Category: DC Machines Objective Questions

DC Machines test on “Characteristics of Separately Excited Generators”.

1. Characteristics of separately excited DC generator are drawn by keeping _____
a) Field current and speed both constant
b) Field current and speed both variable
c) Field current constant and speed variable
d) Field current variable and speed constant
Answer: b
Clarification: The operation considered here assumes that the armature is driven at constant speed (by means of prime mover) and the field excitation (If) is adjusted to give rated voltage at no-load and is then held constant at this value throughout the operation considered.

2. What is the reason behind dropping down of E_a with load?
a) Field resistance
b) Load resistance
c) Internal factors
d) Armature reaction
Answer: d
Clarification: In spite of fixed excitation, E_a drops off with load owing to the demagnetizing effect of the armature reaction. As the voltage drop is caused by magnetic saturation effect, it increases with load non-linearity.

3. External characteristic differ from an internal characteristic in separately excited DC generator by _________
a) I_a *R_a
b) I_f *R_a
c) I_L *R_L
d) I_f *R_f
Answer: a
Clarification: Internal characteristic is drawn across armature generated voltage, it doesn’t account the presence of armature resistance. External characteristic takes into account the presence of armature resistance as it is drawn across terminal voltage.

4. The variation in terminal voltage of DC shunt generator with respect to variation in separately excited DC generator is ___________
a) Much rapid
b) Much slower
c) Remains constant
d) Can’t say
Answer: a
Clarification: The terminal voltage drops off much more rapidly with load in a shunt generator than in a separately-excited generator because of fall in field current with terminal voltage. The external characteristic is a double-valued curve with a certain IL (max).

5. Which of the following characteristic lies above of all others?
a) Differential compound
b) Under compound
c) Level compound
d) Over compound
Answer: d
Clarification: All the graphs when drawn across voltage and load current, start from the same point but with increase in values of load current all machines show different elevation according to the values of series field resisitor.

6. Why differential compound generator is not used in practice?
a) High cost
b) High maintenance
c) High drop down in voltage
d) Difficult construction
Answer: c
Clarification: At a given value of load current, differential compound machine gives lowest voltage output. As the load current increases drop down in terminal voltage of DC differential compound generator is maximum.

7.How compounding level in a compound machine is adjusted?
a) By adding variable resistance in series with series field resistance
b) By adding variable resistance in parallel with series field resistance
c) By adding fixed resistance in parallel with series field resistance
d) By adding fixed resistance in series with series field resistance
Answer: b
Clarification: When fixed resistance is added in parallel with series field resistance we get only one other compounding level. So, by adding variable resistance in parallel with series field we can get various other compounding levels.

8. Which of the following have different external characteristic than other?
a) Self excited DC shunt generator
b) Separately excited generator
c) Compound DC generator
d) Series DC generator
Answer: d
Clarification: Unlike Self excited DC shunt generator, separately excited generator, compound DC generator the voltage at zero load current for DC series motor doesn’t start from some positive non-zero value, instead it starts from origin.

9. External characteristic differ from internal characteristic in DC series motor by ______
a) I_a *(R_a + R_SE)
b) I_a *R_a
c) I_a *R_SE
d) If *R_SE
Answer: a
Clarification: In series DC generator series resistor R_SE is added with armature circuit, thus total drop in terminal voltage from generated armature voltage is equal to I_a *(R_a + R_SE). While drawing internal characteristics presence of armature resistance and series resistance is not taken into account.

10. For series DC generator, internal/external characteristic start from ____________
a) Positive non-zero voltage
b) Zero voltage
c) Negative non-zero voltage
d) Can start from anywhere
Answer: b
Clarification: For all other DC generators other than series DC generators, the external and internal characteristic as well, start from non-zero positive value of voltage. While in series DC generator both internal and external characteristic start from origin.

11. For a DC series generator what is the condition for self-excitation?
a) (R_a+R_se) > RC
b) (R_SE+R_se+R_L) C
c) (R_SE+R_se+R_L) > R_C
d) (R_se+R_L) > R_C
Answer: c
Clarification: Summation of armature resistance, series field resistance, load resistance must be less than the critical resistance. As seen in self-excited DC generator characteristics if this value is greater than critical resistance, voltage build-up will not be possible.

