Digital Electronics - MCQ (Solution) : Engineering Notes
: Digital Electronics - MCQ (Solution) : Engineering Notes
Page 1
DE09 DIGITALS ELECTRONICS
1
TYPICAL QUESTIONS & ANSWERS
PART - I
OBJECTIVE TYPE QUESTIONS
Each Question carries 2 marks.
Choose correct or the best alternative in the following:
Q.1 The NAND gate output will be low if the two inputs are
(A) 00 (B) 01
(C) 10 (D) 11
Ans: D
The NAND gate output will be low if the two inputs are 11
(The Truth Table of NAND gate is shown in Table.1.1)
X(Input) Y(Input) F(Output)
0 0 1
0 1 1
1 0 1
1 1 0
Table 1.1 Truth Table for NAND Gate
Q.2 What is the binary equivalent of the decimal number 368
(A) 101110000 (B) 110110000
(C) 111010000 (D) 111100000
Ans: A
The Binary equivalent of the Decimal number 368 is 101110000
(Conversion from Decimal number to Binary number is given in Table 1.2)
Table 1.2 Conversion from Decimal number to Binary number
2 368
2 184 --- 0
2 92 --- 0
2 46 --- 0
2 23 --- 0
2 11 --- 1
2 5 --- 1
2 2 --- 1
2 1 --- 0
0 --- 1
Page 2
DE09 DIGITALS ELECTRONICS
1
TYPICAL QUESTIONS & ANSWERS
PART - I
OBJECTIVE TYPE QUESTIONS
Each Question carries 2 marks.
Choose correct or the best alternative in the following:
Q.1 The NAND gate output will be low if the two inputs are
(A) 00 (B) 01
(C) 10 (D) 11
Ans: D
The NAND gate output will be low if the two inputs are 11
(The Truth Table of NAND gate is shown in Table.1.1)
X(Input) Y(Input) F(Output)
0 0 1
0 1 1
1 0 1
1 1 0
Table 1.1 Truth Table for NAND Gate
Q.2 What is the binary equivalent of the decimal number 368
(A) 101110000 (B) 110110000
(C) 111010000 (D) 111100000
Ans: A
The Binary equivalent of the Decimal number 368 is 101110000
(Conversion from Decimal number to Binary number is given in Table 1.2)
Table 1.2 Conversion from Decimal number to Binary number
2 368
2 184 --- 0
2 92 --- 0
2 46 --- 0
2 23 --- 0
2 11 --- 1
2 5 --- 1
2 2 --- 1
2 1 --- 0
0 --- 1
DE09 DIGITALS ELECTRONICS
2
Q.3 The decimal equivalent of hex number 1A53 is
(A) 6793 (B) 6739
(C) 6973 (D) 6379
Ans: B
The decimal equivalent of Hex Number 1A53 is 6739
(Conversion from Hex Number to Decimal Number is given below)
1 A 5 3 Hexadecimal
16³ 16² 16¹ 16° Weights
(1A53)
16
= (1X16³) + (10 X 16²) + (5 X 16¹) + (3 X 16º)
= 4096 + 2560 + 80 + 3
= 6739
Q.4 ( ( ( ( ) ) ) ) ( ( ( ( ) ) ) )
16 8
734 = = = =
(A) C 1 D (B) D C 1
(C) 1 C D (D) 1 D C
Ans: D
(734)
8
= (1 D C)
16
0001 ¦ 1101 ¦ 1100
1 D C
Q.5 The simplification of the Boolean expression ( ( ( ( ) ) ) ) ( ( ( ( ) ) ) ) C B A C B A + + + + is
(A) 0 (B) 1
(C) A (D) BC
Ans: B
The Boolean expression is ( ) C B A + ( ) C B A is equivalent to 1
( ) C B A + ( ) C B A = A + B +C + A + B +C = A + B + C + A + B + C
= (A+ A )(B+ B )(C+ C ) = 1X1X1 = 1
Q.6 The number of control lines for a 8 – to – 1 multiplexer is
(A) 2 (B) 3
(C) 4 (D) 5
Ans: B
The number of control lines for an 8 to 1 Multiplexer is 3
(The control signals are used to steer any one of the 8 inputs to the output)
Q.7 How many Flip-Flops are required for mod–16 counter?
