A computer system has a memory capacity of 4K×8 bits. How many address lines are required on the address bus to identify every by…

Question

A computer system has a memory capacity of 4K×8 bits. How many address lines are required on the address bus to identify every byte in this memory uniquely?

1. A. 8 lines
2. B. 12 lines (Correct answer)
3. C. 16 lines
4. D. 10 lines

Correct Answer

Option B — 12 lines

Detailed Solution & Explanation

The correct answer is 12 lines.

Key Points

• The notation 4K × 8 bits means 4K locations, each holding 8 bits (one byte).
• 4K = 4 × 1024 = 4096 locations, so the address bus must uniquely identify 4096 addresses.
• Since n address lines address 2ⁿ locations, solve 2ⁿ = 4096 → n = 12.
• The "× 8" describes the width of each location, which determines the data bus, not the address bus — mistaking it for the address size gives the 8-line distractor.

Additional Information

• The general rule: address lines = log₂(number of locations), and data lines = bits per location.
• Useful powers of two: 2¹⁰ = 1K = 1024, 2¹² = 4K, 2¹⁶ = 64K, 2²⁰ = 1M, 2³⁰ = 1G.
• So 1K needs 10 lines, 64K needs 16, and 1M needs 20 — worth memorising, since these recur constantly.
• For this chip the address bus is 12 bits wide and the data bus 8 bits; total capacity is 4096 × 8 = 32,768 bits = 4 KB.