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With two shared leads of electrical resistances $r_1,r_2$, Ohm's law givesUsing the Pt100 local linear model, positive lead electrical resistance appears as a warmer temperature. At true $25{}^{\circ }\mathrm{C}$, stipulated leads of $0.50\Omega$ each change $109.625\Omega$ into $110.625\Omega$, producing $27.597\dots {}^{\circ }\mathrm{C}$: a calculated bias of about $2.60{}^{\circ }\mathrm{C}$. Subtracting a measured $1.00\Omega$ loop correction works only while that loop electrical resistance remains appropriate; cable temperature changes can invalidate it.

Four-terminal sensing separates the electric current and electric potential difference leads. Negligible sensing electric current suppresses sensing-lead drops; force-lead drops lie outside the interval. Real voltmeter loading, connections, and leakage still need checking.^[ref]

For a stable offset $E$ from the thermoelectric effect, two signed electric potential differences obey $V_{+}=IR+E$ and $V_{-}=-IR+E$. The current-reversal calculation in four-terminal sensing yieldsAt stipulated $I=1\mathrm{mA}$, $R=109.625\Omega$ and $E=3\mu \mathrm{V}$, $V_{+}=109.628\mathrm{mV}$ and $V_{-}=-109.622\mathrm{mV}$ recover $R$ exactly in this model. If the offset or sample temperature changes between readings, cancellation is incomplete. Reversal also leaves Joule heating unchanged because it depends on $I^2$.

A Pt100 has nominal electrical resistance $100\Omega$ at $0{}^{\circ }\mathrm{C}$.^[ref] Here $0{}^{\circ }\mathrm{C}$ is the reference for the nominal element designation; it is distinct from the $0.01{}^{\circ }\mathrm{C}$ water triple point used in SPRT calibration. The following Pt100 local linear model deliberately stipulates one approximate response for calculations.

Choose the illustrative local modelHere $t$ is the Celsius temperature, measured relative to $0{}^{\circ }\mathrm{C}$, so $\alpha t$ is dimensionless. For this chosen affine function, the derivative and inverse relation areCalculated values are $R(0{}^{\circ }\mathrm{C})=100\Omega$, $R(20{}^{\circ }\mathrm{C})=107.70\Omega$, $R(25{}^{\circ }\mathrm{C})=109.625\Omega$, and $R(50{}^{\circ }\mathrm{C})=119.25\Omega$. These are model outputs, not measurements or an assertion that any real Pt100 follows this line exactly. The number $0.00385$ here is the chosen slope parameter; it is not a universal local derivative for all platinum resistance thermometers. Curvature, manufacturing differences, and the fitted calibration interval matter. A wider-range calibration uses a validated response curve rather than extrapolating this line indefinitely.

For a separate synthetic thermometer calibration, stipulate $R_a=100.20\Omega$ at $t_a=0{}^{\circ }\mathrm{C}$ and $R_b=119.60\Omega$ at $t_b=50{}^{\circ }\mathrm{C}$. A two-point affine function has slope $0.388\Omega /{}^{\circ }\mathrm{C}$ and givesThus $R=109.90\Omega$ would indicate $25{}^{\circ }\mathrm{C}$. If an additional synthetic check at reference $25{}^{\circ }\mathrm{C}$ instead supplied $110.10\Omega$, the line would indicate $25.515\dots {}^{\circ }\mathrm{C}$. The $0.20\Omega$ residual exposes disagreement that two fitted points cannot reveal. This separate example changes both offset and slope; it is not a replacement specification for the nominal Pt100 local linear model.

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.