UPPCL-JE’21 EE: Daily Practices Quiz 10-Dec-2021

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UPPCL-JE’21 EE: Daily Practices Quiz

UPPCL-JE’21 EE: Daily Practices Quiz 10-Dec-2021

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

Q1. The average current rating of a semiconductor diode will be maximum for:
(a) Half-wave rectified ac
(b) Full-wave rectified ac
(c) Pure dc
(d) Pure ac

Q2. The maximum efficiency of a half-wave rectifier is …………………
(a) 40.6 %
(b) 81.2 %
(c) 50 %
(d) 25 %

Q3. A Zener diode is used as …………….
(a) an amplifier
(b) a voltage regulator
(c) a rectifier
(d) a multivibrator

Q4. The forward voltage drop across a silicon diode is about ………
(a) 2.5 V
(b) 3 V
(c) 10 V
(d) 0.7 V

Q5. A 6 pole Lap wound dc generator has 300 conductors. Emf induced per conductor is 5 volts. This generator will generate emf of:
(a) 60 V
(b) 250 V
(c) 300 V
(d) 1800 V

Q6. Which type of winding is generally preferred for generating large currents on DC generators?
(a) Progressive wave winding
(b) Retrogressive wave winding
(c) Lap winding
(d) Current depends on design

SOLUTIONS

S1. Ans.(c)
Sol.
For HWR, I_avg=I_m/π=0.3 I_m
For FWR, I_avg=(2I_m)/π=0.6I_m
For pure DC, I_avg=I_m
For pure AC, I_avg=0

S2. Ans.(a)
Sol.

S3. Ans.(b)
Sol. A Zener diode is used as a voltage regulator.

S4. Ans.(d)

Sol.

S5. Ans.(b)
Sol. Emf induced per conductor=5 V
For Lap wound, number of parallel paths(A)=P=6
No. of conductor per parallel path=300/6=50
∴total emf generated by generator=50×5=250 V

S6. Ans.(c)
Sol. LAP winding: large current & low voltage applications
WAVE winding: high voltage & low current applications

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