Semiconductors Question 4
Question 4 - 2024 (27 Jan Shift 2)
The truth table of the given circuit diagram is :
(1) $ \begin{array}{|c|c|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{X} \\ \hline 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \\ \hline \end{array} $
(2) $ \begin{array}{|c|c|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{X} \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \hline \end{array} $
(3) $ \begin{array}{|c|c|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{X} \\ \hline 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \hline \end{array} $
(4) $ \begin{array}{|c|c|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{X} \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\ \hline \end{array} $
Show Answer
Answer: (2)
Solution:
$Y=A \cdot \overline{B}+\overline{A} \cdot B$
This is XOR GATE
