TTL troubleshooting scenario: An inverter feeds one input of a 2-input AND gate. If the inverter’s input pin is left open (no drive), and logic pulses are applied to the other AND input (point B), what appears at the AND output?

Introduction / Context:
In TTL, a floating input tends to read as HIGH. When that floating node is the input to an inverter, the inverter’s output becomes a constant LOW. Understanding this lets you predict downstream gate behavior during faults.

Given Data / Assumptions:

• Standard TTL inverter driving one input of a two-input AND gate.
• The inverter’s input is open (floating).
• Point B (the other AND input) carries logic pulses.

Concept / Approach:
A floating TTL input reads HIGH; therefore the inverter outputs a steady LOW. The AND gate thus sees inputs (LOW, pulses), and LOW AND anything is LOW, independent of the pulses.

Step-by-Step Solution:
1) Floating TTL input → reads HIGH.2) Inverter HIGH in → LOW out.3) AND truth: 0 * X = 0, so output is a steady LOW.

Verification / Alternative check:
Check truth tables and TTL input bias behavior to confirm the constant LOW at the inverter output and, consequently, the AND output.

Why Other Options Are Wrong:
“a steady HIGH” contradicts AND with a constant LOW input; “pulses” would occur only if the other input were a constant HIGH; “an undefined level” reflects CMOS-like floating behavior, not typical TTL input biasing.

Common Pitfalls:
Assuming floating equals undefined in TTL; ignoring the inversion before the AND stage.

Final Answer:
a steady LOW