SSC JE DC Motor & DC Generator Solved Questions (2020)

Ques.71.  Functions of interpoles in a dc machine are to (SSC-2013)

  1. Reduce field winding heating
  2. Improve commutation
  3. Compensate for air-gap variation
  4. Reduce losses

Answer.2. Improve Commutation

Explanation:

  • Interpoles are small poles compared to main poles and placed in the Interpolar region between the main poles and they are connected in series with the armature winding.
  • Their purpose is to counterbalance the armature reaction due to the flow of current in the armature winding.
  • Interpoles primary functions are
    1. It produces a counter flux on the coil which is undergoing commutation to nullify the REACTANCE VOLTAGE.
    2. It also nullifies the armature flux in the inter-polar region automatically.

 

Ques.72. The commutator segments of dc machine are made of (SSC-2013)

  1. Tungsten
  2. Hard-drawn copper
  3. Soft copper
  4. Electrolytic copper

Answer.2. Hard-Drawn copper

Explanation:

Commutators require periodic maintenance such as brush replacement, therefore, to reduce wear and tear commutator segment s of dc machines are made up of hard-drawn copper.

 

Ques.73. An electromagnetic torque is produced as an interaction between flux and current. The angle between flux and current is 45°. If this angle is changed to 30°, flux increases by 100% and current reduce by 25%, then the torque (SSC-2012)

  1. Increases to 183.7% of the original
  2. Reduces to 81.6% of the original
  3. Reduces to 54.4% of the original
  4. Reduces to 66.7% of the original

Answer.1. Increases to 183.7% of the original

Explanation:-

Electromagnetic torque

T ∝ K ΦIcosθ

Where K is designed constant = (PZ/2πA)

Where

P = number of poles

Z = total number of armature conductor

N = rotation speed of the armature in revolution per minute (r.p.m)

A = number of parallel paths

Now the flux Φ is increased by 100% i.e

Φ= Φ1 + 100% Φ1 = 2Φ1

I2 =0.75I1

T1 ∝  ΦI1cos45°  &

T2 ∝  ΦI2cos30° = 2Φx (0.75)Ix cos30°

13

Then T2 = 1.838 T1 = 183.7% of T1

Hence the torque increases to 183.7% of the original.

 

Ques.74. Match List-I (Types of motors) with List-II (Characteristics) and select the appropriate response (SSC-2012)

numerical 74

  1. a → 1, b → 2
  2. a → 2, b → 1
  3. a → 1, b → 1
  4. a →2, b → 2

Answer.2. a → 2, b → 1

Explanation:-

For any dc motor, torque is directly proportional to the flux and armature current. In the case of the series motor, the flux produced is directly proportional to the armature current. Hence torque is directly proportional to the square of armature current.

T ∝ Ia2

So as current during starting is 1.5 times the rated so starting torque is higher for the series motor.

In dc shunt motor flux produced by field winding is proportional to field current Ish. Here the input voltage is constant so field current and flux are also constant. Therefore, a dc shunt motor is also called a constant flux or constant speed motor.

 

Ques.75. The motor characteristics best suited for traction purpose are those of (SSC-2012)

  1. D.C series Motor
  2. Synchronous Motor
  3. Induction Motor
  4. D.C shunt Motor

Answer.1.  D.C series Motor

Explanation:-

Dc series motors are used in traction applications due to their excellent starting torque properties. Torque is directly proportional to the square of the armature current resulting in high starting torque.

 

Ques 79. A lap wound DC generator having 250 armature conductor runs at 1200 rpm. If the generator emf is 200V, then the operating flux of the DC generator is (SSC-2012)

  1. 0.08 wb
  2. 0.04 wb
  3. 0.06 wb
  4. 0.02 wb

Answer.2. 0.04 wb

Explanation:-

The emf induced in a dc machine is given as

Where

E = PΦZN/60A

Φ = flux per pole in Weber

P = number of poles

Z = total number of armature conductor

N = rotation speed of the armature in revolution per minute (r.p.m)

A = number of parallel paths (Number of parallel paths in lap winding, A = P)

Therefore E = ΦZN/60

200 = 1200 × Φ × 250/60

Φ =0.04wb

 

Ques.76. The highest speed attainable by D.C shunt motor is (SSC-2012)

  1. Much higher than the no-load speed
  2. Equal to no-load speed
  3. Much lower than the no-load speed
  4. Ideally infinite

Answer.2. Equal to no-load speed

Explanation:-

In the case of DC shunt motor, flux is practically constant as the field is parallel to the armature. If the load is reduced then the speed of the motor will remain almost constant because the field current remains almost constant. DC shunt motor has full speed at no load and at full load, speed is slightly less (which can be negligible practically).

From the above discussion, it is clear that the highest speed attainable by D.C shunt motor is equal to no-load speed.

 

Ques.77. The direction of rotation of a DC shunt motor can be reversed by interchanging (SSC-2012)

  1. The armature terminal only
  2. Either field or armature terminals
  3. The supply terminals
  4. The field terminals only

Answer.2. Either field or armature terminals

Explanation:-

The direction of rotation of a DC shunt motor can be reversed by interchanging leads of either the field winding or the Armature Winding.

Generally changing the direction of the field is easier, because it carries lesser current as compared to armature current. However, the reversal should not be done while the armature is excited.

 

Ques.78. Pole of d.c machines are often laminated to reduce (SSC-2012)

  1. Eddy current loss
  2. Iron weight
  3. Armature reaction
  4. Hysteresis loss

Answer.1. Eddy current loss

Explanation:-

The pole shoe is laminated to reduce the eddy current losses due to any variable air gap reluctance caused by the rotor slots.

solution 96

In case of open type slot DC machine under teeth position, the air gap is less So reluctance”S, therefore, flux Φ is more

Φ↑ = If x N/S↓

Under slot position, the Air gap is more Reluctance is more hence flux is less

Φ↓= If x N/S↑

So pulsating flux interact with stationary Rotor, as a result, EMF is induced and eddy current will flow so eddy current loss will be there so pole shoe is laminated to reduce eddy current loss.

 

Ques.79. The torque developed in a DC series motor in unsaturated magnetic circuit condition is (SSC-2011)

  1. Almost practically constant at all load currents
  2. Directly Proportional to the load current
  3. Directly proportional to the square of the load current
  4. Inversely Proportional to the square of load current

Answer.3. Directly proportional to the square of the load current

Explanation:-

We know that for D.C series Motor torque is directly proportional to the product of armature current and field flux, Ta ∝ ɸ.Ia. In DC series motors, the field winding is connected in series with the armature, i.e. Ia= If. Therefore, before the magnetic saturation of the field, flux ɸ is directly proportional to Ia. Hence, before magnetic saturation Ta ∝ Ia2.

 

Ques.80. A 220 V shunt motor develops a torque of 60 Nm at an armature current of 10 A. The torque developed when the armature current is 20 A, is (SSC-2011)

  1. 30 Nm
  2. 240 Nm
  3.  84 Nm
  4. 120 Nm

Answer.4. 120 Nm

Explanation:-

For DC shunt motor torque develops

T ∝ Ia

∴ 60 ∝ 10

For the second case

T” = 20

T” = (20 x 60)/10

T” = 120 Nm

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