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JEE Main 2019 (12 Jan, Shift 1) · Logic Gates

Logic Gates question from JEE Main 2019 (12 Jan, Shift 1)

Q146JEE Main 2019 · 12 JAN · SHIFT 1Medium

Logic Gates

The output of the given logic circuit is:

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Answer: (d)

First NAND: $N = \overline{AB}$ . Top NAND: $\overline{A N}$ .
OR gate: $N + B = \bar{A} + \bar{B} + B = 1$ .
Final NAND: $Y = \overline{\overline{AN}\cdot 1} = AN = A(\bar{A} + \bar{B}) = A\bar{B}$ .
So the answer is (d) $A\overline{B}$ .

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