A 2-input AND gate has inputs A and B, plus an active-high inhibit (S). For which input combination is the output logic-1?
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AA = 1, B = 1, S = 1
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BA = 1, B = 1, S = 0
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CA = 1, B = 0, S = 1
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DA = 1, B = 0, S = 0
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EA = 0, B = 0, S = 0
Answer
Correct Answer: A = 1, B = 1, S = 0
Explanation
Introduction / Context:Inhibit inputs are common control features that disable otherwise-valid outputs. Recognizing the polarity of inhibit (active-high vs. active-low) is crucial to correct logic interpretation.
Given Data / Assumptions:
- Gate: AND with inputs A, B.
- Inhibit S is active-high (S = 1 disables output).
Concept / Approach:
Base AND requires A = 1 and B = 1. The active-high inhibit S must be 0 for the output to be enabled. Therefore, the only time output is 1 is when A = 1, B = 1, and S = 0.
Step-by-Step Solution:
AND condition → A * B = 1 requires A = 1 and B = 1.Inhibit condition → S must be 0 to allow output.Combine → A = 1, B = 1, S = 0.Verification / Alternative check:
Truth table inspection confirms that any time S = 1, output is forced low, regardless of A and B.
Why Other Options Are Wrong:
- Any option with S = 1 is inhibited.
- Any option with A = 0 or B = 0 fails AND condition.
Common Pitfalls:
Confusing inhibit polarity; assuming it behaves like an enable (active-high) when it is in fact active-high inhibit.
Final Answer:
A = 1, B = 1, S = 0