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JEE Main 2023 (30 Jan, Shift 2) · Logic Gates

Logic Gates question from JEE Main 2023 (30 Jan, Shift 2)

Q96JEE Main 2023 · 30 JAN · SHIFT 2Medium

Logic Gates

The output $Y$ for the inputs $A$ and $B$ of circuit is given by

Truth table of the shown circuit is:

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Answer: (a)

First NAND: $N = \overline{AB}$ .
Next NANDs: $\overline{A N}$ and $\overline{B N}$ ; final $Y = AN + BN = (A + B)\overline{AB}$ .
$(A + B)(\bar{A} + \bar{B}) = A\bar{B} + \bar{A}B$ , the XOR gate.
Rows: $00 \to 0$ , $01 \to 1$ , $10 \to 1$ , $11 \to 0$ . So the answer is (a).

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