Question

 

 

 

 

 

 

 

For the circuit shown above, the equivalent gate is:

• A. AND gate
• B. OR gate
• C. NOT gate
• D. NAND gate

Hint:

Remember that a NAND gate outputs LOW only when all inputs are HIGH. For other input combinations, the output is HIGH.

Answer 1

The Correct Option is C

Let's break down this logic circuit step by step to figure out the equivalent gate.

1. First Stage:
The inputs $ A $ and $ B $ go into a NAND gate. The output of a NAND gate is the negation of the AND operation. So, the output here is:

$ \overline{A \cdot B} $

2. Second Stage (Top):
Input $ B $ also goes into a NOT gate. The output of the NOT gate is the complement of the input, which is:

$ \overline{B} $

3. Third Stage (Top):
The outputs from the first and second stages, $ \overline{A \cdot B} $ and $ \overline{B} $, are fed into an AND gate. The output of this AND gate is:

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

4. Second Stage (Bottom):
Inputs $ A $ and $ B $ go into an OR gate. The output of an OR gate is:

$ A + B $

5. Third Stage (Bottom):
The output of the OR gate, $ A + B $, goes into a NOT gate. The output of this NOT gate is:

$ \overline{A + B} $

6. Final Stage:
The outputs from the third stage (top) and the third stage (bottom), $ (\overline{A \cdot B}) \cdot (\overline{B}) $ and $ \overline{A + B} $, are fed into an OR gate. The final output $ Y $ is therefore:

$ Y = [(\overline{A \cdot B}) \cdot (\overline{B})] + \overline{A + B} $

Simplifying the Expression Using Boolean Algebra:
Using De Morgan's Laws, we rewrite $ \overline{A \cdot B} $ as $ \overline{A} + \overline{B} $. Substituting this into the expression for $ Y $:

$ Y = [(\overline{A} + \overline{B}) \cdot \overline{B}] + (\overline{A} \cdot \overline{B}) $

Distribute $ \overline{B} $ in the first term:

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

Simplify $ \overline{B} \cdot \overline{B} $ to $ \overline{B} $:

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

Combine like terms:

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

Factor out $ \overline{B} $:

$ Y = \overline{B} (1 + \overline{A}) $

Simplify $ 1 + \overline{A} $ to $ 1 $:

$ Y = \overline{B} \cdot 1 $

$ Y = \overline{B} $

Final Answer:
The equivalent gate for the given circuit is a NOT gate with input $ B $.