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GATE’22 EE: Daily Practices Quiz
GATE’22 EE: Daily Practices Quiz 17-Sep-2021
Each question carries 2 marks
Negative marking: 2/3 mark
Total Questions: 05
Total marks: 10
Time: 12 min.
Q1. The forward resistance of the diode shown in Figure is 5 Ω and the remaining parameters are same at those of an ideal diode. The dc component of the source current is (in Ampere):
(a) Vm/50π
(b) 2Vm/50π
(c) Vm/(50π√2)
(d) Vm/(100π√2)
Q2. A half wave diode rectifier and full wave diode bridge rectifier both have an input frequency of 50 Hz. The frequencies of outputs respectively are:
(a) 100 Hz and 50 Hz
(b) 50 Hz and 100 Hz
(c) 100 Hz each
(d) 50 Hz each
Q3. A concentric cable has a conductor diameter of 1 cm and an insulation thickness of 1.5 cm. when the cable is subjected to a test pressure of 33kV, the maximum field strength will be nearly
(a) 41,000 V
(b) 43,200 V
(c) 45,400 V
(d) 47,600 V
Q4. Radio influence voltage (RIV) generated on a transmission line conductor surface is not affected by
(a) System voltage
(b) Corona discharge on the conductors
(c) Rain
(d) Nearby radio receivers
Q5. A 220 V, 15 kW, 1000 rpm shunt motor with armature resistance of 0.25 Ω, has a rated line current of 68 A and a rated field current of 2.2 A. The change in field flux required to obtain a speed of 1600 rpm while drawing a line current of 52.8 A and a field current of 1.8 A is
(a) 36.36 % decrease
(b) 36.36 % increase
(c) 18.18 % increase
(d) 18.18 % decrease
SOLUTIONS
S1. Ans.(a)
Sol. This is half-wave rectifier circuit.
∴V_dc=V_m/π
∴DC current in the circuit is, Idc=Vdc/(Rf+R)=((Vm/π))/(5+45)=Vm/50π
S2. Ans.(b)
Sol. For HWR, f0=fi=50 Hz
For FWR, f0=2fi=2×50=100 Hz.
S3. Ans.(d)
Sol. Given data
Conductor diameter (2r = d) = 1cm
Insulation thickness (t) = 1.5 cm
Test voltage (V) = 33 kV
Hence D = 2R = 2r + 2t = d + 2t = 4cm
Maximum field strength is given as
E_max=V/(rln R/r )=V/(rln D\/d)
=33000/(0.5 ln 4/1)=47608.93 V\/cm
S4. Ans.(d)
Sol. Radio influence voltage depends on
Operating system voltage
Corona discharge
Surface irregularity
Atmospheric conditions (rain, dust)
Though it does not depend upon nearby radio receivers.
S5. Ans.(a)
Sol. N1/N2 =Eb1/Eb2 ×ϕ2/ϕ1
Ra = 0.25 Ω
Ia1 = 68 – 2.2 = 65.8A;
Ia2 = 52.8 – 1.8 = 51A
∴1000/1600=((220-65.8*0.25))/((220-51*0.25))×ϕ_2/ϕ_1
∴ϕ2/ϕ1 =0.6364
∴% decrease=(1-ϕ2/ϕ1 )×100=36.36 %
