Digital Electronics - Number Systems and Codes - Discussion
Discussion Forum : Number Systems and Codes - General Questions (Q.No. 10)
10.
If a typical PC uses a 20-bit address code, how much memory can the CPU address?
Discussion:
13 comments Page 1 of 2.
VENKATESH said:
1 decade ago
2^20=1MB
Biplaw said:
1 decade ago
2^10=1 MB
Sachin said:
1 decade ago
Memory is byte addressable but request is come in bits
In this question CPU generate 20 bit address so main memory size is 2^20= MB.
In this question CPU generate 20 bit address so main memory size is 2^20= MB.
Sukesh said:
1 decade ago
Address bus is 20bits so 2^20=1Mb used in 8086 microprocessor.
Ashwani said:
1 decade ago
2^20 = 2^10*2^10.
As 2^10 = 1Kb,
Hence 2^20 = 1Kb*1Kb = 1Mb.
As 2^10 = 1Kb,
Hence 2^20 = 1Kb*1Kb = 1Mb.
Lalit said:
1 decade ago
The answer is very simple. We know 1024 bytes is 1 kb and 2^10 is 1024, hence 2^10 * 2^10 =2^20, that is 1024 *1024, 1kb*1024 is equal to 1 Mb.
Niki said:
1 decade ago
Why have we taken 2 as the base, can anyone explain?
Kiran said:
1 decade ago
Because computer works with only binary numbers {0, 1}. So that radix is 2.
That's why base is 2?
That's why base is 2?
Nitish kumar said:
10 years ago
2^20 memory locations can be addressed but it can not measured in mb. Size of memory will depend upon size of data bus.
PETAK said:
8 years ago
2 is used as the base because computer operates in binary, i.e. powers of 2.
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