= Resistance thermometer self-heating
{title2=$P=I^2R$}
Measuring <electric current> creates <Joule heating>, raising the sensing <temperature>. At fixed bath <temperature>, compare settled readings at $I$ and $\sqrt2I$ and extrapolate the approximately linear $I^2$ dependence toward zero current.<Guide on Secondary Thermometry: Industrial Platinum Resistance Thermometers by Jonathan Pearce and others>[§4.1, printed pp. 24–26, equation (4.1.1): self-heating, two-current extrapolation, and medium dependence]{ref}
Under the local model $R(I)=R_{\rm zero}+kI^2$,
$$
R_{\rm zero}=2R(I)-R(\sqrt2I).
$$
For synthetic readings $109.629\,\Omega$ at $1\,\mathrm{mA}$ and $109.633\,\Omega$ at $\sqrt2\,\mathrm{mA}$, the extrapolated value is $109.625\,\Omega$. The lower-current excess $0.004\,\Omega$ corresponds to $0.01039\ldots\,{}^\circ\mathrm C$ in the <Pt100 local linear model>. At $109.625\,\Omega$, $1\,\mathrm{mA}$ dissipates $109.625\,\mathrm{\mu W}$; power alone does not specify a <temperature> rise without a thermal coupling model. These stipulated values are not measurements.
Repeat the comparison after settling and test the $I^2$ approximation if necessary. Changing from liquid to air changes <heat transfer>; a self-heating correction need not transfer unchanged. The <saturation-pressure equilibrium measurement> provides an existing example of controlling this disturbance in resistance thermometry.
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