Physics - Semiconductors Question with Solution | TestHub

The inputs $A$ and $B$ are connected to a NAND gate, producing the output $\overline{A\cdot B}$.

The input $C$ is connected to a NOT gate, producing the output $\bar{C}$.

These two signals are fed into an OR gate to produce the final output $Y$:

$Y=\overline{A\cdot B}+\bar{C}$

Using De Morgan's theorem, $\overline{A\cdot B}=\bar{A}+\bar{B}$.

Substituting this into the expression for $Y$ gives:

$Y=\bar{A}+\bar{B}+\bar{C}$

Answer: $\bar{A}+\bar{B}+\bar{C}$