Priority encoder 74148 — determine outputs from given inputs For an SN74148 8-to-3 priority encoder (active-LOW inputs I0–I7, active-LOW outputs A2 A1 A0; EI active-LOW enable), determine the five output leads (GS, A0, A1, A2, EO) when: I0 = 1, I1 = 1, I2 = 1, I3 = 1, I4 = 0, I5 = 1, I6 = 1, I7 = 1, and EI = 0.

Introduction / Context:
The 74148 is an active-LOW priority encoder: the highest-numbered asserted input (LOW) takes priority and is encoded on active-LOW outputs A2 A1 A0. GS (Group Select) and EO (Enable Output) indicate valid encoding and permit cascading. Interpreting active-LOW conventions correctly is essential.

Given Data / Assumptions:

• Inputs I0–I7 are active-LOW; a logic 0 means “asserted.”
• Outputs A2 A1 A0 are active-LOW codes of the winning input.
• EI = 0 enables the device; GS goes LOW and EO goes HIGH when any input is asserted.

Concept / Approach:
Identify the highest-priority asserted input (lowest logic level among I7..I0). With only I4 = 0 asserted and EI = 0, input 4 wins. Encode decimal 4 (binary 100) onto active-LOW outputs, which means drive A2 LOW, A1 HIGH, A0 HIGH. GS should be LOW (valid), and EO HIGH to indicate a valid output for cascading.

Step-by-Step Solution:

Verification / Alternative check:
Check with a 74148 truth table: single LOW on I4 yields A2 L, A1 H, A0 H and GS L, EO H when EI L.

Why Other Options Are Wrong:

• Options a–c contain incorrect bit patterns for the active-LOW code or misplace H/L levels.

Common Pitfalls:
Forgetting outputs are active-LOW and inverting them mentally; misreading the priority order (I7 highest).

Final Answer:
GS = L, A0 = H, A1 = H, A2 = L, EO = H