The output of the given circuit diagram is

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

Question

(C) The circuit consists of two $OR$ gates whose inputs are modified by $NOT$ gates,and their outputs are fed into a $NOR$ gate.Let the inputs be $A$

Vedclass · Accepted Answer

$A, B, Y$$0, 0, 0$$1, 0, 0$$0, 1, 0$$1, 1, 0$

Vedclass · Answer

(C) The circuit consists of two $OR$ gates whose inputs are modified by $NOT$ gates,and their outputs are fed into a $NOR$ gate.Let the inputs be $A$ and $B$.The upper $OR$ gate has inputs $A$ and $\overline{B}$,so its output is $Y_1 = A + \overline{B}$.The lower $OR$ gate has inputs $B$ and $\overline{A}$,so its output is $Y_2 = \overline{A} + B$.These outputs $Y_1$ and $Y_2$ are fed into a $NOR$ gate,so the final output is $Y = \overline{Y_1 + Y_2} = \overline{(A + \overline{B}) + (\overline{A} + B)}$.Let's construct the truth table:$1$. If $A=0, B=0$: $Y_1 = 0 + \overline{0} = 1$,$Y_2 = \overline{0} + 0 = 1$. $Y = \overline{1+1} = 0$.$2$. If $A=1, B=0$: $Y_1 = 1 + \overline{0} = 1$,$Y_2 = \overline{1} + 0 = 0$. $Y = \overline{1+0} = 0$.$3$. If $A=0, B=1$: $Y_1 = 0 + \overline{1} = 0$,$Y_2 = \overline{0} + 1 = 1$. $Y = \overline{0+1} = 0$.$4$. If $A=1, B=1$: $Y_1 = 1 + \overline{1} = 1$,$Y_2 = \overline{1} + 1 = 1$. $Y = \overline{1+1} = 0$.Thus,for all inputs,the output $Y$ is $0$.

Similar Questions

The logic gate equivalent to the circuit given in the figure is

In the logic circuit given,$A, B$ and $C$ are the inputs and $Y$ is the output. The output $Y$ is $HIGH$ when:

The truth table for the following logic circuit is ...... .

Boolean algebra is essentially based on

Vedclass Test Series

Exam Paper Generator

Online Exam Module