The truth table for the given logic circuit is (a) (b) (c) (d)

Answer: (c). Upper NAND: A · B. Lower NAND gets A and B: A · B. C = A · B · A · B. AB = 00: 1 · 1 = 1; 01: 1 · 0 = 0; 10: 1 · 1 = 1; 11: 0 · 1 = 0. So C = B. Table 1, 0, 1, 0, so the answer is (c).

The truth table for the given logic circuit is | NEET 2022 (Phase 1)

Q109NEET 2022 · PHASE 1Medium

Logic Gates

The truth table for the given logic circuit is

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Answer: (c)

Upper NAND: $\overline{A \cdot B}$ . Lower NAND gets $\bar{A}$ and B: $\overline{\bar{A} \cdot B}$ .
$C = \overline{A \cdot B} \cdot \overline{\bar{A} \cdot B}$ .
$AB = 00$ : $1 \cdot 1 = 1$ ; $01$ : $1 \cdot 0 = 0$ ; $10$ : $1 \cdot 1 = 1$ ; $11$ : $0 \cdot 1 = 0$ . So $C = \bar{B}$ .
Table $1, 0, 1, 0$ , so the answer is (c).

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