Question

The given circuit works as:

• A. NOR gate
• B. OR gate
• C. AND gate
• D. NAND gate

Hint:

This type of gate is sometimes called a "bubbled OR" gate, which by De Morgan's law is equivalent to a NAND gate \( (A' + B') \neq (A'+B')' \). The circuit shown is NOT a bubbled OR, but rather NOT gates feeding a NOR gate. The expression \(Y = (A' + B')' = A \cdot B\) shows that two NOT gates followed by a NOR gate function as an AND gate. It is essential to carefully apply Boolean algebra rules.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a logic circuit diagram and need to determine which fundamental logic gate it is equivalent to.
Step 2: Key Formula or Approach:
We will use Boolean algebra to write the expression for the output `Y` in terms of the inputs `A` and `B`. We need to know the operations of the gates used: \begin{itemize} \item NOT gate (Inverter): Output is the inverse of the input. \( Y = A' \) or \( \overline{A} \). \item NOR gate: Output is the inverse of the OR operation. \( Y = (A+B)' \) or \( \overline{A+B} \). \end{itemize} We will also use De Morgan's Theorems: \( (A+B)' = A' \cdot B' \) and \( (A \cdot B)' = A' + B' \).
Step 3: Detailed Explanation:
1. Trace the inputs through the circuit:
\begin{itemize} \item Input `A` goes through a NOT gate. The output of this gate is \(A'\). \item Input `B` goes through a NOT gate. The output of this gate is \(B'\). \item The outputs \(A'\) and \(B'\) become the two inputs to the final gate. \end{itemize} 2. Analyze the final gate:
The final gate has the shape of an OR gate with a circle at the output, which is the symbol for a NOR gate.
The NOR gate takes inputs \(A'\) and \(B'\). Its output `Y` is therefore:
\[ Y = (A' + B')' \] 3. Simplify the Boolean expression:
Using De Morgan's first theorem, \( (X+Y)' = X' \cdot Y' \). Let \(X = A'\) and \(Y = B'\).
\[ Y = (A')' \cdot (B')' \] The double inversion \( (X')' \) is equal to \(X\).
\[ Y = A \cdot B \] 4. Identify the equivalent gate:
The expression \( Y = A \cdot B \) is the Boolean expression for an AND gate.
Alternative Method: Truth Table \begin{table}[h!] \centering \begin{tabular}{|c|c||c|c||c|} \hline A & B & A' & B' & Y = (A' + B')'
\hline 0 & 0 & 1 & 1 & (1+1)' = 1' = 0
0 & 1 & 1 & 0 & (1+0)' = 1' = 0
1 & 0 & 0 & 1 & (0+1)' = 1' = 0
1 & 1 & 0 & 0 & (0+0)' = 0' = 1
\hline \end{tabular} \end{table} The output column for Y (0, 0, 0, 1) is identical to the truth table for an AND gate.
Step 4: Final Answer:
The given circuit is equivalent to an AND gate. This corresponds to option (C).