Question

The truth table for the following logic circuit is,

• A. $\begin{vmatrix} A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix}$
• B. $\begin{vmatrix} A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{vmatrix}$
• C. $\begin{vmatrix} A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix}$
• D. $\begin{vmatrix} A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{vmatrix}$

Answer 1

Answer: (A)

$Y=\bar{A} \cdot B+A \cdot \bar{B}$

The truth table for the given logic circuit is

$A$	$B$	$\bar{A}$	$\bar{B}$	$\bar{A}.B$	$A.\bar{B}$	$Y=\bar{A} \cdot B + A \cdot\bar{B}$ $$
$0$	$0$	$1$	$1$	$0$	$0$	$0$
$0$	$1$	$1$	$0$	$1$	$0$	$1$
$1$	$0$	$0$	$1$	$0$	$1$	$1$
$1$	$1$	$0$	$0$	$0$	$0$	$0$