WBSETCL-JE’21 EE: Daily Practices Quiz 15-Dec-2021

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WBSETCL JE QUIZ: EE

WBSETCL-JE’21 EE: Daily Practices Quiz 15-Dec-2021

Each question carries 1 mark.
Negative marking: NO NEGATIVE MARKING
Total Questions: 06
Time: 06 min.

Q1. Which of the following is NOT one of the methods to improve string efficiency of insulators?
(a) Guard ring
(b) Grading of insulators
(c) Selection of k
(d) Disc porosity

Q2. The relative permeability of ……………….materials is slightly less than unity.
(a) Ferromagnetic
(b) Paramagnetic
(c) Diamagnetic
(d) Anti-ferrimagnetic

Q3. The difference between the two half-power frequencies is called ……….
(a) Bandwidth
(b) Quality factor
(c) Mid-frequency
(d) Resonant frequency

Q4. If the frequency of a transmission system is changed from 50 Hz to 100 Hz, the string efficiency
(a) Remains unchanged
(b) Will increase
(c) Will decrease
(d) May increase or decrease depending on the line parameters

Q5. What can be the approximate efficiency of a 3-phase, 50 Hz, 4-pole induction motor, which is running at 1470 rpm?
(a) 92 %
(b) 95 %
(c) 96 %
(d) 98 %

Q6. Sumpner’s test is conducted on transformers to study effect of ……………..
(a) Temperature
(b) Stray losses
(c) All-day efficiency
(d) none of these

SOLUTIONS

S1. Ans.(d)
Sol. Methods used for improving string efficiency of insulators:
Use of guard ring
Grading of insulators
Selection of k: k is the ratio of shunt capacitance to mutual capacitance. Lesser the value of k, grater will be efficiency. This can be achieved by using longer cross-arms.

S2. Ans.(c)
Sol. The relative permeability of different materials:
Diamagnetic: μ_r<1
Paramagnetic: μ_r>1(slightly greater than 1)
Ferromagnetic: μ_r≫≫>1(high)

S3. Ans.(a)
Sol. Half power frequencies are the frequencies (ω_1,ω_2) at which power dissipated is one half of the power dissipated at resonate frequency(w_o).
The bandwidth (BW) is defined as the frequency band between half power frequencies;
i.e.,BW=ω2-ω1

S4. Ans.(a)
Sol. The string efficiency is independent of transmission system frequency.

S5. Ans.(d)
Sol. N_s=120f/P=(120×50)/4=1500 rpm
∴efficiency(η)=(1-s)=Nr/Ns =1470/1500×100=98 %

S6. Ans.(a)
Sol. Sumpner’s test is the test which is used to determine the steady temperature rise if the transformer was fully loaded continuously; this is so because under each of these tests the power loss to which the transformer is subjected is either the core-loss or copper-loss but not both.

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