Discussion :: Digital Computer Electronics - Section 3 (Q.No.48)
|Anjali said: (Sep 20, 2016)|
|Please clarify me for this problem. As I have read a number of full adders required is = 2^n, where n = number of bit, then here 15 full adders. How? Please explain.|
|Sameer said: (Nov 6, 2016)|
|The one half-adder can add the least significant bit of the two numbers. Full adders are required to add the remaining 15 bits as they all involve adding carries.|
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