Why Lead Resistance Distorts Small Resistance Measurements

A short metal link has a resistance of \(0.12\ \Omega\), but a two-wire meter reports \(0.52\ \Omega\). Has the link changed, or is the meter measuring something else?

When you’re measuring a small resistance, the test leads and their electrical contacts can contribute as much resistance as the component itself. A two-wire measurement includes those unwanted contributions, so the reported resistance can be much higher than the resistance you’re trying to find.

Here’s a prediction worth making: if the two leads are identical, could their voltage drops cancel because current travels out through one and returns through the other?

No. Current passes through both leads in series. Both produce voltage drops, and those drops add.

01The first experiment: measuring through two leads

Imagine testing a short piece of resistance wire using a DC resistance meter.

A typical two-wire ohmmeter sends a known test current through its probes, measures the potential difference between its own terminals, and calculates resistance from those quantities.

Follow the circuit from the meter:

1. Current leaves the positive meter terminal and travels along the outgoing test lead.
2. It crosses the first electrical contact into the test specimen.
3. It travels through the specimen, then crosses the second contact.
4. It returns to the meter along the other test lead.

The meter measures the voltage associated with that whole path, not just the specimen.

Why the unwanted resistances add

Assume the specimen, leads, and contacts behave as ohmic resistors at constant temperature.

Let:

• \(R_x\) be the specimen resistance, in ohms (\(\Omega\)).
• \(R_a\) and \(R_b\) be the outgoing and return lead resistances, in ohms.
• \(R_c\) be the combined resistance of the two electrical contacts, in ohms.
• \(I\) be the test current, in amperes (A).
• \(V_t\) be the potential difference measured at the meter terminals, in volts (V).
• \(V_s\) be the potential difference directly across the specimen, in volts.

Take conventional current as positive from left to right through the specimen. Voltage drops are positive in that direction, so the left end of the specimen is at a higher potential than the right end.

For the complete series path,

\[ V_t=I(R_x+R_a+R_b+R_c) \]

The resistance reported by the two-wire meter is therefore

\[ \begin{aligned} R_2&=\frac{V_t}{I}\\ &=R_x+R_a+R_b+R_c \end{aligned} \]

Here, \(R_2\) means the apparent resistance obtained using the two-wire method.

The equation explains the problem: the meter cannot distinguish the specimen’s resistance from the resistance of the wires and contacts that connect it.

What should the first experiment show?

Consider this illustrative setup. These values are constructed to explain the measurement method; they are not recorded experimental results.

A metal link has a resistance of \(0.12\ \Omega\). Each test lead has a resistance of \(0.18\ \Omega\), and the two contacts together contribute \(0.04\ \Omega\). The meter supplies \(0.25\ \mathrm{A}\).

The meter would calculate

\[ R_2=\frac{0.130}{0.25}=0.52\ \Omega \]

Yet the voltage across the actual metal link is only \(0.030\ \mathrm{V}\):

\[ R_x=\frac{0.030}{0.25}=0.12\ \Omega \]

The meter is reporting more than four times the specimen’s resistance.

The percentage overestimate, measured relative to the true resistance, is

\[ \begin{aligned} \%\text{ error} &=\frac{0.52-0.12}{0.12}\times100\\ &\approx333\% \end{aligned} \]

That is a large systematic error. Repeating the reading with the same connections won’t remove it.

02Why small resistances are especially vulnerable

The leads don’t somehow become more resistive when you measure a smaller component. The problem is their size relative to the component.

For the simple series model,

\[ \%\text{ error} =\frac{R_{\mathrm{extra}}}{R_x}\times100 \]

Suppose the combined lead and contact resistance remains \(0.40\ \Omega\).

The same unwanted \(0.40\ \Omega\) produces very different relative errors.

This gives a useful decision rule.

If you want the systematic error from lead and contact resistance to be less than \(1\%\), then under this model you need

\[ R_{\mathrm{extra}}\lt 0.01R_x \]

For a \(0.20\ \Omega\) specimen, that requires the total unwanted resistance to be below \(0.002\ \Omega\).

That’s just two milliohms. Ordinary test connections may not meet that requirement reliably.

