PSPCL-JE'21 EE: Daily Practices Quiz 19-Oct-2021 |_00.1
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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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