Question

14. To get the output y = 0 the input A, B and C must be A. B Done c- Da (1) A = 0, B = 1, C = 1 (2) A = 1, B = 0, C = 0 (3) A = 1, B = 1, C = 1 (4) A = 0, B = 0, C = 0​

Answer 1

Answer :

We can see that the the first circuit is OR gate with a NOT gate. So, the summation of A + B will be the complement of A + B.

The 2nd circuit is the NOT gate which means it is the compliment of C.

Let's check out the options :

Option (1) :

• A = 0, B = 1, C = 1

A + B = 1

(A + B)' = 0

C = 1

C' = 0

(A + B)'.C' = 0.0 = 0

[(A + B)'.C']' = 1

Hence, not satisfied.

Option (2) :

• A = 1, B = 0, C = 0

A + B = 1

(A + B)' = 0

C = 0

C' = 1

(A + B)'.C' = 0.1 = 0

[(A + B)'.C']' = 1

Hence, not satisfied.

Option (3) :

• A = 1, B = 1, C = 1

A + B = 1

(A + B)' = 0

C = 1

C' = 0

(A + B)'.C' = 0.0 = 0

[(A + B)'.C']' = 1

Hence, not satisfied.

Option (4) :

• A = 0, B = 0, C = 0

A + B = 0

(A + B)' = 1

C = 0

C' = 1

(A + B)'.C' = 1.1 = 1

[(A + B)'.C']' = 0

Hence, satisfied.

Therefore, Option (4) is the correct option. ✔