Discussion
For a 32KB memory the width of the address bus will be
Answer: C
The address bus needs 15 lines.
Each address line doubles the number of locations that can be selected, so the width is the base-2 logarithm of the number of addressable bytes.
\[ \begin{aligned} 32\ \text{KB} &= 32 \times 1024 = 2^{5} \times 2^{10} \\ &= 2^{15}\ \text{bytes} \end{aligned} \]
So 15 address lines are needed for byte addressing. 10 lines address only 1 KB, 12 lines 4 KB, and 16 lines 64 KB, twice the memory given.
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