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A microprocessor uses RAM chips of 1024 × 1 capacity.(i) How many chips will be required and how many address lines will be connected to provide capacity of 1024 bytes.(ii) How many chips will be required to obtain a memory of capacity of 16 K bytes.Ans. (i) As given available chips = 1024 x 1 capacity andRequired capacity = 1024 x 8 capacityNumber of Chips= (1024X8)/(1024X1) = 8Number of address lines are needed = 10 (that is 1024 = 210)Here the word capacity is similar ( 1024 ) so similar address lines will be connected to all chips. (ii) Number Of Chips Required = (16X1024X8)/(1024X1) = 128
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Q.Prove using Boolean Algebra 1. X (X+Y) = X 2. AB + AC + BC' = AC + BC' 3. (A+B+C) (A+B'+C') (A+B+C') (A+B'+C)=A 4. (A+B'+C) (AB+A'C) = (A+C) (A'+B) 5. XY + XZ + YZ
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