## 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: ?=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