The power dissipated in a resistor is

PHYSICS · Multi-part · Fermi University Physics · Scientific Practice & Experimental Methods · Uncertainty in measurement · Uncertainty in measurement

Measurement and Method, Ch Propagating uncertainty into the quantity you actually want

Problem

The power dissipated in a resistor is P=I2RP = I^2R, with a measured current I=(0.500±0.010) AI = (0.500 \pm 0.010)\,\mathrm{A} and resistance R=(12.0±0.2) ΩR = (12.0 \pm 0.2)\,\Omega, each uncertainty a standard deviation and the two quantities measured independently. ((a)) Find the contribution of II to PP by moving II up to I+u(I)I+u(I) with RR fixed, and the contribution of RR by moving RR up to R+u(R)R+u(R) with II fixed. ((b)) Combine these to give u(P)u(P) and quote P=Pˉ±u(P)P = \bar P \pm u(P).

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