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

The logical operations of this circuit are analyzed sequentially to determine the equivalent gate.

1. Initial Operation:
Inputs $ A $ and $ B $ are processed by a NAND gate. The output of this gate is the negation of the logical AND of $ A $ and $ B $. The expression for this output is:

$ \overline{A \cdot B} $

2. Parallel Operation (Upper Path):
Input $ B $ is also passed through a NOT gate, resulting in:

$ \overline{B} $

3. Intermediate Operation (Upper Path):
The outputs from step 1 ($ \overline{A \cdot B} $) and step 2 ($ \overline{B} $) are combined using an AND gate. The output is:

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

4. Parallel Operation (Lower Path):
Inputs $ A $ and $ B $ are processed by an OR gate. The output is:

$ A + B $

5. Intermediate Operation (Lower Path):
The output from step 4 ($ A + B $) is inverted by a NOT gate. The output is:

$ \overline{A + B} $

6. Final Combination:
The outputs from step 3 ($ (\overline{A \cdot B}) \cdot (\overline{B}) $) and step 5 ($ \overline{A + B} $) are combined using an OR gate to produce the final output $ Y $:

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

Boolean Algebra Simplification:
Applying De Morgan's Law to $ \overline{A \cdot B} $, we get $ \overline{A} + \overline{B} $. Substituting this into the expression for $ Y $:

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

Distributing $ \overline{B} $ in the first term yields:

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

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

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

Combining identical terms:

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

Factoring out $ \overline{B} $:

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

Since $ 1 + \overline{A} $ simplifies to $ 1 $:

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

$ Y = \overline{B} $

Conclusion:
The logic circuit is equivalent to a NOT gate operating on input $ B $.