12. For a given DC series generator with critical resistance equal to 100 Ω, armature resistance is equal to 50 Ω, and series field resistance is equal to 20 Ω, is connected across load of 50 Ω. What will be the load voltage?
a) 20 kV
b) 0 V
c) 2 kV
d) Data insufficient
Answer: b
Clarification: Here, DC series motor fails to excite as addition of armature resistance, load resistance and field resistance is greater than the critical resistance of the machine by 20 Ω. Thus, machine fails to self-excite, as a result we’ll get zero terminal voltage.

all areas of DC Machines for tests, .

DC Machines Multiple Choice Questions on “Ward Leonard Speed Control Method”.

1. 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
Clarification: 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.

2. Which of the following component is not used in Ward Leonard method?
a) AC motor
b) DC generator
c) DC motor
d) AC generator
Answer: d
Clarification: Whole unit of Ward Leonard speed control unit consists of various units like DC generator, DC motor, AC motor, exciter circuit and various pots which are used for carrying out smooth operation.

3. In Ward Leonard speed control method for lowering the speed of the motor ______________
a) Reduce armature voltage
b) Increase armature voltage
c) Increase field current
d) Decrease field current
Answer: a
Clarification: In Ward Leonard speed control method, speed can be reduced under base value by reducing armature voltage. By increasing field current speed can be reduced but this is not employed in Ward Leonard method.

4. 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
Clarification: As seen from speed torque characteristics, reducing armature voltage will reduce the speed of the motor below base value but torque will remain same. Thus, reducing armature voltage will give constant torque speed control.

5. In Ward Leonard speed control method for increasing the speed of the motor ______________
a) Reduce armature voltage
b) Increase armature voltage
c) Increase field current
d) Decrease field current
Answer: d
Clarification: In Ward Leonard speed control method, speed can be increased above base value by weakening of the field, which can be done by lowering field current value. By increasing armature voltage speed can be increased but this is not employed in Ward Leonard method.

6. 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
Clarification: 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.

7. 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
Clarification: 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.

8. 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
Clarification: 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.

9. Ward Leonard method is an ideal choice for motor which undergoes frequent starting, stopping, speed reversal.
a) True
b) False
Answer: a
Clarification: 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 mechanical form and is returned to the mains via the driving ac motor. This aspect makes Ward Leonard method perfect for given application.

10. 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
Clarification: 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.

11. 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
Clarification: 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.

12. Speed control by Ward Leonard method, 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
Clarification: 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.

13. Ward Leonard control is basically a _____________
a) Voltage control method
b) Field diverter method
c) Field control method
d) Armature resistance control method
Answer: a
Clarification: Ward Leonard speed control method is combination of rheostatic series control method and shunted armature control method with field control as well. Thus, it can be called as voltage control method also.

14. In Ward Leonard control of DC motor, the lower limit of speed is imposed by ____________
a) Residual magnetism of the generator
b) Core losses of motor
c) Mechanical losses of motor and generator together
d) Cannot be determined
Answer: a
Clarification: We get the speed below the base value by reducing armature voltage, which is the simple method of reducing back emf which is proportional to the residual magnetism. Thus, lower limit of speed is imposed by residual magnetism of the generator.

15. The disadvantage of the Ward Leonard 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
Clarification: 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.

DC Machines, .

DC Machines Multiple Choice Questions on “Lap Winding”.

1. Resultant pitch in the lap winding is__________
a) Depends on Y_b value
b) Depends on Y_f value
c) Depends on Y_b and Y_f value
d) Always equal to 2
Answer: d
Clarification: In a lap winding the “finish” of one coil is connected to “start” of the adjoining coil. The coil side displacement of the front-end connection is called the front-pitch. The coil side displacement of the back-end connection is called the back-pitch. Resultant-pitch is equal to difference between Y_b and Y_f which is equal to 2, irrespective of Y_b and Y_f value.

