The Potentiometer

Here is a measurement scandal hiding in plain sight: a voltmeter cannot measure the emf of a cell. Connect the best analogue voltmeter you own across a battery and the needle settles always a little too low — the voltmeter draws current, and the moment current flows, the cell's internal resistance quietly pockets part of the voltage. You are measuring the cell while disturbing it, like checking a tyre's pressure with a gauge that leaks air. The potentiometer is the nineteenth century's beautifully sneaky answer. Instead of letting the unknown cell drive a current through a meter, it balances the cell against a known, adjustable voltage — sliding a contact along a long uniform wire until a galvanometer reads exactly zero. At that null point, no current flows through the cell at all: nothing drawn, nothing disturbed, and the cell displays its true, undiminished emf. A measurement made by not measuring — the compensation principle, which kept the potentiometer the precision instrument of electrical science for over a century. This module builds the instrument honestly: the uniform wire and its steady potential fall, the gradient that converts centimetres into volts, the null logic that beats any voltmeter, and the working conditions examiners test. The numericals gym — comparing emfs, extracting internal resistance — trains next door in ARCHETYPE: Potentiometer Numericals; here we make the principle unshakeable.

Key takeaways

A potentiometer is a long uniform resistance wire carrying a steady current from a driver cell; the potential drop along the wire is proportional to length: $V \propto l$ .
The potential gradient $k = V_{\text{wire}}/L$ (volts per metre) is the instrument's conversion rate; a balance length $l$ measures a voltage $kl$ .
An unknown emf is found by sliding a jockey to the null point where the galvanometer reads zero — at balance the cell supplies no current, so its internal resistance drops nothing and the true emf is read.
Working condition: the driver cell's emf (more precisely, the p.d. it maintains across the wire) must exceed the emf being measured, or no balance point exists.
Standardization against a standard cell fixes $k$ experimentally before any unknown is measured.
Smaller gradient (longer wire or added series resistance) spreads the same voltage over more length — a longer, more precisely readable balance length.
Key idea: at the null point the potentiometer draws zero current from the cell under test, so it reads the genuine emf — a feat no ordinary voltmeter can match.

The core insight

A potentiometer balances an unknown emf against the steady potential drop along a current-carrying uniform wire; at the zero-deflection point no current is drawn from the test cell, so the balance length converts directly into the cell's true emf via the potential gradient $k$ , with $\varepsilon = kl$ .

Where this lives today

The word "potentiometer" now mostly names the instrument's descendant: the three-terminal knob inside volume controls and joysticks is a potential divider with a sliding tap — same uniform track, same $V \propto l$ , used to deliver a fraction of a voltage rather than to null one. The measurement idea never retired; it changed clothes. Every digital multimeter on its DC-volts setting presents ten megohms or more of input resistance precisely to approach the draw-no-current ideal the potentiometer achieved exactly, and at the summit of precision, national metrology institutes still measure voltages by comparison against a reference with a null detector — only the reference is now a Josephson junction array whose emf is fixed by quantum mechanics. The slide-wire is gone; Poggendorff's logic — let the unknown reveal itself at the moment it delivers no current — runs the volt itself.

The setup: a ruler made of voltage

Picture the classic school instrument: ten metres of uniform constantan wire, stretched in one-metre runs over a wooden board, with a metre scale beneath. A driver cell, a key, and a rheostat are wired in series across the whole wire, from terminal A to terminal B, pushing a steady current $I$ through it. That current converts the wire into a ruler graduated in volts: each centimetre, carrying the same current through the same resistance-per-centimetre, drops the same small potential. Walk along the wire from A and the potential falls steadily, like a ramp of perfectly even slope. The measuring side hangs off this ramp. The cell under test (emf $\varepsilon$ ) is connected from A — positive terminals together at A, always — through a sensitive galvanometer to a sliding contact called the jockey. Touch the jockey at some point P, and two voltages compete along the loop A-to-P: the wire's drop $V_{AP}$ and the cell's emf $\varepsilon$ . If they disagree, the difference drives current through the galvanometer and the needle kicks. Slide until the kick dies to exactly zero: at that null point, $V_{AP} = \varepsilon$ , and the scale reading $l$ is the measurement.

