You need 12 address lines to access 4K of memory. 212 = 4096. log2 (4096) = 12.
It takes 22 address lines to address 4MB of memory. (222 = 4,194,304)
It takes 16 address lines to address 64 KB of memory. (216 = 65536)
12 address lines are required in 4KBytes pf memory
b'coz 2^12=4K
There are 10 bits in 1 Kb.So 4 Kb requires 12 bits because 2^12=4056 bytes which is equal to 4Kb
2^12 = 4096 = 4k
use bank-switching, or (slower) a shift register.
18
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.
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.
to find the memory from address lines. you just have to make address liness power of 2, as shown below for 12 address lines 212 bits. 212 = 22 * 210 = 4 Kbthat means we can address 4 Kb from 12 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
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
for 16 MB memory has 24 address lines
In the 2k*16 , the 11 address lines are required and the 16 input-output lines are required..