Quiz Electronics Engineering 25 Sep 2020

Quiz Electronics Engineering
Exam: NIC
Topic: Miscellaneous
Date: 25/09/2020

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

Q1. In 8085 processor, which type of instruction does not affect the flag?
(a) Branching Instruction
(b) Data Transfer Instruction
(c) Logical Instruction
(d) Arithmetic Instruction

 

Q2. Digital circuit can be made by the repeated use of
(a) OR gates
(b) NOT Gates
(c) NAND gates
(d) None of these

 

Q3. Insulators have ______ temperature coefficient of
resistance
(a) negative
(b) positive
(c) zero
(d) none of these

 

Q4. In PCM system, if we increase the quantization levels from 2 to 8, the relative bandwidth requirements will:
(a) Be doubled
(b) Remain same
(c) Becomes eight times
(d) Be tripled

 

Q5. A lossless line of 50 ohms is terminated in a load of 100 ohms resistive. The VSWR is
(a) 1 : 2
(b) 2 : 1
(c) 4 : 1
(d) 1 : 4

 

Q6. Which of the following routing algorithms can be used for network layer design?
(a) shortest path algorithm
(b) distance vector routing
(c) link state routing
(d) All of the above

 

Q7. Which network topology requires a central controller or hub?
(a) Ring Topology
(b) Mesh Topology
(c) Star Topology
(d) Bus Topology

 

Q8. For which one of the following ultraviolet light is used to erase the stored contents?
(a) PROM
(b) EPROM
(c) EEPROM
(d) PLA

 

Q9. Capacitors of 2, 4, 6 μF are connected in series, the total capacitance will be
(a) 19μF
(b) 9 μF
(c) 1.09 μF
(d) 99 μF

 

Q10. What is the average power of periodic non-sinusoidal voltage and currents?
(a) The average power of the fundamental component alone.
(b) The sum of the average powers of the sinusoidal components including the fundamental.
(c) The sum of the average powers of the sinusoidal components including the fundamental.
(d) The sum of the root mean square power of the sinusoidal components including the fundamental

 

 

SOLUTIONS

S1. Ans.(b)

S2. Ans.(c)
Sol. NAND and NOR gates are also known as Universal Gates because with the help of these two logic gate any digital circuit can be realized or implemented.

S3. Ans.(a)
Sol. Insulators have Negative Temperature Coefficient of resistance because its resistance decrease with increment of temperature. On increasing temperature insulator may start conducting that means their resistance is decreasing.

S4. Ans.(d)
Sol. It is given that Quantisation Level changes from 2 to 8.
We know that,
Tx BW = Rb = nfs [ where, n is number of bits ]
When, L = 2,
⇒ L = 2n = 2
∴ n = 1
and Tx BW = fs
When, L = 8,
⇒ L = 2n = 8 = 23
∴ n = 3
and Tx BW = 3fs
Therefore, the relative bandwidth requirements will be tripled.

S5. Ans(b)
Sol. Given:
Z_O = 50 ohms
Z_L = 100 ohms
Reflection coefficient
⇒ |Γ| = 〖Z_L+ Z_O〗_ /(Z_L + Z_O )=(100 – 50)/(100 + 50)=50/150 =1/3
Now,
VSWR = (1 + |Γ| )/(1 – |Γ| )
⇒ VSWR =(1 + 1/3)/(1 – 1/3)
⇒ VSWR =(4/3)/(2/3)=4/3×3/2
∴ VSWR = 2∶1

S6. Ans.(d)
Sol. The routing algorithm with the help of routing table decides where a packet should go next. The routing algorithms are shortest path algorithm, static and dynamic routing, distance vector routing, link state routing, etc. The routing algorithms go hand in hand with the operations of all the routers in the networks. These routing table is stored in the routers.

S7. Ans.(c)
Sol. Star topology is a physical network topology in which no computer is connected to another computer directly instead all computers are connected to a central hub. Every message is send from a source device to the hub and the hub then forwards the message to the destination device.

S8. Ans.(b)
Sol. EPROM stands for “Erasable Programmable Read Only Memory”. It is a non-volatile memory. It is programmed by the user and the contents stored in it can be erased repeated times according to the user need. It is erased using Ultraviolet rays. “EEPROM” stands for “Electrically Erasable Programmable Read Only Memory”. The stored content in it is erased using electrical pulses.

S9. Ans.(c)
Sol. When connected in series,
⇒ 1/C_S = 1/C_1 +1/C_2 +1/C_3
⇒ 1/C_S = 1/2+1/4+1/6
⇒ 1/C_S = (6+3+2)/12=11/12
So, C_S=12/11=1.09 μF

S10. Ans.(d)

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