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# UPPSC-LECTURER’21 EE: Daily Practices Quiz 25-NOV-2021

## UPPSC polytechnic lecturer  2021

UPPSC Technical Lecturer Eligibility Criteria 2021. Uttar Pradesh Public Service Commission has released the official notification for the recruitment of PrincipalLecturerLibrarian, and Workshop Superintendent. The Online application has been started on 15th September 2021 and the last date to apply online is 15th October 2021. The recruitment is for a total of 1370 UPPSC Polytechnic Lecturer Vacancies.

Exam Date: 12/12/2021

## UPPSC polytechnic lecturer QUIZ: EE

UPPSC-LECTURER’21 EE: Daily Practices Quiz 25-NOV-2021

Each question carries 3 marks.
Negative marking: 1 mark each or 1/3rd
Total Questions: 06
Time: 08 min.

Q1. The steady-state error of the system can be minimized by
(a) Increasing gain K
(b) Increasing the settling time
(c) Decreasing gain K
(d) Reducing the oscillating frequency

Q2. Diode is a…………device.
(a) Unilateral and linear
(b) Unilateral and non-linear
(c) Bilateral and linear
(d) Bilateral and non-linear

Q3. In a parallel RLC circuit if the lower cut off frequency is 2400 Hz and the upper cut off frequency is 2800 Hz what is the band width?
(a) 400 Hz
(b) 5200 Hz
(c) 2400 Hz
(d) 2800 Hz

Q5. Which one of the following is not a vectored interrupt?
(a) RST 7.5
(b) TRAP
(c) RST 3
(d) INTR

Q6. The synchronous speed for the 7^th space harmonic mmf wave of a 3-phase, 8 pole, 50 Hz induction machine is
(a) 5250 rpm in forward direction
(b) 5250 rpm in reverse direction
(c) 107.14 rpm in forward direction
(d) 107.14 rpm in reverse direction

## SOLUTIONS

S1. Ans.(a)
Sol. ess α 1/K
So, by increasing gain, steady-state error of the system can be minimized.

S2. Ans.(b)
Sol. Diode conducts current in single direction and have non-linear VI characteristics.

S3. Ans.(a)
Sol. Bandwidth=f_H-f_L=2800-2400=400 Hz.

S5. Ans.(d)
Sol. ‘INTR’ is a non-vectored interrupt because it doesn’t have any address of its own. It is the least priority interrupt.
NOTE:
RST7.5, RST6.5, RST5.5, TRAP are vector interrupts.

S6. Ans.(c)
Sol. N_s=120f/P=(120×50)/8=750 rpm
Speed of 7^th harmonic=Ns/7=750/7=107.14 rpm
And, for7^th harmonic=7×120^0=840^0=840-360-360=+120⁰(i.e.forward direction)

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