Question

For the logic gate shown in the diagram, find the output \(Y\) for the given inputs \(A\) and \(B\).

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

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to analyze the circuit gate by gate.
The circuit consists of two AND gates followed by a NOR gate.
Step 2: Key Formula or Approach:
The output of an AND gate with inputs A and B is $A \cdot B$.
The output of a NOR gate with inputs P and Q is $\overline{P + Q}$.
De Morgan's Law states that $\overline{A \cdot B} = \overline{A} + \overline{B}$.
Step 3: Detailed Explanation:
Let the outputs of the two AND gates be $P$ and $Q$.
Both AND gates have inputs A and B.
Therefore, $P = A \cdot B$ and $Q = A \cdot B$.
The final gate is a NOR gate, which takes $P$ and $Q$ as inputs.
The output $Y = \overline{P + Q}$.
Substituting the values:
\[ Y = \overline{(A \cdot B) + (A \cdot B)} \]
Since $X + X = X$ in Boolean algebra, we get:
\[ Y = \overline{A \cdot B} \]
According to De Morgan's Law, this expands to:
\[ Y = \overline{A} + \overline{B} \]
Step 4: Final Answer:
The output $Y$ is $\overline{A} + \overline{B}$.