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

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

Q1. Interpoles are meant for
(a) Increasing counter emf
(b) Increasing the speed of the motor
(c) Reducing sparking at the commutator
(d) Strengthening the main field

Q2. The angle between induced emf and terminal voltage on no-load of a single-phase alternator is
(a) 0⁰
(b) 90⁰
(c) 180⁰
(d) 270⁰

Q3. The sag of a transmission line conductor in summer is……
(a) Same as in winter
(b) More than that in winter
(c) Less than that in winter
(d) None of the above

Q4. The angular velocity of the waveform is given as 440 rad/sec. calculate the frequency (in Hz) of the waveform.
(a) 65
(b) 75
(c) 60
(d) 70

Q5. The power gain in decibels can be expressed as
(a) 20 log_10⁡(PO/Pi )
(b) 10 log_10⁡(VO/Vi )
(c) 10 log_10⁡(PO/Pi )
(d) 20 log_10⁡(PO/Vi )

Q6. Negative feedback used in amplifier results in-
(a) More gain, more bandwidth
(b) More gain, less bandwidth
(c) Less gain, more bandwidth
(d) Less gain, less bandwidth

 

SOLUTIONS

S1. Ans.(c)
Sol. interpoles are used to improve armature reaction and to reduce sparking.

S2. Ans.(a)
Sol. E∠Ө=V∠0^0+Ia∠ϕ×90⁰
At no-load, Ia=0
∴ E∠Ө=V∠0^0
⇒Ө=0⁰.

S3. Ans.(b)
Sol. Sag depends on the temperature and ice loading but effect of temperature is dominating. So, the sag of a transmission line conductor in summer is more than that in winter.

S4. Ans.(d)
Sol. f=ω/2π=440/(2×3.14)=70 Hz.

S5. Ans.(c)
Sol. Power gain =P_o/P_i
in decibel (dB):
dB = 10 log10⁡(Po/Pi )
=10 log10⁡(Vo/Vi )^2
=20 log10⁡(Vo/Vi )

S6. Ans.(c)
Sol. Effects of Negative feedback:
In negative feedback, the feedback energy (voltage or current), is out of phase with the input signal and thus opposes it.
Negative feedback reduces gain of the amplifier. It also reduces distortion, noise and instability.
This feedback increases bandwidth and improves input and output impedances.

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