Question

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

Choose the correct answer from the options given below

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

Hint:

An alternative shortcut is to build a quick truth table with boundary states: If $A=0, B=0 \rightarrow$ inverted inputs are $1, 1 \rightarrow$ NOR output is $0$. If $A=1, B=1 \rightarrow$ inverted inputs are $0, 0 \rightarrow$ NOR output is $1$. The output only goes high when both inputs are high, confirming it behaves exactly like an AND gate.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem requires finding the overall logical function and corresponding Boolean expression for a multi-gate digital circuit layout containing logic gates processing input channels $A$ and $B$.

Step 2: Key Formula or Approach:
We can solve this by constructing a logic truth table or by applying Boolean algebra simplification rules:
1. De Morgan's Laws:
$$\overline{X + Y} = \overline{X} \cdot \overline{Y}$$ $$\overline{X \cdot Y} = \overline{X} + \overline{Y}$$ 2. Double Negation Law: $\overline{\overline{X}} = X$.

Step 3: Detailed Explanation:
Let's trace the configuration from the circuit diagram:
The inputs $A$ and $B$ are passed through independent NOT gates first, producing inverted signals:
$$\text{Output}_1 = \overline{A}$$ $$\text{Output}_2 = \overline{B}$$ These inverted signals are then fed as inputs into a final NOR gate. The mathematical expression for a NOR operation is the inversion of the sum of its inputs:
$$Y = \overline{\text{Output}_1 + \text{Output}_2}$$ Substitute our intermediate values into this expression:
$$Y = \overline{\overline{A} + \overline{B}}$$ Now, apply De Morgan's Law to split the upper inversion bar, changing the internal addition operator into a multiplication operator:
$$Y = \overline{\overline{A}} \cdot \overline{\overline{B}}$$ Applying the double negation rule simplifies the terms completely:
$$Y = A \cdot B$$ This final output expression corresponds to a standard two-input AND logic gate.

Step 4: Final Answer:
The network acts as an AND gate with the expression $A \cdot B$, matching option (B).