Quiz Electrical Engineering

Exam: GATE

Topic: Miscellaneous

Date: 29/06/2021

Each Question carries 2 Mark

Negative Marking: 1/3

Time: 20 Minutes

Q1. For the network shown, the equivalent Thevenin voltage and Thevenin impedance as seen across terminals ‘ab’ is

(a) 50 V in series with 2 Ω

(b) 10 V in series with 2 Ω

(c) 65 V in series with 15 Ω

(d) 10 V in series with 12 Ω

Q2. The Fourier transform of the signum function Sgn (t) defined in the figure is

(a) (-2)/jω

(b) 4/jω

(c) 2/jω

(d) 1/jω+1

Q3. Which one of the following is the correct Fourier transform of the unit step signal?

u(t)=1 for t≥0

=0 for t≤0

(a) πδ(ω)

(b) 1/jω

(c) 1/jω+πδ(ω)

(d) 1/jω+2πδ(ω)

Q4. The Fourier transform of v(t) = cos ω0 t is given by

(a) V(f)=1/2 δ (f-f0 )

(b) V(f)=1/2 δ(f+f0 )

(c) V(f)=1/2 [δ(f-f0 )-δ(f+f0 )]

(d) V(f)=1/2 [δ(f-f0 )+δ(f+f0 )]

Q5. A 400V/100V, 10kVA two winding transformer is reconnected as an auto transformer across a suitable voltage source. The maximum rating of such an arrangement could be

(a) 50 kVA

(b) 15 kVA

(c) 12.5 kVA

(d) 8.75 kVA

Q6. A 10 kVA, 400 kVA/200 kVA, single – phase transformer with a percentage resistance of 3% and percentage reactance of 6% is supplying a current of 50A to a resistive load. The value of the load voltage is

(a) 194 V

(b) 390 V

(c) 192 V

(d) 196 V

Q7. The switch ‘S’ in the circuit has been opened since a very long time. The value of current through resistor R_1 Immediately after switch is closed will be –

(a) zero

(b) V\/R1

(c) V\/R2

(d) V/(R1+R2 )

Q8. In a transistor amplifier, the input resistance is 1KΩ, the collector load resistance is 2KΩ and the transistor β is 10. The voltage gain is

(a) 40

(b) 30

(c) 20

(d) 10

Q9. In the sum of product function

f(x,y,z)=Σ(2,3,4,5)

The prime implicants are-

(a) x ̅y+xy ̅

(b) x ̅y+xy ̅z ̅+xy ̅z

(c) x ̅yz ̅+x ̅yz+xy ̅

(d) x ̅yz ̅+x ̅yz+xy ̅z ̅+xy ̅z

Q10. The output Y of the logic circuit given below is:

(a) 0

(b) 1

(c) X

(d)` X ̅

**SOLUTIONS**

S5. Ans.(a)

Sol. The auto transformer connections can be made as below:

I_P=(10×1000)/400=25A

I_S=400/100×25=100 A

Rating = 500 V × 100 A = 50 kVA

S6. Ans.(a)

Sol. %regulating = ((IRcosϕ+IXsinϕ))/I×100

==(%R cosϕ+%Xsinϕ)

Being a resistive load, 𝟇=0

% regulation = (3×cos0°+6×sin0°)=3%

=(3×200)/100=6V

Hence terminal voltage

= 200 – 6=194 V

S7. Ans.(b)

Sol. S is opened after long time. So, L behaves as short-circuit as soon as switch open.

R_1 ||R_2

∵V_R1=V\/R_1

S8. Ans.(c)

Sol. Since

Av = β×Rc/Rin = 10 × 2/1 = 20