The number of address lines needed to access N-KB is given by log2N Then the number of address lines needed to access 256KB of main memory will be log2256000=18 address lines.
It takes 23 address lines to address 8 mb of memory.
A 2K X 8 memory requires 11 address lines and 8 data lines
How many no of address lines required in 1MB memory 11,16,22 or 24 u haven't specified correct options! 20 address lines will be required because 1 MB is 1024 KB that is 1024*1024 Byte which is equivalent to (2^10)^2 bytes if ur memory is Byte addressable then address lines required will be 20.
2kb=2*1024=2048 2^11=2048 therefore 11 address lines are required
17 address lines and 8 data lines. 2^17=128k
20 address lines are required
You need 12 address lines to access 4K of memory. 212 = 4096. log2 (4096) = 12.
Let N be the number of addresses line 2 megabyte = 2*1024 =2048 N = log (size in bytes) /log 2 N= log 2048/log 2 N=11
The 8086/8088 has 20 address lines. It can access 220, or 1MB, or 1,048,576 bytes of memory.
You need 30 address lines to access 1G of memory. 230 = 1,073,741,824. log2 (1,073,741,824) = 30.
The 8086/8088 has 20 address lines. It can access 220, or 1MB, or 1,048,576 bytes of memory.