A computer system has a memory capacity of 4K x 8 bits. How many address lines are required on the address bus to identify every byte in this memory uniquely?
12 lines
The number of address lines on the address bus is determined by how many separate memory locations must be uniquely identified, following the rule that \(n\) address lines can address \(2^{n}\) locations.
Here the memory is organised as 4K x 8, meaning it has \(4K = 4 \times 1024 = 4096\) locations, each 8 bits (one byte) wide.
To address every one of these locations uniquely we require \(2^{n} = 4096\).
Since \(2^{12} = 4096\), solving gives \(n = 12\), so 12 address lines are needed.
The 8-bit word width only fixes the size of each location (one byte) and does not affect the count of address lines, so the answer is not 8 lines.
Options of 10 or 16 lines would address \(2^{10} = 1024\) or \(2^{16} = 65536\) locations, neither of which equals 4096.
Therefore the address bus needs 12 lines to identify every byte uniquely.
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