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How many address bits are required in an 8K X 8 memory?
15 bits
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 bits are required for accessing 1GB RAM?
It requires 30 address bits to address 1GB of RAM.230 = 1,073,741,824
How many address bits would be required for a memory that stores 64 4-bit words?
It requires 6 bits to address 64 words. It does not matter what the word size is.
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 memory locations can 14 address bits access?
16384
How many bits are required to supernet seven class c address?
3 bits
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.
If a memory has 512 words how many bits should the address registry have?
9
How many lines of the address must be used to access 2048 bytes. How many of these lines are connected to the address input of all chips?
That depends on the memory architecture of the system.if the memory chips are byte wide and not used to create a multibyte bus, 11 address bits are needed.if the memory chips are 32 bits wide, 9 address bits are needed (with the CPU internally selecting which of the 4 bytes it will use).it the memory chips are 64 bits wide, 8 address bits are needed (with the CPU internally selecting which of the 8 bytes it will use.if the memory chips are 4 bits wide, 12 address bits will be needed and the CPU must perform 2 memory cycles per byte that it needs. (yes, I have seen a computer that worked this way!)etc.
A computer has 64MB megabytes of memory Each word is 4 bytes How many bits are needed to address each single word in memory?
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.
How many words are there in an address space specified by 24 bits and the corresponding memory space by 16 bits?
address space=24bits => (2 Power 24)=16M words
