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# GATE’22 EE: Daily Practices Quiz 08-Sep-2021

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GATE’22 EE: Daily Practices Quiz 08-Sep-2021

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

Total Questions: 05

Total marks: 10
Time: 12 min.

Q1. An open loop control system results in a response of e^(-2t) (sin5t + cos5t) for a unit impulse input. The DC gain of the control system is ______.
(a) 241
(b) 0.241
(c)41
(d) 0.21

Q2. For a two-port network the three parameters A, B and C are given by—
A = x1,B = x2,C =1/x2
then D is equal to under the network to be reciprocal—
(a) x1
(b) 0
(c) 2/x(1 )
(d) 2×1

Q3. The incandescent bulbs respectively as P_1 and P_2 for operation at a specified main voltage are connected in series across the mains as shown in the figure, then the total power supplied by the mains to the two bulbs— (a) P1 P2
(b) P1^2+P2^2
(c) (P1 P2)/(P1+P2 )
(d) P1+P2

Q4. The time constant of the network shown in the fig. is— (a) 4RC
(b) 3RC
(c) 2RC
(d) 2RC/3

Q5. Calculate the Voltage regulation of the transformer in which ohmic loss is 1 % of the output and reactance drop is 5 % of the voltage when the power factor is 0.8 lag
(a) 1 %
(b) 3.5 %
(c) 3.8 %
(d) 5 %

SOLUTIONS

S1. Ans.(b)
Sol. g(t)=e^– 2t (sin5t + cos5t)
Taking Laplace transform; G(s)=5/((s+2)^2+5^2 )+(s+2)/((s+2)^2+5^2 )
For DC gain; |G(s)|_(s=0)
∴G(0)=5/((0+2)^2+5^2 )+(0+2)/((0+2)^2+5^2 )=5/29+2/29=7/29=0.241

S2. Ans.(c)
Sol. The condition of reciprocity for ABCD parameters is given by relation
∴x1 D-x2×1/x2 =1
∴x1 D=1+1=2
⇒D=2/x1

S3. Ans.(d)
Sol. Since bulbs are connected in series.
∴PT=I^2 (R1+R2 )=P1+P2

S4. Ans.(a)
Sol. Time constant τ = R_eq C
∵R_eq = 4R | | 4R = 2R
C = 2C
so, τ = 2 R. 2C
= 4RC

S5. Ans.(c)
Sol. here, %R=1 %
%X=5 %
Pf=cos𝟇=0.8 lagging
∴VR=%R cosϕ+%X sinϕ=1*0.8+5*0.6=3.8 %(+ sign because of lagging pf)

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