= Two-wire resistance measurement
A <two-wire resistance measurement> sends a known <electric current> through a specimen and measures the resulting <electric potential difference> using the same two connecting leads. The instrument estimates <electrical resistance> from their ratio. Because the leads carry the test current, their voltage drops enter the reading along with the specimen’s. Keithley’s primary handbook illustrates the two-wire circuit and identifies added lead resistance as its principal low-resistance limitation.<Low Level Measurements Handbook, Seventh Edition by Keithley Instruments>[§3.3.1, Lead Resistance and Four-Wire Method, Figure 3-16]{ref}
For an original linear series-circuit model, let $R_x$ be the specimen resistance, $r_1+r_2$ the total lead resistance, and $r_c$ an included contact resistance. With actual test current $I\ne0$ and additive offset potential difference $E$, the terminal reading is
$$
V_m=I(R_x+r_1+r_2+r_c)+E,
\qquad
\boxed{R_{\rm indicated}=V_m/I=R_x+r_1+r_2+r_c+E/I.}
$$
The division assumes the actual current is known and the specified resistive elements are sufficiently linear at that current. Non-ohmic contacts require a different response model.
For $R_x=0.100\,\Omega$, $r_1=r_2=0.150\,\Omega$, $r_c=0$ and $I=10\,\mathrm{mA}$, the true terminal potential difference is $4.00\,\mathrm{mV}$. The indicated resistance is $0.400\,\Omega$: a $300\%$ positive error relative to the specimen value. Adding $E=20\,\mathrm{\mu V}$ raises the reading to $0.402\,\Omega$. The same $0.300\,\Omega$ lead contribution is only $0.003\%$ of a $10\,\mathrm{k\Omega}$ specimen, illustrating why the acceptable arrangement depends on the measurement scale.
<Lead-resistance subtraction in two-wire measurements> estimates that series contribution using a shorted-lead reading. It can remove a stable offset resistance, but reconnecting clips, moving contacts or changing lead temperature can alter the quantity subtracted. If residual lead resistance changes by $0.020\,\Omega$ after zeroing, the error is already $20\%$ for the $0.100\,\Omega$ specimen above. Display resolution alone cannot establish that subtraction’s accuracy.
Reversing the test current cancels a different effect. If $E$ and the resistances remain unchanged between equal-magnitude positive and negative currents,
$$
V_+=IR_{\rm total}+E,\qquad V_-=-IR_{\rm total}+E,
\qquad \frac{V_+-V_-}{2I}=R_{\rm total}.
$$
The thermoelectric offset cancels, but the lead resistance remains. In the example, $V_+=4.02\,\mathrm{mV}$ and $V_-=-3.98\,\mathrm{mV}$ yield $0.400\,\Omega$, not $0.100\,\Omega$. The handbook treats current reversal separately from the four-wire connection.<Low Level Measurements Handbook, Seventh Edition by Keithley Instruments>[§3.3.2, Figure 3-18, current reversal and changing thermal offsets]{ref}
<Four-terminal sensing> instead separates the current-carrying connections from potential-sensing connections placed at the desired specimen boundaries. A high-input-impedance sense circuit draws sufficiently little current that its lead drops can be much smaller; the exact contact placement still defines what is measured. Raising the two-wire test current to suppress $E/I$ also raises specimen heating, $I^2R_x$. Increasing this example’s current from $10$ to $100\,\mathrm{mA}$ raises specimen dissipation from $10\,\mathrm{\mu W}$ to $1\,\mathrm{mW}$.
A proposed comparison measures the same stable reference specimen with short and long leads, repeats with current reversal, then uses a four-wire connection at fixed specimen terminals. Record current, temperature and connection geometry. Lead-length dependence and reversal dependence distinguish series resistance from additive thermal offsets, while repeating at lower current helps test whether self-heating changed the specimen during the measurement.
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