Q159 / 160 · TS EAMCET 2023 (14 May Shift 2) · Physics

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 Figure

  • Correct answer
    A
    \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}
  • B
    \begin{array}{|c|c|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{Y} \\ \hline 0 & 0 & 1 \\ \hline 1 & 0 & 0 \\ \hline 0 & 1 & 0 \\ \hline 1 & 1 & 1 \\ \hline \end{array}
  • C
    \begin{array}{|c|c|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{Y} \\ \hline 0 & 0 & 0 \\ \hline 1 & 0 & 1 \\ \hline 0 & 1 & 1 \\ \hline 1 & 1 & 1 \\ \hline \end{array}
  • D
    \begin{array}{|c|c|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{Y} \\ \hline 0 & 0 & 1 \\ \hline 1 & 0 & 1 \\ \hline 0 & 1 & 1 \\ \hline 1 & 1 & 1 \\ \hline \end{array}

Solution

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}

Hence, the correct answer is option: 'A'.

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