For the combination of logic gates shown in the figure, the equivalent logic gate is

Options

A. AND
B. NOT
C. NAND
D. NOR

Answer

D

Explanation

`Y_(1) = A + B, Y_(2) = A + B`, as gate 1 and 2 are OR gates.
As gate 3 is NAND gate, its output will be
`Y = bar(Y_(1)Y_(2)) = bar(Y_(1)) + bar(Y_(2)) = bar(A +B.A + B)`
`= bar(A +B) + bar(A +B) = bar(A +B)`
Using De-Morgan.s theorem and identity `A + A =A`.
So, given output is that of NOR gate.