DIGITAL ELECTRONICS - MCQS
Q1. The binary number 10101 is
equivalent to decimal number …………..
19
12
27
21
Answer : 4
Q2. The universal gate is ………………
NAND gate OR gate
AND gate
NOT gate OR gate
AND gate
None of the above Answer :1
Q4. The inputs of a NAND gate are
connected together. The resulting circuit is ………….
OR gate AND gate
NOT gate
None of the above
Answer : 3
Q5. The NOR gate is OR gate
followed by ………………
AND gate NAND
gate NOT gate
None of the above
Answer : 3
Q6. The NAND gate is AND gate
followed by …………………
Q7. Digital circuit can be made by
the repeated use of ………………
OR gates NOT
gates NAND gates
None of the above
Answer : 3
Q8. The only function of NOT gate
is to ……………..
Stop signal
Invert input signal
Act as a
universal gate None of the above Answer : 2
Q9. When an input
signal 1 is applied to a NOT gate, the output is ………………
0
1
Either 0 & 1
None of the above
Answer : 1
Q10. In Boolean algebra, the bar sign
(-) indicates ………………..
OR operation AND operation
NOT operation
None of the above
Answer : 3
Q11. The resolution of an n bit
DAC with a maximum input of 5 V is 5 mV. The value of n is …….
8
9
10
11
Answer : 3
Explanation:
(5/2N-1)1000 = 5 or N = 10
Q12. 2’s complement of binary
number 0101 is ………..
1011
1111
1101
1110
Answer : 1
Explanation: 1’s complement of 0101
is 1010 and 2’s complement is 1010+1 = 1011.
Q13. An OR gate has 4 inputs. One input is high and
the other three are low. The output is …….
Low High
Alternately high and low
May be high or
low depending on relative magnitude of inputs Answer : 2
Explanation: In OR any input high
means high output.
Q14.
Decimal number 10 is equal to binary number ……………
1110
1010
1001
1000
Answer : 2
Explanation: 1010 = 8 + 2 = 10
in decimal.
Q15.
Both OR and AND gates can have only two inputs.
True False
Answer : 2
Explanation: Any numbers of
inputs are possible.
Q16. A device which converts BCD to seven segments is
called ……..
Encoder Decoder
Multiplexer None of these Answer : 2
Explanation: Decoder converts
binary/BCD to alphanumeric.
Q17. In
2’s complement representation the number 11100101 represents the decimal number
……………
+37
-31
+27
-27
Answer : 4
Explanation:
A = 11100101. Therefore Ā =
00011010 and A’ = Ā + 1 = 00011011 = 16 + 8 + 2 + 1 = 27. Therefore A = -
27.
Q18. A
decade counter skips ………..
Binary states 1000 to1111
Binary states 0000 to0011
Binary states
1010 to 1111 Binary states 1111 to higher Answer : 3
Q19. BCD
input 1000 is fed to a 7 segment display through a BCD to 7 segment
decoder/driver. The
segments
which will lit up are ………….
A, b, d
A, b, c All
A, b, g, c, d
Answer : 3
Explanation: 1000 equals decimal 8
Therefore all segments will lit up.
Q20. A ring counter with 5 flip flopswillhave........... States.
5
10
32
Infinite Answer :
1
Q21. For
the gate in the given figure the output will be ………..
0
1
A
Ā
Answer : 4
Explanation: If A = 0, Y = 1 and
A = 1, Y = 0 Therefore Y = Ā.
Q22. In
the expression A + BC, the total number of minterms will be ………
2
3
4
5
Answer : 4
Q23. The circuit in the given figureisa............ gate.
Positive logic OR
gate Negative logic OR gate Negative logic AND gate Positive logic AND gate
Answer : 2
Explanation:
Since V(1) is lower state than V(0) it is a negative logic circuit. Since
diodes are in parallel, it is an OR gate.
Q24. Which
of the following is non-saturating?
TTL CMOS ECL
Both 1 and 2
Answer : 3
Q25. The
number of digits in octal system is ………
Answer : 1 Explanation: The octal
system has 8 digits 0 to 7.
