The output Y of the logic circuit given below | vedclass

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(B) The input $A$ passes through a $NOT$ gate, resulting in $\bar{A}$.This $\bar{A}$ is fed into an $AND$ gate along with input $B$, producing an outp

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$\bar{A}$

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(B) The input $A$ passes through a $NOT$ gate, resulting in $\bar{A}$.This $\bar{A}$ is fed into an $AND$ gate along with input $B$, producing an output of $\bar{A} \cdot B$.This result $(\bar{A} \cdot B)$ and the original $\bar{A}$ are then fed into an $OR$ gate.Therefore, the final output $Y$ is given by:$Y = \bar{A} + (\bar{A} \cdot B)$Using the absorption law of Boolean algebra, which states that $X + (X \cdot Y) = X$, we can simplify the expression:$Y = \bar{A} \cdot (1 + B)$Since $(1 + B) = 1$, we get:$Y = \bar{A} \cdot 1 = \bar{A}$

The output $Y$ of the logic circuit given below is:

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