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Quiz Electronics and Communication Engineering 12 Aug 2020

Quiz Electronics and Communication Engineering
Exam: GATE
Topic: Miscellaneous
Date: 12/08/2020

Each Question carries 2 Mark
Negative Marking: 1/3
Time: 20 Minutes

Q1. Electric field inside a hallow metallic charged sphere is
(a) Zero
(b) Decreasing towards centre
(c) Increasing toward centre
(d) None of these

Q2. An R-L series AC circuit has 3V across resistor and 4V across the inductor, the supply voltage is
(a) 7 volt
(b) 5 volt
(c) 1 volt
(d) zero

Q3. An emitter follower has high input impedance because
(a) Large load resistance may be used
(b) Large biasing resistance is used
(c) large emitter resistance is used
(d) There is negative feedback in the base emitter circuit

Q4. The hexadecimal representation of (657)_8 is:
(a) 1AF H
(b) D78 H
(c) D71 H
(d) 32F H

Q5. On receiving an interrupt from an I/0 device the CPU
(a) Branches off to the interrupt service routine after completion of the current instruction
(b) Halts for a predetermined time
(c) Branches off to the interrupt service routine immediately
(d) Hands over control of address bus and data bus to the interrupting device

Q6. The power factor of DC circuit is
(a) 0
(b) 1
(c) – 1
(d) None of these

Q7. If the response of LTI continuous time system to unit step input is (1/2 – 1/2 e^(-2t)), then impulse response of the system is
(a) (1/2 – 1/2 e^(-2t))
(b) e^(-2t)
(c) (1 – e^(-2t))
(d) Constant

Q8. An FM signal with modulation index mƒ is passed through a frequency tripler. The modulation index of the output signal will be
(a) mƒ
(b) 3mƒ
(c) 9mƒ
(d) 27mƒ

Q9. In Force-Voltage Analogy
(a) Force is analogous to current
(b) Mass is analogous to capacitance
(c) Velocity is analogous to current
(d) Displacement is analogous to magnetic flux linkage

Q10. In a transistor, the base current and collector current are 60 µA and 1.75 mA. The value of α is nearly
(a) 0.97
(b) 0.91
(c) 1.3
(d) 1.7

Solution

S1. Ans.(a)
Sol. Inside a hallow metallic charged sphere, electric field is always zero.

S2. Ans.(b)
Sol. Given:
Voltage across inductor(V_L^ ) = 4 V
Voltage across resistor(V_R^ ) = 3 V
Now, Supply Voltage is
⇒ V = √(V_L^2+V_R^2 )
⇒ V = √(3^2+4²)
So, V = √25=5 volt

S3. Ans.(d)
Sol. Common Collector configuration is also called an Emitter follower. It has high input impedance because there is negative feedback in the base emitter circuit.

S4. Ans.(a)
Sol.
Step 1: Converting Octal number to binary. We know that each octal digit is represented as 3 bit binary number.
⇒ 657_8 = (110 101 111)2
Step 2: Converting Binary number to Hexadecimal. We know that each Hexadecimal digit is represented as 4 bit binary number. If total binary digit are not in multiple of 4 we can include ‘0’ to the left of MSB in binary number.
⇒ (110 101 111)2 = (0001 1010 1111)2
⇒ (0001 1010 1111)2 = 1AF16 [ ∵ 10102 = A16]

Therefore, Hexadecimal representation of (657)_8 = 1AF H

S5. Ans.(b)
Sol. On receiving an interrupt from I/0 device, CPU branch to the interrupt service routine.

S6. Ans.(b)
Sol. In DC circuit, phase difference between current and voltage is ϕ = 0
So, Power Factor is cosϕ = cos0 = 1
OR,
We can also calculate power factor as
cosϕ = (Active power)/(Apparent power )=kw/kVA
For DC circuit, apparent power = active power
Hence, Power factor of DC circuit is unity.

S7. Ans.(b)
Sol. Given:
u(t) = (1/2 – 1/2 e^(-2t))
Impulse response is –
⇒ 𝛿(t) = (du(t))/dt
⇒ 𝛿(t) = d/dt(1/2 – 1/2 e^(-2t))
So, 𝛿(t) = e^(-2t)

S8. Ans. (b)
Sol. The ‘n’ frequency multiplier multiplies the modulation index, carrier frequency, frequency deviation components of FM signal by n times except message signal frequency. Here, frequency tripler is used. So, mƒ changes to 3mƒ .

S9. Ans.(c)

S10. Ans.(a)
Sol. Given:
I_C = 1.75 mA = 1.75 × 10^(-3)
I_B = 60 µA = 60 × 10^(-6)

Now, β = I_C/I_B
⇒ β = (1.75 × 10^(-3))/(60 × 10^(-6) )
So, β ≈ 29

and α = β/(β+1)
⇒ α = 29/(29+1)
⇒ α = 29/30
So, α ≈ 0.966

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