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5 logic gates are connected as shown in the figure. If $\mathrm{A}$ and $\mathrm{B}$ are the inputs, $\mathrm{Y}$ is the output then the truth table of the circuit is

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Verified Answer
The correct answer is:
\begin{array}{|c|c|c|}
\hline \mathrm{A} & \mathrm{B} & \mathrm{Y} \\
\hline 0 & 0 & 0 \\
\hline 1 & 0 & 0 \\
\hline 0 & 1 & 0 \\
\hline 1 & 1 & 0 \\
\hline
\end{array}
\hline \mathrm{A} & \mathrm{B} & \mathrm{Y} \\
\hline 0 & 0 & 0 \\
\hline 1 & 0 & 0 \\
\hline 0 & 1 & 0 \\
\hline 1 & 1 & 0 \\
\hline
\end{array}
Boolean expression for the combination of logic gates $\mathrm{Y}=(\mathrm{A} \cdot \overline{\mathrm{B}}) \cdot(\overline{\mathrm{A}} \cdot \mathrm{B})$
The truth table of the circuit is
\begin{array}{|l|l|l|l|l|l|l|}
\hline \mathrm{A} & \mathrm{B} & \overline{\mathrm{A}} & \overline{\mathrm{B}} & \mathrm{A} \cdot \overline{\mathrm{B}} & \overline{\mathrm{A}} \cdot \mathrm{B} & \mathrm{Y}=(\mathrm{A} \cdot \overline{\mathrm{B}}) \cdot(\overline{\mathrm{A}} \cdot \mathrm{B}) \\
\hline 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
\hline 0 & 1 & 1 & 0 & 0 & 1 & 0 \\
\hline 1 & 0 & 0 & 1 & 1 & 0 & 0 \\
\hline 1 & 1 & 0 & 0 & 0 & 0 & 0 \\
\hline
\end{array}
The truth table of the circuit is
\begin{array}{|l|l|l|l|l|l|l|}
\hline \mathrm{A} & \mathrm{B} & \overline{\mathrm{A}} & \overline{\mathrm{B}} & \mathrm{A} \cdot \overline{\mathrm{B}} & \overline{\mathrm{A}} \cdot \mathrm{B} & \mathrm{Y}=(\mathrm{A} \cdot \overline{\mathrm{B}}) \cdot(\overline{\mathrm{A}} \cdot \mathrm{B}) \\
\hline 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
\hline 0 & 1 & 1 & 0 & 0 & 1 & 0 \\
\hline 1 & 0 & 0 & 1 & 1 & 0 & 0 \\
\hline 1 & 1 & 0 & 0 & 0 & 0 & 0 \\
\hline
\end{array}
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