Q: P=I^2R

If the percentage error in I is 0.75% and R is 0.25%

What is the maximum Percentage change in P?

Could anyone please help me out in this one?

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- November 8th 2007, 10:38 AMstuwyPercentage Change
Q: P=I^2R

If the percentage error in I is 0.75% and R is 0.25%

What is the maximum Percentage change in P?

Could anyone please help me out in this one? - November 8th 2007, 01:21 PMTKHunny
P = I^2 R

Are we to assume that P, I, and R are all functions of a common variable, maybe 't'?

If so, we have by the Product Rule dP = I^2 dR + 2*I*dI*R

Relative error:

I has 0.75%, or dI/I = 0.0075 ==> I = dI/*0.0075

R has 0.25%, or dR/R = 0.0025 ==> R = dR/0.0025

Putting it all together

dP = I^2 dR + 2*I*dI*R = (dI/*0.0075)^2 dR + 2*(dI/*0.0075)*dI*dR/0.0025 = [dR * dI^2 / 0.0075]*[1/0.0075 + 2/0.0025]

For Relative Error, calculate dP/P and see all the stuff that cancels out. - November 19th 2007, 06:02 AMstuwy