Question

A computer has 24-bit address bus and an instruction format providing 12 bits in the address part. What is the maximum addressable range?

• A. \(2^{24}\) locations
• B. \(2^{12}\) locations
• C. \(2^{36}\) locations
• D. \(2^{48}\) locations

Hint:

The number of addressable locations is determined by the width of the address bus, where the number of locations is \(2^{\text{number of bits in address bus}}\).

Answer 1

The Correct Option is A

The total addressable range is determined by the address bus. Since the address bus is 24 bits wide, the system can address \(2^{24}\) unique locations.
Therefore, the maximum addressable range is \(2^{24}\) locations.
Therefore, the correct answer is 1. \(2^{24}\) locations.