a computer memory is composed of 16k words of 8 bits

• 1 byte = 8 bits
  1 nybble = 4bits
  1 word = 32bits
  1 halfword =16bits
  1 doubleword = 64bits
  16k words = 512 bits(ie 512/8 = 64bits)
  
• 9 years agoHelpfull: Yes(0) No(0)
• Bits can represent only two number 0 & 1. so,
  
  A 1-bit address can address two words (0, 1). [ 2^1 = 2 ]
  
  A 2-bit address can address four words (00, 01, 10, 11). [ 2^2 = 4 ]
  
  A 3-bit address can address eight words (000, 001, 010, 011, 100, 101, 110, 111). [ 2^3 = 8 ]
  
  now 16K words in total = 16000 words => each single word can thus be addressed in 14 bits [ AS 2^13 = 8192 & 2 ^ 14 = 16384 ] => so 14 bits are needed to address each word in memory.
• 8 years agoHelpfull: Yes(0) No(0)
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• 5 years agoHelpfull: Yes(0) No(0)