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# UPPSC-LECTURER’21 EE: Daily Practices Quiz 06-Oct-2021

## UPPSC Technical Education Teaching Exam Notification 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.

## UPPSC-LECTURER’21 EE:  Daily Practices Quiz

UPPSC-LECTURER’21 EE: Daily Practices Quiz 06-Oct-2021

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

Q1. For a JFET IDSS = 8 mA and peak voltage Vp = –8V, what will be the drain current for gate to source voltage of –2V?
(a) 4.5 mA
(b) 8 mA
(c) 16 mA
(d) 12 mA

Q2. The plot of modulation index versus carrier amplitude for an amplitude modulated signal yields
(a) Horizontal line
(b) Vertical line
(c) Parabola
(d) Hyperbola

Q3. Gain cross-over frequency is defined as one at which |G(jω) H(jω)| equals
(a) 0
(b) 1
(c) √2
(d) 1/√2

Q4. A MOSFET has………terminals.
(a) Two
(b) Five
(c) Four
(d) Three

Q5. Which of the following is not Maxwell’s equation?
(a) ∆×E= -∂B/∂t
(b) ∆.E= -B
(c) ∆.D=ρ
(d) ∆.B= 0

Q6. A transformer has a core loss of 64 W and copper loss of 144 W, when it is carrying 20 % overload current. The load at which this transformer will operate at the maximum efficiency……
(a) 120 %
(b) 80 %
(c) 44 %
(d) 66 %

## SOLUTIONS

S1. Ans.(a)
Sol. ID=IDSS (1-VGS/VP )^2=8×10^(-3) (1-(-2)/(-8))^2=4.5 mA

S2. Ans.(d)
Sol. For an amplitude modulated signal, modulation index(μ)=Am/Ac
So, plot of μ v/s Ac will be hyperbola. (Graph similar to y=1/x).

S3. Ans.(b)
Sol. At gain cross-over frequency, the magnitude of G(jω) is 1.
Or at ω=ω_gc; |G(jω) H(jω)|=1
Phase margin is calculated at gain cross-over frequency.

S4. Ans.(d)
Sol.

Schematic symbol of n-channel MOSFET

S6. Ans.(b)
Sol. At 20% overload current (i.e.,I=1.2I_fl) & copper loss=144 W
As copper loss(P_cu ) α I^2
∴Copper loss at full-load, Pcu(fl) =144/(1.2)^2 =100 W
Core loss or constant loss=64 W
Load at which efficiency is maximum,
∴η_max=rated KVA×√(Pi/Pcu(fl) )=rated KVA×√(64/100)=0.8×rated KVA
∴At 80% load, efficiency is maximum.

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