(A) 5 (B) 6
(C) 3 (D) 4
Ans: D
The number of flip-flops is required for Mod-16 Counter is 4.
Page 3
DE09 DIGITALS ELECTRONICS
1
TYPICAL QUESTIONS & ANSWERS
PART - I
OBJECTIVE TYPE QUESTIONS
Each Question carries 2 marks.
Choose correct or the best alternative in the following:
Q.1 The NAND gate output will be low if the two inputs are
(A) 00 (B) 01
(C) 10 (D) 11
Ans: D
The NAND gate output will be low if the two inputs are 11
(The Truth Table of NAND gate is shown in Table.1.1)
X(Input) Y(Input) F(Output)
0 0 1
0 1 1
1 0 1
1 1 0
Table 1.1 Truth Table for NAND Gate
Q.2 What is the binary equivalent of the decimal number 368
(A) 101110000 (B) 110110000
(C) 111010000 (D) 111100000
Ans: A
The Binary equivalent of the Decimal number 368 is 101110000
(Conversion from Decimal number to Binary number is given in Table 1.2)
Table 1.2 Conversion from Decimal number to Binary number
2 368
2 184 --- 0
2 92 --- 0
2 46 --- 0
2 23 --- 0
2 11 --- 1
2 5 --- 1
2 2 --- 1
2 1 --- 0
0 --- 1
DE09 DIGITALS ELECTRONICS
2
Q.3 The decimal equivalent of hex number 1A53 is
(A) 6793 (B) 6739
(C) 6973 (D) 6379
Ans: B
The decimal equivalent of Hex Number 1A53 is 6739
(Conversion from Hex Number to Decimal Number is given below)
1 A 5 3 Hexadecimal
16³ 16² 16¹ 16° Weights
(1A53)
16
= (1X16³) + (10 X 16²) + (5 X 16¹) + (3 X 16º)
= 4096 + 2560 + 80 + 3
= 6739
Q.4 ( ( ( ( ) ) ) ) ( ( ( ( ) ) ) )
16 8
734 = = = =
(A) C 1 D (B) D C 1
(C) 1 C D (D) 1 D C
Ans: D
(734)
8
= (1 D C)
16
0001 ¦ 1101 ¦ 1100
1 D C
Q.5 The simplification of the Boolean expression ( ( ( ( ) ) ) ) ( ( ( ( ) ) ) ) C B A C B A + + + + is
(A) 0 (B) 1
(C) A (D) BC
Ans: B
The Boolean expression is ( ) C B A + ( ) C B A is equivalent to 1
( ) C B A + ( ) C B A = A + B +C + A + B +C = A + B + C + A + B + C
= (A+ A )(B+ B )(C+ C ) = 1X1X1 = 1
Q.6 The number of control lines for a 8 – to – 1 multiplexer is
(A) 2 (B) 3
(C) 4 (D) 5
Ans: B
The number of control lines for an 8 to 1 Multiplexer is 3
(The control signals are used to steer any one of the 8 inputs to the output)
Q.7 How many Flip-Flops are required for mod–16 counter?
(A) 5 (B) 6
(C) 3 (D) 4
Ans: D
The number of flip-flops is required for Mod-16 Counter is 4.
DE09 DIGITALS ELECTRONICS
3
(For Mod-m Counter, we need N flip-flops where N is chosen to be the smallest number
for which 2N is greater than or equal to m. In this case 24 greater than or equal to 1)
Q.8 EPROM contents can be erased by exposing it to
(A) Ultraviolet rays. (B) Infrared rays.
(C) Burst of microwaves. (D) Intense heat radiations.
Ans: A
EPROM contents can be erased by exposing it to Ultraviolet rays
(The Ultraviolet light passes through a window in the IC package to the EPROM chip
where it releases stored charges. Thus the stored contents are erased).
