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PSPCL-JE’21 EE: Daily Practices Quiz 13-Oct-2021

PSPCL RECRUITMENT 2021

Punjab State Power Corporation Ltd. (PSPCL), a power generating and distributing company of the Government of Punjab state, has released the notification for recruitment to the post of Clerk, Revenue Accountant, Junior Engineer (Electrical), Assistant Lineman & Assistant Sub Station Attendant Posts.
A total of 75 vacancies have been announced for the recruitment process of JE ELECTRICAL Engineering.

Exam Dates: 10 N0V-21 to 17 Nov-21

PSPCL-JE’21 EE: Daily Quiz

PSPCL-JE’21 EE: Daily Practices Quiz 13-Oct-2021

Each question carries 1 mark.
Negative marking: 1/4 mark
Total Questions: 06
Time: 06 min.

Q1. If the area, current and number of turns of an ac circuit are doubled, then the new inductance will be:
(a) Twice the former
(b) Four times the former
(c) Same
(d) Half the former

Q2. The unit of magnetic flux density is
(a) Weber
(b) Weber per meter
(c) Ampere per meter
(d) Tesla

Q3. Two long parallel conductors are placed 10 mm apart from each other carrying current of 150 A. what will be the force per meter length of each one?
(a) 0.45 N
(b) 0.1 N
(c) 4.5 N
(d) 9 N

Q4. In a material an application of magnetic field of 1.75 ×10^5 A/m causes a magnetic density of 0.2182 wb/m^2. Calculate its permeability (µH/m).
(a) 0.6
(b) 0.4
(c) 1
(d) 1.2

Q5. The reluctance of a straight magnetic path is:
(a) Directly proportional to area
(b) Inversely proportional to area
(c) Directly proportional to permeability
(d) Inversely proportional to length

Q6. Which of the following statement is correct?
Spiral springs are used in instruments
(a) Provide controlling torque
(b) Provide damping torque
(c) Lead the current to moving coil as well as to provide the controlling torque
(d) Pivot and jewel bearings

SOLUTIONS

S1. Ans.(a)
Sol. We know that, L=Nϕ/I=NBA/I.
According to given question:
N’= 2N
I’= 2I
A’=2A
∴L^’= (2N×B×2A)/2I=2×NBA/I=2L

S2. Ans.(d)
Sol. Magnetic flux density(B) =ϕ/A(weber per meter square) or Tesla

S3. Ans.(a)
Sol. F/l=(µ_0 µ_r I_1 I_2)/2πd= (4π×10^(-7)×(150)^2×1)/(2π×〖10×10〗^(-3) )=0.45 N.

S4. Ans.(d)
Sol. Permeability(µ) =B/H=0.2182/(1.75×10^5 )=1.2 µH/m.

S5. Ans.(b)
Sol. Reluctance(S) =l/(µ_0 µ_r A) AT/Wb
∴S ⍺ l/A

S6. Ans.(c)
Sol. Spiral springs are used in instruments to lead the current to moving coil as well as to provide the controlling torque.

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