Would increasing the test current solve the problem?

You might expect a larger current to produce a clearer voltage reading and improve accuracy.

A larger current does increase the voltage across the specimen. Unfortunately, it also increases the voltage drop across the leads.

If all resistances stay constant, doubling the current doubles every series voltage drop. The ratio \(V_t/I\) remains unchanged, including the unwanted lead resistance.

Higher current can also heat the specimen, making its resistance change. So increasing current is not a reliable cure.

03The revised experiment: four-wire Kelvin measurement

The first design has a fundamental weakness: the same two wires are used to deliver current and define where voltage is measured.

A better design separates those jobs.

A four-wire resistance measurement, also called a Kelvin measurement, uses two force leads to supply current and two sense leads to measure voltage.

Here’s how to arrange it.

Starting at the current source, connect the positive force lead to one end of the specimen. Connect the negative force lead to the other end and back to the source.

Then attach two separate voltage-sense probes directly to the specimen, as close as practical to the ends of the section whose resistance you want. Place those sense contacts inside the main current-carrying contacts when the goal is to exclude their voltage drops.

Connect the sense probes to a high-input-resistance voltmeter.

Why four wires remove the lead-voltage error

The current still passes through the force leads, so those leads still produce voltage drops.

But the voltmeter now measures only the potential difference between its two sense contacts.

An ideal voltmeter draws no current. A real high-input-resistance voltmeter draws very little.

Consequently, negligible current passes through the sense leads. Using \(V=IR\), their voltage drops are also negligible, even though the sense leads themselves have resistance.

That gives

\[ R_4\approx\frac{V_s}{I} \]

where \(R_4\) is the four-wire resistance estimate.

If the sense contacts span only the intended specimen, the voltmeter draws negligible current, and all the supplied current passes through that specimen, then

\[ R_4\approx R_x \]

Notice what actually improved: the experiment measures voltage at the correct locations.

The force leads haven’t magically lost their resistance. Their voltage drops have simply been excluded from the measured potential difference.

Comparing the two methods directly

Return to our original \(0.12\ \Omega\) specimen, with \(0.40\ \Omega\) of total lead and contact resistance and a current of \(0.25\ \mathrm{A}\).

The four-wire result uses the specimen’s voltage drop rather than the whole circuit’s voltage drop.

This is useful for measuring low-resistance connections, short conductors, and metal current-sense resistors. Those components are intended to have very small resistance, so the voltage across them can be comparable to the voltage lost in ordinary test leads.

04Can you just short the probes and subtract the reading?

Sometimes. Many meters offer a relative or zero function that lets you short the probe tips, measure the unwanted resistance, and subtract it from subsequent readings.

This is called lead compensation or zero compensation.

The idea is reasonable: if the leads contribute a constant \(0.15\ \Omega\), subtracting \(0.15\ \Omega\) should remove their effect.

But a shorted-probe test doesn’t necessarily reproduce the contact resistance when the probes touch the actual specimen.

Suppose a shorted-probe reading is \(0.18\ \Omega\), while the leads and probe-to-probe contact together really do contribute that amount.

You zero the meter, then move the probes onto a specimen. If the new specimen contacts introduce an additional \(0.05\ \Omega\) that wasn’t present during zeroing, the corrected result still overestimates the specimen’s resistance by \(0.05\ \Omega\).

For a \(20\ \Omega\) resistor, that may be unimportant. For a \(0.10\ \Omega\) specimen, it’s a 50% error.

Zero compensation is useful when the connections are stable and repeatable. Four-wire sensing is more robust because it excludes voltage drops outside the sense contacts rather than trying to estimate and subtract them.

05Practical extension: four-wire measurements aren’t automatically perfect

The simple Kelvin model assumes an ideal voltmeter and a single current path through the specimen.

Real instruments and complicated specimens can break those assumptions.

These effects are practical extensions to the standard ideal-meter model, rather than requirements you should automatically assume in every HSC circuit problem.

A real voltmeter can draw current

A real voltmeter has a finite input resistance, \(R_V\), measured in ohms.

