Question

According to De Morgan's first theorem a NOR Gate is equivalent to a __________ Gate.

• A. Bubbled NOR
• B. XNOR
• C. XAND
• D. Bubbled AND

Hint:

Rewrite NOT(A+B) as (NOT A)(NOT B), an AND with inverted inputs.

Answer 1

The Correct Option is D

Step 1: Write the NOR output. For inputs A and B, a NOR gate gives \(\overline{A + B}\).

Step 2: Apply De Morgan's first theorem, which states \(\overline{A + B} = \overline{A} \cdot \overline{B}\).

Step 3: Read the right side. \(\overline{A} \cdot \overline{B}\) is an AND operation performed on the complemented (inverted) inputs.

Step 4: An AND gate whose inputs each carry an inversion bubble is called a bubbled AND gate. So a NOR gate equals a bubbled AND gate. The answer is Bubbled AND (option D).