Q.9 The hexadecimal number ‘A0’ has the decimal value equivalent to
(A) 80 (B) 256
(C) 100 (D) 160
Ans: D
The hexadecimal number ‘A0’ has the decimal value equivalent to 160
( A 0
16
1
16
0
= 10X16
1
+ 0X16
0
= 160)
Q.10 The Gray code for decimal number 6 is equivalent to
(A) 1100 (B) 1001
(C) 0101 (D) 0110
Ans: C
The Gray code for decimal number 6 is equivalent to 0101
(Decimal number 6 is equivalent to binary number 0110)
+ + +
0 1 1 0
0 1 0 1
Q.11 The Boolean expression B A B A B A . . . + + is equivalent to
(A) A + B (B) B A.
(C) B A+ (D) A.B
Ans: A
The Boolean expression A .B + A. B + A.B is equivalent to A + B
( A .B + A. B + A.B = B( A + A ) + A. B
= B + A. B {Q( A + A ) = 1}
= A + B {Q(B + A. B ) = B + A}
Q.12 The digital logic family which has minimum power dissipation is
Page 4
DE09 DIGITALS ELECTRONICS
1
TYPICAL QUESTIONS & ANSWERS
PART - I
OBJECTIVE TYPE QUESTIONS
Each Question carries 2 marks.
Choose correct or the best alternative in the following:
Q.1 The NAND gate output will be low if the two inputs are
(A) 00 (B) 01
(C) 10 (D) 11
Ans: D
The NAND gate output will be low if the two inputs are 11
(The Truth Table of NAND gate is shown in Table.1.1)
X(Input) Y(Input) F(Output)
0 0 1
0 1 1
1 0 1
1 1 0
Table 1.1 Truth Table for NAND Gate
Q.2 What is the binary equivalent of the decimal number 368
(A) 101110000 (B) 110110000
(C) 111010000 (D) 111100000
Ans: A
The Binary equivalent of the Decimal number 368 is 101110000
(Conversion from Decimal number to Binary number is given in Table 1.2)
Table 1.2 Conversion from Decimal number to Binary number
2 368
2 184 --- 0
2 92 --- 0
2 46 --- 0
2 23 --- 0
2 11 --- 1
2 5 --- 1
2 2 --- 1
2 1 --- 0
0 --- 1
DE09 DIGITALS ELECTRONICS
2
Q.3 The decimal equivalent of hex number 1A53 is
(A) 6793 (B) 6739
(C) 6973 (D) 6379
Ans: B
The decimal equivalent of Hex Number 1A53 is 6739
(Conversion from Hex Number to Decimal Number is given below)
1 A 5 3 Hexadecimal
16³ 16² 16¹ 16° Weights
(1A53)
16
= (1X16³) + (10 X 16²) + (5 X 16¹) + (3 X 16º)
= 4096 + 2560 + 80 + 3
= 6739
Q.4 ( ( ( ( ) ) ) ) ( ( ( ( ) ) ) )
16 8
734 = = = =
(A) C 1 D (B) D C 1
(C) 1 C D (D) 1 D C
Ans: D
(734)
8
= (1 D C)
16
0001 ¦ 1101 ¦ 1100
1 D C
Q.5 The simplification of the Boolean expression ( ( ( ( ) ) ) ) ( ( ( ( ) ) ) ) C B A C B A + + + + is
(A) 0 (B) 1
(C) A (D) BC
Ans: B
The Boolean expression is ( ) C B A + ( ) C B A is equivalent to 1
( ) C B A + ( ) C B A = A + B +C + A + B +C = A + B + C + A + B + C
= (A+ A )(B+ B )(C+ C ) = 1X1X1 = 1
Q.6 The number of control lines for a 8 – to – 1 multiplexer is
(A) 2 (B) 3
(C) 4 (D) 5
Ans: B
The number of control lines for an 8 to 1 Multiplexer is 3
(The control signals are used to steer any one of the 8 inputs to the output)
Q.7 How many Flip-Flops are required for mod–16 counter?
