Question

When three NAND logic gates are connected as shown in the figure, then the logic gate equivalent to the circuit is

• A. NOT
• B. AND
• C. OR
• D. NOR

Hint:

- NAND gate output: \( \overline{A \cdot B} \). - A NAND gate with inputs tied together (\(A,A\)) acts as a NOT gate: \( \overline{A \cdot A} = \overline{A} \). - De Morgan's Theorems: 1. \( \overline{A \cdot B} = \overline{A} + \overline{B} \) 2. \( \overline{A + B} = \overline{A} \cdot \overline{B} \) - \( \overline{\overline{A}} = A \) (Double negation). Trace the logic signals step-by-step through the circuit.

Answer 1

The Correct Option is C

Let's analyze the output of each gate.
NAND gate 1: Inputs are A and A.
Output \( X = \overline{A \cdot A} = \overline{A} \).
(A NAND gate with inputs tied together acts as a NOT gate).
NAND gate 2: Inputs are B and B.
Output \( Y = \overline{B \cdot B} = \overline{B} \).
(Acts as a NOT gate).
NAND gate 3: Inputs are X and Y.
Output \( Z = \overline{X \cdot Y} \).
Substitute X and Y: \[ Z = \overline{(\overline{A}) \cdot (\overline{B})} \] Using De Morgan's theorem \( \overline{P \cdot Q} = \overline{P} + \overline{Q} \): \[ Z = \overline{(\overline{A})} + \overline{(\overline{B})} \] Since \( \overline{\overline{P}} = P \): \[ Z = A + B \] The expression \( A+B \) represents the OR logic operation.
Therefore, the equivalent logic gate is OR.
This matches option (3).