A computer with N address lines can address 2N different objects in memory. 221 is 2,097,152; in a byte oriented computer, otherwise known as two megabytes.
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
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.
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.
8086 has 20 address lines. Therefore it can address 220 bits or 1,048,576 bits of memory, or roughly 1 MB (mega byte).
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.
A 2K X 8 memory requires 11 address lines and 8 data lines
20 address lines are required
There are 24 address lines required for 16 mb. That covers 12 mb. The next step down is 23 address lines, which is 8 mb. The 8085 and 8086/8088 cannot address 12 mb. Only the 80286 and higher can.
The 8086/8088 microprocessor has a 20 bit address bus, so the number of memory locations it can address is 220 or 1,048,576.
2kb=2*1024=2048 2^11=2048 therefore 11 address lines are required