Q26. The access time of a word in 4 MB main
memory is 100 ms. The access time of a word in a 32 kb datacachememoryis10ns.Theaveragedatacachebit
ratiois0.95.Theefficiencyofmemoryaccess time is………
9.5 ns
14.5 ns 20 ns 95
ns
Answer : 2
Explanation: Access time = 0.95
x 10 + 0.05 x 100.
Q27. The
expression Y = pM (0, 1, 3, 4) is …………..
Explanation: This is a product of
sums expression.
Q28. An
8 bit DAC has a full scale output of 2 mA and full scale error of ± 0.5%. If
input is 10000000 the
range of
outputs is ………….
994 to 1014μA
990 to 1020μA
800 to 1200μA
None of the above
Answer : 1
Q29. Decimal 43 in hexadecimal and BCD number system
is respectively……. And ……..
2B and00110100
B2 and01000100
Answer : 2
Q30. An
AND gate has two inputs A and B and one inhibit input 3, Output is 1 if
Answer: 2
Explanation: All AND inputs must
be 1 and inhibit 0 for output to be 1.
Q31. The greatest negative number which can be stored is 8
bit computer using 2’s complement arithmetic is ……..
-256
-128
-255
-127
Answer: 2
Explanation: The largest negative
number is 1000 0000 = -128.
Q33. A
JK flip flop has tpd= 12 ns. The largest modulus of a ripple counter using
these flip flops and
operating
at 10 MHz is ……..
16
64
128
256
Answer: 4
Q34. The
basic storage element in a digital system is ………….
Flipflop Counter
Multiplexer
Encoder Answer :
1
Explanation:
Storing can be done only in memory and flip-flop is a memory element.
Q35. In a ripple counter,
Whenever a
flipflop sets to 1, the next higher FF toggles
Whenever a flipflop sets to 0, the next higher FF remains unchanged
Whenever a flipflop sets to 1, the next higher FF faces race condition Whenever
a flipflop sets to 0, the next higher FF faces race condition Answer : 1
Explanation: In a ripple counter the
effect ripples through the counter.
24.4mV
1.2 V
None of these
Answer : 1
Q37. For the truth table of the
given figure Y = ………….
A + B + C
Ā +BC Ā
B¯
Answer : 4
Q38. A full adder can be made
out of …………
Two half adders
Two half adders
and a OR gate Two half adders and a NOT gate Three half adders
Answer : 2
Q39. If the functions w, x, y, z
are as follows
W = z x = z W = z, x = y W = y
Q40. The output of a half adder is
……….
Sum
Sum and Carry Carry
None of these
Answer: 2
Q41. Minimum number of 2-input
NAND gates required to implement the function F = (x + y) (Z + W) is
………..
3
4
5
6
Answer : 2
Q42. Which device has one input
and many outputs?
Multiplexer
Demultiplexer Counter
Flip flop Answer:
2
Explanation:
Demultiplexer takes data from one line and directs it to any of its N output
depending on the status of its select lines.
Q43. A carry look ahead adder is
frequently used for addition because
It costs less It isfaster
It is more accurate Uses fewer gates Answer:2
Explanation:
Q44. The counter in the given figure
is ………….
Mod3
Mod6
Mod8
Mod7
Answer : 2
Explanation: When counter is 110
the counter resets. Hence mod 6.
Q45. In register index
addressing mode the effective address is given by ……..
Index register value
Sum of the index
register value and the operand Operand
Difference of the index register
value and the operand Answer : 2
4 = 22, in up scaling digit will
be shifted by two bit in right direction.
Q46.7BF16= 2
0111 10111110
0111 10111111
0111 10110111
0111 10110011
Answer : 2
Explanation:
7BF16 = 7 x 162 + 11 x 161 + 15
x 160 = 1983 in decimal = 0111 1011 1111 in binary.
Q48. Zero suppression is not used
in actual practice.
True False
Answer:2
Explanation: Zero suppression is
commonly used.
