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A resistance thermometer using platinum as its sensing material. It provides the temperature readout in the saturation-pressure equilibrium measurement.

Transduction turns electrical resistance into a temperature indication. A uniform wire has $R={\rho }_{\mathrm{el}}\ell /A$, with electrical resistivity ${\rho }_{\mathrm{el}}$, length $\ell$, and cross-sectional area $A$. Stronger atomic vibrations at higher temperature scatter Electrons more, increasing platinum's electrical resistivity; impurities and crystallographic defects also contribute.^[ref]

A small known electric current $I$ and measured electric potential difference $V$ give $R=V/I$ through Ohm's law; a resistance bridge can instead compare the element with a reference resistor. Inverting the calibrated response then gives the sensing element's temperature. This makes electrical readout convenient, but the platinum resistance thermometer lead resistance error, resistance thermometer self-heating, and resistance thermometer immersion test explain why a stable display alone is insufficient. The gas thermometer supplies a contrasting route through an equation of state; the saturation-pressure equilibrium measurement shows why a trustworthy independent temperature readout matters.

A thermometer that infers temperature from a calibrated electrical resistance. Measuring current can heat its sensing element; extrapolation toward zero current controls that bias in the saturation-pressure equilibrium measurement.

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.