5 logic gates are connected as shown in the f | vedclass

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Question

(A) The circuit consists of two $NOT$ gates, two $AND$ gates, and one final $AND$ gate. The inputs to the first $AND$ gate are $A$ and $\overline{B}$,

Vedclass · Accepted Answer

$A, B$$Y$$0, 0$$0$$1, 0$$0$$0, 1$$0$$1, 1$$0$

Vedclass · Answer

(A) The circuit consists of two $NOT$ gates, two $AND$ gates, and one final $AND$ gate. The inputs to the first $AND$ gate are $A$ and $\overline{B}$, so its output is $A \cdot \overline{B}$. The inputs to the second $AND$ gate are $\overline{A}$ and $B$, so its output is $\overline{A} \cdot B$. The final output $Y$ is the $AND$ operation of these two outputs: $Y = (A \cdot \overline{B}) \cdot (\overline{A} \cdot B)$. Using the associative and commutative properties of Boolean algebra: $Y = A \cdot \overline{A} \cdot B \cdot \overline{B}$. Since $A \cdot \overline{A} = 0$ and $B \cdot \overline{B} = 0$, the output $Y = 0 \cdot 0 = 0$ for all input combinations. Therefore, the output $Y$ is always $0$.

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

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