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# PSPCL-JE’21 EE: Daily Practices Quiz 19-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 19-Oct-2021

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

Q1. The secondary coil of an ideal 2:1 transformer has a 2 F capacitor connected across its terminals. The referred impedance on the primary side is of an element:
(a) L=8 H
(b) C=8 F
(c) L=0.5 H
(d) C=0.5 F

Q2. A 3 H inductor has 2000 turns. How many turns must be added to increase the inductance to 5 H?
(a) 582
(b) 2582
(c) 2500
(d) 1000

Q3. In the indirect method of lighting about how much percentage of light falls on the surface?
(a) 10 %
(b) 50 %
(c) 75 %
(d) 90 %

Q4. Ripple factor of half wave rectifier is _________
(a) 1.414
(b) 1.21
(c) 1.3
(d) 0.48

Q5. Quadrature axis synchronous reactance is the ratio of….
(a) Vmax/Imax
(b) Vmax/Imin
(c) Vmin/Imax
(d) Vmin/Imin

Q6. Slip of an induction motor ……………. when the rotor starts gaining speed.
(a) Increases
(b) Decreases
(c) Remains same
(d) Increase or decrease

## SOLUTIONS

S1. Ans.(d)
Sol. Referred element on primary side will be capacitor.
As Xc=1/2πfC i.e., C α 1/Xc
So, C2^’=C2/(N1/N2 )^2 =2/(2)^2 =0.5 F

S2. Ans.(a)
Sol. As L α N^2
∴(N2/N1 )^2=L2/L1
∴N2=N1×√(L2/L1 )=2000×√(5/3)=2582
∴Number of turns to be added=2582-2000=582.

S3. Ans.(a)
Sol. indirect lighting method of lighting, 90 % of the light is reflected upwards and only 10 % falls on the surface. They are used for mainly decoration purposes in cinemas, hotels, club etc.

S4. Ans.(b)
Sol. Ripple factor of a rectifier is the measure of the effectiveness of a power supply filter in reducing the ripple voltage. It is calculated by taking ratio of RMS AC component of output voltage to DC component of output voltage.
r = √(Irms^2 – Idc^2)/Idc
For a half wave rectifier, it is 1.21.

S5. Ans.(c)
Sol. Quadrature axis synchronous reactance: Xq=Vmin/Imax
Direct axis synchronous reactance: Xd=Vmax/Imin

S6. Ans.(b)
Sol. Slip(s)=(Ns-Nr)/Ns
So, when the rotor starts gaining speed, due to increase of rotor speed, slip of an induction
motor decreases.

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