Two NOT gates are connected at the two inputs of a NAND gate. This combination will behave like

Options

A. NAND gate
B. AND gate
C. OR gate
D. NOR gate

Answer

To solve the problem of determining what the combination of two NOT gates connected to the inputs of a NAND gate behaves like, we can follow these steps:

### Step 1: Understand the Components
We have two NOT gates and one NAND gate. The NOT gate inverts its input, while the NAND gate produces a low output only when all its inputs are high.

### Step 2: Define the Inputs
Let’s denote the inputs to the NOT gates as A and B. Therefore, the outputs of the NOT gates will be:
- Output of the first NOT gate: A' (which is NOT A)
- Output of the second NOT gate: B' (which is NOT B)

### Step 3: Connect the Outputs to the NAND Gate
The outputs of the NOT gates (A' and B') are fed into the NAND gate. The NAND gate will then take these inverted inputs.

### Step 4: Write the Output Expression for the NAND Gate
The output Y of the NAND gate can be expressed as:
\[ Y = \overline{A' \cdot B'} \]
This means Y is the negation of the product of A' and B'.

### Step 5: Apply De Morgan's Theorem
Using De Morgan's theorem, we can simplify the expression:
\[ Y = \overline{A' \cdot B'} = \overline{A'} + \overline{B'} \]
This simplifies to:
\[ Y = A + B \]
because the negation of A' is A, and the negation of B' is B.

### Step 6: Identify the Resulting Logic Gate
The expression \( Y = A + B \) is the output of an OR gate. Therefore, the combination of two NOT gates connected to the inputs of a NAND gate behaves like an OR gate.

### Conclusion
The final answer is that the combination behaves like an OR gate.