Engineering Jobs   »   Electrical Engineering quizs   »   UPPCL 2021, quiz: UPPCL-JE ELECTRICAL,

UPPCL-JE’21 EE: Daily Practices Quiz 14-Dec-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 14-Dec-2021

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

Q1. Which one of the following methods gives voltage regulation higher than the actual value in an alternator?
(a) ZPF method
(b) MMF method
(c) EMF method
(d) ASA method

Q2. The purpose of coating on arc welding electrodes is to
(a) stabilize the arc
(b) provide a protecting atmosphere
(c) provide slag to protect the molten metal
(d) all of the above

Q3. Thin laminations are used in a machine in order to reduce
(a) Eddy current losses
(b) Hysteresis losses
(c) Both a and b
(d) None of the above

Q4. Conductors, which connects the consumer’s terminal to distribution line is known as…
(a) Feeders
(b) Service-mains
(c) Distributors
(d) General cables

Q5. One unit of electrical energy equals
(a) 1 KWH
(b) 1 WH
(c) 10 WH
(d) 100 WH

Q6. If the applied voltage of a DC motor is 230 V, then back emf, for maximum power developed is
(a) 460 V
(b) 230 V
(c) 200 V
(d) 115 V


S1. Ans.(c)
Sol. EMF method of voltage Regulation is known as “Pessimistic Method” because gives VR higher than actual.

S2. Ans.(d)
Sol. The primary purpose of a light coating is to increase arc stability, provide a protective
gaseous atmosphere to prevent oxygen, hydrogen, and nitrogen picks up by the molten
metal and Protective slag over hot metal etc.

S3. Ans.(a)
Sol. Thin laminations are used in order to reduce the eddy current losses only. Due to laminations the area of the eddy currents loops is minimized and the losses due to eddy current losses are minimized.

S4. Ans.(b)
Sol. The service mains conductors form connecting links between distributors and metering points at the consumer terminal.

S5. Ans.(a)
Sol. One unit of electrical energy equals 1KWH.

S6. Ans.(d)
Sol. For maximum power output, the back emf is equal to half of the applied voltage.
Hence, Eb = V/2 = 230/2 ⇒115 V. where V is the terminal voltage.

Sharing is caring!

Leave a comment

Your email address will not be published. Required fields are marked *