JEE Physics 2023 Question with Solution

Given: A digital logic circuit with inputs $A$ and $B$ is shown.

Find: The equivalent logic operation performed by the circuit.

From the solution, the final conclusion is that the output simplifies to an AND operation.

Let the output of the first NAND gate be:

$$X_1 = \overline{A \cdot B}$$

The second NAND stage gives:

$$X_3 = \overline{X_1 \cdot X_2}$$

Using $X_1 = \overline{A \cdot B}$ and $X_2 = \overline{A \cdot B}$,

$$X_3 = \overline{\overline{A \cdot B} \cdot \overline{A \cdot B}}$$

So,

$$X_3 = A \cdot B$$

Hence the final output represents the logic of an AND gate.

Therefore, the correct option is D.

Given: The circuit is to be reduced to an equivalent single logic gate.

Find: Which gate among the given options matches the output $Y$.

1. Analyze the first NAND gate:

$$X_1 = \overline{A \cdot B}$$

2. The third NAND gate also provides:

$$X_2 = \overline{A \cdot B}$$

3. These two outputs are fed into the next NAND gate:

$$X_3 = \overline{X_1 \cdot X_2}$$

Substituting,

$$X_3 = \overline{\overline{A \cdot B} \cdot \overline{A \cdot B}}$$

Since the two terms are identical,

$$X_3 = A \cdot B$$

4. The solution concludes that the final output remains:

$$Y = A \cdot B$$

Thus the circuit is equivalent to an AND gate. Hence, the correct option is D.