In the following circuit,the output Y for all | vedclass

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(C) The circuit consists of a $NOR$ gate followed by a $NOR$ gate acting as a $NOT$ gate (since both inputs are tied together).Let the output of the f

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

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(C) The circuit consists of a $NOR$ gate followed by a $NOR$ gate acting as a $NOT$ gate (since both inputs are tied together).Let the output of the first $NOR$ gate be $Y^{\prime} = \overline{A + B}$.The second gate is a $NOR$ gate with both inputs as $Y^{\prime}$,so its output is $Y = \overline{Y^{\prime} + Y^{\prime}} = \overline{Y^{\prime}} = \overline{\overline{A + B}} = A + B$.This is the Boolean expression for an $OR$ gate.The truth table for $Y = A + B$ is:$A$$B$$Y$$0$$0$$0$$0$$1$$1$$1$$0$$1$$1$$1$$1$Comparing this with the given options,option $C$ is correct.

$A$. 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 $< 1000$

In the following circuit,the output $Y$ for all possible inputs $A$ and $B$ is expressed by the truth table.

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