Question

Consider the following logic circuit.

The output is Y = 0 when :

• A. A = 1 and B = 1
• B. A = 0 and B = 1
• C. A = 1 and B = 0
• D. A = 0 and B = 0

Hint:

To analyze logic circuits, systematically determine the output of each logic gate based on its inputs and the truth table of the gate. For complex circuits, writing the Boolean expression for the final output in terms of the inputs can be helpful. Then, evaluate this expression for each given input combination.

Answer 1

The Correct Option is A

To determine when the output \( Y = 0 \) for the given logic circuit, let's analyze the circuit step-by-step.

1. The circuit consists of an AND gate, an OR gate, and a NOT gate.
2. Inputs: A and B
3. The AND gate receives inputs A and the output of the NOT gate (which inverts B).
4. The OR gate receives inputs B and the output of the AND gate.
5. The final output Y is obtained by passing the output of the OR gate through another AND gate with input A.

Let's analyze the circuit logic:

• The NOT gate inverts B, so if \( B = 0 \), the NOT gate outputs 1, and if \( B = 1 \), the NOT gate outputs 0.
• The output of the AND gate is given by \( A \cdot \overline{B} \). This is 1 if \( A = 1 \) and \( B = 0 \).
• The output of the OR gate is \( B + (A \cdot \overline{B}) \). This is 1 if either \( B = 1 \) or if both \( A = 1 \) and \( B = 0 \).
• The final AND gate gives output \( Y = A \cdot (B + (A \cdot \overline{B})) \).

We need to determine when \( Y = 0 \):

• From the expression \( Y = A \cdot (B + (A \cdot \overline{B})) \), \( Y \) becomes 0 only if \( A = 0 \).
• However, if \( A = 0 \), the first input to the final AND gate is 0, making \( Y = 0 \) irrespective of the other term.

Examining the options, we find:

• For \( \textbf{A = 1 and B = 1} \):
  • NOT gate output: \( \overline{B} = 0 \)
  • AND gate output: \( A \cdot \overline{B} = 1 \cdot 0 = 0 \)
  • OR gate output: \( B + (A \cdot \overline{B}) = 1 + 0 = 1 \)
  • Final AND gate output: \( Y = 1 \cdot 1 = 1 \)
• By verifying the calculations for all combinations, we find that the output \( Y = 0 \) when
• Correct Option:

A = 1 and B = 1

Answer 2

Step 1: Analyze the logic circuit and identify the gates.
The circuit consists of an AND gate, a NOT gate, an OR gate, and a final NAND gate. 
Step 2: Write the Boolean expression for the output of each gate.
Let the output of the AND gate be \( X \), the output of the NOT gate be \( \bar{B} \), and the output of the OR gate be \( Z \). 
The final output is \( Y \). \[ X = A \cdot B \] \[ \bar{B} = \text{NOT}(B) \] \[ Z = A + \bar{B} \] The final output \( Y \) is the NAND of \( X \) and \( Z \): \[ Y = \overline{X \cdot Z} = \overline{(A \cdot B) \cdot (A + \bar{B})} \] 
Step 3: Evaluate the output \( Y \) for each given input combination.
Case 1: A = 1 and B = 1
\[ X = 1 \cdot 1 = 1 \] \[ \bar{B} = \text{NOT}(1) = 0 \] \[ Z = 1 + 0 = 1 \] \[ Y = \overline{1 \cdot 1} = \overline{1} = 0 \] So, \( Y = 0 \) when \( A = 1 \) and \( B = 1 \). 
Case 2: A = 0 and B = 1
\[ X = 0 \cdot 1 = 0 \] \[ \bar{B} = \text{NOT}(1) = 0 \] \[ Z = 0 + 0 = 0 \] \[ Y = \overline{0 \cdot 0} = \overline{0} = 1 \] So, \( Y = 1 \) when \( A = 0 \) and \( B = 1 \). 
Case 3: A = 1 and B = 0
\[ X = 1 \cdot 0 = 0 \] \[ \bar{B} = \text{NOT}(0) = 1 \] \[ Z = 1 + 1 = 1 \] \[ Y = \overline{0 \cdot 1} = \overline{0} = 1 \] So, \( Y = 1 \) when \( A = 1 \) and \( B = 0 \). 
Case 4: A = 0 and B = 0
\[ X = 0 \cdot 0 = 0 \] \[ \bar{B} = \text{NOT}(0) = 1 \] \[ Z = 0 + 1 = 1 \] \[ Y = \overline{0 \cdot 1} = \overline{0} = 1 \] So, \( Y = 1 \) when \( A = 0 \) and \( B = 0 \). 
Step 4: Identify the input combination for which \( Y = 0 \).
From the evaluations above, the output \( Y = 0 \) only when \( A = 1 \) and \( B = 1 \).