2. What is the condition of retrogressive winding?
a) Y_b > Y_f
b) Y_bf
c) Y_b = Y_f
d) No condition in bterms of Y_b and Y_f
Answer: b
Clarification: The coil side displacement of the front-end connection is called the front-pitch. The coil side displacement of the back-end connection is called the back-pitch. The direction in which the winding progresses depends upon which is more, Y_b or Y_f. For retrogressive winding Y_bf .

3. What is the value of Y_b for a lap winding with a 4-pole, 12-slot armature with two coil sides/slot. Assume single-turns coils.
a) 3
b) 5
c) 7
d) 9
Answer: c
Clarification: Coil span is defined as a ratio of number of slots in the armature winding which are also equal to the number of commutator segments to the number of poles. Here, Slots in the armature winding= 12, Number of poles= 4.
Y_CS= 12/4= 3.
Y_b = 2_YCS +1= 7.

4. What is the value of Yf for a lap winding with a 4-pole, 12-slot armature with two coil sides/slot? (Assume single-turns coils – progressive winding)
a) 3
b) 5
c) 7
d) 9
Answer: b
Clarification: Coil span is defined as a ratio of number of slots in the armature winding which are also equal to the number of commutator segments to the number of poles. Here, Slots in the armature winding= 12, Number of poles= 4.
Y_CS= 12/4= 3.
Y_b = 2Y_CS +1= 7
Y_f =Y_b – 2= 7-2= 5.

5. Equalizer rings are needed in lap winding.
a) False
b) True
Answer: b
Clarification: Each parallel path in lap winding is under the influence of one pair of poles, so if a machine consists of multiple pairs of poles then dissimilarities occurs, due to which unequal voltages may be induced in the paths and a circulating current may flow. In wave winding each path is under the influence of all poles, so voltages are induced in each path causing no such dissimilarities like lap winding. Equalizers in lap windings are used to remove this dissimilarity, they’re not needed in wave winding.

6. What is the symmetry requirement of lap winding?
a) 2C/P= 0
b) 2C/P= integer
c) 2C/P= non-integer
d) Can’t express mathematically
Answer: b
Clarification: To avoid no-load circulating currents and certain consequential commutation problems, all the parallel paths must be identical so as to have the same number of coil-sides. Symmetry thus requires ratio of 2C/P is equal to the integer. Also, US/P equal to integer represents the same.

7. What is the relation between number of parallel paths(A) and number of poles(P)?
a) A = P
b) A < P
c) A > P
d) No relation exists
Answer: a
Clarification: Complex winding can be divided into different parallel paths lying under different pole pairs. It is, therefore, concluded that the number if parallel paths is equal to the number of poles. In wave winding number of parallel paths is equal to 2.

8. Current flowing through the armature conductors Ic is related to total current Ia by_______
a) Ic = A Ia
b) Ic = Ia/A
c) Ic = A² Ia
d) Ic = A/Ia
Answer: b
Clarification: Two positive and two negative brushes are respectively connected in parallel for feeding the external circuit. As per the ring diagram Ia splits into the number of poles equally. Poles = Parallel paths. Thus, Ic = Ia /A.

9. Value of commutator pitch in lap winding is_____
a) +2
b) +1 or -1
c) -2
d) Different for different parameters
Answer: b
Clarification: Two ends of coil are connected across the adjacent commutator segments. Depending on the type of winding that is, retrogressive or progressive, we have two values for commutator pitch.
For progressive winding, commutator pitch = +1. For retrogressive winding, commutator pitch = -1.

10. What is the value of Yf for a lap winding with a 4-pole, 12 commutator segments, with two coil sides/slot.?
(Assume single-turns coils -retrogressive winding).
a) 9
b) 3
c) 11
d) 5
Answer: a
Clarification: Coil span is defined as a ratio of number of slots in the armature winding which are also equal to the number of commutator segments to the number of poles. Here, Slots in the armature winding= Number of commutator segments= 12, Number of poles= 4.
Y_CS= 12/4= 3.
Y_b = 2Y_CS +1= 7
Y_f =Y_b + 2= 7+2= 9.

