(B) is correct.

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Lakshman Patel RJIT
asked
in Digital Logic
Mar 31, 2020
recategorized
Aug 24, 2020
by Lakshman Patel RJIT

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2 votes

Range of $n-bit$ $2's$ complement numbers is : $-2^{n-1}$ to $2^{n-1}-1$ (Refer: Range of integers in 2's complement)

For $n=8$, the smallest number that can be represented is : $-2^{8-1}=-2^{7}=-128$

Option B is correct.