Three-input OR gate behavior For a digital OR gate with three inputs A, B, and C (positive-logic system), identify the specific input condition(s) that will force the gate's output to be LOW. Explain using the OR truth rule that the output is 1 if any input is 1, and 0 only when all inputs are 0.

Introduction / Context:
An OR gate implements the Boolean sum. In positive logic, a logic HIGH represents 1 and a logic LOW represents 0. Understanding when an OR gate outputs LOW is essential for designing enable lines, fault-detection logic, and combining asynchronous flags in digital electronics.

Given Data / Assumptions:

• Three inputs: A, B, C.
• Positive logic (HIGH = 1, LOW = 0).
• Ideal gate with standard OR truth behavior.

Concept / Approach:
The Boolean OR operation yields 1 if any input is 1. Symbolically for three inputs: Y = A + B + C. The only case that produces Y = 0 is A = 0 and B = 0 and C = 0. Any single HIGH (or more) forces the output HIGH.

Step-by-Step Solution:

Verification / Alternative check:
Create a short truth table subset. All-zero inputs produce LOW; any nonzero input vector (001, 010, 100, 011, 101, 110, 111) produces HIGH. This directly confirms OR behavior.

Why Other Options Are Wrong:

• A = 0, B = 0, C = 1: At least one input is HIGH, so output is HIGH.
• A = 0, B = 1, C = 1: Two inputs are HIGH, so output is HIGH.
• All of the above: Incorrect because only the all-zero case gives LOW.

Common Pitfalls:
Confusing OR with NOR (which inverts the result), or assuming that multiple LOW inputs are required to keep the output LOW even when a single HIGH is present. For OR, any HIGH dominates to produce a HIGH output.

Final Answer:
A = 0, B = 0, C = 0