The correct truth table for the following log | vedclass

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Question

(B) The given logic circuit consists of an $AND$ gate,a $NOT$ gate,and an $OR$ gate.Input $A$ is connected to a $NOT$ gate,so its output is $\bar{A}$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The given logic circuit consists of an $AND$ gate,a $NOT$ gate,and an $OR$ gate.Input $A$ is connected to a $NOT$ gate,so its output is $\bar{A}$.Inputs $A$ and $B$ are connected to an $AND$ gate,so its output is $A \cdot B$.These two outputs are then fed into an $OR$ gate.Therefore,the final output $Y$ is given by the Boolean expression: $Y = \bar{A} + (A \cdot B)$.Using the distributive law of Boolean algebra,$Y = (\bar{A} + A) \cdot (\bar{A} + B)$.Since $\bar{A} + A = 1$,we get $Y = 1 \cdot (\bar{A} + B) = \bar{A} + B$.Now,let's construct the truth table for $Y = \bar{A} + B$:- If $A=0, B=0$: $Y = \bar{0} + 0 = 1 + 0 = 1$.- If $A=0, B=1$: $Y = \bar{0} + 1 = 1 + 1 = 1$.- If $A=1, B=0$: $Y = \bar{1} + 0 = 0 + 0 = 0$.- If $A=1, B=1$: $Y = \bar{1} + 1 = 0 + 1 = 1$.Comparing this with the given options,option $B$ matches the calculated truth table.

The correct truth table for the following logic circuit is:

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

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

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

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