Physics - Semiconductors Question with Solution | TestHub

The given logic circuit consists of a NOT gate and a NAND gate.

The input $A$ passes through the NOT gate, so its output is $\overline{A}$.

The inputs to the NAND gate are $\overline{A}$ and $B$.

The Boolean expression for the output $Y$ of the NAND gate is:
$Y=\overline{\overline{A}\cdot B}$

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

Evaluating $Y$ for all possible combinations of inputs $A$ and $B$:
If $A=0,B=0$, then $Y=0+\overline{0}=0+1=1$
If $A=0,B=1$, then $Y=0+\overline{1}=0+0=0$
If $A=1,B=0$, then $Y=1+\overline{0}=1+1=1$
If $A=1,B=1$, then $Y=1+\overline{1}=1+0=1$

This corresponds to the truth table (R).

Answer: R