In the following combination of logic gates,t | vedclass

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(C) $1$. The upper branch consists of an $AND$ gate followed by a $NOT$ gate ($NAND$ gate). The inputs to the $AND$ gate are $A$ and $B$,so its output

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) $1$. The upper branch consists of an $AND$ gate followed by a $NOT$ gate ($NAND$ gate). The inputs to the $AND$ gate are $A$ and $B$,so its output is $(A \cdot B)$. After the $NOT$ gate,the output becomes $(\overline{A \cdot B})$.$2$. The lower branch consists of a $NOT$ gate on input $A$ (giving $\overline{A}$) followed by an $AND$ gate with input $B$. Thus,the output of this branch 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: $Y = (\overline{A \cdot B}) + (\overline{A} \cdot B)$.$4$. Comparing this with the given options,option $C$ matches our derived expression.

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

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Inputs	Inputs	Outputs
$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$1$
$1$	$1$	$1$