Question

In the following circuit, the output $\text{Y}$ for all possible inputs $\text{A}$ and $\text{B}$ is expressed by a truth table:

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

The flow of input in the circuit is shown below


The output from the $2^{nd}$ gate will be

$\text{Y}=\overline{\overline{(\text{A}+\text{B})}+\overline{(\text{A}+\text{B})}}$

Using Demorgan's theorem, $\text{A}+\text{A}=\text{A}$

$\Rightarrow \text{Y}=\overline{\overline{(\text{A}+\text{B})}}$

$\Rightarrow \text{Y}=\text{A}+\text{B}$

This is a representation of $\text{OR}$ logic function.

Therefore, the truth table is,

$\begin{array}{ccc}\text{A}& \text{B}& \text{Y}\\ 0& 0& 0\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 1\end{array}$

Hence, option $\left(\text{C}\right)$ is correct.