The combination of gates shown below produces

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

Question

(D) The given circuit consists of two $NOT$ gates ($G_1$ and $G_2$) at the inputs $A$ and $B$,followed by a $NOR$ gate $(G_3)$ and another $NOT$ gate

Vedclass · Accepted Answer

$NAND$ gate

Vedclass · Answer

(D) The given circuit consists of two $NOT$ gates ($G_1$ and $G_2$) at the inputs $A$ and $B$,followed by a $NOR$ gate $(G_3)$ and another $NOT$ gate $(G_4)$.$1$. The outputs of $G_1$ and $G_2$ are $\bar{A}$ and $\bar{B}$ respectively.$2$. These are fed into the $NOR$ gate $G_3$,which produces the output: $Y' = \overline{\bar{A} + \bar{B}}$.$3$. Using De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.$4$. This output $Y'$ is then passed through the final $NOT$ gate $G_4$,resulting in $Y = \overline{Y'} = \overline{A \cdot B}$.$5$. The expression $\overline{A \cdot B}$ represents the output of a $NAND$ gate.Therefore,option $(d)$ is correct.

Similar Questions

Write the truth table for the circuit given in the figure below,which consists of $NOR$ gates,and identify the logic operation $(OR, AND, NOT)$ that this circuit performs.

The circuit shown here can be used as a logic gate. The logic gate is

Which of the following represents a universal gate?

The output $Y$ of the combination of gates shown is equal to :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module