The truth table for the given logic circuit i | vedclass

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(C) The circuit consists of two $AND$ gates and one $OR$ gate. Let the inputs be $A$ and $B$.$1$. The upper $AND$ gate receives inputs $A$ and $B$. It

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$A$$B$$E$$0$$0$$0$$0$$1$$1$$1$$0$$0$$1$$1$$1$

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(C) The circuit consists of two $AND$ gates and one $OR$ gate. Let the inputs be $A$ and $B$.$1$. The upper $AND$ gate receives inputs $A$ and $B$. Its output is $C = A \cdot B$.$2$. The lower $AND$ gate receives inputs $\bar{A}$ (from the $NOT$ gate) and $B$. Its output is $D = \bar{A} \cdot B$.$3$. The final output $E$ is the $OR$ combination of $C$ and $D$,so $E = C + D = (A \cdot B) + (\bar{A} \cdot B)$.$4$. Factoring out $B$,we get $E = B(A + \bar{A})$. Since $A + \bar{A} = 1$,the expression simplifies to $E = B \cdot 1 = B$.$5$. Constructing the truth table:- If $A=0, B=0$,then $E=0$.- If $A=0, B=1$,then $E=1$.- If $A=1, B=0$,then $E=0$.- If $A=1, B=1$,then $E=1$.This matches the truth table where $E$ is identical to $B$.

The truth table for the given logic circuit is: (Inputs $A, B$ and Output $E$)

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$A$	$B$	$E$
$0$	$0$	$1$
$0$	$1$	$0$
$1$	$0$	$1$
$1$	$1$	$0$

$A$	$B$	$E$
$0$	$0$	$1$
$0$	$1$	$0$
$1$	$0$	$0$
$1$	$1$	$1$

$A$	$B$	$E$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$0$
$1$	$1$	$1$

$A$	$B$	$E$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$0$