The truth table for the gate shown in the fol | vedclass

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(D) The given circuit consists of two $NOT$ gates (as the inputs are shorted) followed by a $NOR$ gate.Let the inputs be $A$ and $B$.The output of the

Vedclass · Accepted Answer

$A, B, X$$0, 0, 0$$0, 1, 0$$1, 0, 0$$1, 1, 1$

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(D) The given circuit consists of two $NOT$ gates (as the inputs are shorted) followed by a $NOR$ gate.Let the inputs be $A$ and $B$.The output of the first $NOT$ gate is $\bar{A}$.The output of the second $NOT$ gate is $\bar{B}$.These are the inputs to the $NOR$ gate.The output $X$ of the $NOR$ gate is given by $X = \overline{\bar{A} + \bar{B}}$.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:If $A=0, B=0$,then $X=0$.If $A=0, B=1$,then $X=0$.If $A=1, B=0$,then $X=0$.If $A=1, B=1$,then $X=1$.Comparing this with the given options,option $D$ is correct.

$A$	$X$
$0$	$1$
$1$	$0$

The truth table for the gate shown in the following figure will be

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