5 μsec
Answer : 1
Explanation:
Q50. The hexadecimal number
(3E8)16 is equal to decimal number ………
1000
982
768
323
Answer : 1
Explanation: 3 x 162 + 14 x 161
+ 8 = 1000
Q51. The number of distinct Boolean
expression of 4 variables is …….
1024
65536
Answer : 4
Explanation:
Q53. A memory system of size 16 k
bytes is to be designed using memory chips which have 12 address
lines and 4 data lines each. The
number of such chips required to design the memory system is ……….
2
4
8
18
Answer : 3
Explanation:
(16×1024×8)/(4096×4) = 8
Q54. In a 7 segment display, LEDs b
and c lit up. The decimal number displayed is ……….
3
1
Answer : 1
Q55. In a BCD to 7 segment decoder
the minimum and maximum number of outputs active at any time
is ….
2 and7
3 and7
1 and 6
3 and 6
Answer: 1
Explanation:
Minimum number
of outputs when input is decimal 1 and maximum number of outputs when input is
decimal 8.
Q56. A three state switch has
three outputs. These are …….. , …….. , ……….
Low, low and high
Low, high, high Low. Floating, low Low, high, floating Answer: 4
Q57. Maxterm designation for A + B +
C is ……….
M0 M1 M3 M4
Answer: 1
Explanation: A + B + C = 000 =
M0
Q58. 1’s complement of 11100110 is
……………….
00011001
10000001
00011010
00000000
Answer:1
Explanation: By replacing 1 by 0
and 0 by 1.
BASIC QUESTIONS
1. Which
of the following is UniversalGate?
OR gate NAND gate
AND gate NOR gate Answer
The Universal Gate is NAND gate.
2.
Which of the
following isInverter?
OR gate NOT gate
AND gate NAND GATE
Answer
Inverter
is NOT gate.
3. The
NOR gate is OR gate followed by which gate?
AND gate NOT gate
NAND gate
None of the
mentioned Answer
The NOR gate is OR gate followed by NOT gate.
4. In
the boolean algebra, avariablehas _
differentstate(s)/value(s).
3
1
2
4
Answer
2
5. AND
operation is equivalent to–
None of the above Answer:
Intersection
6. A
+ Ā is =?
0
1
A
Ā
View
Answer 1
7. Which
is the example of digital device from the given option?
Record
players Microprocessors Sensors Thermistors View Answer Microprocessors
8. numbers are
used in the decimal numbersystem?
0 to 9
0 to10
1 to10
None of the above
View Answer
9. Combinations that not listed for the
input variables are –
Borrow Don’t Care
Overflow Carry
View Answer
Don’t Care
10.
A full adder have–
2 inputs, 2 outputs
2 inputs,
1output
3 inputs,
2outputs
3 inputs, 1
output View Answer
3 inputs, 2 outputs
11.
In Positive logic, logic gate 1 corresponds to–
Zero Voltage
Level Positive Voltage Lower Voltage Level Higher Voltage Level View Answer
Higher Voltage Level
12.
An X-OR Gate Produces an output only when it’s
two inputs are–
View Answer Different
13. The
Only Function of a Not gate is to–
Invert an
inputsignal Stop Asignal
Act an universal
set Recomplement a signal ViewAnswer
Invert an input signal
14.
ASCII codeisa.............................. bitcode.
8
7
2
1
View
Answer 7
15. Multiplexer
means…………………
Many in to One
Many in to Many One in to Many None of the Above View Answer
16.
Binary
Numbersystemhas........... symbols.
10
2
View
Answer 2
17. The
Steps required for the analysis of combinational circuits are–
Obtain the
functions of intermediate points and outputs Label the inputs and outputs
Draw the truth
table All of the Above View Answer:All of the Above
18.
There are two types of parity–
Odd Even
Both 1 & 2
None of the above
View Answer
Both 1 & 2
19.
The Four common and useful MSI circuitsare
Decoder Encoder Demultiplexer All of the above View
Answer
20.Multiplexers can be constructed from
smallerones.
Larger Small
Dimultiplexers
None of the above View Answer Larger
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