This set of Switching Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Boolean Algebra – Exclusive-OR and Equivalence Operations”.
1. The symbol for Exclusive-OR operation is ⊕.
a) True
b) False
View Answer
Explanation: The symbol for Exclusive-OR operation is ⊕.
The Exclusive OR operation is given by: A⊕B = A’.B + A.B’. The Exclusive-OR operation results in a LOW when both inputs are equal.
2. Given below is the circuit symbol for Exclusive-OR.
a) True
b) False
View Answer
Explanation: That is the circuit symbol for Exclusive-OR gate.
The Exclusive OR operation is given by: A⊕B = A’.B + A.B’. The circuit symbol for Exclusive-Or gate is similar to an OR gate, but the difference is an extra line before the input.
3. Identify the logic gate which has the following truth table.
| A | B | Y |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
a) AND gate
b) OR gate
c) NAND gate
d) Exclusive-OR
View Answer
Explanation: The Exclusive OR operation is given by: A⊕B = A’.B + A.B’. By making a truth table of that expression, we get the truth tale in the question.
| A | B | A’.B+A.B’ |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
4. Evaluate the boolean expression A⊕0.
a) A
b) A’
c) 1
d) 0
View Answer
Explanation: The steps to derive the solution are given below.
A ⊕ 0
= A’.0 + A.(0’)
= 0 + A.1
= A
5. Evaluate the Boolean expression A⊕1.
a) A
b) A’
c) 1
d) 0
View Answer
Explanation: The steps to derive the solution are given below.
A ⊕ 1
= A’.1 + A.(1’)
= A’ + A.0
= A’
6. Evaluate the Boolean expression A⊕A.
a) A
b) A’
c) 1
d) 0
View Answer
Explanation: The steps to derive the solution are given below.
A ⊕ A
= A’.A + A.A’
Using A.A’ = 0, we get
= 0
7. Evaluate the Boolean expression A⊕A’.
a) A
b) A’
c) 1
d) 0
View Answer
Explanation: The steps to derive the solution are given below.
A⊕A’
= A’.A’ + A.(A’)’
= A’ + A
Using A + A’ = 1, we get
= 1
8. Complement the Boolean expression (A⊕B).
a) A’⊕B = A⊕B’
b) AB
c) A + B
d) A’⊕B’
View Answer
Explanation: The steps to derive the solution are given below.
(A⊕B)’
= A’.A’ + A.(A’)’
= A’ + A
Using A + A’ = 1, we get
= 1
9. Identify the logic gate used to implement the equivalence operation (≡).
a) NOR gate
b) NAND gate
c) AND gate
d) Exclusive-NOR gate
View Answer
Explanation: The Exclusive-NOR gate is used to implement the Equivalence operation.
This can be concluded by observing the following truth table.
| A | B | A’.B+A.B’ (Exclusive-OR) | (A’B+A.B’)’ = (A’+B).(A+B’) = A.B+A’.B+A’ (Exclusive-NOR) |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
Thus, output is 1 only when A ≡ B for Exclusive-NOR Gate. Hence equivalence operation is performed by Exclusive-NOR gate.
10. Calculate the minimum number of AND gates, OR gates and NOT gates required to realize Exclusive-OR operation if only using AND, OR and NOT gates are allowed.
a) 2 AND, 1 OR and 2 NOT gates or 1 AND, 2 OR and 2 NOT gates
b) 2 AND, 2 NOT and 2 OR gates
c) 3 AND, 1 NOT and 2 OR gates
d) 2 AND and 2 OR gates
View Answer
Explanation: The Exclusive-OR operation is given by A⊕B = A’.B + A.B’ = (A + B).(A’ + B’).
Thus, it can be realized using 2 AND, 1 OR and 2 NOT gates for the Sum of Products expression and 1 AND, 2 OR and 2 NOT gates for the Product of Sums expression.
Sanfoundry Global Education & Learning Series – Switching Circuits.
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