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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.

  • Physics -

    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

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