Consider the following logic circuit. The output is Y = 0 when: (a) A = 1 and B = 0 (b) A = 0 and B = 0 (c) A = 1 and B = 1 (d) A = 0 and B = 1

Answer: (c). AND gate: AB. OR gate gets B and A: A + B. Y = AB·(A + B) (NAND). Y = 0 only when both NAND inputs are 1: AB = 1 ⇒ A = B = 1, and then A + B = 1 too. So the answer is (c) A = 1 and B = 1.

Consider the following logic circuit. The output is Y = 0 when: | JEE Main 2025 (7 Apr, Shift 2)

Q43JEE Main 2025 · 7 APR · SHIFT 2Medium

Logic Gates

Consider the following logic circuit.

The output is Y = 0 when:

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Answer: (c)

AND gate: $AB$ . OR gate gets $B$ and $\bar{A}$ : $\bar{A} + B$ .
$Y = \overline{AB\cdot(\bar{A} + B)}$ (NAND).
$Y = 0$ only when both NAND inputs are 1: $AB = 1 \Rightarrow A = B = 1$ , and then $\bar{A} + B = 1$ too.
So the answer is (c) $A = 1$ and $B = 1$ .

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