When connected across a specimen, it forms a parallel path. If the current meter is positioned before that branch, its reading includes both the specimen current and the voltmeter current.

Writing \(I_x\) for current through the specimen and \(I_V\) for current through the voltmeter,

\[ I=I_x+I_V \]

If the voltage across both branches is \(V_s\),

\[ I_V=\frac{V_s}{R_V} \]

The specimen’s actual resistance is therefore

\[ R_x=\frac{V_s}{I-I_V} \]

For a very high-input-resistance voltmeter, \(I_V\) is usually negligible relative to \(I_x\). That restores the ordinary four-wire approximation.

A real ammeter also has some internal resistance. This can change the current delivered by a fixed-voltage source. However, its resistance does not automatically become part of the specimen’s measured resistance when the voltage is correctly sensed across the specimen.

Four-wire sensing cannot fix a wrong current path

Imagine a second conductor accidentally connected in parallel with your specimen.

Your voltage-sense probes might be perfectly positioned, but the current supplied by the source now splits between two paths.

If you divide the sensed voltage by the total current, the result describes the combined parallel network, not the resistance of the specimen alone.

Similarly, if the sense probes are attached partway along a long specimen, the measured voltage belongs to the section between those probes, not automatically to the entire specimen.

Four-wire sensing solves a particular problem: unwanted series voltage drops outside the sensed region. It does not remove every possible experimental error.

A diagnostic guide

Before deciding which correction to use, mark two things on the circuit: the path taken by the measured current and the exact points between which voltage is measured.

For a refresher on how those measurements are connected in a basic circuit, see How to Measure Resistance with a Voltmeter and Ammeter.

06Questions and solutions

The questions become progressively more demanding. All numerical data are illustrative, and any assumptions needed for a unique result are stated in the question.

Question 1

A metal test link has a resistance of \(0.080\ \Omega\). It is measured with a two-wire ohmmeter whose outgoing and return leads each have a resistance of \(0.060\ \Omega\). Contact resistance is negligible.

The test current is \(0.50\ \mathrm{A}\).

Calculate:

a) The resistance displayed by the ohmmeter.

b) The percentage overestimate.

c) The voltage across the link and the voltage measured at the meter terminals.

Assume constant resistance and a single series current path.

Solution 1

The ohmmeter displays \(0.20\ \Omega\), a 150% overestimate. The link has \(0.040\ \mathrm{V}\) across it, but the meter measures \(0.100\ \mathrm{V}\) across the complete path.

a) Displayed resistance

All three resistances are in series:

\[ \begin{aligned} R_2&=0.080+0.060+0.060\\ &=0.200\ \Omega \end{aligned} \]

b) Percentage overestimate

\[ \begin{aligned} \%\text{ error} &=\frac{0.200-0.080}{0.080}\times100\\ &=150\% \end{aligned} \]

c) Voltage measurements

Using \(V=IR\), the voltage across the specimen is

\[ \begin{aligned} V_s&=(0.50)(0.080)\\ &=0.040\ \mathrm{V} \end{aligned} \]

The meter-terminal voltage is

\[ \begin{aligned} V_t&=(0.50)(0.200)\\ &=0.100\ \mathrm{V} \end{aligned} \]

The extra \(0.060\ \mathrm{V}\) is dropped across the two leads.

The tempting mistake is to treat the complete terminal voltage as though it belongs to the small link. Only \(0.040\ \mathrm{V}\) is across the link itself.

Question 2

A laboratory measures the resistance of a small specimen using two identical test leads, each of length \(L\).

The same experiment is repeated after replacing both leads with wires of length \(2L\), made from the same material and with the same cross-sectional area.

A regulated source maintains a current of \(0.40\ \mathrm{A}\).

The specimen and its electrical contacts remain unchanged between measurements. Resistance is proportional to lead length, all resistances remain constant, and the four-wire sense probes measure the specimen alone.

The following readings are obtained.

Determine:

a) The combined lead resistance in the original arrangement.

b) The actual specimen resistance.

c) The combined contact resistance.

d) Why doubling both lead lengths does not double the terminal voltage.

