Sequential logic basics: The S–R (Set–Reset) flip-flop is a bistable element used for storing 1 bit of information. Counting only its stable memory conditions, how many logic states does an S–R flip-flop have?

Introduction:
Flip-flops are the building blocks of registers and state machines. The S–R flip-flop is the simplest latch-type circuit, demonstrating how digital systems store a single binary value.

Given Data / Assumptions:

• Bistable device, meaning two stable outputs.
• Inputs labeled S (set) and R (reset).
• Invalid or metastable conditions are not counted as normal stable states.

Concept / Approach:
A bistable circuit has exactly two stable operating points. For the S–R flip-flop, those are Q = 1 (set) and Q = 0 (reset). When inputs are driven to certain combinations (e.g., S = 1, R = 1 in the basic NOR implementation), the state becomes invalid or leads to uncertainty, but that does not create an additional stable memory condition.

Step-by-Step Solution:
1) Identify stable states: stored 1 and stored 0.2) Recognize that state persists when inputs return to the hold condition.3) Exclude invalid/metastable scenarios from the count of stable states.4) Conclude that the device has two stable logic states.

Verification / Alternative check:
Truth tables for NOR- or NAND-based S–R latches show two valid stable outputs and one forbidden input combination, reinforcing the count of two stable states.

Why Other Options Are Wrong:
3: would imply an additional stable or tri-stable state; not present in S–R latches.4 and 5: exceed the storage capacity of one flip-flop, which holds exactly one bit.

Common Pitfalls:
Misinterpreting the forbidden input condition as a third state; it is not a valid stable memory condition.

Final Answer:
2