Question

The resultant gate and its Boolean expression for the given circuit is

• A. OR, $A + B$
• B. NAND, $\overline{A \cdot B}$
• C. NOR, $\overline{A + B}$
• D. AND, $A \cdot B$

Hint:

A NOR gate following two NOT gates behaves as an AND gate due to De Morgan’s law.

Answer 1

The Correct Option is D

Step 1: Identify individual gates.
The first two gates $G_1$ and $G_2$ are NOT gates, so they invert inputs $A$ and $B$.

Step 2: Analyze the final gate.
The outputs of the NOT gates are fed into an OR gate with an inverted output, i.e., a NOR gate.

Step 3: Write Boolean expression.
\[ Y = \overline{(\overline{A} + \overline{B})} \]

Step 4: Apply De Morgan’s theorem.
\[ Y = A \cdot B \]

Step 5: Conclusion.
The resultant gate is AND gate with Boolean expression $A \cdot B$.