Digital Electronics - Memory and Storage - Discussion
Discussion Forum : Memory and Storage - General Questions (Q.No. 1)
1.
How many address bits are needed to select all memory locations in the 2118 16K × 1 RAM?
Discussion:
21 comments Page 2 of 3.
Abshir abdirizak said:
9 years ago
The answer is 10 + 4 = 14.
Dhanaji said:
9 years ago
Sir but how to use this 10+4.
Pankaj sethia said:
1 decade ago
2^14=(16*(1024))=16K
A Purohit said:
10 years ago
Here K means Kilobytes.
So, 1 K = 1024 bytes.
So, 1 K = 1024 bytes.
Ravi said:
1 decade ago
1024 bits is equal to the 1 k
Ravikiran said:
1 decade ago
Why did you use 1024 as 1 K?
Nandhini said:
1 decade ago
Why did you use 1024?
Jeffrey maghinay said:
1 decade ago
2^4= 2 x 2 x 2 x 2 = 16.
2^10= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1024.
2 to the power of 4 = 16.
2 to the power of 10 = 1024.
So: add the two power 4 + 10 = 14.
2^10= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1024.
2 to the power of 4 = 16.
2 to the power of 10 = 1024.
So: add the two power 4 + 10 = 14.
Sri said:
1 decade ago
How to factorising 16384 by 2?
Priyanka said:
1 decade ago
16k = 16*1024 = 2^4*2^10.
So total 14 address lines require.
So total 14 address lines require.
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