?
a) γ = r_n/r_n-1
b) γ = r_n/r_n+1
c) γ = r_n*r_n-1
d) γ = r_n*r_n+1
Answer: a
Clarification: For given values of armature currents upper and lower, corresponding equivalent resistances are calculated. So, by the induction formula we calculate the value of γ and the step resistance as well, where n denotes the number of steps.

11. Which of the following starter can sufficiently start the DC series motor?
a) 3-point starter
b) 4-point starter
c) 2-point starter
d) Cannot be determined
Answer: c
Clarification: 2-point starter is enough to start the DC series motor. Since armature and field winding are in series already high armature current will not flow. No requirement of 3 or 4-point starter in DC series motor.

DC Machines for online Quizzes, .

DC Machines Assessment Questions and Answers on “No Load with On Load Characteristics of DC Generator”.

1. Which of the following is not the operating characteristics of Dc generator?
a) No-load characteristics
b) Load characteristics
c) External characteristics
d) Internal characteristics
Answer: d
Clarification: The relationship between various parameters has to be presented graphically because of the magnetic saturation effect. Four characteristics of importance are the following: 1) No load characteristics 2) Load characteristics 3) External characteristics 4) Armature characteristics.

2. Characteristics drawn at Ia = 0 is also called as ____________
a) Magnetization characteristics
b) Non-magnetization characteristics
c) Anti-magnetization characteristics
d) Cannot be determined
Answer: a
Clarification: With I_a = 0 (no load) at constant n, it is the presentation of Vt (=Ea) vs If. This is the most important characteristic as it reveals the nature of the magnetization of the machine. It is easy to determine as the generator is on no load and so only low rated prime mover will serve the purpose. It is commonly called the open–circuit/magnetization characteristic.

3. Open circuit characteristics (OCC) is generally drawn across __________
a) E_a vs I_f, I_a=constant (not equal to rated)
b) E_a vs I_f, I_a=0
c) E_a vs I_f, I_a=constant
d) E_a vs I_f, I_a=constant (rated)
Answer: b
Clarification: Open circuit characteristics is also called as no-load characteristics or magnetization characteristics. No load clearly states that armature current will equal to 0. Thus, OCC is drawn at E_a vs If, I_a=0.

4. Characteristics of a DC generator drawn across Vt vs If at rated armature current and constant speed, is called as ____________
a) Load characteristics
b) No-load characteristics
c) External characteristics
d) Armature characteristics
Answer: a
Clarification: Since we have I_a value which is equal to rated i.e. non-zero, it is indeed not a no-load characteristic. Axes given are V_t and I_f, hence it is not an armature characteristic. Thus, it’s called as load characteristic or magnetization characteristic on load.

5. In an OCC at If =0, graph starts from origin.
a) True
b) False
Answer: b
Clarification: As the machine would have been previously subjected to magnetization, a small residual voltage would be present with field unexcited. As will be seen practically, this is necessary for generator to self-excite. So, graph will start from just above the origin on Voc axis.

6. While conducting OCC, in order to avoid hysteresis loop, in which direction If should be increased?
a) -ve direction
b) +ve direction
c) In any direction
d) In both direction there exists hysteresis loop
Answer: b
Clarification: In conducting the OCC test, If must be raised gradually only in the forward direction otherwise the curve would exhibit local hysteresis loops. In OCC at If =0 there exists small residual voltage shown by non-zero Voc.

7. Air gap line is drawn at iron _________
a) Saturated
b) Unsaturated
c) Moderately saturated
d) Variable saturation
Answer: b
Clarification: The extension of the liner portion of the magnetization curve, is known as the air-gap line as it represents mainly the magnetic behaviour of the machine’s air-gap, the iron being unsaturated in this region consumes negligible ampere-turns; in any case the effect of iron is also linear here.

8. If suppose OCC is conducted at speed n1, where n1< nrated, OCC will lie ____________
a) Above OCC at nrated
b) On OCC at nrated
c) Below OCC at nrated
d) Can’t comment by only speed information
Answer: c
Clarification: For a less speed than the rated one, residual voltage appearing at terminal call Voc will also be less than Voc at rated value, it will vary in parallel manner but will never intersect OCC at rated speed.

9. E_a can be determined using __________
a) No-load characteristics
b) Load characteristics
c) Cannot be determined
d) Above OCC
Answer: a
Clarification: Under load conditions Ea cannot be determined from the OCC for If in the saturation region because of the demagnetizing effect of armature reaction. We must therefore determine experimentally the equivalent demagnetizing ampere-turns ATd due to armature reaction under actual load conditions.

10. If load characteristics are drawn on OCC itself, we get curve ________
a) Above OCC
b) On OCC
c) Below OCC
d) Intersecting OCC
Answer: c
Clarification: Since on load operation of a DC machine, we’ll get terminal voltage less than the terminal voltage obtained in OCC, graph will start from below OCC. On load, the effect of armature reaction will draw load characteristics parallel to the OCC below it, causing no intersection.

11. Load characteristics drawn at Ra =0 and Ra not equal to 0, will lie _____
a) Above
b) On
c) Below
d) Intersecting
Answer: c
Clarification: Load characteristics with at Ra =0 will lie below the load characteristics drawn at Ra not equal to 0. To the load characteristic we add IaRa drop to get Ea induced emf with load. Thus, it will lie above.

12. OCC is drawn at two different speeds both less than rated speed. OCC drawn at speed N1 lies below OCC drawn at speed N2. Which of the following relation is correct?
a) N2 = N1
b) N2 < N1
c) N2 >> N1
d) Can’t comment
Answer: c
Clarification: As a speed on which OCC is taken decreases, the residual voltage appearing on Voc axis also decrease and OCC starts from below, compare to first one. Thus, N1 is less comparatively, as its OCC lies below than the OCC drawn at other speed.

13. Why No-load or load characteristics are also called as magnetization characteristics?
a) E_a α I_f
b) E_a α φ
c) I_f α φ
d) Cannot be determined
Answer: b
Clarification: As the generated voltage in the armature in the case of DC generator is proportional to terminal voltage, which also proportional to magnetic flux, as seen by residual voltage appearing at 0 field current. No-load and load characteristics are called as magnetization curves.

DC Machines Assessment Questions, .

DC Machines Multiple Choice Questions 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
Answer: b
Clarification: 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
Answer: c
Clarification: The motor will accelerate the mechanical load connected during this period but no increase in the mechanical load as P_load = T₁W₁ = T₂W₂ where W₂ >W₁. 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
Answer: a
Clarification: 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
Answer: a
Clarification: 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
Answer: b
Clarification: Speed regulation is defined as a ratio of difference of no-load speed with full load speed with no-load speed. Here, no-load speed is more than the full load speed. Thus, we divide difference by no-load value and not by full load value.

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
Answer: b
Clarification: 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 %
Answer: b
Clarification: 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
Answer: b
Clarification: 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
Answer: b
Clarification: 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
Answer: a
Clarification: 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_f1f2
b) I_f1 >I_f2
c) I_f1 =I_f2
d) Cannot comment on I_f1, I_f2
View Answer

Answer: b
Clarification: 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
Answer: d
Clarification: 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
Answer: a
Clarification: 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.

DC Machines, .