Question

The following configuration of gate equivalent to:

• A. NAND
• B. XOR
• C. OR
• D. None of these

Answer 1

The correct option is B

XOR

Evaluating the gate from the given figure:

In the given configuration, the input is $A$ and $B$. So, the output will be $\left(A+B\right)$ for the OR gate
In the given configuration, the input is $A$ and $B$. So, the output will be $\overline{\left(A.B\right)}=\overline{A}+\overline{B}$ for the NAND gate, according to De Morgan's theorem.

De Morgan's theorem states that $\overline{\left(A.B\right)}=\overline{A}+\overline{B}$ and $\overline{A+B}=\overline{A}.\overline{B}$.
Here, the input is $\left(A+B\right)$ and $\left(\overline{A}+\overline{B}\right)$. So, the output will be $\left(A+B\right).\left(\overline{A}+\overline{B}\right)$
So, the given output is $Y=(A+B).(\overline{A}+\overline{B})$
We can write the above expression as,
$Y=A\overline{A}+A\overline{B}+B\overline{A}+B\overline{B}$
$⟹Y=0+A\overline{B}+B\overline{A}+0\left[\because A\overline{A}=0\right]$
$⟹Y=A\overline{B}+B\overline{A}$
This is the expression of an XOR gate.

Hence option B is the correct answer.