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Processor A has a clock rate of 3.6 GHz and voltage 1.25V. Assume that, on average, it consumes 90W of dynamic power. Processor B has a clock rate of 3.4 GHz and voltage of 0.9V. Assume that, on average, it consumes 40W of dynamic power. For each processor find the average capacitive loads.

User JensOlsen
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1 Answer

1 vote

Answer:

Processor A :
C=3.2* 10^(-8)\ F

Processor B :
C=2.98* 10^(-8)\ F

Step-by-step explanation:

We know that


Dynamic\ power=(1)/(2)* C* V^2* f

C=Average capacitive loads.

Processor A :

f= 3.6 GHz

V= 1.25 V

P=90 W


Dynamic\ power=(1)/(2)* C* V^2* f


90=(1)/(2)* C* 1.25^2* 3.6* 10^9


C=(90* 2)/(1.25^2* 3.6* 10^9)\ F


C=3.2* 10^(-8)\ F

Processor B :

f= 3.4 GHz

V= 0.9 V

P=40 W


Dynamic\ power=(1)/(2)* C* V^2* f


40=(1)/(2)* C* 0.9^2* 3.4* 10^9


C=(40* 2)/(0.9^2* 3.4* 10^9)\ F


C=2.98* 10^(-8)\ F

User Josmarie
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