I get that the capacitive load is determined which is also indicated by this question: $$\frac{Power}{Voltage^2 \times frequency} = Capacitive Load$$

However, I am not sure how power (which is given in Watts) is determined when given both the dynamic and static power consumption and unfortunately my textbook doesn't really give any suggestions either. My guess is power would be equal to either dynamic - static or dynamic + static, but I don't which one would be correct, if either are correct, and more importantly why that statement is correct.

  • \$\begingroup\$ Just an fyi I am an intro Comp Sci student so I don't have background in any Engineering so if possible write my level. :P \$\endgroup\$
    – Asterisk
    Jul 4, 2015 at 19:47
  • \$\begingroup\$ In the title you ask one thing, while your question asks something different. Please fix that. \$\endgroup\$ Jul 4, 2015 at 20:11

1 Answer 1


Both dynamic power consumption and static power consumption are consumptions, so the total consumption is the sum of both. AFAIK for modern CPUs the dynamic power consumption dominates at practical frequencies.


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