SSC Junior Engineer Electrical Online Exam 2018 CPWD/CWC/MES Electrical Engineering 22-01-2018 Morning3

21. Which of the following is the dimensional formula for mutual inductance?

A) ML2T 2A –2
B) ML2T 2A 2
C) ML2T –2A –2
D) ML–2T –2A 2

... Answer is C)
Mutual inductance dimensinal formula ML2T –2A –2
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22. Which of the following is the CORRECT expression for hysteresis loss occuring in a material?

A) η$Bm^2$$f^2$V
B) η$Bm^2$$f^2$$V^2.5$
C) η$Bm^1.6$fV
D) η$Bm^2$$f^1.6$V

... Answer is C)
Pb = η$Bm^1.6$fV

Pb = η * $Bm^n$ * f * V

Pb = hysteresis loss (W)
η = Steinmetz hysteresis coefficient, depending on material (J/m3)
Bm = maximum flux density (Wb/m2)
n = Steinmetz exponent, ranges from 1.5 to 2.5, depending on material
f = frequency of magnetic reversals per second (Hz)
V = volume of magnetic material (m3)
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23. Which of the following expression satisfies the Faraday's law of electromagnetic induction?

A) e = -Ndф/dt
B) e = -N Integral dф dt
C) e = N$d^2$ф/$dt^2$
D) e = N|dф/dt

... Answer is A)
Faraday's law of electromagnetic induction
e = -Ndф/dt
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24. Which property of a material opposes the passage of magnetic flux through it?

A) Permeance
B) Capacitance
C) Inductance
D) Reluctance

... Answer is D)
Reluctance.

25. Determine the intensity of magnetization (in A/m) of a magnet when the pole magnet is 30 A-m and the pole of the magnet is 2 sq.m.?

A) 60
B) 30
C) 25
D) 15

... Answer is D)
intensity of magnetization (in A/m) of a magnet = Pole strenght/Area
H = M/A
H = 30/2 = 15 Amp/meter
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26. What will be the produced mmf (in Ampturns) in a coil, if the coil has 160 turns and carries a current of 0.15 A?

A) 32
B) 24
C) 16
D) 8

... Answer is B)
N = 160 turns
I = 0.15 Amp

produced mmf = NI = 160 * 0.15 = 24 Amp-turn
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27. Determine the reluctance (in Amp-turns/Wb) of a coil, when the flux through the coil is 25 Wb and the value of produced mmf is 50 Ampturns.

A) 2
B) 4
C) 6
D) 8

... Answer is A)
Reluctance = mmf/flux
Reluctance = 50 /25 = 2 Amp-turns/Wb

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28. Determine the magnetic field intensity (in Amp-turns/m) of 5 meter long coil when the coil has 100 turns and carries a current of 0.6 A?

A) 15
B) 12
C) 10
D) 8

... Answer is B)
N = 100 turns
I = 0.6 A
l = 5 mtr
magnetic field intensity H = NI /l
magnetic field intensity H = 100*0.6/5 = 12 Amp-turns/m
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29. Determine the self-inductance (in mH) of a 3m long air-cored solenoid, when the coil has 300 turns and the diameter of the coil 12 cm?

A) 0.41
B) 0.35
C) 0.32
D) 0.24

... Answer is A)
N = 300 turns
l = 3 meter
d = 12 cm
r =d/2 =6 cm = 0.06 meter

A = π$r^2$ = 3.14*.06*.06

M = (µ0µr $N^2$ A)/l
where μ0 is the permeability of free space and has the value of µ0 = 4π *$10^-7$
for air Value of µr = 1
M = (4π *$10^-7$ *1*300*300*3.14*.06*.06)/3

M = 0.41mH

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30. Determine the current (in A) through a 60 cm long solenoid when the solenoid has 400 turns and the value of magnetic field at the centre of the solenoid is 6 mT?

A) 4.4
B) 5.6
C) 7.2
D) 8.4

... Answer is A)
flux density = permeability x intensity
B = µ0µr x H
H = B/µ0µr ------------(1)

where μ0 is the permeability of free space and has the value of µ0 = 4π *$10^-7$

for air Value of µr = 1

And we also know
H*l=NI
H = NI/l------------------(2)
Compare intensity from both equation

B/µ0µr = NI/l
I = (B/µ0µr) * (l/N)

So now we have
l = 60 cm = 0.6 meter
turn N =400
B = 6*$10^-3$ tesla

Apply above formula
I = (B/µ0µr) * (l/N)
I = (6*$10^-3$ / 4π *$10^-7$) * (0.6/400)
I = 7.2 Amp
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