Computer Science - Digital Computer Electronics - Discussion
Discussion Forum : Digital Computer Electronics - Section 2 (Q.No. 2)
2.
A microprocessor has memory locations from 0000 to 3FFF. Each memory location stores 1 byte. How bytes can the memory store? Express this in kilobytes?
Discussion:
8 comments Page 1 of 1.
Sachin meshram said:
1 decade ago
3FFF binary is 0011 1111 1111 1111.
(note:0011 1111 1111 =2^10=1KB (here 10 is last 10 binary bit)).
In 3FFF their is total 14 binary bit.
i.e. 2^4*2^10=2^4KB=16KB=16*1024=16384.
(note:0011 1111 1111 =2^10=1KB (here 10 is last 10 binary bit)).
In 3FFF their is total 14 binary bit.
i.e. 2^4*2^10=2^4KB=16KB=16*1024=16384.
(2)
Richa said:
1 decade ago
Convert into decimal form : 3FFF --- gives 16383. add 1 for location 0000. => 163384
(1)
Chandni said:
1 decade ago
Convert 3fff into binary and then into decimal. Which gives 16383 and add 1 for 0000 location. So 16384is correct answer.
(1)
Pandees said:
1 decade ago
Answer is:
Hex = 3fff.
Decimal = 3x163+15x162+15x161+15x160 = 16383.
Signed decimal = 16383.
Binary = 11111111111111.
Hex = 3fff.
Decimal = 3x163+15x162+15x161+15x160 = 16383.
Signed decimal = 16383.
Binary = 11111111111111.
(1)
Ajit said:
1 decade ago
Can anyone explain the method ?
Richa said:
1 decade ago
Please explain the method for the solution.
Anjali said:
9 years ago
Thank you all for explaining it.
Anubhav said:
4 years ago
Starting Address = 0 x 0000.
Last Address = 0 x 3FFF,
Address Space = 0 x 3FFF = 2^14 Addresses = 16KiB = 16384 Bytes.
Last Address = 0 x 3FFF,
Address Space = 0 x 3FFF = 2^14 Addresses = 16KiB = 16384 Bytes.
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