Question:
When three NAND logic gates are connected as shown in the figure, then the logic gate equivalent to the circuit is
When three NAND logic gates are connected as shown in the figure, then the logic gate equivalent to the circuit is
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- 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.
Updated On: Jun 5, 2025
- NOT
- AND
- OR
- NOR
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The Correct Option is C
Solution and Explanation
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).
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).
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