Question 183 of 209

The logic circuit shown above is equivalent to :

ABCD

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Answer: (C)

Truth table of the given gate:
$\begin{matrix}A & B & C \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 0\end{matrix}$
Truth table of option $\left(1\right)$
$\begin{matrix}A & B & C \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1\end{matrix}$
Truth table of option $\left(2\right)$
$\begin{matrix}A & B & C \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 1\end{matrix}$
Truth table of option $\left(3\right)$
$\begin{matrix}A & B & C \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 0\end{matrix}$
Truth table of option $\left(4\right)$
$\begin{matrix}A & B & C \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 0\end{matrix}$
Since option $\left(1\right)$ has same truth table, hence answer is option $\left(4\right)$ is correct.
Given Boolean expression can be written as
$\bar{A+\bar{B}}=C$
$∴C=\bar{A}\cdot \bar{B}=\bar{A}\cdot B$

JEE Main Physics practice question on Semiconductors (Logic gates), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.