Semiconductor Electronics Materials Devices and Simple Circuits - Result Question 101
103. In the following circuit, the output $Y$ for all possible inputs $A$ and $B$ is expressed by the truth table
[2007]
(a) A B Y
(b) A B Y $\begin{matrix} 0 & 1 & 1\end{matrix} $ $\begin{matrix} 0 & 0 & 1\end{matrix} $ $\begin{matrix} 0 & 1 & 1\end{matrix} $ $\begin{matrix} 0 & 1 & 0\end{matrix} $ $\begin{matrix} 1 & 0 & 1\end{matrix} $ $\begin{matrix} 1 & 0 & 0\end{matrix} $ $\begin{matrix} 1 & 1 & 0\end{matrix} $ 110
(c) A B Y
(d) A B Y 000 $\begin{matrix} 0 & 0 & 0\end{matrix} $ $\begin{matrix} 0 & 1 & 1\end{matrix} $ 010 101 100 $\begin{matrix} 1 & 1 & 1\end{matrix} $ $\begin{matrix} 1 & 1 & 1\end{matrix} $
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Answer:
Correct Answer: 103. (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 |
