Question

The output Y of following circuit for given inputs is :

• A. \(A \cdot B (A + B)\)
• B. \(A \cdot B\)
• C. 0
• D. \(\overline{A} \cdot B\)

Answer 1

The Correct Option is C

To solve this problem, we need to analyze the logic circuit depicted in the given image. The circuit consists of AND, OR, and NOT gates.

Let's break down the circuit step-by-step:

1. The input A is connected to both an OR gate and an AND gate.
2. The input B is connected to a NOT gate. Thus, it outputs \(\overline{B}\).
3. The OR gate takes inputs A and B, resulting in the output \(A + B\).
4. The AND gate combines input A and NOT B, giving the output \(A \cdot \overline{B}\).
5. Finally, these two outputs from the OR and the AND gates are inputs to another AND gate, thus:

The final output Y from the AND gate is:

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

We analyze this expression:

1. The term \(A \cdot \overline{B}\) means A is true and B is false.
2. The expression \((A + B)\) evaluates to true if either A or B is true.
3. Combining \((A + B)\) with \((A \cdot \overline{B})\) in an AND operation results in:
   • If A is true and B is false, \((A \cdot \overline{B})\) is true, but for the AND operation with \((A + B)\) to hold, B must be false.
   • If B is true, \((A \cdot \overline{B})\) becomes false, leading the overall expression to false.

Therefore, the output Y will always be zero because the condition for it to be true is not exclusively satisfied.

Conclusion: The correct answer is \(0\).

Answer 2

Using the truth table:

A	B	Y
0	0	0
0	1	0
1	0	0
1	1	0

Thus, \( Y = 0 \).

Final Answer: 0.