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 $B$.

Similar Questions

The output of a $NOR$ gate is $HIGH$ when

How many $NAND$ gates are required to construct an $AND$ gate?

The figure shows a combination of two $NOT$ gates and a $NOR$ gate. The combination is equivalent to a

$A$ logic gate which gives output '$HIGH$' only when its two input terminals are at different logic levels with respect to each other is

Consider the following logic circuit. The output is $Y = 0$ when

Vedclass Test Series

Exam Paper Generator

Online Exam Module