For the logic circuit shown,the truth table i | vedclass

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(B) The given circuit consists of two $NOT$ gates followed by a $NOR$ gate.Let the inputs be $A$ and $B$.The outputs of the two $NOT$ gates are $\bar{

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

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(B) The given circuit consists of two $NOT$ gates followed by a $NOR$ gate.Let the inputs be $A$ and $B$.The outputs of the two $NOT$ gates are $\bar{A}$ and $\bar{B}$.These are the inputs to the $NOR$ gate.The output $Y$ of the $NOR$ gate is given by the Boolean expression:$Y = \overline{\bar{A} + \bar{B}}$Using De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.Thus,the circuit acts as an $AND$ gate.The truth table for an $AND$ gate is:$A$$B$$Y$$0$$0$$0$$0$$1$$0$$1$$0$$0$$1$$1$$1$Therefore,the correct option is $B$.

For the logic circuit shown,the truth table is:

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

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

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

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