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

Assertion: The logic gate $NOT$ can be built using a diode.
Reason: The output voltage and the input voltage of the diode have a $180^o$ phase difference.

Assertion : $NAND$ or $NOR$ gates are called digital building blocks.
Reason : The repeated use of $NAND$ (or $NOR$) gates can produce all the basic or complicated gates.

In the Boolean algebra,the expression $\left( {\overline {\bar A \cdot \bar B} } \right) \cdot A$ is equal to:

If $A = 100101$ and $B = 110110$ are applied to $X-NOR$ gate,then the output will be

Which one of the following will be the output of the given circuit?

Vedclass Test Series

Exam Paper Generator

Online Exam Module