In the following circuit, the output Y for all possible inputs A and B is expressed by the truth table. (a) A B Y 0 0 0 1 0 0 0 1 0 1 1 1 (b) A B Y 0 0 1 0 1 0 1 0 0 1 1 0 (c) A B Y 0 0 0 0 1 1 1 0 1 1 1 1 (d) A B Y 0 0 1 1 0 1 0 1 1 1 1 0

Answer: (c). First NOR gives A + B. Second NOR has both inputs joined, so it acts as NOT: Y = A + B = A + B. OR table: Y = 0, 1, 1, 1 for AB = 00, 01, 10, 11. So the answer is (c).

In the following circuit, the output Y for all possible inputs A and… | NEET 2007 (AIPMT)

Q178NEET 2007 · AIPMTMedium

Logic Gates

In the following circuit, the output Y for all possible inputs A and B is expressed by the truth table.

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Answer: (c)

First NOR gives $\overline{A + B}$ .
Second NOR has both inputs joined, so it acts as NOT: $Y = \overline{\overline{A + B}} = A + B$ .
OR table: $Y = 0, 1, 1, 1$ for $AB = 00, 01, 10, 11$ .
So the answer is (c).

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