Question

For the logic gates shown below, the correct output is

• A. \(A+B+C\)
• B. \(\overline{A}\cdot \overline{B}\cdot \overline{C}\)
• C. \(\overline{A}+\overline{B}+\overline{C}\)
• D. \(\overline{AB}+\overline{BC}\)

Hint:

For NAND gates, always use De Morgan's law: \[ \overline{AB}=\overline{A}+\overline{B}. \] This helps simplify logic gate circuits quickly.

Answer 1

The Correct Option is C

Step 1: Identify the outputs of the NAND gates.
From the circuit, the upper NAND gate gives \[ \overline{AB}. \] The lower NAND gate gives \[ \overline{BC}. \] Also, the upper and lower side gates act as inverters, giving \[ \overline{A} \] and \[ \overline{C}. \]

Step 2: Combine the middle outputs.
The outputs \[ \overline{AB} \] and \[ \overline{BC} \] are fed into an OR gate.
Therefore, the middle output is \[ \overline{AB}+\overline{BC}. \] Using De Morgan's law, \[ \overline{AB}=\overline{A}+\overline{B} \] and \[ \overline{BC}=\overline{B}+\overline{C}. \] So, \[ \overline{AB}+\overline{BC} = (\overline{A}+\overline{B})+(\overline{B}+\overline{C}). \] \[ = \overline{A}+\overline{B}+\overline{C}. \]

Step 3: Final output.
The final OR gate combines the available outputs, so the final output remains \[ Y=\overline{A}+\overline{B}+\overline{C}. \]

Step 4: Final conclusion.
Therefore, \[ \boxed{Y=\overline{A}+\overline{B}+\overline{C}} \] Hence, the correct option is \[ \boxed{(3)} \]