Semiconductor Electronics Material Devices and Simple Circuits
151250
For the given logic gates combination the correct truth table will be
1 \(\begin{array}{|c|c|c|}
\hline \mathrm{A} & \mathrm{B} & \mathrm{X} \\
\hline 0 & 0 & 0 \\
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \\
\hline
\end{array}\)
2 \(\begin{array}{|c|c|c|}
\hline \mathrm{A} & \mathrm{B} & \mathrm{X} \\
\hline 0 & 0 & 0 \\
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 1 \\
\hline
\end{array}\)
3 \(\begin{array}{|c|c|c|}
\hline \mathrm{A} & \mathrm{B} & \mathrm{X} \\
\hline 0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
1 & 1 & 0 \\
\hline
\end{array}\)
4 \(\begin{array}{|c|c|c|}
\hline \mathrm{A} & \mathrm{B} & \mathrm{X} \\
\hline 0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 1 \\
1 & 1 & 0 \\
\hline
\end{array}\)
Explanation:
B 
The Boolean expression of this circuit
\(\mathrm{X}=\overline{\mathrm{A}} \mathrm{B}+\overline{\mathrm{B}} \mathrm{A} \\
\mathrm{X}=\mathrm{A} \oplus \mathrm{B}\)
Symbol of XOR gate

\(\begin{array}{|c|c|c|}
\mathrm{A} &\mathrm{B} &\mathrm{X}=\mathrm{A} \oplus \mathrm{B} \\
\hline 0 & 0 0 \\
\hline 0 & 1 & 1 \\
\hline 1 & 0 & 1 \\
\hline 1 & 1 & 0 \\
\hline
\end{array}\)