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UPPSC-LECTURER’21 EE: Daily Practices Quiz 07-Oct-2021

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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+tan⁡ϕ2)
(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 %


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 cos⁡ϕ+X sin⁡ϕ)×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 %

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