Engineering Jobs   »   Electrical Engineering quizs   »   UPPCL JE 2021, UPPCL JE ELECTRICAL...

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

Table of Contents

## UPPCL RECRUITMENT 2021

Uttar Pradesh Power Corporation Limited has rolled out the official notification for the recruitment of Assistant Engineers and Junior Engineers for a total of 286 UPPCL Vacancy 2021. The Online submission of the application form for UPPCL Recruitment 2021 starts on 12th November 2021 and the last date to apply for the same is 2nd December 2021.

## 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.

 Rectifier Max. Efficiency HWR 40.6 % FWR 81.2 %

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

S4. Ans.(d)

Sol.

 Diodes Si (silicon) 0.7 V Ge(germanium) 0.3 V

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

Sharing is caring!

Thank You, Your details have been submitted we will get back to you.

Leave a comment