MCQOPTIONS
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MCQOPTIONS
This section includes 223 Mcqs, each offering curated multiple-choice questions to sharpen your Digital Electronics knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
Identify the following logic gate |
| A. | AND gate |
| B. | OR gate |
| C. | NAND gate |
| D. | NOT gate |
| Answer» D. NOT gate | |
| 102. |
Match the following list : List – I(Boolean logic function) List – II(Inverse of function) (a) ab + bc + ca(i)a̅ (b̅ + c̅) (b)ab + a̅ b̅ + c̅(ii)a̅ b̅ + b̅ c̅ + c̅ a̅ (c)a + bc(iii)(a ⊕ b) c (d)(a̅ + b̅ + c̅) (a + b̅ + c̅) (a̅ + b̅ + c) (iv)abc + a̅bc + abc̅ |
| A. | (a) – (iii), (b) – (ii), (c) – (i), (d) – (iv) |
| B. | (a) – (ii), (b) – (iii), (c) – (i), (d) – (iv) |
| C. | (a) – (iii), (b) – (ii), (c) – (iv), (d) – (i) |
| D. | (a) – (ii), (b) – (iii), (c) – (iv), (d) – (i) |
| Answer» C. (a) – (iii), (b) – (ii), (c) – (iv), (d) – (i) | |
| 103. |
A + A’ has the logic value |
| A. | 0 |
| B. | 1 |
| C. | A |
| D. | A’ |
| Answer» C. A | |
| 104. |
How many AND, OR and XOR gates are required for implementation of full adder? |
| A. | 1, 2, 2 |
| B. | 2, 2, 1 |
| C. | 3, 2, 2 |
| D. | 3, 0, 1 |
| Answer» D. 3, 0, 1 | |
| 105. |
Find out the equivalent of A + A' + B'. |
| A. | 1 |
| B. | AB |
| C. | A |
| D. | B |
| Answer» B. AB | |
| 106. |
Assertion (A): It is possible that a digital circuit gives the same output for different input voltages.Reason (R): A digital circuit is also called a logic circuit. |
| A. | Both (A) and (R) are true and (R) is the correct explanation of (A) |
| B. | Both (A) and (R) are true, but (R) is not the correct explanation of (A) |
| C. | (A) is true, but (R) is false |
| D. | (A) is false, but (R) is true |
| Answer» B. Both (A) and (R) are true, but (R) is not the correct explanation of (A) | |
| 107. |
If a signal passing through a gate is initiated by sending low into one of the inputs and the output is high, the gate is |
| A. | NOR |
| B. | NAND |
| C. | AND |
| D. | OR |
| Answer» C. AND | |
| 108. |
In the circuit shown, what are the values of F for EN = 0 and EN = 1, respectively? |
| A. | 0 and D |
| B. | Hi-Z and D |
| C. | 0 and 1 |
| D. | Hi-Z and \(\bar D\) |
| Answer» C. 0 and 1 | |
| 109. |
Consider the following sum of products expression, F\(F=ABC+\bar A\bar B C+A\bar BC+\bar ABC+\bar A\bar B\bar C\)The equivalent product of sums expression is |
| A. | \(F = \left( {A + \bar B + C} \right)\left( {\bar A + B + C} \right)\left( {\bar A + \bar B + C} \right)\) |
| B. | \(F = \left( {A + \bar B + \bar C} \right)\left( {A + B + C} \right)\left( {\bar A + \bar B + \bar C} \right)\) |
| C. | \(F = \left( {\bar A + B + \bar C} \right)\left( {A + \bar B + \bar C} \right)\left( {A + B + C} \right)\) |
| D. | \(F = \left( {\bar A + \bar B + C} \right)\left( {A + B + \bar C} \right)\left( {A + B + C} \right)\) |
| Answer» B. \(F = \left( {A + \bar B + \bar C} \right)\left( {A + B + C} \right)\left( {\bar A + \bar B + \bar C} \right)\) | |
