Physics - Semiconductors Question with Solution | TestHub

Let us analyze the output of each logic gate given in the options:

For gate (A): The symbol represents a NAND gate. The inputs are $1$ and $1$.
The Boolean expression for a NAND gate is $Y=\overline{A\cdot B}$.
Output $Y=\overline{1\cdot 1}=\overline{1}=0$.

For gate (B): The symbol represents a NOR gate. The inputs are $0$ and $0$.
The Boolean expression for a NOR gate is $Y=\overline{A+B}$.
Output $Y=\overline{0+0}=\overline{0}=1$.

For gate (C): The symbol represents an AND gate. The inputs are $1$ and $0$.
The Boolean expression for an AND gate is $Y=A\cdot B$.
Output $Y=1\cdot 0=0$.

For gate (D): The symbol represents an XOR gate. The inputs are $0$ and $1$.
The Boolean expression for an XOR gate is $Y=A\oplus B$.
Output $Y=0\oplus 1=1$.

Therefore, the gates which give an output of '0' are (A) and (C).

Answer: (A) and (C)