The output of the circuit is low (zero) for: | vedclass

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

Question

(C) The given circuit consists of an $OR$ gate followed by a $NOR$ gate. Let the output of the $OR$ gate be $Z = X + Y$. This $Z$ is fed into both inp

Vedclass · Accepted Answer

$(B), (C)$ and $(D)$ only

Vedclass · Answer

(C) The given circuit consists of an $OR$ gate followed by a $NOR$ gate. Let the output of the $OR$ gate be $Z = X + Y$. This $Z$ is fed into both inputs of the $NOR$ gate. The output of the $NOR$ gate is $Out = \overline{Z + Z} = \overline{Z}$.Substituting $Z = X + Y$,we get $Out = \overline{X + Y}$. This is the Boolean expression for a $NOR$ gate.The truth table for the circuit is:$X$$Y$$X+Y$$Out = \overline{X+Y}$$0$$0$$0$$1$$0$$1$$1$$0$$1$$0$$1$$0$$1$$1$$1$$0$From the table,the output is low (zero) for cases $(B)$,$(C)$,and $(D)$.

The output of the circuit is low (zero) for: $(A) \ X = 0, Y = 0$,$(B) \ X = 0, Y = 1$,$(C) \ X = 1, Y = 0$,$(D) \ X = 1, Y = 1$. Choose the correct answer from the options given below:

Similar Questions

Which of the following logic gates is called a Universal gate?

In the given circuit as shown,two input waveforms $A$ and $B$ are applied simultaneously. The resultant waveform $Y$ is:

What are basic gates and universal gates?

The truth table for the given circuit is

Identify the correct truth table of the given logical circuit.

Vedclass Test Series

Exam Paper Generator

Online Exam Module