The figure shows a combination of two NOT gat | vedclass

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

Question

(C) The inputs to the $NOR$ gate are $\bar{A}$ and $\bar{B}$ because they pass through $NOT$ gates. The output $Y$ of the $NOR$ gate is given by the B

Vedclass · Accepted Answer

$AND$ gate

Vedclass · Answer

(C) The inputs to the $NOR$ gate are $\bar{A}$ and $\bar{B}$ because they pass through $NOT$ gates. The output $Y$ of the $NOR$ gate is given by the Boolean expression: $Y = \overline{\bar{A} + \bar{B}}$.According to De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.This is the Boolean expression for an $AND$ gate.Truth table:$A$$B$$\bar{A}$$\bar{B}$$\bar{A} + \bar{B}$$Y = \overline{\bar{A} + \bar{B}}$$0$$0$$1$$1$$1$$0$$0$$1$$1$$0$$1$$0$$1$$0$$0$$1$$1$$0$$1$$1$$0$$0$$0$$1$Thus,the combination is equivalent to an $AND$ gate.

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

Similar Questions

The following figure shows the variation of inputs $A$ and $B$ with time. If these inputs are given to a $NAND$ gate,select the correct output signal $Y$ from the given options.

The combination of logic gates shown below becomes:

In the following circuit,the correct relation between output $(Y)$ and inputs $A$ and $B$ will be

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

The following configuration of logic gates is equivalent to

Vedclass Test Series

Exam Paper Generator

Online Exam Module