DC Machines, .

DC Machines Multiple Choice Questions on “Graphical Representation of External Characteristics”.

1. No load point of DC generator is __________
a) Intersection of OCC and Rf line
b) Point on the Y axis at rated field current
c) Point on the X axis at rated terminal voltage
d) Can’t find through graphical interpretation
Answer: a
Clarification: Intersection of OCC with field resistance line gives the no-load point. All the value so obtained from x and y axis respectively gives the terminal voltage and field current of a DC generator at no-load.

2. Rf line intersects with OCC in ________
a) 1^st quadrant
b) 2^nd quadrant
c) 3^rd quadrant
d) 4^th quadrant
Answer: a
Clarification: Rf line is a straight line passing through the origin and having constant slope, so rising always in a positive direction. OCC starts from some positive value on Y axis and increases till maximum point, afterwards it starts becoming constant, where generally it intersects with Rf line.

3. How armature resistance effect is shown graphically?
a) By adding I_aR_a product horizontally with R_f line
b) By subtracting I_aR_a product vertically with R_f line
c) By adding I_aR_a product vertically with R_f line
d) By subtracting I_aR_a product horizontally with R_f line
Answer: c
Clarification: For representing the voltage drop in armature resistance, we add product I_aR_a vertically with R_f line at minimum 2 points and draw line parallel to R_f to get line with V + I_aR_a = constant.

4. For determining IaRa maximum _____________
a) Distance between R_f line and v+ I_aR_a line is taken
b) Distance between OCC and v+ I_aR_a line is taken
c) Maximum Distance between OCC and v+ I_aR_a line is taken
d) Can’t calculated graphically
Answer: c
Clarification: OCC when starts from some residual voltage on y axis, goes on increasing till some maximum value and then starts reducing slightly. The bulk is formed where maximum distance between OCC and v+ I_aR_a line is taken, to get effective frop in armature.

5. How demagnetization effect of armature reaction is shown graphically?
a) By shifting origin towards +ve y axis
b) By shifting origin towards +ve x axis
c) By shifting origin towards -ve y axis
d) By shifting origin towards -ve x axis
Answer: d
Clarification: The demagnetization caused by armature reaction can be quantified by equivalent field current Ifd which can be taken as proportional to the armature current Ia. So, by shifting the origin towards -ve x axis by Ifd we can show armature reaction graphically.

6. External characteristics of DC shunt motor lies in ______________
a) 1^st quadrant
b) 2^nd quadrant
c) 1^st and 2^nd quadrant
d) 4^th quadrant
Answer: c
Clarification: When external characteristics are plotted on graphs, and if effect of armature reaction is not considered graph lies in 1st quadrant only. When effect of armature reaction is considered we shift the origin thus, characteristics lies in 2^nd quadrant as well.

7. From magnetization characteristic at I_f = 7.1 A, E_a = 225 V at 1000 rpm. What will be the terminal voltage at speed 950 rpm?
a) 225 V
b) 235 V
c) 214 V
d) 220 V
Answer: c
Clarification: Speed is directly proportional to the back emf of a machine. So, E_a = 225 V at 1000 rpm
E_a (950 rpm) = 225 x 950 / 1000, this will give speed of given DC machine at 950 rpm. Upon calculations we get, E_a = 213.7 V= 214 V approx.

8. For a given compound DC machine, Net field current obtained from characteristic is equal to 7.5 A, where shunt field current is equal to 5 A, armature current is 505 A, demagnetizing current equal to 0.95 A and shunt field winding of 1000 turns at rated speed of 1000 rpm. What will be the series field turns?
a) 7
b) 8
c) 5
d) 9
Answer: a
Clarification: From the excitation balance equation, I_f + [N_se/N_f] I_a – I_fd =I_f (net).
5 + 505(N_se /1000) – 0.95 = If (net).
Calculating for N_se by substituting If (net), we get N_se = 6.8 that is 7 turns.

DC Machines, .

DC Machines Multiple Choice Questions on “Braking of DC Motors – 1”.