| 110. |
In the sum of products function \(\left( {X,Y,Z} \right) = \mathop \sum \nolimits \left( {2,3,4,5} \right)\), the prime implicants are |
| A. | \(\bar XY,X\bar Y\) |
| B. | \(\bar XY,X\bar Y\bar Z,X\bar YZ\) |
| C. | \(\bar XY\bar Z,\bar XYZ,X\bar Y\) |
| D. | \(\bar XY\bar Z,\bar XYZ,X\bar Y\bar Z,X\bar YZ\) |
| Answer» B. \(\bar XY,X\bar Y\bar Z,X\bar YZ\) | |
| 111. |
A 3 - input majority gate is defined by the logic function \({\rm{M}}\left( {{\rm{a}},{\rm{b}},{\rm{c}}} \right) = {\rm{\;ab\;}} + {\rm{\;bc\;}} + {\rm{\;ac}}\) . Which one of the following gate is represented by the function \({\rm{M}}\left( {\overline {{\rm{M}}\left( {{\rm{a}},{\rm{b}},{\rm{c}}} \right)} ,{\rm{\;M}}\left( {{\rm{a}},{\rm{b}},\overline {{\rm{c\;}}} } \right),{\rm{c}}} \right)?\) |
| A. | 3 input NAND gate |
| B. | 3 input EX-OR gate |
| C. | 3 input NOR gate |
| D. | 3 input OR gate |
| Answer» C. 3 input NOR gate | |
| 112. |
For the gates shown in Fig. (a) and Fig. (b), the x and y inputs are respectively, |
| A. | 0 and 0 |
| B. | 0 and 1 |
| C. | 1 and 0 |
| D. | 1 and 1 |
| Answer» E. | |
| 113. |
Determine the logical operation of the given circuit. |
| A. | \(X = \overline {A + B + C + D} \) |
| B. | \(X = \overline {ABCD} \) |
| C. | X = A + B + C + D |
| D. | X = ABCD |
| Answer» E. | |
| 114. |
All digital circuits can be realized by using only |
| A. | Exclusive OR gates |
| B. | NOR gate |
| C. | Multiplexers |
| D. | OR gate |
| Answer» C. Multiplexers | |
| 115. |
Gate is a circuit with one or more input, but output is |
| A. | two |
| B. | one |
| C. | three |
| D. | more than 1 |
| Answer» C. three | |
| 116. |
A K map of 3 variables contains _______ cells. |
| A. | 8 |
| B. | 3 |
| C. | 6 |
| D. | 9 |
| Answer» B. 3 | |
| 117. |
Commutative law states |
| A. | A.B = B.A |
| B. | A.(B.C) = (A.B).C |
| C. | A + (B + C) = (A + B) + C |
| D. | A.(B + C) = A.B + A.C |
| Answer» B. A.(B.C) = (A.B).C | |
| 118. |
If we group four 1’s from the adjacent cells of a K-map, then the group is called: |
| A. | quad |
| B. | byte |
| C. | nibble |
| D. | word |
| Answer» B. byte | |
| 119. |
If the output of a logic gate is ‘1’ when all its inputs are at logic ‘0’, the gate is either |
| A. | a NAND or a NOR |
| B. | an AND or an EX–NOR |
| C. | an OR or a NAND |
| D. | an EX–OR or an EX–NOR |
| Answer» B. an AND or an EX–NOR | |
| 120. |
AND operation of (79)10 and (-56)10 result in |
| A. | 50 H |
| B. | 48 H |
| C. | 42 H |
| D. | 08 H |
| Answer» C. 42 H | |
| 121. |
Find Z |
| A. | (A̅ + B̅) (C̅ + D̅) (E̅ + F̅) |
| B. | AB + CD + EF |
| C. | (A + B) (C + D) (E + F) |
| D. | \(\overline {AB} + \overline {CD} + \overline {EF} \) |
| Answer» C. (A + B) (C + D) (E + F) | |
| 122. |
_______ is an Universal gate. |
| A. | XOR |
| B. | AND |
| C. | XNOR |
| D. | NAND |
| Answer» E. | |
| 123. |
If the input signals (A & B) and output signals are as below then the circuit element is |
| A. | AND Gate |
| B. | OR Gate |
| C. | NOR Gate |
| D. | XOR Gate |
