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(A) The given circuit consists of an $AND$ gate and an $OR$ gate.Let the inputs be $A$ and $B$.The output of the $AND$ gate is $A \cdot B$.This output

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

Vedclass · Answer

(A) The given circuit consists of an $AND$ gate and an $OR$ gate.Let the inputs be $A$ and $B$.The output of the $AND$ gate is $A \cdot B$.This output is one of the inputs to the $OR$ gate,while the other input to the $OR$ gate is $A$ directly.Therefore,the final output $Y$ is given by the Boolean expression: $Y = A + (A \cdot B)$.Using the Boolean algebra identity $A + (A \cdot B) = A(1 + B) = A \cdot 1 = A$,we find that $Y = A$.Let's verify this with a truth table:- If $A=0, B=0$: $Y = 0 + (0 \cdot 0) = 0 + 0 = 0$.- If $A=0, B=1$: $Y = 0 + (0 \cdot 1) = 0 + 0 = 0$.- If $A=1, B=0$: $Y = 1 + (1 \cdot 0) = 1 + 0 = 1$.- If $A=1, B=1$: $Y = 1 + (1 \cdot 1) = 1 + 1 = 1$.Comparing this with the given options,option $A$ matches this truth table.

What is the truth table for the given logic gate combination?

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

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

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

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