1. Which of the following is the best braking method?
a) Friction
b) Electromechanical action
c) Eddy-currents
d) Electric braking
Answer: d
Clarification: Braking methods based on friction, electromechanical action, eddy-currents, etc. are independent of the motor but sometimes electric braking is better justified owing to its greater economy and absence of brake wear.

2. DC motor is still widely used in tractions due to its excellent braking properties.
a) True
b) False
Answer: b
Clarification: Dc motor is used in tractions because of its excellent braking characteristics and ability of smooth transition from the motor to the generator mode and vice versa. Also, characteristics suit perfectly for traction application.

3. Which of the following is not the method of electrical braking?
a) Plugging or counter-current
b) Dynamic or rheostatic
c) Regenerative
d) Eddy current
Answer: d
Clarification: Eddy current is the electrical effect or response of the system, which is reflected mechanically at brakes to reduce the speed of the motor. Thus, eddy current is not an electrical brake, it is mechanical one.

4. 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
Clarification: Plugging is method where connections are reversed at a given instant. Because of the problem of interrupting highly inductive field current and the time needed for the field current to build up in opposite direction, it is a common practice to reverse armature connections.

5. Which of the following is correct formula for braking torque in plugging?
a) n (ka²/R_b)
b) n² (ka²/Rb)
c) n^-1 (ka²/Rb)
d) (ka²/Rb)
Answer: a
Clarification: Braking torque is equal to braking power divided by speed of the motor.
T= [(nka )²/Rb]/n. As, Braking power is equal to Ea²/Rb/n. By solving for the braking torque from the above equation, we get n (ka²/Rb).

6. 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
Clarification: Braking torque of the DC machine is given by n (ka2/Rb). Here, braking torque is directly proportional to the speed of the motor, so as the speed decreases the efficiency of electrical brakes which is dependent on braking torque decreases.

7. Plugging is applied in a motor, if we don’t make the switch OFF what will happen?
a) Motor will come to rest as a result of plugging
b) Motor will come to rest and will start rotating in another direction
c) Motor will burn
d) Nothing will happen
Answer: b
Clarification: If the switch is kept ON near to zero speed, motor will have braking torque acting in opposite direction greater than the electromechanical torque. Thus, motor will come to rest and for the next instant motor will start rotating in opposite direction.

8. Plugging is used in ____________
a) Small motors only
b) Small and medium powered
c) Only in large heavy machines
d) Everywhere
Answer: a
Clarification: Plugging is used in small scale applications only. The large initial current and high mechanical stress restrict the application of plugging in large machines. So, in order to balance stress this method is used in small machines only.

9. Which of the following is dynamic braking?
a) Reversal of field connections
b) Reversal of armature connections
c) Addition of equal and opposite field
d) Removal of armature circuit from current machine circuit
Answer: d
Clarification: Reversal of the connections of armature is the method called plugging. In dynamic braking we remove the armature circuit and connect it to different resistor, with field circuit still connected to the external supply.

10. 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
Clarification: In dynamic braking, when brakes are applied the armature is disconnected from machine circuit and connected to the braking resistor. Now, at this point motor is driven by kinetic energy gained earlier, dissipating power in braking resistor.

11. 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
Clarification: The armature is disconnected from the supply and then a braking resistor Rb is immediately connected across it. The motor acts as a generator, driven by the inertia and stored kinetic energy dissipating power in Rb. This is a simple method of bringing a motor nearly to a standstill.

12. In which of the following electrical braking method, energy is supplied back to the supply?
a) Plugging
b) Dynamic braking
c) Regenerative braking
d) In all electrical braking
Answer: c
Clarification: In plugging energy is wasted in braking resistance which is equal to starting resistance while running as a motor. In dynamic braking energy is generated but it is not fed back to supply. In regenerative method energy is sent back for reuse.

DC Machines, .

DC Machines Multiple Choice Questions on “Wave Winding”.

