SSC JE 2014 Electrical question paper with solution(Evening-shift)

Ques 51. In a 3- phase induction motor crawling happens at

  1. Any speed
  2. No-load speed
  3. Odd multiples of fundamental
  4. Even multiple of fundamental

An induction motor is a single excited machine. The crawling word it self-suggest crawl means moving at low speed.
This characteristic is the result of improper functioning of the motor which means either motor is running at a very slow speed or it is not taking the load.

The resultant speed is nearly 1/7th of its synchronous speed. This action is due to the harmonics fluxes produced in the gap of the stator winding of odd harmonics like 3rd, 5th, 7th, etc.

 

Ques 52. A 4-pole, 3-phase induction motor runs at 1440 RPM on a 50 Hz supply. Find the slip speed

  1. 2940 rpm
  2. 1500 rpm
  3. 1440 rpm
  4. 60 rpm

Synchronous speed Ns is

Ns = 120f/P

Ns = 120 × 50/4 = 1500 rpm

Slip speed = Ns – Nr

= 1500 – 1440

= 60 rpm

 

Ques 53. Low voltage windings are placed nearer to the core in the case of concentric windings because

  1. It reduces hysteresis loss
  2. It reduces eddy current loss
  3. It reduces insulation requirement
  4. It reduces leakage fluxes

Why does low voltage winding of the transformer placed near to core?

The LV winding of the Transformer is placed near the core in order to reduce the cost of insulation and the size of the Transformer. The insulation is directly proportional to the voltage so, If the HV winding of the Transformer is placed near the core, the insulation would have to be thicker which leads to higher cost.

Placing the HV winding after the LV winding makes the much lesser thickness of insulation for the HV winding.

 

Ques 54. If K is the phase-to-phase voltage ratio, then the line-to-line voltage ratio in a 3-phase Y-Δ transformer is

  1. K
  2. K /√3
  3. √3K
  4. √3/K

In star connection

VL = √3VPh

IL = Iph

In delta connection

VL = Vph

IL = √3Iph

VL = line voltage

Vph = phase voltage

IL = line current

Iph = phase current

Line-to-line voltage VL  = √3 × Vp

VL = √3K

 

Ques 55. In an auto-transformer of voltage ratio V1/V2, V1 > V2, the fraction of power transferred inductively is proportional to

  1. V1 / (V1 +V2 )
  2. V2/V1
  3. (V1 – V2 )/(V1 +V2 )
  4. (V1 – V2 )/V1 

Auto transformer

In an Auto-transfer the power is transferred both inductively and conductively

For inductive power transfer in auto-transformer is

Total power input × (V1 -V2)/V1

And power transfer conductively is given as

total power input × V2/V1

 

Ques 56. The stepped core is used in transformers in order to reduce

  1. Volume of Iron
  2. Volume of copper
  3. Iron loss
  4. Reluctance of core

For medium and large capacity transformers circular coils are used because they are mechanically stronger.It is more economical to use circular shape coils around the stepped core.

For the same area of the iron core, required by magnetic flux the diameter of the circumscribing circle (d) get reduced with an increase in the number of steps.

The reduction of diameter “d” will reduce the mean length turn of the winding around the core. Thus lesser copper (mean turn wise) will be used and the cost will reduce.

The core area also can be better utilized for cooling purposes and mechanically the core structure will be stable with stackings intact.

 

Ques 57. Commutation conditions at full load for large DC machines can be efficiently checked by the

  1.  Brake test
  2. Swinburne’s test
  3. Hopkinson’s test
  4.  Field test

Hopkinson’s test is also referred to as the regenerative or Back-to-Back test.  It is a full load test and required two identical machines are connected parallel, whereas the first machine gets excited from the source so it acts as a motor and is mechanically coupled with the next machine and makes it run as a generator.

The electrical output power from the generator (second machine) is fed to the motor(First machine) and vice-versa. Hopkinson test is performed to measure the efficiency of the DC machine.

 

Ques 58. The emf induced in a DC shunt generator is 230 V. The armature resistance is 0.1Ω. If the armature current is 200 A, the terminal voltage will be

  1.  200V
  2. 230V
  3. 210 V
  4. 250 V
Terminal voltage of DC shunt generator is

V = E – IaRa

Where E is generated EMFI

Ia is armature current

Ra Armature resistance

V = 230 – 0.1 x 200

V = 230 – 20

V = 210 v

 

Ques 59. The commutator of a DC generator acts as

  1. An amplifier
  2. A rectifier
  3. A load
  4. A multiplier
  • The commutator is a mechanical rectifier, so the commutator collects induced EMF or current developed in the armature.
  • The commutator converts the alternating current generated in the armature into the unidirectional current

 

Ques 60. Fleming’s left-hand rule is applicable to

  1.  DC generator
  2.  DC motor
  3.  Alternator
  4.  Transformer
Fleming left-hand rule is applicable to DC motor because direction of rotation of motor or direction of force experienced i.e motoring action can be determined by Fleming’s left hand rule.
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