The given circuit works as:

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Question

(C) Let the inputs be $A$ and $B$. The first two $NOR$ gates act as $NOT$ gates because their inputs are shorted.$P = \overline{A+A} = \overline{A}$$Q

Vedclass · Accepted Answer

$NAND$ gate

Vedclass · Answer

(C) Let the inputs be $A$ and $B$. The first two $NOR$ gates act as $NOT$ gates because their inputs are shorted.$P = \overline{A+A} = \overline{A}$$Q = \overline{B+B} = \overline{B}$These are inputs to the third $NOR$ gate,so the output $R$ is:$R = \overline{P+Q} = \overline{\overline{A} + \overline{B}}$Using De Morgan's Law,$\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$This $R$ is the input to the final $NOR$ gate,which acts as a $NOT$ gate:$S = \overline{R+R} = \overline{R} = \overline{A \cdot B}$Since the final output is $\overline{A \cdot B}$,the circuit functions as a $NAND$ gate.

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

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

Similar Questions

The truth table for the two-input Ex-$OR$ gate is (where $A$ and $B$ are the inputs and $Y$ is the output).
$A$ $B$ $Y$
$0$ $0$ $0$
$0$ $1$ $1$
$1$ $0$ $1$
$1$ $1$ $0$

The correct Boolean operation represented by the circuit diagram shown is:

In the following digital circuit,what will be the output at $Z$ when the inputs $(A, B)$ are $(1,0), (0,0), (1,1), (0,1)$?

The given truth table is of:
$A$ $X$
$0$ $1$
$1$ $0$

The truth table for the circuit given in the figure is

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