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 $B$.

$A$. 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 $< 1000$

For the logic circuit shown,the truth table is:

Similar Questions

The given truth table is correct for:
$P: 1, 1, 0, 0$
$Q: 1, 0, 1, 0$
$R: 1, 0, 0, 0$

The correct truth table for the following logic circuit is:

The truth table corresponding to the circuit given below is

If ${V}_{A}$ and ${V}_{B}$ are the input voltages (either $5\, V$ or $0\, V$) and ${V}_{0}$ is the output voltage,then the two gates represented in the following circuit $(A)$ and $(B)$ are:

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

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

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

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