Question

Truth table for the given circuit is:

• A. \[ \begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\ \hline \end{array} \]
• B. \[ \begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ \hline \end{array} \]
• C. \[ \begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\ \hline \end{array} \]
• D. \[ \begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ \hline \end{array} \]

Hint:

For logic circuits, first identify the logic gates, derive Boolean expressions, and verify results using truth tables.

Answer 1

The Correct Option is C

Step 1: Identify Logic Gates in the Circuit
The circuit consists of:
- An OR gate taking inputs \( A \) and \( B \).
- An AND gate taking inputs \( A \) and \( B \).
- The outputs of these gates are passed through another OR gate.
Step 2: Derive Boolean Expression The intermediate outputs are: \[ X = A + B \quad \text{(OR gate)} \] \[ Z = A \cdot B \quad \text{(AND gate)} \] The final output \( Y \) is obtained as: \[ Y = X + Z = (A + B) + (A \cdot B) \] Using Boolean algebra: \[ Y = A + B \] Step 3: Construct Truth Table \[ \begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\ \hline \end{array} \] Final Answer: The correct truth table matches option (3).