The output (X) of the logic circuit shown in | vedclass

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(C) The first gate is a $NAND$ gate with inputs $A$ and $B$. Its output is $\overline{A.B}$.This output is fed into a $NOT$ gate (a $NAND$ gate with b

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$X = A.B$

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(C) The first gate is a $NAND$ gate with inputs $A$ and $B$. Its output is $\overline{A.B}$.This output is fed into a $NOT$ gate (a $NAND$ gate with both inputs connected together acts as a $NOT$ gate). If the input to a $NOT$ gate is $Y$,the output is $\overline{Y}$.Here,$Y = \overline{A.B}$,so the final output $X = \overline{\overline{A.B}}$.Using the Boolean algebra property $\overline{\overline{Z}} = Z$,we get $X = A.B$.

$A$. Small Scale Integration $(SSI)$	$I$. Logic gates $< 100$
$B$. Medium Scale Integration $(MSI)$	$II$. Logic gates $> 1000$
$C$. Large Scale Integration $(LSI)$	$III$. Logic gates $\leq 10$
$D$. Very Large Scale Integration $(VLSI)$	$IV$. Logic gates $< 1000$

The output $(X)$ of the logic circuit shown in the figure will be:

Similar Questions

The truth table for this given circuit is:

The combination of the gates shown will produce:

The Boolean expression for a two-input Ex-$OR$ gate is (where $A$ and $B$ are inputs,and $Y$ is the output):

The truth table given below is for which gate?
$A: 0, 0, 1, 1$
$B: 0, 1, 0, 1$
$C: 1, 1, 1, 0$

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