SSC JE Magnetic-Circuit Solved Questions (2018-2009)

SSC Topic wise Solved paper / March 15, 2022

Ques.61. Two couple coil of L1 = 0.8 H and L2 = 0.2 H have a coupling coefficient of K = 0.9. The mutual inductance M is (SSC-2016)

  1. 0.144 H
  2. 0.23 H
  3. 0.36 H
  4. 0.43 H

Answer.3. 0.36

Explanation:

Mutual inductance of two coupling coil is

M = K√L1L2

= 0.9√0.16

= 0.36 H

 

Ques.62. The relative permittivity of paramagnetic material is (SSC-2016)

  1. Greater than one
  2. Less than one
  3. 1
  4. Infinite

Answer.1. Greater than one

Explanation:

Paramagnetism is a form of magnetism which occurs only in the presence of an externally applied magnetic field. Paramagnetic materials are attracted to magnetic fields, hence have a relative magnetic permeability greater than one (or, equivalently, a positive magnetic susceptibility).

 

Ques.63. Magnetic reluctance depends upon

  1. Cross-sectional area
  2. Length of material
  3. Material properties
  4. All of the above

Answer.4. All of the above

Explanation:

Flux in a magnetic circuit also depends on the opposition that the circuit presents to it. Reluctance (RM), is the opposition a magnetic circuit offers to the formation of magnetic flux.

The opposition depends on the dimensions of the core and the material of which it is made. As the resistance of a wire, reluctance is directly proportional to length (L) and inversely proportional to the cross-sectional area (A). In equation form,

RM = L/A (Ampere-Weber)

The unit of Reluctance is Ampere-Weber or A/Wb

To change the ‘proportional’ sign into an ‘equals’ sign we must introduce a constant. This constant is called ‘magnetic reluctivity‘ and is directly equivalent to ‘resistivity’ in an electric circuit. However, it’s far more common to use it’s reciprocal instead, which we call ‘absolute permeability(μ)‘ Which is equivalent to conductivity.

Permeability is a measure of how easy it is to establish the flux in a material. Ferromagnetic materials have high permeability and hence low RM, while non-magnetic materials have low permeability and high RM.

RM = L/μA

 

Ques.63. Two coils have an inductance of 8 mH and 18 mH and a coefficient of coupling of 0.5. If the two coils are connected in series aiding, the total inductance will be (SSC-2016)

  1. 32 mH
  2. 38 mH
  3. 40 mH
  4. 48 mH

Answer.2. 38 mH

Explanation:

Total inductance is given as

Leq = L1 + L2 + 2M……………..1

Where M = K√L1L2

= 0.5√8 x 18

= 6

Putting the value of M in equation number 1

= 8 + 18 + 2 x 6

= 8 + 18 + 12

= 38 mH

 

Ques.64.  Which material is preferred in permanent magnets (SSC-2016)

  1. Cast Iron
  2. Steel
  3. Pure Iron
  4. None of these

Answer.2. Steel

Explanation:

The material which we choose for making magnet should have high coercivity. The high coercivity ensures that magnetism in the material stays for a longer time. As steel and alnico have high coercivity, therefore, they are used for making permanent magnets.

 

Ques.65. In order to minimize loss due to hysteresis, the magnetic material should have (SSC-2016)

  1. High resistivity
  2. Low Hysteresis co-efficient
  3. The smaller B-H loop area
  4. High retentivity

Answer.3. The smaller B-H loop area

Explanation:

  • Hysteresis loss is associated with the cyclic magnetization and demagnetization of the material as the magnetic field changes with the sinusoidally varying current.
  • The hysteresis loss is proportional to the area of the hysteresis loop and therefore in order to reduce the loss the material of the core should have a smaller hysteresis loop area.

 

Ques.66.  The unit of retentivity is (SSC-2016)

  1. Weber
  2. Weber/sq.m
  3. Amper turn/meter
  4. Amper turn

Answer.2. Weber/sq.m

The Unit of retentivity is Weber/sq.m.

 

Ques.67. An air gap is usually inserted in magnetic circuits to (SSC-2016)

  1. Increase M.M.F
  2. Increase the flux
  3. Prevent Saturation
  4. None of the above

Answer.3. Prevent Saturation

Explanation:

  • Saturation is the state reached when an increase in applied external magnetic field H cannot increase the magnetization of the material further, so the total magnetic flux density B levels off.
  • The air gap is mostly used in the magnetic circuit because magnetic saturation causes loss of inductance, increasing current, power loss in the circuit.
  • implementation of the air gap in a magnetic circuit extends the range before magnetic saturation occurs and increasing the saturation current of the magnetic inductor. 

 

Ques.68. The magnetising force (H) and magnetic flux density (B) are connected by the relation (SSC-2016)

  1. B = μrH/μο
  2. B = μοH
  3. B = H/μομr
  4. B = μrHμο/μr

Answer.2. B = μοH

Explanation:

In free space,

B=μoH

Where,

B is the magnetic flux density,

H is the magnetic field intensity

μo is the permeability of free space.

 

Ques.69. The self-inductance of a long cylindrical conductor due to its internal flux linkages is 1mH/m. If the diameter of the conductor is doubled, then the self-inductance of the conductor due to internal flux linkage would be (SSC-2016)

  1. 0.5 KH/m
  2. 1mH/m
  3. 1.414 KH/m
  4. 4 KH/m

Answer.2. 1mH/m

Explanation:

Self-inductance is independent of the shape and size of the conductor. Hence the self-inductance will remain the same.

 

Ques.70. Two coils in differential connection have self-inductance of 2 mH and 4 mH and a mutual inductance 0f 0.15 mH. The equivalent inductance of the combination is (SSC-2016)

  1. 5.7 mH
  2. 5.85 mH
  3. 6 mH
  4. 6.15 mH

Answer.1. 5.7 mH

Explanation:

Total inductance for differential connection is given as

Leq = L1 + L2 + 2M……………..1

= 2 + 4 – 2 x 0.15

= 5.7 mH

Pages: 1 2 3 4 5 6 7 8 9 10 11 12

Related Posts

Scroll to Top