From the circuit of the following logic gates | vedclass

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

Question

(D) Let the inputs be $A$ and $B$. The outputs of the two $NOT$ gates are $\overline{A}$ and $\overline{B}$.These are fed into a $NOR$ gate,so the out

Vedclass · Accepted Answer

Vedclass · Answer

(D) Let the inputs be $A$ and $B$. The outputs of the two $NOT$ gates are $\overline{A}$ and $\overline{B}$.These are fed into a $NOR$ gate,so the output of the $NOR$ gate is $\overline{\overline{A} + \overline{B}}$.By De Morgan's theorem,$\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$.This output $(A \cdot B)$ and the input $B$ are fed into a $NAND$ gate.The final output $Y = \overline{(A \cdot B) \cdot B} = \overline{A \cdot (B \cdot B)}$.Since $B \cdot B = B$,we have $Y = \overline{A \cdot B}$.This is the Boolean expression for a $NAND$ gate. However,looking at the provided circuit image,the final gate is a $NAND$ gate,and the expression simplifies to $Y = \overline{A \cdot B}$,which is a $NAND$ gate.

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

From the circuit of the following logic gates,the basic logic gate obtained is

Similar Questions

The truth table given below corresponds to which logic gate?

$A$ $B$ $X$

$0$ $0$ $0$

$0$ $1$ $1$

$1$ $0$ $1$

$1$ $1$ $1$

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:

In the given logic circuit diagram,$A$,$B$,and $C$ are the inputs,and $Y$ is the output. The output $Y$ is ' $HIGH$ ' (i.e.,$Y=1$):

In the adjacent circuit,$A$ and $B$ represent two inputs and $C$ represents the output. The circuit represents:

Vedclass Test Series

Exam Paper Generator

Online Exam Module