Engineering Jobs   »   Quiz Electronics Engineering 17 Oct 2020

Quiz Electronics Engineering 17 Oct 2020

Quiz Electronics Engineering
Exam: NIC
Topic: Miscellaneous
Date: 17/10/2020

Each Question carries 1 Mark
Negative Marking: 1/4
Time: 10 Minutes

Q1. Consider the function F(s) = 1/(s(s^2+3s+2)) where F(s) is Laplace Transform of function f(t). The initial value of f(t) is:
(a) 5
(b) 5/2
(c) 5/3
(d) 0

Q2. In a discrete time complex exponential sequence of frequency ω_o = 1, the sequence is
1. Periodic with period 2π/ω_o
2. Non-periodic
3. Periodic for some value of period N
Which of the above statements is/are correct?
(a) 1 only
(b) 2 only
(c) 3 only
(d) 1 and 3

Q3. Resonant converters are basically used to
(a) reduce the switching losses
(b) eliminates harmonics
(c) generates large peaky voltage
(d) convert a square wave into a sine wave

Q4. In which of the following applications usually tuned voltage amplifiers are not used?
(a) Television receivers
(b) Radio receivers
(c) Public address system
(d) All of the above.

Q5. The unit of conductivity is
(a) ohm/m
(b) mho/m
(c) ohm. m
(d) mho. m

Q6. Which one of the following is not a function of network layer?
(a) routing
(b) inter-networking
(c) congestion control
(d) error control

Q7. Microcontroller contains:
(a) Processor and Timers
(b) Processor and I/O devices
(c) Processor Memory Timers and I/O devices
(d) Processor only

Q8. The cumulative addition of the four binary bits (1 + 1 + 1 + 1) gives
(a) 100
(b) 111
(c) 1001
(d) 1111

Q9. When a semiconductor bar is heated at one end, a voltage across the bar is developed. If the heated end is positive, the semiconductor is
(a) p-type
(b) n-type
(c) intrinsic
(d) highly degenerate

Q10. Consider the following:
In a Parallel Plate Capacitor, let the charge be held constant while the dielectric material is replaced by a different dielectric. Consider
1. Stored energy.
2. Electric field intensity.
3. Capacitance.
Which of these changes?
(a) 1 only
(b) 1 and 2 only
(c) 2 and 3 only
(d) 1, 2 and 3


S1. Ans.(d)
Sol. Given:
F(s) = 1/(s(s^2+3s+2))
Initial Value Theorem is –
⇒ f(0^+) = lim┬(t → 0 )⁡〖f(t)〗 = lim┬(s → ∞)⁡〖sF(s)〗
So, lim┬(t → 0 )⁡〖f(t)〗 = 0

S2. Ans.(b)
Sol. Given:
ω_o = 1
In a discrete time for complex exponential sequence to be periodic the period must be rational.
Period = 2π/ω_o = 2π
The value obtained is irrational so the discrete time complex exponential sequence is non-periodic.

S3. Ans.(a)
Sol. Resonant converters eliminate much of the switching losses encountered in Pulse Width Modulation (PWM) converters.

S4. Ans.(c)
Sol. Tuned voltage amplifiers are not used in Public address system.

S5. Ans.(b)
Sol. We know that,
Conductivity = 1/Resistivity=1/(ohm.m)
Hence, unit of conductivity is mho/m.

S6. Ans.(d)
Sol. In the OSI network model, network layer provides data routing paths for communications with different devices at different network i.e. inter-networking. It also helps in congestion control. Error and flow control is a function of the data link layer and the transport layer.

S7. Ans.(c)
Sol. Microcontroller contains Processor Memory Timers and I/O devices. I/O devices perform the operation based on the instruction stored in the memory.

S8. Ans.(a)
Sol. Converting each binary bit to decimal and then adding we get,
1 + 1 + 1 + 1 = 4
Converting back to binary digit we get,
4_10 = (0100)_2

S9. Ans.(b)
Sol. Highly doped semiconductor is called Degenerate Semiconductor. It may be of N – Type
or P – Type. Thermoelectric Phenomenon causes diffusion of flow of charge carriers from hot
to a cold part of a crystal.
Based on Thermoelectric Phenomenon, we get that –
The electron charge carriers present at the hot end of the semiconductor moves towards the cold end due to diffusion, creating negative excessive charge carriers on the cold contact. So, the heated end of the semiconductor becomes positive and it is called N – Type semiconductor.
In P – Type semiconductor holes are majority charge carriers so it takes part in diffusion. The heated end of the semiconductor is negative so it is of P – Type.

S10. Ans.(d)
Sol. We know that,
W_E = Q^2/2C
E = εD
C = Aε/d
All the three are proportional to dielectric material ε so when dielectric material changes these three value also changes.

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