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An 8085 microprocessor based system uses a 4 k x 8 bit RAM whose address in AAOOH. The address of the last byte in this RAM is
- a)OFFFH
- b)1000H
- c)B9FFH
- d)BA00H
Correct answer is option 'D'. Can you explain this answer?
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Explanation:
The given RAM is a 4 k x 8 bit RAM which means it has a capacity of storing 4 kilobytes of data, and each location can store 8 bits of data. The starting address of this RAM is AAOOH.
To find the address of the last byte in this RAM, we need to calculate the ending address of this RAM. The ending address can be calculated as follows:
Ending address = Starting address + Size of RAM - 1
Here, the size of the RAM is 4 kilobytes, which is equal to 4096 bytes. Hence, the ending address can be calculated as:
Ending address = AAOOH + 4096 - 1
Ending address = BA00H
Therefore, the address of the last byte in this RAM is BA00H. Hence, the correct answer is option D.
The given RAM is a 4 k x 8 bit RAM which means it has a capacity of storing 4 kilobytes of data, and each location can store 8 bits of data. The starting address of this RAM is AAOOH.
To find the address of the last byte in this RAM, we need to calculate the ending address of this RAM. The ending address can be calculated as follows:
Ending address = Starting address + Size of RAM - 1
Here, the size of the RAM is 4 kilobytes, which is equal to 4096 bytes. Hence, the ending address can be calculated as:
Ending address = AAOOH + 4096 - 1
Ending address = BA00H
Therefore, the address of the last byte in this RAM is BA00H. Hence, the correct answer is option D.
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An 8085 microprocessor based system uses a 4 k x 8 bit RAM whose address in AAOOH. The address of the last byte in this RAM isa)OFFFHb)1000Hc)B9FFHd)BA00HCorrect answer is option 'D'. Can you explain this answer?
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