(A) 5 (B) 6
(C) 3 (D) 4
Ans: D
The number of flip-flops is required for Mod-16 Counter is 4.
DE09 DIGITALS ELECTRONICS
3
(For Mod-m Counter, we need N flip-flops where N is chosen to be the smallest number
for which 2N is greater than or equal to m. In this case 24 greater than or equal to 1)
Q.8 EPROM contents can be erased by exposing it to
(A) Ultraviolet rays. (B) Infrared rays.
(C) Burst of microwaves. (D) Intense heat radiations.
Ans: A
EPROM contents can be erased by exposing it to Ultraviolet rays
(The Ultraviolet light passes through a window in the IC package to the EPROM chip
where it releases stored charges. Thus the stored contents are erased).
Q.9 The hexadecimal number ‘A0’ has the decimal value equivalent to
(A) 80 (B) 256
(C) 100 (D) 160
Ans: D
The hexadecimal number ‘A0’ has the decimal value equivalent to 160
( A 0
16
1
16
0
= 10X16
1
+ 0X16
0
= 160)
Q.10 The Gray code for decimal number 6 is equivalent to
(A) 1100 (B) 1001
(C) 0101 (D) 0110
Ans: C
The Gray code for decimal number 6 is equivalent to 0101
(Decimal number 6 is equivalent to binary number 0110)
+ + +
0 1 1 0
0 1 0 1
Q.11 The Boolean expression B A B A B A . . . + + is equivalent to
(A) A + B (B) B A.
(C) B A+ (D) A.B
Ans: A
The Boolean expression A .B + A. B + A.B is equivalent to A + B
( A .B + A. B + A.B = B( A + A ) + A. B
= B + A. B {Q( A + A ) = 1}
= A + B {Q(B + A. B ) = B + A}
Q.12 The digital logic family which has minimum power dissipation is
DE09 DIGITALS ELECTRONICS
4
(A) TTL (B) RTL
(C) DTL (D) CMOS
Ans: D
The digital logic family which has minimum power dissipation is CMOS.
(CMOS being an unipolar logic family, occupy a very small fraction of silicon Chip
area)
Q.13 The output of a logic gate is 1 when all its inputs are at logic 0. the gate is either
(A) a NAND or an EX-OR (B) an OR or an EX-NOR
(C) an AND or an EX-OR (D) a NOR or an EX-NOR
Ans: D
The output of a logic gate is 1 when all inputs are at logic 0. The gate is either a NOR or an
EX-NOR .
(The truth tables for NOR and EX-NOR Gates are shown in fig.1(a) & 1(b).)
Fig.1(a) Truth Table for NOR Gate Fig.1(b) Truth Table for EX-NOR Gate
Q.14 Data can be changed from special code to temporal code by using
(A) Shift registers (B) counters
(C) Combinational circuits (D) A/D converters.
Ans: A
Data can be changed from special code to temporal code by using Shift Registers.
(A Register in which data gets shifted towards left or right when clock pulses are
applied is known as a Shift Register.)
Q.15 A ring counter consisting of five Flip-Flops will have
(A) 5 states (B) 10 states
(C) 32 states (D) Infinite states.
Ans: A
A ring counter consisting of Five Flip-Flops will have 5 states.
Q.16 The speed of conversion is maximum in
(A) Successive-approximation A/D converter.
(B) Parallel-comparative A/D converter.
(C) Counter ramp A/D converter.
(D) Dual-slope A/D converter.
Input
A B
Output
Y
0 0 1
0 1 0
1 0 0
1 1 0
Input
A B
Output
Y
0 0 1
0 1 0
1 0 0
1 1 1
Page 5
DE09 DIGITALS ELECTRONICS
1
TYPICAL QUESTIONS & ANSWERS
PART - I
OBJECTIVE TYPE QUESTIONS
Each Question carries 2 marks.