Solution 2

The original lead pair has a combined resistance of \(0.16\ \Omega\), the specimen is \(0.20\ \Omega\), and the contacts contribute \(0.020\ \Omega\). Doubling the lead lengths increases only the lead resistance, not the complete circuit resistance.

a) Original lead resistance

First calculate the two terminal-resistance readings:

\[ \begin{aligned} R_{2,L}&=\frac{0.152}{0.40}\\ &=0.38\ \Omega \end{aligned} \]

\[ \begin{aligned} R_{2,2L}&=\frac{0.216}{0.40}\\ &=0.54\ \Omega \end{aligned} \]

The difference is

\[ \Delta R=0.54-0.38 =0.16\ \Omega \]

Because both leads doubled in length, the added resistance equals the resistance of the original pair.

Thus,

\[ R_{\mathrm{leads},L}=0.16\ \Omega \]

The doubled pair has resistance \(0.32\ \Omega\).

b) Specimen resistance

The sense probes measure the specimen’s voltage directly:

\[ R_x=\frac{0.080}{0.40} =0.20\ \Omega \]

c) Contact resistance

Subtract the specimen and original leads from the first two-wire result:

\[ \begin{aligned} R_c&=0.38-0.20-0.16\\ &=0.020\ \Omega \end{aligned} \]

We can verify this using the longer leads:

\[ 0.20+0.32+0.020=0.54\ \Omega \]

At \(0.40\ \mathrm{A}\), that gives

\[ V_t=(0.40)(0.54) =0.216\ \mathrm{V} \]

This agrees with the supplied reading.

d) Why the terminal voltage doesn’t double

Only the lead resistance doubled. The specimen and contact resistances remained unchanged.

The original circuit resistance was \(0.38\ \Omega\), while the new resistance is \(0.54\ \Omega\), not \(0.76\ \Omega\).

The important experimental limitation is that the two terminal readings alone would have revealed the original combined lead resistance, but not separated the specimen resistance from its contacts.

The extra four-wire measurement supplied the information needed to distinguish those contributions.

Question 3

A uniform resistance wire is stretched horizontally between two current-carrying clips located at positions \(x=0\ \mathrm{cm}\) and \(x=60\ \mathrm{cm}\).

Two high-resistance voltage-sense probes touch the wire at \(x=10\ \mathrm{cm}\) and \(x=50\ \mathrm{cm}\).

A current of \(0.80\ \mathrm{A}\) flows from left to right. The sense voltmeter reads \(0.096\ \mathrm{V}\), while the complete terminal voltage is \(0.200\ \mathrm{V}\).

Assume the wire is uniform, ohmic, and at constant temperature, with no parallel current paths. The sense probes draw negligible current.

a) Find the resistance between the sense probes.

b) Calculate the resistance of the full 60 cm wire.

c) Determine the combined resistance of the force leads and their contacts outside the wire.

d) Predict the four-wire voltage if the sense probes are moved to the extreme ends of the 60 cm wire, without moving the current clips.

e) Explain which conclusions would no longer be justified if the wire were non-uniform.

Solution 3

The sense probes measure \(0.12\ \Omega\) across the middle 40 cm, but the full wire has resistance \(0.18\ \Omega\). The force leads and contacts contribute another \(0.070\ \Omega\), and sensing the full wire would produce \(0.144\ \mathrm{V}\).

a) Resistance between the sense probes

\[ \begin{aligned} R_{\mathrm{middle}} &=\frac{0.096}{0.80}\\ &=0.12\ \Omega \end{aligned} \]

This is the resistance of the 40 cm section, not the whole specimen.

b) Full wire resistance

The wire is uniform, so resistance is proportional to length.

\[ \begin{aligned} R_{\mathrm{wire}} &=0.12\times\frac{60}{40}\\ &=0.18\ \Omega \end{aligned} \]

c) External lead and contact resistance

The complete measured resistance is

\[ R_2=\frac{0.200}{0.80} =0.25\ \Omega \]

Therefore,

\[ \begin{aligned} R_{\mathrm{extra}} &=0.25-0.18\\ &=0.070\ \Omega \end{aligned} \]

d) New sense voltage

With the sense contacts at the extreme ends of the wire, the sensed region now includes all 60 cm.

\[ \begin{aligned} V_s&=(0.80)(0.18)\\ &=0.144\ \mathrm{V} \end{aligned} \]

The specimen hasn’t changed. Only the endpoints used for voltage sensing have changed.

e) If the wire is non-uniform

The measured middle-section resistance remains \(0.12\ \Omega\), but scaling it by \(60/40\) would no longer be justified.

