JEE Physics 2023 Question with Solution

Given: The logic circuit has inputs $A$ and $B$. Find: The correct truth table for output $Y$.

Given circuit represent XOR.

The circuit is an implementation of the XOR gate using NAND gates. Therefore its output is

$$Y = A \oplus B$$

The XOR truth table is

$$\begin{array}{c|c|c} A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}$$

Therefore, the correct option is D.

Given: The circuit is made of NAND gates. Find: Which option matches the output $Y$.

To solve the given circuit and determine the output $Y$ for the inputs $A$ and $B$, analyze the logic gates used in the circuit.

The circuit provided in the image represents an XOR gate. The intermediate outputs are:

$$\overline{A \cdot B}$$

The next two NAND gate outputs become

$$\overline{A \cdot \overline{A \cdot B}}$$

and

$$\overline{B \cdot \overline{A \cdot B}}$$

The final NAND gate gives

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

Evaluating for all input combinations:

1. When $A = 0$ and $B = 0$, $Y = 0$.
2. When $A = 0$ and $B = 1$, $Y = 1$.
3. When $A = 1$ and $B = 0$, $Y = 1$.
4. When $A = 1$ and $B = 1$, $Y = 0$.

So the truth table is

$$\begin{array}{c|c|c} A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}$$

Hence, the given circuit represents an XOR gate and the correct option is D.