Ques 111. Following graph shows the loss characteristics of a sheet of ferromagnetic material against varying frequency f. Pi is the iron loss at frequency f, Hysteresis and eddy current losses of the sheet at 100 Hz are (SSC-2012)
10 W, 100 W
10 W, 50 W
1 W, 5 W
1 W, 10 W
Answer.4. 1 W, 10 W
Explanation:-
Iron loss is given as
Pi = Pe + Ph
Pi = Kef2 + Khf
Pi/f = Kef + Kh
where Kh = 0.01
Ke = 0.001
Ph = Khf = 0.01 x 100 = 1W
Pe = Kef2 = 0.001 x (100)2 = 10W.
Ques 112. Hysteresis losses are present in iron core coil when (SSC-2011)
The current in the coil is sinusoidal only
The current in the coil is alternating
The current is unsymmetrical alternation only
The current in the coil is DC only
Answer.2. The current in the coil is alternating
Explanation:-
Hysteresis losses are also known as Iron Loss or Core Loss and it is always constant.
Hysteresis loss is due to the reversal of magnetization of the transformer core whenever it is subjected to alternating nature of magnetizing force. Whenever the core is subjected to an alternating magnetic field, the domain present in the material will change their orientation after every half cycle. The power consumed by the magnetic domains for changing the orientation after every half cycle is called Hysteresis loss.
Ques 113. Eddy current loss in the ferromagnetic core is proportional to (SSC-2011)
Square of frequency
The square root of frequency
Frequency
Reciprocal of the frequency
Answer.1. Square of frequency
Explanation:-
Eddy current is given by
Pe = f2B2max
∴ Eddy current losses are directly proportional to the square of the frequency.
Ques 114. Two inductors have self-inductances of 9 mH and 25 mH. The mutual inductance between the two is 12 mH. The coefficient of inductive coupling between the two inductors is (SSC-2011)
18.75
0.25
0.8
1.25
Answer.3. 0.8
Explanation:-
Mutual inductance
Ques 115. The magnetic materials that are used to prepare permanent magnets should have (SSC-2011)
The steeply rising magnetization curve
Small hysteresis loop
Highly retentivity
Low coercive force
Answer.3. Highly retentivity
Explanation:-
Properties of the material of a permanent magnet :
(1) It should have high retentivity so that it remains magnetized in the absence of the magnetizing field.
(2) It should have a high saturation magnetization.
(3) It should have high coercivity so that it does not get demagnetized easily.
Ques 116. The resonant frequency of the AC series circuit shown in the figure given below, in Hz is (SSC-2011)
1/4π√3
1/4π√2
1/4π
1/4π√10
Answer.3. 1/4π
Explanation:-
When the coils are joined in series with additive flux the equivalent inductance is
Leq = L1 + L2 – 2M = 2 + 2 – 2 x 1 = 2H
The resonant frequency of the AC series circuit in Herz is given as
F = 1/2π√LC
Where L = 2H & C = 2F
F = 1/2π√LC = 1/2π√4
= 1/4π
Ques 117. Permeance is analogous to (SSC-2010)
Conductance
Reluctance
Inductance
Resistance
Answer.1. Conductance
Explanation:-
In a magnetic circuit, permeance is a measure of the quantity of magnetic flux for a number of current-turns. A magnetic circuit almost acts as though the flux is ‘conducted’, therefore permeance is larger for large cross-sections of a material and smaller for longer lengths. This concept is analogous to electrical conductance in the electric circuit.
Ques 118. Which of the following will improve the mutual coupling between primary and secondary circuits? (SSC-2009)
Transformer oil of high break down voltage
High reluctance magnetic core
Winding material of high resistivity
Low reluctance magnetic core
Answer.4. Low reluctance magnetic core
Explanation:-
Magnetic reluctance, or magnetic resistance, is a concept used in the analysis of magnetic circuits. It is analogous to resistance in an electrical circuit, but rather than dissipating electric energy it stores magnetic energy.
Low the reluctance, less the opposition to flux therefore more flux can pass through the transformer core.
Therefore the transformer core is made magnetic steel that has high permeability (low reluctance) to the magnetic flux.
Ques 119. Two coupled coils with L1 = L2 = 0.6H have the coupling coefficient of K = 0.8. The turn ratio N1/N2 is (SSC-2009)
4
2
1
0.5
Answer.3. 1
Explanation:-
The self-inductance is given as
L = μN2A/I
L ∝ N2
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
L1/L2 = N21/N22
0.6/0.6 = N21/N22
N1/N2 = 1
Ques 120. Which among the following magnetic materials has the highest energy-product to make it a permanent magnet? (SSC-2009)
Alnico
Ferrite
Samarium cobalt
Cobalt – Iron alloy
Answer.1. Alnico
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.
For SSC JE Measurement & Instrumentation Solved Questions (Part-2) Click Here
For SSC JE Measurement & Instrumentation Solved Questions Click Here
For SSC JE Basic Electrical Questions Solved (2009 – 2018) (Part-3) Click Here
For SSC JE Basic Electrical Questions Solved (2009 – 2018) (Part-2) Click Here
For SSC JE Basic Electrical Questions Solved (2009 – 2018) (Part-1) Click Here
For SSC JE Electrical Transformer Questions Solved (2009 – 2018) Click Here
