Question 164 of 209

The output $Y$ for the inputs $A$ and $B$ of circuit is given by

Truth table of the shown circuit is :

A$\left[\begin{matrix}A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{matrix}\right]$B$\left[\begin{matrix}A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1\end{matrix}\right]$C$\left[\begin{matrix}A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1\end{matrix}\right]$D$\left[\begin{matrix}A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{matrix}\right]$

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Answer: (D) $\left[\begin{matrix}A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{matrix}\right]$

Output using DeMorgan's theorem can be written as,
$Y=\left(\left(A\cdot A\cdot B\right)\cdot \left(A\cdot B\cdot B\right)\right) =\left(A\cdot A\cdot B\right)+\left(A\cdot B\cdot B\right) =\left(A\cdot A\cdot B\right)+\left(A\cdot B\cdot B\right) =\left(A\cdot \left(A+B\right)\right)+\left(\left(A+B\right)\cdot B\right) =A\cdot A+A\cdot B+A\cdot B+B\cdot B =\left(A+B\right)\cdot \left(A+B\right)$
Which represents XOR gate.
Therefore, required truth table will be
$\left[\begin{matrix}A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{matrix}\right]$

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.