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PSPCL-JE’21 EE: Daily Practices Quiz 27-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 27-Oct-2021

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

Q1. Find the number of strands of ACSR conductor for 3-layer transmission line?
(a) 1
(b) 3
(c) 19
(d) 37

Q2. At standstill, in a three-phase induction motor, the ratio of slip speed to speed at which stator magnetic field rotates is equal to……….
(a) 3
(b) 0
(c) 2
(d) 1

Q3. Which gate is used when an output is desired to be complement of the input?
(a) NOT Gate
(b) AND Gate
(c) XOR gate
(d) OR Gate

Q4. In a 4-pole, 25 KW, 200 V wave wound dc shunt generator, the current in each parallel path will be:
(a) 125 A
(b) 31.25 A
(c) 250 A
(d) 62.5 A

Q5. Which of the following is not a Maxwell’s equation?
(a) ∆×E= -∂B/∂t
(b) ∆.E= -B
(c) ∆.D=ρ
(d) ∆.B= 0

Q6. In a dynamometer type wattmeter, the pressure coil connected across the load terminal is
(a) Highly inductive
(b) Highly capacitive
(c) Highly resistive
(d) Non inductive

SOLUTIONS

S1. Ans.(c)
Sol. Number of strands for ‘n’ layers=3n^2-3n+1=3×3^2-(3×3)+1=19.

S2. Ans.(d)
Sol. The slip speed of an induction motor is defined as the difference between synchronous speed and rotor speed.
i.e., N_slip=N_S-N_r
stator magnetic field rotates at synchronous speed.
At standstill condition, N_r=0
∴required ratio=N_s/N_s =1.

S3. Ans.(a)
Sol. A NOT gate is a logic gate that inverts the digital input signal. For this reason, a NOT gate is sometimes is referred to as an inverter.

S4. Ans.(d)
Sol. I_total=P/V=(25×1000)/200=125 A
For WAVE wound: number of parallel paths=2
∴ the current in each parallel path=I_total/2=125/2=62.5 A

S5. Ans.(b)
Sol. Max well’s equations
(i) ∇. D ⃗=ρ or ∇. E = ρ/ϵ_o → Gauss law
(ii) ∇.B ⃗ = O → Gauss’ law of magnetism
(iii) ∇ × E = – ∂B/∂t = – B’ → Faraday’s law
(iv) ∇ × H = ∂D/∂t + J → Amperes circuital law

S6. Ans.(c)
Sol. In a dynamometer type wattmeter, the pressure coil connected across the load terminal is highly Resistive.

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