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UPPSC Technical Education Teaching Exam Notification 2021
Uttar Pradesh Public Service Commission has released the official notification for the recruitment of Principal, Lecturer, Librarian, 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.
UPPSC-LECTURER’21 EE: Daily Practices Quiz
UPPSC-LECTURER’21 EE: Daily Practices Quiz 07-Oct-2021
Each question carries 3 marks.
Negative marking: 1 mark each or 1/3rd
Total Questions: 06
Time: 08 min.
Q1. The unit step response of a network is (1-e^(-αt)), then its unit impulse response is
(a) αe^(-αt)
(b) (1-α^(-1) ) e^(-αt)
(c) α^(-1) e^(-αt)
(d) (1-α) e^(-αt)
Q2. In thyristor, holding current is:
(a) More than the latching current
(b) Less than the latching current
(c) Equal to latching current
(d) None of the above
Q3. A load draws an active power P at a lagging p.f. of cosϕ1. If the p.f. is improved to
cosϕ2, the leading kVAR supplied by p.f. correction equipment will be
(a) P (cosϕ2-cosϕ1)
(b) P (sinϕ2-sinϕ1)
(c) P (tanϕ1+tanϕ2)
(d) P (tanϕ1-tanϕ2)
Q4. In TV, video signals are transmitted through
(a) Frequency modulation
(b) Pulse modulation
(c) Amplitude modulation
(d) Phase modulation
Q5. A 10 kVA, 400 V/ 200 V, single-phase transformer with a percentage resistance of 3% and percentage reactance of 6% is supplying a current of 50 A to a resistive load. The value of the load voltage is
(a) 192 V
(b) 194 V
(c) 196 V
(d) 390 V
Q6. A 40 KVA transformer has a core loss of 400 W and full-load copper loss of 800 W. The fraction of rated load at maximum efficiency is
(a) 50 %
(b) 62.3 %
(c) 70.7 %
(d) 100 %
SOLUTIONS
S1. Ans.(a)
Sol. To find out impulse response, simply differentiate unit step response:
i.e., d/dt (1-e^(-αt) )=-(-e^(-αt) )=αe^(-αt).
S2. Ans.(b)
Sol. In thyristor, holding current is less than latching current.
Generally, IL/IH ≃2.5
S3. Ans.(d)
Sol. Leading kVAR supplied = P (tanϕ1 – tanϕ2)
S4. Ans.(c)
Sol. Frequency modulation is used for Radio TV sound transmissions while amplitude modulation is used for video transmission.
S5. Ans.(b)
Sol. % regulation= (R cosϕ+X sinϕ)×100
For resistive load: ?=0⁰
∴% regulation=3×1+6×0=3%=200×3/100=6 V
∴Terminal voltage=200-6=194 V
S6. Ans.(c)
Sol. η=√(Pi/Pcufl )=√(400/800) ×100=70.7 %