Physics - Semiconductors Question with Solution | TestHub

The given logic circuit consists of a NOR gate, an AND gate, and a NAND gate.

The output of the upper NOR gate is $Y_1=\overline{A+B}$.

The output of the lower AND gate is $Y_2=A\cdot B$.

These two outputs are fed into a NAND gate. The final output $Y$ is given by:

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

$Y=\overline{(\overline{A+B})\cdot (A\cdot B)}$

Using De Morgan's law, $\overline{A+B}=\overline{A}\cdot \overline{B}$. Substituting this into the expression for $Y$:

$Y=\overline{(\overline{A}\cdot \overline{B})\cdot (A\cdot B)}$

Rearranging the terms, we get:

$Y=\overline{(\overline{A}\cdot A)\cdot (\overline{B}\cdot B)}$

Since $A\cdot \overline{A}=0$ and $B\cdot \overline{B}=0$, the expression simplifies to:

$Y=\overline{0\cdot 0}=\overline{0}=1$

Thus, the output $Y$ is always $1$ regardless of the input values of $A$ and $B$.

This corresponds to the truth table (P).

Answer: P