The Following file consists of 16 Presentations which covers almost all the portions and also included the solved 2 marks and 16 marks of the digital electronics....
Q-1 Convert the following decimal numbers to their binary equivalents:
(a) 37
(b) 14
(c) 167
(d) 72.45
(e) 0.4475
(f) 52
(g) 4097.188
(h) 2048.625
A-1 (a) 37
Q-1 A certain memory has capacity of 16K X 32. How many words does it store?
A-1 This memory consists of 16K words of 32 bits each. Therefore, the total number
of words contained in the memory is 16384 because 16 X 1024 = 16384. The bit
storage capacity of this memory is 16384 X 32 = 524288 bits.
Q-2 How many different addresses are required by the memory that contains 16K
words?
A-2 16384 addresses are required by the memory that contains 16K words.
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Q-1 Convert the following decimal numbers to their binary equivalents:
(a) 37
(b) 14
(c) 167
(d) 72.45
(e) 0.4475
(f) 52
(g) 4097.188
(h) 2048.625
A-1 (a) 37
Q-1 A certain memory has capacity of 16K X 32. How many words does it store?
A-1 This memory consists of 16K words of 32 bits each. Therefore, the total number
of words contained in the memory is 16384 because 16 X 1024 = 16384. The bit
storage capacity of this memory is 16384 X 32 = 524288 bits.
Q-2 How many different addresses are required by the memory that contains 16K
words?
A-2 16384 addresses are required by the memory that contains 16K words.
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