Three-input NOR gate condition for HIGH output From the truth table of a 3-input NOR gate, which input combination (A, B, C) produces a HIGH (logic 1) output X?

Introduction / Context:
NOR gates are universal elements. Understanding their truth behavior is essential for designing active-LOW control signals and wired-OR types of logic in digital systems.

Given Data / Assumptions:

• 3-input NOR with inputs A, B, C, output X.
• NOR is NOT(OR) across all inputs.
• Logic levels are ideal (no analog thresholds considered).

Concept / Approach:
The NOR function is X = NOT(A + B + C). The OR of any input being HIGH yields 1, and after inversion, the output becomes 0. Therefore, the only way for X to be 1 is for the OR term to be 0, which requires A = B = C = 0.

Step-by-Step Solution:
Compute OR: S = A + B + C.If any of A, B, or C equals 1, then S = 1.NOR output: X = NOT(S). Thus, if S = 1 → X = 0; if S = 0 → X = 1.S = 0 only when A = 0, B = 0, and C = 0.

Verification / Alternative check:
Truth table enumeration shows that the line with A = 0, B = 0, C = 0 is the only row producing X = 1; all other rows produce X = 0.

Why Other Options Are Wrong:

• Any case with at least one 1 (like 100, 010, 001, 110, 101, 011, 111) makes the OR term 1 and the NOR output 0.

Common Pitfalls:

• Thinking NOR is the same as NAND; they complement different base operations.
• Forgetting that the presence of even a single HIGH at the input of OR forces the OR result to HIGH.

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