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A computer with 32-bit wide data bus uses 4 K x 8 static RAM memory chips. The smallest memory this computer can have is:

(a) 32 kb (b) 16 kb (c) 8 kb (d) 24 kb
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is the ans 16KB ?
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yes, but how, plz explain?
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Its a 32bit data bus so you need minimum 4 chips because each chip has 8 bits of data Input/Output (8*4) = 32 bits .

 

Since with 4 chips the smallest memory is  :  4 * ( 4kB * 8 ) / 8 =  2 ^14 = 16 KB
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From my question, where did you get the information: "each chip has 8 bits of Data I/O" ?

What does "4K*8" actually mean?

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See they said that 4K*8 so the 8 here is no of data lines coming out and 4k is the addressable lines. Since you have 32bit bus you need to have 4 chips minimum.
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32 bit data bus means 4B data is transferred per sec (means memory word size)

Given ram - 4K* 8 = 4K*2*4= 8K*4

So memory size = 8Kb

Address bus bits= 13

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Hi, please explain a bit,

1] Why did you break 4K*8 into 4K*2*4 ?

2] How did you get 13 bit address bus? Where is 2^13 coming from?
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