GATE'22 EE: Daily Practices Quiz 30-Oct-2021 |_00.1
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GATE’22 EE: Daily Practices Quiz 30-Oct-2021

GATE 2022

IIT Kharagpur has released the GATE 2022 exam dates. As per the GATE 2022 exam date, the exam will be conducted on February 5, 6, 12 & 13. Candidates preparing for GATE 2022 must attempt this Civil Engineering Quiz to boost your GATE exam preparation.

GATE’22 EE QUIZ

GATE’22 EE: Daily Practices Quiz 30-Oct-2021

Each question carries 2 marks
Negative marking: 2/3 mark

Total Questions: 10
Total marks: 20
Time: 20 min.

Q1. The impulse response of a system is h(n)=a^n u(n). The condition for the system to be BIBO stable is:
(a) |a|>1
(b) |a|<1
(c) a is real and negative
(d) a is real and positive

Q2. For a “two-port reciprocal” network the output open-circuit voltage divided by the input current is equal to:
(a) h12
(b) 1/h12
(c) z12
() 1/y12

Q3. Match the statements in right column with the Left column and select the correct answer using the code given below the lists

(i)                Lead compensator A.      Maximum peak or overshoot decreases
(ii)              Lag compensator B.      Improves transient response
(iii)           Proportional controller C.      Steady state error reduces
(iv)            Integral controller D.     Improves the steady state response

(a) (i)-D, (ii)-B, (iii) – A, (iv)-C
(b) (i)-C, (ii)-D, (iii) – B, (iv)-A
(c) (i)-B, (ii)-D, (iii) – C, (iv)-A
(d) (i)-A, (ii)-C, (iii) – D, (iv)-B

Q4. The maximum phase shift that can be provided by a lead compensator with transfer function G(S)=(1+6s)/(1+2s) is
(a) 15°
(b) 30°
(c) 45°
(d) 60°

Q5. Consider a parallel RLC circuit having inductance of 10 mH, capacitance of 100 μF.
Determine the value of resistance that would lead to a critically damped response?
(a) 5 Ω
(b) 10 Ω
(c) 20 Ω
(d) 15 Ω

Q6. Neglecting losses, if the power transformed inductively is equal to power transformed conductively in an auto-transformer, then the secondary to primary ratio of transformer is:
(a) 1
(b) 2
(c) 0.5
(d) 1.25

Q7. What is the power transmitted inductively in an auto-transformer which supplies a load at 161 Volts with an applied primary voltage of 230 Volts?
(a) 15 % of input
(b) 30 % of input
(c) 35 % of input
(d) 70 % of input

Q8. A current of 10+6 sin⁡(ωt+30)A is passed through two meters. They are a PMMC meter and a moving iron instrument. The respective reading in (A) will be:
(a) 10 A and 10 A
(b) 10 A and 16 A
(c) 10 A and 10.86 A
(d) 0 A and 6 A

Q9. The energy gap between valance band and conduction band for semi-conductor is:
(a) 1 eV
(b) 7 eV
(c) 0.1 eV
(d) 0.7 eV

Q10. In case of class A amplifiers, the ratio of efficiency of transformer coupled amplifier to efficiency of a transformer less amplifier is
(a) 0.5
(b) 1.0
(c) 1.36
(d) 2.0

SOLUTIONS

S1. Ans.(b)
Sol. h(n)=a^n u(n)
∴H(z)=1/(1-az^(-1) )=z/(z-a); |z|>|a|
Thus, system is stable for |a|<1 .

S2. Ans.(c)
Sol. ├ V12/I1 ┤(I2=0)=Z12=Z21 (reciprocal network)

S3. Ans.(c)
Sol.
Lead compensator – improves transient response
Lag compensator – improves the steady state response
Proportional controller – Steady state error reduces
Integral controller – Maximum peak or overshoot decrease

S4. Ans.(b)
Sol. Given, G(s) = (1+6s)/(1+2s)=(1+T1 s)/(1+αT1 s)
Here, T1=6
And, αT1=2 ∵ α = 2/3=1/3
∵ sin ϕm=(1-α)/(1+α)=(1-1/3)/(1+1/3)=2/4=1/2
ϕm=sin^(-1) (1/2)
ϕm=30°= maximum phase shift

S5. Ans.(a)
Sol. For parallel RLC: ξ=1/2R √(L/C)
For critically damped: 𝜉=1
∴1=1/2R √((10×10^(-3))/(100×10^(-6) ))
⇒R=10/2 =5Ω

S6. Ans.(c)
Sol. Power transfer inductive=V2 I2 (1-K)
Power transfer conductive=V2 I2 K
ACQ: Power transfer inductive= Power transfer conductive
∴1-K=K⇒K=0.5

S7. Ans.(b)
Sol. Power transfer inductive=Sin (1-VL/VH )=Sin (1-161/230)=30 % of S_in

S8. Ans.(c)
Sol. PMMC instrument reads only DC value. Moving iron instrument will read rms value of the current.
∴PMMC reading=10 A
MI instrument reading=√(10^2+(6/√2)^2 )=√118=10.86 A

S9. Ans.(a)
Sol. A semiconductor material has almost filled valence band and nearly empty conduction band with a very small energy gap (≃ 1 eV) between them.

S10. Ans.(d)
Sol. efficiency of class-A amplifier: –
With transformer coupled = 50%
Without transformer coupled = 25%
∴ Required ratio = 50/25=2

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