Semiconductor Electronics Materials Devices And Simple Circuits Ques 6
6. In the following circuit, the output $Y$ for all possible inputs $A$ and $B$ is expressed by the truth table
[2007]
(a) $A \quad B \quad Y$
$\begin{matrix} \quad 0 & 1 & 1\end{matrix} $
$\begin{matrix} \quad 0 & 1 & 1\end{matrix} $
$\begin{matrix} \quad 1 & 0 & 1\end{matrix} $
$\begin{matrix}\quad 1 & 1 & 0\end{matrix} $
(b) $\begin{matrix} A & B & Y\end{matrix} $
$\begin{matrix} \quad 0 & 0 & 1\end{matrix} $
$\begin{matrix} \quad 0 & 1 & 1\end{matrix} $
$\begin{matrix} \quad 1 & 0 & 0\end{matrix} $
$\begin{matrix} \quad 1 & 1 & 0\end{matrix} $
(c) $\begin{matrix} A & B & Y\end{matrix} $
$\begin{matrix} \quad 0 & 0 & 0\end{matrix} $
$\begin{matrix} \quad 0 & 1 & 1\end{matrix} $
$\begin{matrix} \quad 1 & 0 & 1\end{matrix} $
$\begin{matrix} \quad 1 & 1 & 1\end{matrix} $
(d) $\begin{matrix} A & B & Y\end{matrix} $
$\begin{matrix} \quad 0 & 0 & 0\end{matrix} $
$\begin{matrix} \quad 0 & 1 & 0\end{matrix} $
$\begin{matrix} \quad 1 & 0 & 0\end{matrix} $
$\begin{matrix} \quad 1 & 1 & 1\end{matrix} $
Show Answer
Answer:
Correct Answer: 6.(c)
Solution:
- (c)
$Y^{\prime}=\overline{A+B} ; Y=\overline{\overline{A+B}}=A+B$
Therefore truth table:
| $A$ | $B$ | $Y$ |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
BETA
