The resultant gate and its Boolean expression in the given circuit is
Choose the correct answer from the options given below
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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.
Step 1: Read the circuit. Inputs $A$ and $B$ are each passed through a NOT gate, and the two inverted outputs feed a NOR gate. Step 2: Write the inverted inputs. After the NOT gates the signals are $\overline{A}$ and $\overline{B}$. Step 3: Apply the NOR operation. A NOR gate outputs the complement of the sum of its inputs, so $Y = \overline{\overline{A} + \overline{B}}$. Step 4: Use De Morgan's law. The bar over a sum becomes a product of bars: $Y = \overline{\overline{A}}\cdot\overline{\overline{B}}$. Step 5: Remove the double bars. Since $\overline{\overline{A}} = A$ and $\overline{\overline{B}} = B$, we get $Y = A\cdot B$. Step 6: Identify the gate. The expression $A\cdot B$ is the AND function, so the resultant gate is AND, which is option (2). \[ \boxed{\text{AND},\ Y = A\cdot B} \]
