Question

Which of the following is a correct truth table for the circuit shown in figure?

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

Answer 1

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

Here, Input to the $\text{NAND}$ gate is $A$ and $A$

Thus,
$Y=\overline{A\cdot A}$

$\Rightarrow Y=\overline{A}(\because A\cdot A=A)$

For
$A=1,Y=0$
and $A=0,Y=1$

$\therefore$ option $\left(\text{A}\right)$ is the correct answer.