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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.

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Related Questions

How many address lines are needed to access 256KB of main memory?

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.


How many nuMBer of address lines required for 8 MB of memory?

It takes 23 address lines to address 8 mb of memory.


How many address lines and data lines are required for a2Kx8 memory?

A 2K X 8 memory requires 11 address lines and 8 data lines


How many no of address lines required in 1MB memory 111622 or 24?

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.


How many address lines are required if micro processor has to access 2kb memory?

2kb=2*1024=2048 2^11=2048 therefore 11 address lines are required


How many address lines and data lines are required for a 128k x 8 memory system?

17 address lines and 8 data lines. 2^17=128k


How do you calculate number of address lines needeed for 1 MB?

20 address lines are required


How many address lines are required to address 64 kbytes of memory?

You need 12 address lines to access 4K of memory. 212 = 4096. log2 (4096) = 12.


How many memory address does this number of address lines allow the 8086 to access directly?

The 8086/8088 has 20 address lines. It can access 220, or 1MB, or 1,048,576 bytes of memory.


How many address lines are necessary to address two megabytes 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


How many address lines and data lines are required from 256K 16 memory system?

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.


Minimum number of address lines needed to access a physical memory of 1GB?

You need 30 address lines to access 1G of memory. 230 = 1,073,741,824. log2 (1,073,741,824) = 30.

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