Physics - Semiconductors Question with Solution | TestHub

From the given logic circuit, the output $Y$ is the result of an OR gate with inputs $A$ and the output of gate $G$.

Let the output of gate $G$ be $Y_G$. The Boolean expression for the circuit is:

$Y=A+Y_G$

From the given truth table, we can analyze the required output of gate $G$ for different inputs:

For $A=0,B=0$, the output is $Y=1$.

$0+Y_G(0,0)=1\Rightarrow Y_G(0,0)=1$

For $A=0,B=1$, the output is $Y=1$.

$0+Y_G(0,1)=1\Rightarrow Y_G(0,1)=1$

Now let us check the given options for gate $G$:

If $G$ is an AND gate: $Y_G(0,0)=0\cdot 0=0\ne 1$.

If $G$ is an OR gate: $Y_G(0,0)=0+0=0\ne 1$.

If $G$ is a NOR gate: $Y_G(0,1)=\overline{0+1}=0\ne 1$.

If $G$ is a NAND gate: $Y_G(0,0)=\overline{0\cdot 0}=1$ and $Y_G(0,1)=\overline{0\cdot 1}=1$.

Let us verify the remaining cases for a NAND gate:

For $A=1,B=0$: $Y_G(1,0)=\overline{1\cdot 0}=1\Rightarrow Y=1+1=1$.

For $A=1,B=1$: $Y_G(1,1)=\overline{1\cdot 1}=0\Rightarrow Y=1+0=1$.

This perfectly matches the given truth table. Therefore, the logic gate $G$ is a NAND gate.

Answer: NAND