Answer:
Processor A :[tex]C=3.2\times 10^{-8}\ F[/tex]
Processor B :[tex]C=2.98\times 10^{-8}\ F[/tex]
Explanation:
We know that
[tex]Dynamic\ power=\dfrac{1}{2}\times C\times V^2\times f[/tex]
C=Average capacitive loads.
Processor A :
f= 3.6 GHz
V= 1.25 V
P=90 W
[tex]Dynamic\ power=\dfrac{1}{2}\times C\times V^2\times f[/tex]
[tex]90=\dfrac{1}{2}\times C\times 1.25^2\times 3.6\times 10^9[/tex]
[tex]C=\dfrac{90\times 2}{1.25^2\times 3.6\times 10^9}\ F[/tex]
[tex]C=3.2\times 10^{-8}\ F[/tex]
Processor B :
f= 3.4 GHz
V= 0.9 V
P=40 W
[tex]Dynamic\ power=\dfrac{1}{2}\times C\times V^2\times f[/tex]
[tex]40=\dfrac{1}{2}\times C\times 0.9^2\times 3.4\times 10^9[/tex]
[tex]C=\dfrac{40\times 2}{0.9^2\times 3.4\times 10^9}\ F[/tex]
[tex]C=2.98\times 10^{-8}\ F[/tex]
