Truth table for the given circuit is

Truth table for the given circuit is
  1. $\begin{array}{lll}\mathrm{A} & \mathrm{B} & \mathrm{Y} \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 1\end{array}$
  2. $\begin{array}{lll}\mathrm{A} & \mathrm{B} & \mathrm{Y} \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1\end{array}$
  3. $\begin{array}{lll}A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1\end{array}$
  4. $\begin{array}{lll}A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1\end{array}$

Solution

The output of the logic gate circuit is $\begin{aligned} & \mathrm{Y}=\overline{\overline{(\mathrm{A}+\mathrm{B})}) \cdot(\mathrm{A} \cdot \mathrm{~B})}=\overline{\overline{\mathrm{A}+\mathrm{B}}}+\overline{\mathrm{A} \cdot \mathrm{~B}} \\ & =\mathrm{A}+\mathrm{B}+\overline{\mathrm{A}}+\overline{\mathrm{B}}=(\mathrm{A}+\overline{\mathrm{A}})+(\mathrm{B}+\overline{\mathrm{B}})=1+1=1 \end{aligned}$ $\therefore$ For any value of input, the output of the circuit is 1 .

Asked in: AP EAMCET 2024 (22 May Shift 2)

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