 Resistance thermometer self-heating

Measuring electric current creates Joule heating, raising the sensing temperature. At fixed bath temperature, compare settled readings at $I$ and $\sqrt{2}I$ and extrapolate the approximately linear $I^2$ dependence toward zero current.^[ref]

Under the local model $R(I)=R_{\mathrm{zero}}+k{I}^2$,For synthetic readings $109.629\Omega$ at $1\mathrm{mA}$ and $109.633\Omega$ at $\sqrt{2}\mathrm{mA}$, the extrapolated value is $109.625\Omega$. The lower-current excess $0.004\Omega$ corresponds to $0.01039\dots {}^{\circ }\mathrm{C}$ in the Pt100 local linear model. At $109.625\Omega$, $1\mathrm{mA}$ dissipates $109.625\mu \mathrm{W}$; 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.