Physics - Semiconductors Question with Solution | TestHub

From the given logic circuit, the output $Y$ is given by the Boolean expression:

$Y=A+G(A,B)$

From the given truth table, we can observe that the output $Y$ is identical to the input $A$ for all combinations of $A$ and $B$.

Thus, $Y=A$.

Equating the two expressions for $Y$, we get:

$A+G(A,B)=A$

Let us check the given options for gate G:

If G is an AND gate, $G(A,B)=A\cdot B$.

Then, $Y=A+A\cdot B=A(1+B)=A\cdot 1=A$.

This perfectly matches the required condition.

Checking other options:
If G is an OR gate, $Y=A+(A+B)=A+B\ne A$.
If G is a NAND gate, $Y=A+\overline{A\cdot B}$. For $A=0,B=0$, $Y=0+1=1\ne 0$.
If G is a NOR gate, $Y=A+\overline{A+B}$. For $A=0,B=0$, $Y=0+1=1\ne 0$.

Therefore, the gate G must be an AND gate.

Answer: AND gate.