posted by Robert .
An electric current of 50 +/- 1 mA flows through an electric bulb filament that has a resistance of 10 +/- 2 ohm. Estimate the power dissipated by the filament and the associated error in the power.
The power dissipated is
I^2*R = (.05^2)(10) = 0.025 W
That is pretty feeble for a light bulb.
The relative current error is negligible compared to the relative resistance error of 20%.
The power measurement uncertainty is therefore 20% or +/- 0.005 W