A memory unit has a capacity of 4K × 8 bits. How many address lines are required in the address bus to uniquely access each memory location (byte)?

Explanation

A 4K × 8-bit memory has 4 × 1024 = 4096 addressable locations, with 8 bits (one byte) at each location. An address bus with n binary lines can select 2^n locations. Since 4096 = 2^12, 12 address lines are required. The 8-bit width describes the data stored at each location, not the number of address lines.