4. For a new version of processor, suppose the capacitive load remains, how much more energy will the processor consume if we increase voltage by 20% and increase clock rate by 20%?

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4. For a new version of processor, suppose the capacitive load remains, how much more energy will the processor consume if we increase voltage by 20% and increase clock rate by 20%?

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Answer 1

Answer #1

`Hey,

Note: Brother in case of any queries, just comment in box I would be very happy to assist all your queries

The power dissipation of a processor is given by the formula:

$$P = C V^2 f$$

Where : C = Capacitive load, V = Voltage, f = Clock rate.

So clearly when C is held constant:

P2/P1=(V2/V1)^2*f2/f1=(1.2)^2*(1.2)=(1.2)^3=1.7280

So, the processor will now consume approximately 1.7280 times (72.80% more) the energy as compared to the previous version.

Kindly revert for any queries

Thanks.

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Answer 2

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4. For a new version of processor, suppose the capacitive load remains, how much more energy will the processor consume if we increase voltage by 20% and increase clock rate by 20%?

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4. For a new version of processor, suppose the capacitive load remains, how much more energy will the processor consume if we increase voltage by 20% and increase clock rate by 20%?

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