1. What will be the value of “Y_f + Y_b” for a wave winding?
a) Equal to Y_c
b) Half of the Y_c value
c) Double of the Y_c value
d) Four times Y_c value
Answer: c
Clarification: In the wave winding, as the number of coil-sides is double the number of segments, the top coil-side of the second coil will be numbered as (1+2*Y_c). After numbering other coil sides,
1 + 2*Y_c – Y_f = 1+ Y_b
So Y_f + Y_b = 2Y_c.

2. For a progressive wave winding Y_c = ______
a) 2C/P
b) 2(C+1)/P
c) 2(C-1)/P
d) 2C/(P+1)
Answer: b
Clarification: Starting at segment 1 and after going through P/2 coils or Yc (P/2) segments, the winding should end in segment 2 for progressive winding or segment (C) for retrogressive winding. That is mathematically,
Y_c (P/2) = (C+1)
Y_c = 2(C+1)/P

3. Number of parallel paths in wave winding are ______
a) Equal to P
b) Equal to P/2
c) 2
d) Depends on other parameters
Answer: c
Clarification: In wave winding all coils are divided into 2 groups- all coils carrying clockwise current are series connected and so are all coils with counter-clockwise current- and these 2 groups are in parallel because the winding is closed. Thus, a wave winding has 2 parallel paths irrespective of number of poles.

4. What is the spacing between the brushes for a wave winding when a machine is 6-pole DC armature with 16 slots having 2-coil sides per slot and single-turn coils.
a) 4 segments
b) 8 segments
c) 16 segments
d) 12 segments
Answer: b
Clarification: Only 2 brushes are required in this case as the number of poles in wave winding is equal to 2. So, spacing between the brushes is equal to total number of segments i.e. total slots divided by 2. Spacing between brushes = C/A = 16/2 = 8 segments.

5. What is the relation between conductor current and armature current in wave winding?
a) Ic = Ia
b) Ic = 2Ia
c) Ic = 4Ia
d) Ic = Ia/2
Answer: d
Clarification: the number of parallel paths in the in a wave winding is equal to 2. So, armature current will get divided equally into total number of conductors/paths. Conductor current in a wave wounded machine is half of the Ia.

6. For a conductor current equal to 4mA, Current carried by a particular brush in a 2-pole machine will be _____
a) 16mA
b) 8mA
c) 2mA
d) 10mA
Answer: b
Clarification: Conductor current in a wave wounded machine is half of the Ia. So, Ia= 8mA. All positive and all negative brushes are respectively connected in parallel to feed the external circuit. Thus, IBRUSH = Ia /(P/2). Solving we get Brush current = 8mA.

7. Equalizer rings are needed in the wave winding.
a) True
b) False
Answer: b
Clarification: The armature coil forms 2 parallel paths under the influence of all pole-pairs so that the effect of the magnetic circuit asymmetry is equally present in both the parallel paths resulting in equal parallel-path voltages. Thus, equalizer rings are not needed in wave winding.

8. For a wave winding when a machine is 6-pole DC armature with 16 slots having 2-coil sides per slot and single-turn coil, Yf value is ____
a) 5
b) 3
c) 2
d) 7
Answer: a
Clarification: Y_cs = 16/6 = 2 slots (nearest lower integral value)
Y_b= 2*2+1 = 5
Y_c= 2(16-1)/6 = 5 segments
Y_f = 2Y_c – Yb = 5.

9. Wave winding machines are used in ______ currents applications.
a) High
b) Moderate
c) Low
d) Can be used anywhere
Answer: c
Clarification: Lap winding machine has the advantage of large number of parallel paths and lower conductor current and is therefore used in low voltage and high current applications. Wave winding has fixed number of parallel paths so, wave wounded machine is used in low currents application.

10. For a wave wounded machine number of brushes for small, large machines respectively is ________ _________
a) 2, 2
b) 4, 2
c) 2, P
d) Both values depend on the given conditions
Answer: c
Clarification: For a small wave wounded machine number of parallel paths are 2, thus 2 brushes are used. For a large machine total number of brushes is equal to the total number of poles. The spacing between adjacent brushes is C/P commutator segments.

DC Machines, here is complete set of
1000+ Multiple Choice Questions and Answers.