Question

For the logic circuit shown, the truth table is?

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

Answer 1

The correct option is B $\begin{array}{ccc}\text{A}& \text{B}& \text{Y}\\ 0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\\ 1& 1& 0\end{array}$

$Y=\overline{\overline{(\overline{A}.\overline{B})}}=(\overline{A}.\overline{B})=\overline{A+B}$
Truth table is
$\begin{array}{ccc}\text{A}& \text{B}& \text{Y}\\ 0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\\ 1& 1& 0\end{array}$