= Platinum resistance thermometer lead resistance error
{title2=$\Delta t=(r_1+r_2)/(R_0\alpha)$}
With two shared leads of <electrical resistances> $r_1,r_2$, <Ohm's law> gives
$$
V_{\rm two}=I[R(t)+r_1+r_2],\qquad
R_{\rm indicated}=R(t)+r_1+r_2.
$$
Using 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 $27.597\ldots\,{}^\circ\mathrm C$: 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.
\Image[/platinum-resistance-two-wire-four-wire-sensing.png]
{title=Two shared leads include cable voltage drops; four leads sense at the platinum element}
{id=platinum-resistance-two-wire-four-wire-sensing}
{description=Red paths carry <electric current>; blue paths connect an ideal high-input-<electrical resistance> <voltmeter>. In the two-wire circuit the sensing terminals span both leads and the element. In <four-terminal sensing> they span the element alone. The stated lead values and <temperature> bias are illustrative calculations.}
{source=platinum-resistance-two-wire-four-wire-sensing.py}
<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.<Guide on Secondary Thermometry: Industrial Platinum Resistance Thermometers by Jonathan Pearce and others>[§§4.4.1 and 4.4.3, printed pp. 30 and 32–33: two-wire inclusion of leads and four-wire exclusion]{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> yields
$$
R=\frac{V_+-V_-}{2I}.
$$
At stipulated $I=1\,\mathrm{mA}$, $R=109.625\,\Omega$ and $E=3\,\mathrm{\mu V}$, $V_+=109.628\,\mathrm{mV}$ 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 $I^2$.
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