In the combination of the following gates,the | vedclass

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(B) The given circuit consists of two $AND$ gates,two $NOT$ gates,and one $OR$ gate.$1$. The upper $AND$ gate receives inputs $A$ and $\overline{B}$ (

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$A\cdot \overline {B} + \overline {A} \cdot B$

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(B) The given circuit consists of two $AND$ gates,two $NOT$ gates,and one $OR$ gate.$1$. The upper $AND$ gate receives inputs $A$ and $\overline{B}$ (since $B$ passes through a $NOT$ gate). Its output is $A \cdot \overline{B}$.$2$. The lower $AND$ gate receives inputs $\overline{A}$ (since $A$ passes through a $NOT$ gate) and $B$. Its output is $\overline{A} \cdot B$.$3$. These two outputs are fed into an $OR$ gate. Therefore,the final output $Y$ is the sum of these two expressions: $Y = A \cdot \overline{B} + \overline{A} \cdot B$.This is the Boolean expression for an $XOR$ (Exclusive-$OR$) gate.

In the combination of the following gates,the output $Y$ can be written in terms of inputs $A$ and $B$ as

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