| Answer» D. XOR Gate | |
| 124. |
If X = 0 in the logic equation, [X + Z{Y + (Z + X̅ Y)}] [Y + X̅ (Z + Y)] = 0 then |
| A. | Z = 0 |
| B. | Z = Y |
| C. | Z = 1 |
| D. | Z = Y̅ |
| Answer» B. Z = Y | |
| 125. |
A + A’B gives |
| A. | 1 |
| B. | A |
| C. | B |
| D. | A + B |
| Answer» E. | |
| 126. |
A Circuit that operates in such a way that its output is high only when all its inputs are high |
| A. | OR |
| B. | NAND |
| C. | NOR |
| D. | AND |
| Answer» E. | |
| 127. |
Find the function represented by the following circuit |
| A. | A + B + C |
| B. | AB |
| C. | AB + C |
| D. | B + C |
| Answer» C. AB + C | |
| 128. |
In Boolean algebra, (A.A̅) + A =? |
| A. | A |
| B. | 0 |
| C. | A̅ |
| D. | 1 |
| Answer» B. 0 | |
| 129. |
How many cells will be there in the K-Map to solve in a 4 variable Boolean expression using K-Map? |
| A. | 64 |
| B. | 16 |
| C. | 24 |
| D. | 32 |
| Answer» C. 24 | |
| 130. |
For a four-input OR gate the number of input condition, that will produce HIGH output are |
| A. | 1 |
| B. | 3 |
| C. | 15 |
| D. | 0 |
| Answer» D. 0 | |
| 131. |
A universal logic gate can implement any Boolean function by connecting sufficient number of them appropriately. Three gates are shown:Which one of the following statements is TRUE? |
| A. | Gate 1 is a universal gate. |
| B. | Gate 2 is a universal gate. |
| C. | Gate 3 is a universal gate. |
| D. | None of the gates shown is a universal gate. |
| Answer» D. None of the gates shown is a universal gate. | |
| 132. |
DeMorgan’s Theorem is |
| A. | (AB)’ = A’ + B’ |
| B. | (AB)’ = A' + B |
| C. | (AB)’ = A + B’ |
| D. | (AB)’ = A + B |
| Answer» B. (AB)’ = A' + B | |
| 133. |
A room has a fan with 2 switches. The fan can be turned on by any switch regardless of the position of the second switch. Logic for a set of switches is equal to ______. |
| A. | XNOR gate |
| B. | NOR gate |
| C. | XOR gate |
| D. | NAND gate |
| Answer» D. NAND gate | |
| 134. |
If the input to the digital circuit of the below figure consisting of a cascade of 20 XOR gates is X, then what is the output Y? |
| A. | 0 |
| B. | 1 |
| C. | X̅ |
| D. | X |
| Answer» C. X̅ | |
| 135. |
In Exclusive–Or gate, when output is zero, the inputs has to be |
| A. | 1, 0 |
| B. | 0, 1 |
| C. | 1, 1 |
| D. | 1, X |
| Answer» D. 1, X | |
| 136. |
A staircase light is controlled by two switches, one at the top of the stairs and the other at the bottom of the stairs, such that the light is ON when and only when one of the switches is ON and the other switch is OFF. What will be the logic equation in SOP form? |
| A. | A + B |
| B. | \(\overline {A + B} \) |
| C. | A̅B + AB̅ |
| D. | AB |
| Answer» D. AB | |
| 137. |
Find the output Boolean function for the logic circuit. |
| A. | X = ABC |
| B. | X = A̅BC |
| C. | X = A̅ B̅ C |
| D. | X = A̅ BC̅ |
| Answer» C. X = A̅ B̅ C | |
| 138. |
From the given logic circuit, determine the expression for Z. |
| A. | P̅ + R |
| B. | \(\overline {PQ} + R\) |
| C. | Q̅ + P |
| D. | Q̅ + R |
