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 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.
The 8086/8088 has 20 address lines. It can access 220, or 1MB, or 1,048,576 bytes of memory.
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 a 256K x 16 memory system, the memory has 256K (256 * 1024 = 262,144) addressable locations and each location holds 16 bits of data. To calculate the number of address lines needed, we find the base-2 logarithm of 256K, which is 18 (since 2^18 = 262,144). For the data lines, since each location holds 16 bits, 16 data lines are required. Thus, the system requires 18 address lines and 16 data lines.
You need 30 address lines to access 1G of memory. 230 = 1,073,741,824. log2 (1,073,741,824) = 30.