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GATE’22 EE: Daily Practices Quiz 13-Sep-2021
Each question carries 2 marks
Negative marking: 2/3 mark
Total Questions: 05
Total marks: 10
Time: 12 min.
Q1. A sensor requires 30 s to indicate 90% of the response to a step input. If the sensor is a first – order system, the time constant is
[given,log_e (0.1)= -2.3]
(a) 15 s
(b) 13 s
(c) 21 s
(d) 28 s
Q2. Consider the time response of a second – order system with damping coefficient less than 1 to a unit step input:
1. It is over damped.
2. It is periodic function.
3. Time duration between any two consecutive values of 1 is the same.
Which of the above statements is/are correct?
(a) 1, 2 and 3
(b) 1 only
(c) 2 only
(d) 3 only
Q3. Buses for load flow studies are classified as:
1. load bus
2. PV bus
3. Slack bus
Which one of the following is the correct combination of the pair of quantities specified having their usual meaning for different buses?
Q4. If the applied voltage of a certain transformer is increased by 50 % and the frequency is reduced to 50 % (assuming that the magnetic circuit remains unsaturated), the maximum core flux will
(a) Remains the same as original
(b) Change to three times the original value
(c) Change to 0.5 times the original value
(d) Change to 1.5 times the original value
Q5. What happened to the speed when the flux is reduced by 10 % in a 200 V DC shunt motor having an armature resistance of 0.2 Ω carrying a current of 50 A and running at 960 rpm prior to weakening of field. The total torque may be assumed constant & neglect losses….
(a) 1250 rpm
(b) 576 rpm
(c) 920 rpm
(d) 1066 rpm
SOLUTIONS
S1. Ans.(b)
Sol. Step response of first order system,
c(t)=1-e^(-t\/τ)
0.9=1-e^(-30\/τ)
⇒ e^(-30\/τ)=0.1;
⇒ τ= -30/(log_e (0.1) )=30/2.3 ▭(τ=13 sec)
S2. Ans.(d)
Sol.
For an underdamped system ξ < 1; oscillation is damped by frequency.
ωd=ωn √(1-ξ^2 )
S4. Ans.(b)
Sol. ϕ α V/f
ACQ: the applied voltage of a certain transformer is increased by 50 % and the frequency is reduced to 50 %.
∴ϕ^’=(1.5 V_1)/(0.5 f_1 )=3ϕ.
S5. Ans.(d)
Sol. Torque (T)=kϕN=constant
ACQ: T_=T2
∴ϕ1 N1=ϕ2 N2
∴N2=(ϕ1 N1)/ϕ2 =(ϕ1×960)/(0.9ϕ1 ) (Flux is reduced by 10%)
⇒N2=1066 rpm.
