The combination of logic gates shown below be | vedclass

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

Question

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

Vedclass · Accepted Answer

$OR$ gate

Vedclass · Answer

(D) The given circuit consists of two $NAND$ gates used 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 (acting as $NOT$) is $\bar{A}$.The output of the second $NAND$ gate (acting as $NOT$) is $\bar{B}$.These two outputs are fed into a third $NAND$ gate.The final output $Y$ is given by $Y = \overline{(\bar{A} \cdot \bar{B})}$.Using De Morgan's theorem,$\overline{(\bar{A} \cdot \bar{B})} = \overline{\bar{A}} + \overline{\bar{B}} = A + B$.This is the Boolean expression for an $OR$ gate.Therefore,the combination acts as an $OR$ gate.

The combination of logic gates shown below becomes:

Similar Questions

The truth table given below is for which gate?
$A: 0, 0, 1, 1$
$B: 0, 1, 0, 1$
$C: 1, 1, 1, 0$

Identify the operation performed by the circuit given below.

Which of the following relations is correct according to Boolean algebra?

The circuit diagram shown here corresponds to which logic gate?

The output $Y$ of the following digital logic circuit will be '$1$' (one) for the inputs:

Vedclass Test Series

Exam Paper Generator

Online Exam Module