Choose correct or the best alternative in the following:
Q.1 The NAND gate output will be low if the two inputs are
(A) 00 (B) 01
(C) 10 (D) 11
Ans: D
The NAND gate output will be low if the two inputs are 11
(The Truth Table of NAND gate is shown in Table.1.1)
X(Input) Y(Input) F(Output)
0 0 1
0 1 1
1 0 1
1 1 0
Table 1.1 Truth Table for NAND Gate
Q.2 What is the binary equivalent of the decimal number 368
(A) 101110000 (B) 110110000
(C) 111010000 (D) 111100000
Ans: A
The Binary equivalent of the Decimal number 368 is 101110000
(Conversion from Decimal number to Binary number is given in Table 1.2)
Table 1.2 Conversion from Decimal number to Binary number
2 368
2 184 --- 0
2 92 --- 0
2 46 --- 0
2 23 --- 0
2 11 --- 1
2 5 --- 1
2 2 --- 1
2 1 --- 0
0 --- 1
DE09 DIGITALS ELECTRONICS
2
Q.3 The decimal equivalent of hex number 1A53 is
(A) 6793 (B) 6739
(C) 6973 (D) 6379
Ans: B
The decimal equivalent of Hex Number 1A53 is 6739
(Conversion from Hex Number to Decimal Number is given below)
1 A 5 3 Hexadecimal
16³ 16² 16¹ 16° Weights
(1A53)
16
= (1X16³) + (10 X 16²) + (5 X 16¹) + (3 X 16º)
= 4096 + 2560 + 80 + 3
= 6739
Q.4 ( ( ( ( ) ) ) ) ( ( ( ( ) ) ) )
16 8
734 = = = =
(A) C 1 D (B) D C 1
(C) 1 C D (D) 1 D C
Ans: D
(734)
8
= (1 D C)
16
0001 ¦ 1101 ¦ 1100
1 D C
Q.5 The simplification of the Boolean expression ( ( ( ( ) ) ) ) ( ( ( ( ) ) ) ) C B A C B A + + + + is
(A) 0 (B) 1
(C) A (D) BC
Ans: B
The Boolean expression is ( ) C B A + ( ) C B A is equivalent to 1
( ) C B A + ( ) C B A = A + B +C + A + B +C = A + B + C + A + B + C
= (A+ A )(B+ B )(C+ C ) = 1X1X1 = 1
Q.6 The number of control lines for a 8 – to – 1 multiplexer is
(A) 2 (B) 3
(C) 4 (D) 5
Ans: B
The number of control lines for an 8 to 1 Multiplexer is 3
(The control signals are used to steer any one of the 8 inputs to the output)
Q.7 How many Flip-Flops are required for mod–16 counter?
(A) 5 (B) 6
(C) 3 (D) 4
Ans: D
The number of flip-flops is required for Mod-16 Counter is 4.
DE09 DIGITALS ELECTRONICS
3
(For Mod-m Counter, we need N flip-flops where N is chosen to be the smallest number
for which 2N is greater than or equal to m. In this case 24 greater than or equal to 1)
Q.8 EPROM contents can be erased by exposing it to
(A) Ultraviolet rays. (B) Infrared rays.
(C) Burst of microwaves. (D) Intense heat radiations.
Ans: A
EPROM contents can be erased by exposing it to Ultraviolet rays
(The Ultraviolet light passes through a window in the IC package to the EPROM chip
where it releases stored charges. Thus the stored contents are erased).
Q.9 The hexadecimal number ‘A0’ has the decimal value equivalent to
(A) 80 (B) 256
(C) 100 (D) 160
Ans: D
The hexadecimal number ‘A0’ has the decimal value equivalent to 160
( A 0
16
1
16
0
= 10X16
1
+ 0X16
0
= 160)
Q.10 The Gray code for decimal number 6 is equivalent to
(A) 1100 (B) 1001
(C) 0101 (D) 0110
Ans: C
The Gray code for decimal number 6 is equivalent to 0101
(Decimal number 6 is equivalent to binary number 0110)
+ + +
0 1 1 0
0 1 0 1
Q.11 The Boolean expression B A B A B A . . . + + is equivalent to
(A) A + B (B) B A.
