This set of Switching Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Complementing Boolean Expressions”.
1. The Boolean expression (A + B).(A + C’) when complemented results in A’.(B’ + C).
a) True
b) False
View Answer
Explanation: The steps to derive the solution are given below.
= ((A + B).(A + C’))’
Using distributive law, we get,
= (A + B.C’)’
Using DeMorgan’s law, we get
= A’.(B.C’)’
Using DeMorgan’s law, we get
= A’.(B’ + C)
2. The Boolean expression (A + B.C’.D) when complemented results in (A + B).(A + C’).(A + D).
a) True
b) False
View Answer
Explanation: The steps to derive the solution are given below.
Using DeMorgan’s law,
= (A + B.C’.D)’
= A’.(B.C’.D)’
= A’.(B’ + C + D’)
Hence the answer is false.
3. Complement A.B + C.D.
a) (A + C).(B + C).(A + D).(B + D)
b) B.(A + B +C)
c) (A’ + B’).(C’ + D’)
d) A.B.C.D
View Answer
Explanation: The steps to derive the solution are given below.
Using DeMorgan’s law, we get
= (A.B + C.D)’
= (A.B)’.(C.D)’
Applying DeMorgan’s law again,
= (A’ + B’).(C’ + D’)
4. Complement A.B + A.C’ + D’.E’.
a) (A’ + B’.C).(D + E)
b) (A + D’).( A + E’).B.C ’
c) (A + D’).( A + E’).(B + C’)
d) A.(B + C’ + D’).(B + C’ + E’)
View Answer
Explanation: The steps to derive the solution are given below.
Using the ordinary distributive law, X.Y + X.Z = X.(Y + Z)
= (A.(B + C’) + D’.E’)’
Using DeMorgan’s law twice, we get
= (A.(B +C’))’.(D’.E’)’
= (A’ + (B + C’)’).(D + E)
= (A’ + B’.C).(D + E)
5. Complement the Boolean expression A + B.C.D.E
a) A’.(B’ + C’ + D’ + E’)
b) (A + B).(B + D).(C +E)
c) (A + B).(C + D).(E + B)
d) (A + B).(C + D).(D + E)
View Answer
Explanation: The steps to derive the solution are given below.
Using DeMorgan’s law twice, we get
= (A + B.C.D.E)’
= A’.(B.C.D.E)’
= A’.(B’ + C’ + D’ + E’)
6. Find the complement of the Boolean expression (A’ + B).(A’ + C).(A’ + D).
a) A.(B’ + C’ + D’)
b) A’ + B.C.D’
c) A’ + B.C + D
d) A’ + B + C + D
View Answer
Explanation: The steps to derive the solution are given below.
Using Distributive law, (X + Y).(X + Z) = X + YZ
= ((A’ + B).(A’ + C).(A’ +D))’
= (A’ + B.C.D)’
Using DeMorgan’s law twice we get,
= A.(B.C.D)’
= A.(B’ + C’ + D’)
7. Find the complement of the Boolean expression (A + B).(A + C).(A + D).
a) A’.(B’ + C’ + D’)
b) A + B.C.D’
c) A + B.C + D
d) A + B + C + D
View Answer
Explanation: The steps to derive the solution are given below.
Using Distributive law, (X + Y).(X + Z) = X + YZ
= ((A + B).(A + C).(A +D))’
= (A + B.C.D)’
Using DeMorgan’s law twice we get,
= A’.(B.C.D)’
= A’.(B’ + C’ + D’)
8. Find the complement of the Boolean expression (A + B’).(A + C’).(A + D’).
a) A’.(B + C + D)
b) A + B.C.D
c) A + B.C’ + D’
d) A + B + C + D
View Answer
Explanation: The steps to derive the solution are given below.
Using Distributive law, (X + Y).(X + Z) = X + YZ
= ((A + B’).(A + C’).(A + D’))’
= (A + B’.C’.D’)’
Using DeMorgan’s law twice,
= A’.(B’.C’.D’)’
= A’.(B + C + D)
9. Find the complement of the Boolean expression (A + B).(A + C).(A + D’).
a) A’.(B’ + C’ + D)
b) A + B.C.D
c) A + B.C’ + D’
d) A + B + C + D
View Answer
Explanation: The steps to derive the solution are given below.
Using Distributive law, (X + Y).(X + Z) = X + YZ
= ((A + B).(A + C).(A + D’))’
= (A + B.C.D’)’
Using DeMorgan’s law twice, we get
= A’.(B.C.D’)’
= A’.(B’ + C’ + D)
10. Find the complement of the Boolean expression (A + B).(A + C’).(A + D).
a) A’.(B’ + C + D’)
b) A + B.C.D
c) A + B.C’ + D’
d) A + B + C + D
View Answer
Explanation: The steps to derive the solution are given below.
Using Distributive law, (X + Y).(X + Z) = X + YZ
= ((A + B).(A + C’).(A + D’))’
= (A + B.C’.D)’
Using DeMorgan’s law twice, we get
= A’.(B.C’.D)’
= A’.(B’ + C + D’)
Sanfoundry Global Education & Learning Series – Switching Circuits.
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