74148 (8-to-3 priority encoder): When data input I5 is the asserted (active) input on a 74148 priority encoder, the three data outputs (A2, A1, A0) are claimed to be A0 = 1, A1 = 0, A2 = 0. Decide if this statement is correct (assume standard active-low inputs and outputs).

Introduction / Context:
The 74148 is an 8-to-3 priority encoder with active-low inputs (I0–I7) and active-low outputs (A2, A1, A0). It encodes the highest-priority asserted input (I7 highest) into a 3-bit code. Many learners stumble over the active-low convention and inadvertently report the non-inverted code.

Given Data / Assumptions:

• Active-low inputs: a logic 0 on Ix means “active.”
• Active-low outputs: the binary code appears inverted at A2, A1, A0.
• Assume I5 is the only active input and higher-priority I6, I7 are inactive.

Concept / Approach:
If input 5 (binary 101) is the asserted line, the encoder outputs the binary index of the highest asserted input—but as active-low signals. That means the output pattern is the bitwise inversion of 101, which is 010. Therefore, A2 = 0, A1 = 1, A0 = 0. Any claim that A0 = 1, A1 = 0, A2 = 0 swaps and misinterprets the active-low result.

Step-by-Step Solution:

Verification / Alternative check:
Check the manufacturer’s truth table: With I5 low and I6, I7 high, the encoded outputs read 0 1 0 (active-low). Enable/GS behavior does not change the fundamental code pattern.

Why Other Options Are Wrong:

• Correct: Would require A2 A1 A0 = 0 0 1 (miscode), which is not the 74148 behavior.
• Only correct if active-high: The 74148 is defined with active-low interfaces; converting to active-high requires inversion.
• Ambiguous: Even with enables handled, the code mapping remains as above.

Common Pitfalls:
Forgetting that both inputs and outputs are active-low; mixing up bit order; not accounting for priority (I7 highest).

Final Answer:
Incorrect