Address bit calculation: Because 4096 = 2^12, determine how many address bits are required for a 4K × 1 RAM to uniquely address all storage locations.

Introduction / Context:
Address lines select unique memory locations. The number of address bits determines the maximum addressable depth according to a power-of-two relationship. Designers must compute this quickly when planning memory maps and decoding logic.

Given Data / Assumptions:

• Total locations: 4K = 4096 addresses.
• Each address selects one 1-bit cell (width is irrelevant to address count).
• Binary addressing: N address lines can select 2^N unique locations.

Concept / Approach:
Find N such that 2^N = 4096. Recognize that 4096 is 2^12. Therefore N = 12 address bits are required, commonly labeled A0 through A11. This is independent of the data width; a 4K × 8 device also uses 12 address lines for the same depth.

Step-by-Step Solution:
1) Note 4K = 4096 = 2^12.2) Use formula: number_of_locations = 2^N → solve for N = log2(4096) = 12.3) Thus 12 address inputs are needed to uniquely select each location.4) Labeling convention: A0 is LSB, A11 is MSB for 4K depth.

Verification / Alternative check:
Build a quick table: 2^10 = 1024, 2^11 = 2048, 2^12 = 4096; confirms 12 lines.

Why Other Options Are Wrong:
“4096” confuses count with number of address lines. “10” and “1024” correspond to 1K. Only 12 maps to 4096 addresses.

Common Pitfalls:
Mixing decimal K (1000) with binary K (1024); forgetting that width does not change address line count.

Final Answer:
12