The outer 20 cm could have a different resistance per unit length.

Without knowing those outer sections, we also cannot determine the force-lead and contact resistance separately.

What we can establish is that the resistance outside the sensed middle section totals

\[ \begin{aligned} R_{\mathrm{outside}} &=0.25-0.12\\ &=0.13\ \Omega \end{aligned} \]

That value includes the outer sections of the wire, plus the force leads and contacts.

The misconception is that four-wire sensing always measures an entire component. It measures the region between its sense contacts. Converting that result into the resistance of a larger region requires additional information about the specimen.

Question 4

Practical extension: finite voltmeter resistance

A regulated current source supplies \(1.0\ \mathrm{A}\) through two force leads, each with resistance \(0.20\ \Omega\), to an unknown ohmic resistor.

A voltmeter is connected directly across the unknown resistor using sense leads of negligible resistance.

Unlike an ideal voltmeter, this instrument has an input resistance of \(9.0\ \Omega\). It reads \(0.90\ \mathrm{V}\).

The source ammeter is placed before the current splits between the specimen and the voltmeter, so it measures their total current.

Assume constant resistance, no additional contact resistance, and the stated current direction through the specimen.

a) Calculate the current through the voltmeter and the current through the specimen.

b) Determine the specimen’s actual resistance and the naive four-wire resistance estimate.

c) Calculate the voltage required across the regulated current source’s output terminals.

d) The voltmeter is replaced by an ideal voltmeter while the current source continues supplying \(1.0\ \mathrm{A}\). Predict the new specimen voltage and source-terminal voltage. Explain why both change.

Solution 4

The specimen has a true resistance of \(1.0\ \Omega\), even though the naive four-wire calculation gives \(0.90\ \Omega\). The difference comes from current entering the finite-resistance voltmeter, not from the force leads.

a) Branch currents

The voltmeter is in parallel with the specimen, so each has \(0.90\ \mathrm{V}\) across it.

Using Ohm’s law for the voltmeter,

\[ \begin{aligned} I_V&=\frac{0.90}{9.0}\\ &=0.10\ \mathrm{A} \end{aligned} \]

The remaining current flows through the specimen:

\[ \begin{aligned} I_x&=1.0-0.10\\ &=0.90\ \mathrm{A} \end{aligned} \]

b) Actual resistance versus apparent resistance

The true specimen resistance is

\[ \begin{aligned} R_x&=\frac{0.90}{0.90}\\ &=1.0\ \Omega \end{aligned} \]

If we incorrectly divide the voltage by the total source current,

\[ \begin{aligned} R_4&=\frac{0.90}{1.0}\\ &=0.90\ \Omega \end{aligned} \]

The naive reading is 10% too low.

c) Current-source terminal voltage

The force leads have a combined resistance of

\[ R_{\mathrm{leads}}=0.20+0.20 =0.40\ \Omega \]

They carry the full \(1.0\ \mathrm{A}\), so their combined voltage drop is

\[ V_{\mathrm{leads}}=(1.0)(0.40) =0.40\ \mathrm{V} \]

The source must therefore provide

\[ \begin{aligned} V_t&=0.90+0.40\\ &=1.30\ \mathrm{V} \end{aligned} \]

d) Replacing the voltmeter with an ideal meter

An ideal voltmeter draws no current. The entire \(1.0\ \mathrm{A}\) now flows through the specimen.

Its voltage becomes

\[ \begin{aligned} V_s&=(1.0)(1.0)\\ &=1.0\ \mathrm{V} \end{aligned} \]

The force-lead voltage drop remains \(0.40\ \mathrm{V}\), so

\[ \begin{aligned} V_t&=1.0+0.40\\ &=1.40\ \mathrm{V} \end{aligned} \]

Improving the voltmeter actually increases the required source voltage because the parallel loading path has been removed while total current remains fixed.

