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

Ques 101.  An inductor is supplied from a sinusoidal voltage source. The magnetic field energy in the inductor changes from the peak value to the minimum value in 10 msec. The supply frequency is (SSC- 2013)

  1. 50 Hz
  2. 25 Hz
  3. 1 Hz
  4. 100 Hz

Answer.2. 25 Hz

Explanation:

For a quarter part of wave pulse energy going down to the maximum value to minimum value. i,e

The magnetic field energy in the inductor changes from the peak value to the minimum value in 10 msec. 2

 

Ques 102. The magnetic field energy in an inductor changes from maximum value to minimum value in 5 msec when connected to an ac source. The frequency of the source (SSC- 2013)

  1. 500
  2. 200
  3. 50
  4. 20

 Answer.3. 50

Explanation:
For a quarter part of wave pulse energy going down to the maximum value to minimum value. i,e
T/4 = 5ms
T =20ms
f = 1/T  =1/20ms = 50Hz

Ques 103.  Two coupled coils, connected in series, have an equivalent inductance of 16 mH or 8 mH depending on the connection. The mutual inductance between the coils is (SSC- 2013)

  1. 12 mH
  2. 8 √2 mH
  3. 4 mH
  4. 2 mH

Answer.4. 2mH

Explanation:

The overall inductance of 2 coil L1 and L2 connected in series with mutual inductance aiding self-inductance L1 with mutual inductance opposing  self-inductance L2  then the mutual inductance M is given as

1/2 (L1 – L2)

1/2 (16 – 8)

= 2mH

 

Ques 104. Which one of the following is a valid value of the coefficient of coupling between two inductors? (SSC- 2013)

  1. 1.414
  2. 0.9
  3. 1.732
  4. 17.32

Answer.2. 0.9

Explanation:

The value of the coefficient of coupling is always greater than 0 and less than 1, or 0% and 100% respectively. A coefficient of coupling of 0 would represent no coupling, and 1 would represent perfect coupling.

 

Ques 105. Given two coupled inductors L 1 and L 2 having their mutual inductance M. The relationship among them must satisfy (SSC-2012)

  1. M > L1L2
  2. M ≤ L1L2
  3. M = L1L2
  4. M > (L1 + L2)/2

Answer.2. M ≤ L1L2

Expalanation:-

The expression for mutual inductance

M = K√L1L2

Now the value of the coefficient of coupling K lies between 0 and 1 i.e Mutual inductance is maximum when K = 1 and mutual inductance is zero when K = 0

So K ≤ 1

∴ M ≤ √L1L2

 

Ques 106. If the length of a bar of magnetic material is increased by 20% and the cross-sectional area is decreased by 20% then the reluctance is (SSC-2012)

  1. Increased by 50%
  2. Decreased by 33%
  3. Increased by 67%
  4. Remaining same

Answer.1. Increased by 50%

Explanation:-

The reluctance of a uniform magnetic circuit can be calculated as:

S = L/μA

S2/S1 = A1 L2/A2 L1

Since length l2 of the bar is increased by 20% = 120/100 = 1.2  i.e

L2 = 1.2L1

And area is decreased by 20% = 80/100 = 0.8 i.e

A2 = 0.8A1

18

 

Ques 107. Two coupled inductors L1 = 0.2 H and L2 = 0.8 H, have a coefficient of coupling K = 0.8, The mutual inductance M is (SSC-2012)

  1. 0.16 H
  2. 0.02 H
  3. 0.32 H
  4. 0.24 H

Answer.3. 0.32 H

Expalantion:-

The expression for mutual inductance

19

 

Ques 108. A coil with a certain number of turns has a specified time constant. If the number of turns is doubled, its time constant would (SSC-2012)

  1. Become doubled
  2. Get halved
  3. Remain Unaffected
  4. Become Fourfold

Answer.4. Become Fourfold

Explanation:-

The self-inductance of a solenoid is given as

L = μN2A/I = L1

where

N is the number of turns of the solenoid

A is the area of each turn of the coil

l is the length of the solenoid

and μ is the permeability constant

so, if the number of turns was to be doubled the self-inductance would be

L2 = u (2N)2A/l

or

L2 = 4L1

it would be quadrupled or increase fourfold.

 

Ques 109. The iron loss per unit frequency in a ferromagnetic core, when plotted against frequency, is a (SSC-2012)

  1.  Constant
  2.  Straight-line with positive slope
  3.  Straight-line with a negative slope
  4.  Parabola

Answer.2. Straight-line with positive slope

Explanation:-

Both hysteresis loss and eddy current loss give rise to heat in a magnetic circuit. The two losses are usually taken together and are called ‘iron loss’.

Pi = PE + PH

Pi = f2B2max + f B1.6max

Iron losses thus vary with both frequency and magnetic flux density. The power transformer is likely to be fed at the constant frequency and at the constant voltage, which means that magnetic flux density is almost constant. Thus iron losses in a transformer are constant from no load to full load.

 

Ques 110. The mutual inductance between two closely coupled coils is 1 H. If the turns of one coil is decreased to half and those of the other is doubled, the new value of the mutual inductance would be (SSC-2012)

  1. 1/4 H
  2. 1 H
  3. 2 H
  4. 1/2 H

Answer.2. 1 H

Explanation:-

The expression for mutual inductance

20

∴ M1 = M2 = 1H

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