The combination of gates shown below yields:

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The circuit consists of two $NAND$ gates acting as $NOT$ gates (since their inputs are shorted) followed by a $NAND$ gate.Let the inputs be $A$ an

Vedclass · Accepted Answer

$OR$ gate

Vedclass · Answer

(B) The circuit consists of two $NAND$ gates acting as $NOT$ gates (since their inputs are shorted) followed by a $NAND$ gate.Let the inputs be $A$ and $B$.The output of the first $NAND$ gate is $\bar{A}$.The output of the second $NAND$ gate is $\bar{B}$.These two outputs are fed into the third $NAND$ gate.The final output $X$ is given by the boolean expression:$X = \overline{\bar{A} \cdot \bar{B}}$Using De Morgan's theorem,$\overline{\bar{A} \cdot \bar{B}} = \overline{\bar{A}} + \overline{\bar{B}} = A + B$.Since the expression $X = A + B$ represents an $OR$ gate,the correct option is $B$.

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

Similar Questions

Which of the following logic gates is called a Universal gate?

In the digital circuit shown in the figure,for the given inputs,the $P$ and $Q$ values are:

The output $(Y)$ of the given logic gate is similar to the output of an/a

The truth table given below is for ($A$ and $B$ are the inputs,$Y$ is the output).
$A$ $B$ $Y$
$0$ $0$ $1$
$0$ $1$ $1$
$1$ $0$ $1$
$1$ $1$ $0$

The resultant gate and its Boolean expression for the given circuit is

Vedclass Test Series

Exam Paper Generator

Online Exam Module