Which of the following is the correct truth t | vedclass

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(C) The given circuit consists of two $NOT$ gates, two $AND$ gates, and one $OR$ gate.Let the inputs be $A$ and $B$.The output of the first $AND$ gate

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The given circuit consists of two $NOT$ gates, two $AND$ gates, and one $OR$ gate.Let the inputs be $A$ and $B$.The output of the first $AND$ gate is $Y_1 = (	ext{NOT } A) 	ext{ AND } B = \bar{A} \cdot B$.The output of the second $AND$ gate is $Y_2 = A 	ext{ AND } (	ext{NOT } B) = A \cdot \bar{B}$.The final output $Y$ is the $OR$ combination of $Y_1$ and $Y_2$: $Y = Y_1 + Y_2 = \bar{A}B + A\bar{B}$.This is the Boolean expression for an $XOR$ gate.The truth table for an $XOR$ gate is:- If $A=0, B=0$, then $Y = \bar{0} \cdot 0 + 0 \cdot \bar{0} = 1 \cdot 0 + 0 \cdot 1 = 0 + 0 = 0$.- If $A=0, B=1$, then $Y = \bar{0} \cdot 1 + 0 \cdot \bar{1} = 1 \cdot 1 + 0 \cdot 0 = 1 + 0 = 1$.- If $A=1, B=0$, then $Y = \bar{1} \cdot 0 + 1 \cdot \bar{0} = 0 \cdot 0 + 1 \cdot 1 = 0 + 1 = 1$.- If $A=1, B=1$, then $Y = \bar{1} \cdot 1 + 1 \cdot \bar{1} = 0 \cdot 1 + 1 \cdot 0 = 0 + 0 = 0$.Thus, the correct truth table is option $C$.

Which of the following is the correct truth table for the logic gate circuit shown in the figure?

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

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

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

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