The logic gate equivalent to the combination | vedclass

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Question

(A) 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 $\ba

Vedclass · Accepted Answer

$AND$

Vedclass · Answer

(A) 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 $y = \overline{\bar{A} + \bar{B}}$. Using De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$. The expression $y = A \cdot B$ represents an $AND$ gate. Therefore,the combination is equivalent to an $AND$ gate.

Input $A$	Input $B$	Output $Y$
$0$	$1$	$C$
$0$	$0$	$D$
$1$	$0$	$E$
$1$	$1$	$F$

The logic gate equivalent to the combination of logic gates shown in the figure is

Similar Questions

In the given digital circuit,if the inputs are $A=1, B=1$ and $C=1$,then the values of $y_1$ and $y_2$ are respectively:

For a $NAND$ gate, the inputs and outputs are given in the table below. The values taken by $C, D, E, F$ are respectively:
Input $A$ Input $B$ Output $Y$
$0$ $1$ $C$
$0$ $0$ $D$
$1$ $0$ $E$
$1$ $1$ $F$

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

The Boolean equation of a logic gate is $Y = \bar{A} \cdot B + A \cdot \bar{B}$. Identify the logic gate and describe its construction using $OR$,$AND$,and $NOT$ gates.

The output $D$ for the given circuit is:

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