In the following circuit, the output `Y` for all possible inputs `A` and `B` is expressed by the truth table:

Options

A. `|(A,B,Y),(0,0,0),(0,1,0),(1,0,0),(1,1,1)|`
B. `|(A,B,Y),(0,0,1),(0,1,1),(1,0,1),(1,1,0)|`
C. `|(A,B,Y),(0,0,1),(0,1,0),(1,0,0),(1,1,0)|`
D. `|(A,B,Y),(0,0,0),(0,1,1),(1,0,1),(1,1,1)|`

Answer

D

Explanation

Gates`-I` and `II` are `NOR` gates. We can simplify the gate circuit as

Here, gates `I` and `II` are `NOR` gates. The output `bar(A+B)`of gate `I` will be appeared as input of gate `II`. The final output is `Y=bar(bar(A+B))=A+B`
This is the Boolean expression of `OR` gate whose truth table is given below
`|(A,B,Y),(0,0,0),(0,1,1),(1,0,1),(1,1,1)|`