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: Trace NAND gate circuit using De Morgan's law.
A NAND gate on a single input A gives \( \overline{A\cdot A} = \bar{A} \) (NOT gate).

Step 2: NAND of the inverted inputs.
If each input A, B, C is individually inverted, then fed into a NAND: output \( = \overline{\bar{A}\cdot\bar{B}\cdot\bar{C}} = A+B+C \). But if the three NOT-outputs \( \bar{A},\bar{B},\bar{C} \) are the final output directly (OR via De Morgan on the combined NAND), the output is \( \bar{A}+\bar{B}+\bar{C} \).

\[ \boxed{\bar{A}+\bar{B}+\bar{C}} \]