Figure

The potentiometer: driver circuit powers the wire, jockey hunts the null point

Two design choices deserve their reasons. The wire is an alloy like constantan because its resistivity barely changes with temperature (Temperature Dependence, Thermistors & Superconductors-Lite explains why alloys sit still), so the volt-ruler's calibration doesn't wander as the room warms. And the wire is long on purpose: the same voltage spread over ten metres instead of one gives ten times more millimetres per millivolt — the instrument's precision. Gradient down, readable length up: that intuition returns throughout.

The potential gradient: converting length to volts

Everything the potentiometer does routes through one conversion factor. If the driver circuit maintains a potential difference $V_{\text{wire}}$ across the full length $L$ , then the drop per unit length is the potential gradient:

k = \frac{V_{\text{wire}}}{L} \qquad (1)

where: $k$ — the potential gradient in volts per metre (exam answers often quote V/cm — carry the unit explicitly); $V_{\text{wire}}$ — the p.d. between the wire's ends A and B, which is less than the driver's emf whenever a rheostat or the driver's internal resistance takes its share; $L$ — the wire's full length. Read $k$ as the exchange rate between the scale and the volt: at $k = 0.05\ \mathrm{V/m}$ , every metre of balance length testifies to one-twentieth of a volt. A point P at distance $l$ from A then sits at potential drop

V_{AP} = k\,l \qquad (2)

where: $l$ — the balance length read off the scale. This linear relation $V \propto l$ is the potentiometer principle, and it leans on two conditions worth naming aloud in any answer: the wire must be uniform (same cross-section and material everywhere, so resistance per metre is constant) and the driver current must be steady (same $I$ during the whole experiment). Break either and the ruler's markings effectively smear.

Figure

The volt-ruler: potential drop grows linearly with distance along the wire

In practice nobody computes $k$ from first principles — the wire's exact resistance and the driver's true emf are never known well enough. Instead the instrument is standardized: a standard cell of precisely known emf (historically a Weston cell, $1.0186\ \mathrm{V}$ ) is balanced first, its balance length fixes $k$ experimentally, and only then is the unknown connected. Standardization turns a comparative gadget into an absolute instrument; repeating it at the end checks that the driver current never drifted.

Derivation AWhy the potential falls uniformly along the wire

We want the heart of the instrument: that a uniform wire carrying a steady current drops potential in exact proportion to length, $V_{AP} \propto l$ . The strategy is to build the wire's resistance from its geometry — the same argument as Resistivity & Geometry, applied to a fraction of the wire — and then let Ohm's law convert steady current into steady voltage drop. Watch where each of the two working conditions (uniform wire, steady current) enters, because exam questions probe exactly those joints.

Step 1. A wire of resistivity $\rho$ , cross-sectional area $A_w$ and length $L$ has total resistance $R_{\text{wire}} = \rho L/A_w$ . The stretch from A to a point P at distance $l$ is identical wire, just shorter: $R_{AP} = \rho l/A_w$ . Why uniformity matters: a varying cross-section would make resistance-per-metre vary, breaking $R_{AP} \propto l$ — a ruler with unevenly spaced markings.

Step 2. Divide the two resistances: $R_{AP}/R_{\text{wire}} = l/L$ . The fraction of resistance equals the fraction of length — geometry alone, independent of the material's actual resistivity.

Step 3. The driver pushes current $I$ through the whole wire. Ohm's law on the segment: $V_{AP} = I R_{AP} = I \rho l / A_w$ . Everything in $I\rho/A_w$ is constant along the wire, so $V_{AP} = k\,l$ with $k = I\rho/A_w$ — the gradient of equation (1), since $V_{\text{wire}} = I\rho L/A_w = kL$ . Why the steady current matters: k contains I. If the driver cell sags or the rheostat is nudged between the standardization and the measurement, k changes silently and every subsequent reading converts length to volts at the wrong rate.

Step 4. The graph of $V_{AP}$ against $l$ is a straight line through the origin with slope $k$ . Verified: with $I = 0.25\ \mathrm{A}$ , $R_{\text{wire}} = 15\ \Omega$ , $L = 8\ \mathrm{m}$ : $V_{\text{wire}} = 3.75\ \mathrm{V}$ and $k = 0.469\ \mathrm{V/m}$ ; the segment at $l = 4\ \mathrm{m}$ has $R_{AP} = 7.5\ \Omega$ and drop $1.875\ \mathrm{V} = k \times 4$ — proportionality confirmed.

