For a 32 KB byte-addressable memory, the width of the address bus will be:
2026
For a 32 KB byte-addressable memory, the width of the address bus will be:
- A.
10
- B.
12
- C.
15
- D.
16
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Correct answer: C
Address-bus width depends on the number of addressable locations.
32 KB = 32 × 1024 bytes = 32768 bytes
32768 = 215 addressable byte locations
To select 215 locations, the address bus needs 15 lines
Correct answer: 15