QJEEM21Phd78b: JEE Main 2021 Physics - Class 12 JEEM21Phd78b Solution: The following logic gate is equivalent to :

Question

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The following logic gate is equivalent to :

JEE Main 2021 (Online) 16th March Evening Shift Physics - Semiconductor Question 115 English

Accepted Answer

Output of first NAND gate with input A

\begin{aligned} = \overset{―}{A . A} \\ = \overset{―}{A} + \overset{―}{A} \\ = \overset{―}{A} \end{aligned}

Similarly, Output of first NAND gate with input B

= \overset{―}{B}

\begin{aligned} Y_{1} = \overset{―}{\overset{―}{A} . \overset{―}{B}} \\ = \overset{―}{\overset{―}{A}} + \overset{―}{\overset{―}{B}} \\ = A + B \end{aligned}

\begin{aligned} Y = \overset{―}{(A + B) . (A + B)} \\ = \overset{―}{(A + B)} + \overset{―}{(A + B)} \\ = \overset{―}{(A + B)} \end{aligned}

So, given logic gates circuit is a NOR gate.

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