(C) B A+ (D) A.B
Ans: A
The Boolean expression A .B + A. B + A.B is equivalent to A + B
( A .B + A. B + A.B = B( A + A ) + A. B
= B + A. B {Q( A + A ) = 1}
= A + B {Q(B + A. B ) = B + A}
Q.12 The digital logic family which has minimum power dissipation is
DE09 DIGITALS ELECTRONICS
4
(A) TTL (B) RTL
(C) DTL (D) CMOS
Ans: D
The digital logic family which has minimum power dissipation is CMOS.
(CMOS being an unipolar logic family, occupy a very small fraction of silicon Chip
area)
Q.13 The output of a logic gate is 1 when all its inputs are at logic 0. the gate is either
(A) a NAND or an EX-OR (B) an OR or an EX-NOR
(C) an AND or an EX-OR (D) a NOR or an EX-NOR
Ans: D
The output of a logic gate is 1 when all inputs are at logic 0. The gate is either a NOR or an
EX-NOR .
(The truth tables for NOR and EX-NOR Gates are shown in fig.1(a) & 1(b).)
Fig.1(a) Truth Table for NOR Gate Fig.1(b) Truth Table for EX-NOR Gate
Q.14 Data can be changed from special code to temporal code by using
(A) Shift registers (B) counters
(C) Combinational circuits (D) A/D converters.
Ans: A
Data can be changed from special code to temporal code by using Shift Registers.
(A Register in which data gets shifted towards left or right when clock pulses are
applied is known as a Shift Register.)
Q.15 A ring counter consisting of five Flip-Flops will have
(A) 5 states (B) 10 states
(C) 32 states (D) Infinite states.
Ans: A
A ring counter consisting of Five Flip-Flops will have 5 states.
Q.16 The speed of conversion is maximum in
(A) Successive-approximation A/D converter.
(B) Parallel-comparative A/D converter.
(C) Counter ramp A/D converter.
(D) Dual-slope A/D converter.
Input
A B
Output
Y
0 0 1
0 1 0
1 0 0
1 1 0
Input
A B
Output
Y
0 0 1
0 1 0
1 0 0
1 1 1
DE09 DIGITALS ELECTRONICS
5
Ans: B
The speed of conversion is maximum in Parallel-comparator A/D converter
(Speed of conversion is maximum because the comparisons of the input voltage are
carried out simultaneously.)
Q.17 The 2’s complement of the number 1101101 is
(A) 0101110 (B) 0111110
(C) 0110010 (D) 0010011
Ans: D
The 2’s complement of the number 1101101 is 0010011
(1’s complement of the number 1101101 is 0010010
2’s complement of the number 1101101is 0010010 + 1 =0010011)
Q.18 The correction to be applied in decimal adder to the generated sum is
(A) 00101 (B) 00110
(C) 01101 (D) 01010
Ans: B
The correction to be applied in decimal adder to the generated sum is 00110.
When the four bit sum is more than 9 then the sum is invalid. In such cases, add +6(i.e.
0110) to the four bit sum to skip the six invalid states. If a carry is generated when adding 6,
add the carry to the next four bit group .
Q.19 When simplified with Boolean Algebra (x + y)(x + z) simplifies to
(A) x (B) x + x(y + z)
(C) x(1 + yz) (D) x + yz
Ans: D
When simplified with Boolean Algebra (x + y)(x + z) simplifies to x + yz
[(x + y) (x + z)] = xx + xz + xy + yz = x + xz + xy + yz (
Q
xx = x)
= x(1+z) + xy + yz = x + xy + yz {
Q
(1+z) = 1}
= x(1 + y) + yz = x + yz {
Q
(1+y) = 1}]
Q.20 The gates required to build a half adder are
(A) EX-OR gate and NOR gate (B) EX-OR gate and OR gate
(C) EX-OR gate and AND gate (D) Four NAND gates.
Ans: C
The gates required to build a half adder are EX-OR gate and AND gate
Fig.1(d) shows the logic diagram of half adder.
S
C
A
B
Fig.1(d) Logic diagram of Half Adder
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