The trap is assuming that a four-wire connection automatically guarantees that the measured current all passes through the specimen.

Ordinary high-input-resistance voltmeters make this effect much smaller, but the governing principle is unchanged.

Question 5

A low-resistance metal strip consists of two ohmic sections, \(AB\) and \(BC\), connected end to end.

A removable metal brace is connected directly between points \(B\) and \(C\), providing a second conducting path parallel to section \(BC\).

A regulated current source drives \(0.50\ \mathrm{A}\) from \(A\) to \(C\) through external force leads.

Three high-input-resistance sense probes touch the strip directly at \(A\), \(B\), and \(C\). They draw negligible current, and their voltage measurements exclude all external force-lead and contact resistance.

A separate terminal-voltage measurement includes the force leads and their contacts.

The experiment gives the following readings.

The current source maintains \(0.50\ \mathrm{A}\) throughout.

Assume the sections, brace, leads, and contacts all remain at constant temperature. Removing the brace does not disturb the contacts, and its removal leaves no other conducting path between \(B\) and \(C\).

a) Determine the resistance of each section of the strip and the resistance of the removable brace.

b) Find the total resistance of the strip when the brace is removed.

c) With the brace fitted, calculate the four-wire resistance estimate across \(A\) and \(C\), the two-wire resistance estimate at the instrument terminals, and the combined external lead and contact resistance.

d) A student notices that the two-wire result with the brace fitted equals the actual resistance of the strip with the brace removed. They conclude that the two-wire connection was accurate and that the four-wire measurement was faulty. Explain precisely why this conclusion is wrong.

e) Predict both the four-wire and two-wire results after the brace is removed.

Solution 5

The isolated strip has a total resistance of \(0.48\ \Omega\), the removable brace is \(0.20\ \Omega\), and the external leads and contacts contribute \(0.18\ \Omega\). With the brace connected, the four-wire estimate is \(0.30\ \Omega\), while the two-wire estimate is \(0.48\ \Omega\).

The apparent agreement between the two-wire reading and the isolated strip’s resistance is accidental. Two different measurement effects cancel numerically.

a) Resistances of the two sections and the brace

Section \(AB\) carries the full \(0.50\ \mathrm{A}\), because the current has not yet reached the branching point \(B\).

Therefore,

\[ \begin{aligned} R_{AB}&=\frac{0.090}{0.50}\\ &=0.18\ \Omega \end{aligned} \]

When the brace is removed, all the source current must pass through section \(BC\).

\[ \begin{aligned} R_{BC}&=\frac{0.150}{0.50}\\ &=0.30\ \Omega \end{aligned} \]

Now consider the brace-fitted condition.

The voltage across section \(BC\) is \(0.060\ \mathrm{V}\). The brace connects the same two points, so it must have exactly the same potential difference.

Current through the strip section is

\[ \begin{aligned} I_{BC}&=\frac{0.060}{0.30}\\ &=0.20\ \mathrm{A} \end{aligned} \]

Only \(0.20\ \mathrm{A}\) of the supplied \(0.50\ \mathrm{A}\) travels through section \(BC\).

The rest flows through the brace:

\[ \begin{aligned} I_{\mathrm{brace}} &=0.50-0.20\\ &=0.30\ \mathrm{A} \end{aligned} \]

The brace resistance is

\[ \begin{aligned} R_{\mathrm{brace}} &=\frac{0.060}{0.30}\\ &=0.20\ \Omega \end{aligned} \]

The key is that the current splits at \(B\). You cannot use the full source current to calculate the resistance of section \(BC\) while the brace is connected.

b) Total resistance of the isolated strip

With the brace removed, sections \(AB\) and \(BC\) are in series.

\[ \begin{aligned} R_{\mathrm{strip}} &=0.18+0.30\\ &=0.48\ \Omega \end{aligned} \]

This is the resistance of the actual strip between \(A\) and \(C\), with no bypass path.

c) Four-wire and two-wire readings with the brace fitted

The sensed voltage from \(A\) to \(C\) is the sum of the two section voltages:

\[ \begin{aligned} V_{AC}&=0.090+0.060\\ &=0.150\ \mathrm{V} \end{aligned} \]

The four-wire estimate is therefore

\[ \begin{aligned} R_4&=\frac{0.150}{0.50}\\ &=0.30\ \Omega \end{aligned} \]

That is the equivalent resistance of section \(AB\) in series with the parallel combination of section \(BC\) and the brace.

