Physics - Semiconductors Question with Solution | TestHub

The given circuit consists of three NAND gates.

The first NAND gate has its two inputs tied together and connected to $A$. It acts as a NOT gate. Its output is $Y_1=\overline{A\cdot A}=\overline{A}$.

The second NAND gate has its two inputs tied together and connected to $B$. It also acts as a NOT gate. Its output is $Y_2=\overline{B\cdot B}=\overline{B}$.

The third NAND gate takes $Y_1$ and $Y_2$ as inputs. Its output $Y$ is given by:
$Y=\overline{Y_1\cdot Y_2}$

Substituting the values of $Y_1$ and $Y_2$:
$Y=\overline{\overline{A}\cdot \overline{B}}$

Using De Morgan's theorem ($\overline{X\cdot Y}=\overline{X}+\overline{Y}$):
$Y=\overline{\overline{A}}+\overline{\overline{B}}$
$Y=A+B$

This is the Boolean expression for an OR gate. The truth table for an OR gate is:
If $A=0,B=0$, then $Y=0+0=0$
If $A=0,B=1$, then $Y=0+1=1$
If $A=1,B=0$, then $Y=1+0=1$
If $A=1,B=1$, then $Y=1+1=1$

This matches the truth table in the third option.

Answer: