This set of Switching Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Commutative, Associative and DeMorgan’s Laws”.
1. The AND operation is commutative.
a) True
b) False
View Answer
Explanation: The answer to the above question can be verified using the following truth table.
| A | B | A.B | B.A |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 0 |
| 1 | 1 | 1 | 1 |
From the above truth table, it can be verified that A.B = B.A. Hence, the AND operation is commutative.
2. The OR operation is commutative.
a) True
b) False
View Answer
Explanation: The answer to the above question can be verified using the following truth table.
| A | B | A+B | B+A |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 |
From the above truth table, it can be verified that A+B = B+A. Hence, the OR operation is commutative.
3. Which among the below Boolean expressions is equal to (A.B).C?
a) A + B + C
b) A.(B.C)
c) A + B.C
d) A.C + B
View Answer
Explanation: The answer to the above question can be verified using the following truth table.
| A | B | C | A.B | B.C | (A.B).C | A.(B.C) |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 | 0 | 0 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
From the above table it can be verified that (A.B).C = A.(B.C). Thus, the AND operation has associative property.
4. Which among the below Boolean expressions is equal to (A + B) + C?
a) A.B + C
b) A + (B + C)
c) A + B.C
d) A.C + B
View Answer
Explanation: The answer to the above question can be verified using the following truth table.
| A | B | C | A+B | B+C | (A+B).C | A+(B+C) |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 1 |
| 1 | 0 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
From the above table it can be verified that (A + B) + .C = A + (B + C). Thus, the OR operation has associative property.
5. Simplify the Boolean expression (A + B)’.
a) A’.B’
b) A + B’
c) (A.B)’
d) A’.B
View Answer
Explanation: The answer to the above question can be verified using the following truth table.
| A | A’ | B | B’ | A+B | A’.B’ | (A+B)’ |
|---|---|---|---|---|---|---|
| 0 | 1 | 0 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 | 0 | 0 |
| 1 | 0 | 1 | 0 | 1 | 0 | 0 |
From the above table it can be verified that (A + B)’ = A’.B’. Thus, DeMorgan’s law is verified.
6. Simplify the Boolean expression (A.B)’.
a) A’ + B’
b) A.B’
c) A + B’
d) A’ + B
View Answer
Explanation: The answer to the above question can be verified using the following truth table.
| A | A’ | B | B’ | A.B | (A.B)’ | A’+B’ |
|---|---|---|---|---|---|---|
| 0 | 1 | 0 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 0 | 0 |
From the above table it can be verified that (A.B)’ = A’ + B’. Thus, DeMorgan’s law is verified.
7. Simplify the Boolean expression A.(B + C).
a) A.B + C
b) B.(A + C)
c) A.B + A.C
d) A.B + C’
View Answer
Explanation: The answer to the above question can be verified using the following truth table.
| A | B | C | A.B | A.C | B+C | A.(B+C) | A.B+A.C |
|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
From the above table it can be verified that A.(B + C) = A.B + A.C. Thus, the first Distributive Law of Boolean Algebra is verified.
8. Simplify the Boolean expression A + B.C.
a) A.C + B
b) (A + B).(A + C)
c) A.B + C
d) A.B.C’
View Answer
Explanation: The answer to the above question can be verified using the following truth table.
| A | B | C | A+B | A+C | B.C | A+(B.C) | (A+B).(A+C) |
|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |
| 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
From the above table it can be verified that A + B.C = (A +B).(A+C). Thus, the second Distributive Law of Boolean Algebra is verified.
9. Which operation is the dual of AND operation?
a) OR
b) NOR
c) NOT
d) NAND
View Answer
Explanation: The dual of a Boolean Algebra expression is obtained by interchanging the constants 0 and 1 and interchanging the operations of AND and OR. Variables and complements are left unchanged. The dual of AND is OR and the dual of OR is AND.
Sanfoundry Global Education & Learning Series – Switching Circuits.
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