Question:
The combination of the gates shown in the following figure yields
The combination of the gates shown in the following figure yields
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A Bubbled NAND gate (NAND with inverted inputs) acts as an OR gate. Similarly, a Bubbled AND gate acts as a NOR gate. Remembering these "bubbled" equivalencies saves time in logic circuit analysis!
Updated On: Apr 16, 2026
- NAND gate
- OR gate
- NOT gate
- XOR gate
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
According to De Morgan's Laws, the combination of inverted inputs followed by a specific gate can be simplified into a single basic logic gate.
Step 2: Key Formula or Approach
De Morgan's Theorem states: 1. \( \overline{A \cdot B} = \bar{A} + \bar{B} \) 2. \( \overline{A + B} = \bar{A} \cdot \bar{B} \)
Step 3: Detailed Explanation
1. The inputs are first passed through NOT gates, giving \( \bar{A} \) and \( \bar{B} \). 2. These are then fed into a NAND gate. The output \( Y \) of a NAND gate is the inversion of the product of its inputs: \[ Y = \overline{(\bar{A}) \cdot (\bar{B})} \] 3. Using De Morgan's Law (\( \overline{X \cdot Y} = \bar{X} + \bar{Y} \)): \[ Y = \overline{(\bar{A})} + \overline{(\bar{B})} \] 4. Since the double inversion cancels out (\( \bar{\bar{A}} = A \)): \[ Y = A + B \] 5. The expression \( A + B \) represents an OR gate.
Step 4: Final Answer
The combination yields an OR gate.
According to De Morgan's Laws, the combination of inverted inputs followed by a specific gate can be simplified into a single basic logic gate.
Step 2: Key Formula or Approach
De Morgan's Theorem states: 1. \( \overline{A \cdot B} = \bar{A} + \bar{B} \) 2. \( \overline{A + B} = \bar{A} \cdot \bar{B} \)
Step 3: Detailed Explanation
1. The inputs are first passed through NOT gates, giving \( \bar{A} \) and \( \bar{B} \). 2. These are then fed into a NAND gate. The output \( Y \) of a NAND gate is the inversion of the product of its inputs: \[ Y = \overline{(\bar{A}) \cdot (\bar{B})} \] 3. Using De Morgan's Law (\( \overline{X \cdot Y} = \bar{X} + \bar{Y} \)): \[ Y = \overline{(\bar{A})} + \overline{(\bar{B})} \] 4. Since the double inversion cancels out (\( \bar{\bar{A}} = A \)): \[ Y = A + B \] 5. The expression \( A + B \) represents an OR gate.
Step 4: Final Answer
The combination yields an OR gate.
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