The truth table for the circuit shown in the | vedclass

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(D) The circuit consists of a $NAND$ gate and an $OR$ gate feeding into a final $OR$ gate. Let the inputs be $A$ and $B$. The output of the $NAND$ gat

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

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(D) The circuit consists of a $NAND$ gate and an $OR$ gate feeding into a final $OR$ gate. Let the inputs be $A$ and $B$. The output of the $NAND$ gate is $\overline{AB}$. The output of the $OR$ gate is $A+B$. The final output $Y$ is the $OR$ combination of these two: $Y = (A+B) + \overline{AB}$. Using Boolean algebra: $Y = A + B + \bar{A} + \bar{B}$. Since $A + \bar{A} = 1$ and $B + \bar{B} = 1$,the expression simplifies to $Y = 1 + 1 = 1$. Thus,for all input combinations $(A, B)$,the output $Y$ is always $1$. Therefore,the correct truth table is option $D$.

The truth table for the circuit shown in the figure is

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

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

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

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