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
In order to address 2 megabytes of memory, you need 21 address lines.
It requires 21 address lines to address two megabytes of memory.
221 = 2,097,152
It requires 31 address lines for 2 GB of memory.
You would need at least 22 address lines. If you take 2 to the 22nd power, you get a bit over 4.1 million, and that is the minimum number of addresses you'd need for four megabytes of data.
Two megabytes of memory requires 21 address lines. 221 = 2097152.
21 address lines are required to address two mb of memory. (221 = 2097152)
21
It takes 23 address lines to address 8 mb of memory.
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.
for 16 MB memory has 24 address 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.
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).
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.
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.
32 bit address bus can access more than 4 gigabytes (232) of memory. Sandeep Kr. Singh (MCA)
The minimum number of address lines required to address 4k of memory is 12.To reach this number, remember that each line has two possibilities and keep doubling as you count off. So one line can be used for two possibilities. Two lines represent four possibilities. Three represent eight. When you get to ten, you have 1024 possibilities. So double to 2048 at 11 and again to 4096 at 12. Or for the shortcut, if you take two to the 12th power, you get 4096.