Semiconductors Question 8
Question 8 - 30 January - Shift 2
The output $Y$ for the inputs $A$ and $B$ of circuit is given by
Truth table of the shown circuit is :
| $A$ | $B$ | $Y$ |
|---|---|---|
| 0 | 0 | 1 |
(1) $0 \begin{matrix} 0 & 1 & 1\end{matrix} $
| 1 | 0 | 1 |
|---|
| 1 | 1 | 0 |
|---|
| $A$ | $B$ | $Y$ |
|---|---|---|
| 0 | 0 | 1 |
(2) $0 \begin{matrix} 0 & 1 & 0\end{matrix} $
| 1 | 0 | 0 |
|---|
$\begin{matrix} 1 & 1 & 1\end{matrix} $
| $A$ | $B$ | $Y$ |
|---|---|---|
| 0 | 0 | 0 |
(3) $0 \begin{matrix} 0 & 1 & 1\end{matrix} $
| 1 | 0 | 1 |
|---|
$\begin{matrix} 1 & 1 & 1\end{matrix} $
| $A$ | $B$ | $Y$ |
|---|---|---|
| 0 | 0 | 0 |
(4) $0 \begin{matrix} 0 & 1 & 1\end{matrix} $
$\begin{matrix} 1 & 0 & 1\end{matrix} $
| 1 | 1 | 0 |
|---|
Show Answer
Answer: (4)
Solution:
Given circuit represent XOR
