RAM chips arranged in 4X6 array and of 8kX4bit capacity
each. How many address lines reqd. to access each byte
a. 12 b. 16 c.15 d. 17
Answer Posted / subhash
d. 17
8k*4 8k*4 8k*4 8k*4 8k*4 8k*4 - 24KB
8k*4 8k*4 8k*4 8k*4 8k*4 8k*4 - 24KB
8k*4 8k*4 8k*4 8k*4 8k*4 8k*4 - 24KB
8k*4 8k*4 8k*4 8k*4 8k*4 8k*4 - 24KB
------
96kB
To address it byte by byte you need 17 address lines
16 lines - 2^16 = 64kB
17 lines - 2^17 = 128kB
| Is This Answer Correct ? | 20 Yes | 0 No |
Post New Answer View All Answers
When does stack (process stack) overflow? What are the remedies?
What is the most used operating system in the world?
Can you install a 32bit program on a 64 bit computer?
Which program files folder is 64 bit?
Define page fault interrupt?
When is a system in a safe state?
What is the programs x86 folder?
What is external fragmentation?
How long does it take to run chkdsk?
When does page fault error occur?
When would you choose bottom-up methodology?
What are the ways in which fault tolerance can be ensured in systems? (Different redundancy techniques)?
What do you understand by the term ‘transistors’?
How do I know chkdsk is finished?
State the advantages of segmented paging over pure segmentation?