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?

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:

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