Question

The output $Y$ of the gate shown in the figure will be represented by :

• A. $Y=A\cdot B$
• B. $Y=A+B$
• C. $Y=\overline{A+B}$
• D. $Y=A\overline{B}+B\overline{A}$

Answer 1

The correct option is B $Y=A+B$

Let the output from the $1^{st}$ gate and the $2^{nd}$ gate be $Y_1\text{and}{Y}_2$

$Y_1=\overline{A\cdot A}=\overline{A}$

Similarly, $Y_2=\overline{B\cdot B}=\overline{B}$

Now the inputs to the final $\text{NAND}$ gate is $Y_1\text{and}{Y}_2$.

$\therefore Y=\overline{Y_1\cdot Y_2}$

$\Rightarrow Y=\overline{\overline{A}\cdot \overline{B}}$

By DeMorgan's theorem,

$Y=\overline{\overline{A}}+\overline{\overline{B}}$

$\Rightarrow Y=A+B$

Hence, option $\left(\text{B}\right)$ is correct.