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

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 \( B = 1 \).