The terminal measurement instead gives

\[ \begin{aligned} R_2&=\frac{0.240}{0.50}\\ &=0.48\ \Omega \end{aligned} \]

The voltage across the external force leads and contacts is

\[ \begin{aligned} V_{\mathrm{extra}} &=0.240-0.150\\ &=0.090\ \mathrm{V} \end{aligned} \]

Hence,

\[ \begin{aligned} R_{\mathrm{extra}} &=\frac{0.090}{0.50}\\ &=0.18\ \Omega \end{aligned} \]

d) Why the apparent agreement proves nothing

The bare strip’s resistance is \(0.48\ \Omega\).

Adding the brace creates a parallel path across section \(BC\). This reduces the resistance of the connected network.

With the brace fitted, the resistance between \(B\) and \(C\) becomes

\[ \begin{aligned} R_{BC,\mathrm{eq}} &=\frac{0.30(0.20)}{0.30+0.20}\\ &=0.12\ \Omega \end{aligned} \]

The effective resistance between \(A\) and \(C\) is therefore

\[ \begin{aligned} R_{AC,\mathrm{eq}} &=0.18+0.12\\ &=0.30\ \Omega \end{aligned} \]

Compared with the isolated strip, the brace reduces the effective resistance by

\[ 0.48-0.30=0.18\ \Omega \]

But the external lead and contact resistance adds exactly \(0.18\ \Omega\) to the two-wire reading.

The two effects cancel:

\[ \begin{aligned} R_2&=0.30+0.18\\ &=0.48\ \Omega \end{aligned} \]

This happens to equal the isolated strip’s resistance.

The two-wire measurement is not accurately measuring the strip alone. It is measuring the shunted strip plus the external leads and contacts.

The four-wire method isn’t faulty either. It correctly measures the equivalent resistance of the electrical network between its sense points, provided the total supplied current is used to define that network’s equivalent resistance.

What it cannot do is distinguish the strip’s current from current flowing through a parallel brace.

This is the conceptual hinge: a measurement can give the expected numerical answer for the wrong physical reason.

e) Measurements after removing the brace

Once the brace is removed, all \(0.50\ \mathrm{A}\) passes through both sections of the strip.

The sensed voltage becomes

\[ \begin{aligned} V_{AC}&=0.090+0.150\\ &=0.240\ \mathrm{V} \end{aligned} \]

The four-wire reading is

\[ \begin{aligned} R_4&=\frac{0.240}{0.50}\\ &=0.48\ \Omega \end{aligned} \]

The external lead and contact resistance remains \(0.18\ \Omega\).

The two-wire reading is therefore

\[ \begin{aligned} R_2&=0.48+0.18\\ &=0.66\ \Omega \end{aligned} \]

Its expected terminal voltage is

\[ \begin{aligned} V_t&=(0.50)(0.66)\\ &=0.330\ \mathrm{V} \end{aligned} \]

Removing the brace makes the four-wire measurement equal the isolated strip’s resistance, while the two-wire measurement now overestimates it.

The experiment demonstrates two separate requirements for a correct resistance measurement: voltage must be measured across the intended specimen, and the current used in \(R=V/I\) must be the current through that specimen.

07What this changes about your next measurement

When measuring a small resistance, don’t begin by asking whether the meter’s displayed number looks reasonable. First identify where the voltage is measured, where the current flows, and which unwanted resistances or parallel paths are included.

Once those are correct, a changing resistance reading becomes useful physical evidence rather than just a measurement problem. You can then investigate whether the specimen itself is changing with current or temperature, using the distinction explained in Ohm’s Law in HSC Physics: When Resistance Is Constant.