Semiconductors Question 4
Question 4 - 2024 (27 Jan Shift 2)
The truth table of the given circuit diagram is :
A $\quad$ B $Y$
$\begin{array}{lll}0 & 0 & 1\end{array}$
(1) $0 \begin{array}{lll}0 & 1 & 0\end{array}$
$\begin{array}{lll}1 & 0 & 0\end{array}$
$\begin{array}{lll}1 & 1 & 1\end{array}$
A $\quad$ B $\quad Y$
0 a 0 o 0
(2) $0 \quad 1 \quad 1$
1 1 0 1
$\begin{array}{lll}1 & 1 & 0\end{array}$
A $\quad$ B $\quad Y$
$\begin{array}{lll}0 & 0 & 0\end{array}$
(3) $0 \quad 1 \quad 0$
100
$\begin{array}{lll}1 & 1 & 1\end{array}$
$A+B \quad Y$
$\begin{array}{lll}0 & 0 & 1\end{array}$
(4) 0 กat 1
$\begin{array}{lll}1 & 0 & 1\end{array}$
$\begin{array}{lll}1 & 1 & 0\end{array}$
Show Answer
Answer: (2)
Solution:
$\mathrm{Y}=\mathrm{A} \cdot \overline{\mathrm{B}}+\overline{\mathrm{A}} \cdot \mathrm{B}$
This is XOR GATE
