How many address lines are needed for a 64Kb segment with each register having 512 b

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kumar_eee

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How many address lines are needed to address a 64Kb segment with each register storing upto 512 bytes
 

Re: Solution plz


64k=2^6*2^10=2^16
now you should devide it by 512 to get the no of addressed locations
2^16/2^9=2^7=128 addressed location of size 512
which will need 7 address lines
TeE ThE EdE
 

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