If you are addressing bytes, then 512K words (16 bit words) requires 20 address lines.
I gave that answer because the question was categorized 8086/8088. If you are addressing words, then the answer is 19 address lines.
512 bytes=4096 bits . so 2^12 bits..therefore 12 address lines are required.
MB = Mega Bytes = 2^20
512 Locations = 2^9
So, total memory location are 2^(20+9) = 2^29
So, total number of address lines are 29.
29 address lines..
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.
It takes 23 address lines to address 8 mb of memory.
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.
2kb=2*1024=2048 2^11=2048 therefore 11 address lines are required
17 address lines and 8 data lines. 2^17=128k
You need 12 address lines to access 4K of memory. 212 = 4096. log2 (4096) = 12.
In the 2k*16 , the 11 address lines are required and the 16 input-output lines are required..
for 16 MB memory has 24 address lines
Firstly we need to convert Mb's into bits i.e 1Mb=1024x1024 = 210x210 =220 That means there are 220 memory locations and we will need 20 address lines.
The 8086/8088 has 20 address lines. It can access 220, or 1MB, or 1,048,576 bytes of memory.
ANSWER There are 2128 combinations of addresses. This is about 3.4 x 1038 locations. Assuming each address holds a 32-bit word, that's 1.2 x 1039 bytes. That's a LOT of memory.