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.
It takes 12 address bits to address 4K of memory. 212 = 4096.
8 bits to the byte = 32,000
6363
You need 20 bits of address bus to address 1 Mb of memory.
It takes 23 address lines to address 8 mb of memory.
1,374,389,534,720 bits
The memory address space is 64 MB, which means 226. However, each word is 4 bytes, which means that you have 224 words. This means you need log2 224 or 24 bits, to address each word.
There are 8388608 bits in 1MB, as there are 8bits in a Byte and 1024Bytes in a kilobyte and then there are 1024 kilobytes in a megabyte.There are 8388608 bits in 1 megabyte.
8,388,608 bits = 1,048,576 bytes = 1.048 megabytes (MB) = 1 mebibyte (MiB).
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.
There are 1,000,000 or 106 bits in a megabit.
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.
4096000
4800 mb