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.

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