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Physics
The truth table for the following logic circuit is
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Q. The truth table for the following logic circuit is
KEAM
KEAM 2009
Semiconductor Electronics: Materials Devices and Simple Circuits
A
$ \left| \begin{matrix} A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{matrix} \right| $
0%
B
$ \left| \begin{matrix} A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\ \end{matrix} \right| $
50%
C
$ \left| \begin{matrix} A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{matrix} \right| $
25%
D
$ \left| \begin{matrix} A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ \end{matrix} \right| $
0%
E
$ \left| \begin{matrix} A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{matrix} \right| $
0%
Solution:
Output, $ Y=A\overline{B}+\overline{A}B $ $ \left. \begin{matrix} A=0 & B=0 \\ \overline{A}=1 & \overline{B}=0 \\ \end{matrix} \right]Y=0 $
$ \left. \begin{matrix} A=0 & B=1 \\ \overline{A}=1 & \overline{B}=0 \\ \end{matrix} \right]Y=1 $
$ \left. \begin{matrix} A=1 & B=0 \\ \overline{A}=0 & \overline{B}=1 \\ \end{matrix} \right]Y=1 $
$ \left. \begin{matrix} A=1 & B=1 \\ \overline{A}=0 & \overline{B}=0 \\ \end{matrix} \right]Y=0 $
Hence, option is correct.