| Answer» E. | |
| 139. |
If a signal passing through a gate is inhibited by sending a LOW into one of the inputs, and the output is HIGH, the gate is: |
| A. | AND |
| B. | NAND |
| C. | NOR |
| D. | OR |
| Answer» C. NOR | |
| 140. |
Identify the Boolean function for the given switching circuit. |
| A. | (A + B) C |
| B. | A (B + C) |
| C. | A + B + C |
| D. | ABC |
| Answer» C. A + B + C | |
| 141. |
Considering X and Y as binary variables, the Boolean expression XY’ + X’Y’ is equivalent to |
| A. | X |
| B. | X’ |
| C. | Y |
| D. | Y’ |
| Answer» E. | |
| 142. |
Considering X as a binary variable, the Boolean expression X + 0 is equivalent to |
| A. | X |
| B. | 1 |
| C. | 0 |
| D. | X' |
| Answer» B. 1 | |
| 143. |
Consider the following expression:A. B. C. D + A. B. C̅ . D̅ + A. B. C. D̅ + A. B. C̅ . D + A. B. C. D. E + A. B. C̅ . D̅. E̅ + A. B. C̅. D. EThe simplification of this by using theorems of Boolean algebra will be |
| A. | A + B |
| B. | A ⊕ B |
| C. | (A + B) (A. B) |
| D. | A . B |
| Answer» E. | |
| 144. |
_______ are universal logic gates. |
| A. | NAND and NOR gates |
| B. | NOT Gates and EX-OR Gates |
| C. | AND Gates and NOT Gates |
| D. | OR Gates and EX-OR Gates |
| Answer» B. NOT Gates and EX-OR Gates | |
| 145. |
A function (A, B, C) defined by three boolean variables A, B, and C when expressed as the sum of products is given by:F = A̅.B̅.C̅ + A̅.B.C̅ + A.B̅.C̅where, A̅, B̅, and C̅ are the complements of the respective variables. The product of sums (POS) form of the function F is |
| A. | F = (A + B + C) . (A + B̅ + C) . (A̅ + B + C) |
| B. | F = (A̅ + B̅ + C̅) . (A̅ + B + C̅) + (A + B̅ + C̅) |
| C. | F = (A + B + C̅) . (A + B̅ + C̅) . (A̅ + B + C̅) . (A̅ + B̅ + C) . (A̅ + B̅ + C̅) |
| D. | F = (A̅ + B̅ + C) . (A̅ + B + C) . (A + B̅ + C) . (A + B + C̅) . (A + B + C) |
| Answer» D. F = (A̅ + B̅ + C) . (A̅ + B + C) . (A + B̅ + C) . (A + B + C̅) . (A + B + C) | |
| 146. |
A Product-of-sums (POS) expression leads to what kind of logic circuit? |
| A. | OR-AND circuit |
| B. | NOR-NOR circuit |
| C. | AND-OR_INVERT circuit |
| D. | NAND-NAND circuit |
| Answer» B. NOR-NOR circuit | |
| 147. |
In K-map reduction for 4-variables POS expression, the cell with address 0000 indicates: |
| A. | ABCD |
| B. | A + B + C + D |
| C. | A̅B̅C̅D̅ |
| D. | A̅ + B̅ + C̅ + D̅ |
| Answer» C. A̅B̅C̅D̅ | |
| 148. |
Identify the gates G1 and G2 respectively in the figure given below to get Y = ABC |
| A. | NOR & AND |
| B. | AND & AND |
| C. | NAND & OR |
| D. | OR & AND |
| Answer» B. AND & AND | |
| 149. |
Find X |
| A. | 0 |
| B. | AB̅ |
| C. | AB |
| D. | A̅B |
| Answer» B. AB̅ | |
| 150. |
Consider the logic circuit as shown in the figure below, the function f1, f2 and f (in canonical sum of products form in decimal notation) aref1(w, x, y, x ) = ∑ (8, 9, 10)f2 (w, x, y, x ) = ∑ (7, 8, 12, 13, 14, 15)f(w, x, y, x ) = ∑ (8, 9)The function f3 is |
| A. | ∑ (9, 10) |
| B. | ∑ (9) |
| C. | ∑ (1, 8, 9) |
| D. | ∑ (8, 10, 18) |
| Answer» C. ∑ (1, 8, 9) | |