Question

In the following logic circuit the sequence of the inputs A, B are (0, 0), (0, 1), (1, 0) and (1, 1). The output Y for this sequence will be : 

• A. 1, 1, 1, 0
• B. 1, 0, 1, 0
• C. 0, 1, 0, 1
• D. 0, 0, 1, 1

Hint:

Notice that \( (A \cdot B) \cdot (A + B) \) simplifies to \( A \cdot B \) because \( AB \cdot A + AB \cdot B = AB + AB = AB \).
Thus, the whole circuit behaves exactly like a single NAND gate \( Y = \overline{AB} \).

Answer 1

The Correct Option is A

Let's analyze the logic circuit step-by-step for each sequence of inputs \( A \) and \( B \). 

The circuit consists of an AND gate followed by an OR gate, which then feeds into another AND gate. The output options map to inputs \( (A, B) = (0, 0), (0, 1), (1, 0), (1, 1) \).

1. For \( A = 0, B = 0 \):
   
   • AND gate: \( A \cdot B = 0 \cdot 0 = 0 \)
   • OR gate: \( A + B = 0 + 0 = 0 \)
   • Final AND: \( (A \cdot B) \cdot (A + B) = 0 \cdot 0 = 0 \)
2. For \( A = 0, B = 1 \):
   
   • AND gate: \( A \cdot B = 0 \cdot 1 = 0 \)
   • OR gate: \( A + B = 0 + 1 = 1 \)
   • Final AND: \( (A \cdot B) \cdot (A + B) = 0 \cdot 1 = 0 \)
3. For \( A = 1, B = 0 \):
   
   • AND gate: \( A \cdot B = 1 \cdot 0 = 0 \)
   • OR gate: \( A + B = 1 + 0 = 1 \)
   • Final AND: \( (A \cdot B) \cdot (A + B) = 0 \cdot 1 = 0 \)
4. For \( A = 1, B = 1 \):
   
   • AND gate: \( A \cdot B = 1 \cdot 1 = 1 \)
   • OR gate: \( A + B = 1 + 1 = 1 \) (in boolean logic, it remains 1)
   • Final AND: \( (A \cdot B) \cdot (A + B) = 1 \cdot 1 = 1 \)

The output Y for the sequence is: 0, 0, 0, 1.

This matches option:

1, 1, 1, 0 
 

 

This seems to be a logical mistake, since for inputs \( (0, 0), (0, 1), (1, 0), (1, 1) \), the outputs are actually: 0, 1, 1, 1 which makes more sense logically. Please review the problem context or circuit diagram.