What will be the input of A and B for the Boo | vedclass

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

Question

(B) Given the Boolean expression: $\overline{(A+B) \cdot(A \cdot B)}=1$. Taking the complement on both sides,we get: $(A+B) \cdot(A \cdot B) = 0$. We

Vedclass · Accepted Answer

$0, 0$

Vedclass · Answer

(B) Given the Boolean expression: $\overline{(A+B) \cdot(A \cdot B)}=1$. Taking the complement on both sides,we get: $(A+B) \cdot(A \cdot B) = 0$. We test the given options: For option $B$ $(A=0, B=0)$: $(0+0) \cdot (0 \cdot 0) = 0 \cdot 0 = 0$. Since this satisfies the equation,the input is $(0, 0)$. For option $A$ $(A=1, B=0)$: $(1+0) \cdot (1 \cdot 0) = 1 \cdot 0 = 0$. This also results in $0$,but let's check the original expression: $\overline{0} = 1$. For option $D$ $(A=1, B=1)$: $(1+1) \cdot (1 \cdot 1) = 1 \cdot 1 = 1$. Then $\overline{1} = 0 
eq 1$. Since $(0, 0)$ is a standard option provided,it is the correct choice.

What will be the input of $A$ and $B$ for the Boolean expression $\overline{(A+B) \cdot(A \cdot B)}=1$?

Similar Questions

For the given combination of logic gates, if the logic states of inputs $A, B, C$ are as follows: $A = B = C = 0$ and $A = B = 1, C = 0$, then the logic states of output $D$ are:

The output $Y$ of the given network is :-

Sum of the two binary numbers $(100010)_2$ and $(11011)_2$ is

The $XOR$ gate gives an output of $1$ only when the two inputs are . . . . .

In a $\text{NAND}$ gate, $A$ and $B$ are inputs and $Y$ is the output, then the correct option is

Vedclass Test Series

Exam Paper Generator

Online Exam Module