The null point: measuring by not measuring

Now the payoff. Around the test loop — A → cell → galvanometer → jockey → back along the wire to A — two sources push in opposition: the cell's emf $\varepsilon$ one way, the wire's drop $V_{AP}$ the other. The galvanometer current is driven by their difference. Where $V_{AP} < \varepsilon$ , the cell wins and the needle deflects one way; where $V_{AP} > \varepsilon$ , the wire wins and the needle reverses. Between them lies the point of perfect stalemate:

\varepsilon = k\,l \qquad (3)

where: $l$ — the balance length at zero deflection. Equation (3) is the entire measurement: read $l$ , multiply by the standardized $k$ , done. The two-direction deflection is also the search strategy — tap near each end, confirm opposite deflections, bisect toward the null.

Go deeper · Derivation B · 1 minWhy a null reading measures the true emf

Derivation BWhy a null reading measures the true emf

We want the claim that makes the potentiometer superior to any voltmeter: at balance, the reading equals the cell's genuine emf $\varepsilon$ , unpolluted by its internal resistance $r$ . The strategy is to write the potential drop around the test loop with an arbitrary galvanometer current $i_g$ , then watch what happens to every $r$ -dependent and galvanometer-dependent term when $i_g = 0$ . The thing to watch: internal resistance only does anything when current passes through it.

Step 1. Let the test loop carry current $i_g$ (through cell, galvanometer of resistance $G$ , and jockey contact). Traversing from A through cell and galvanometer to the jockey, the loop condition reads $\varepsilon - i_g r - i_g G = V_{AP}$ .

Step 2. Off balance, $i_g \neq 0$ , and the wire drop $V_{AP}$ differs from $\varepsilon$ by $i_g(r + G)$ — an error depending on two quantities we generally do not know. Any instrument that reads while drawing current inherits this error; that is the voltmeter's incurable disease (Real Meters — Loading Effects & Reading Errors dissects the damage).

Step 3. At the null point, $i_g = 0$ . The error term $i_g(r+G)$ vanishes identically — not approximately, not for small $r$ , but exactly, whatever the values of $r$ and $G$ . The loop condition collapses to $\varepsilon = V_{AP} = kl$ . Why this is the compensation idea: the wire's drop has been adjusted to compensate the cell's emf precisely, so the cell sits in the circuit yet delivers nothing. A source delivering zero current suffers zero internal drop and shows its emf at its terminals — the open-circuit condition, arranged deliberately inside a closed loop.

Step 4. Note what the argument never required: a calibrated galvanometer (it is only a zero-detector), or a clean jockey contact — no current, no drop, either way. Verified: for $\varepsilon = 1.5\ \mathrm{V}$ , $r = 0.9\ \Omega$ — a voltmeter of resistance $9\ \Omega$ would read $\varepsilon \times 9/9.9 = 1.36\ \mathrm{V}$ (a 9% understatement), while the potentiometer at balance reads $1.50\ \mathrm{V}$ exactly; setting $i_g = 0$ in Step 1 reproduces $\varepsilon = kl$ independent of $r$ and $G$ .

The working conditions

One requirement is absolute: the wire's full-scale voltage must exceed the emf under test. The wire offers drops between $0$ (at A) and $kL = V_{\text{wire}}$ (at B); if $\varepsilon > kL$ , the stalemate point lies beyond the end of the ruler and the galvanometer deflects the same direction everywhere — the classic no-balance symptom. Since $V_{\text{wire}}$ is at most the driver's emf (less after the rheostat and internal resistance take their cuts), the driver must comfortably out-vote the test cell. The habits: both positive terminals at A (reversed polarity gives the same one-way-deflection symptom); standardize before and after; tap the jockey rather than dragging it (dragging scrapes the wire non-uniform); and set the rheostat so the balance lands long — the same ±1 mm reading uncertainty is a smaller fraction of a longer length. That is the precision dial of the instrument: reduce $k$ and every measured voltage stretches over more centimetres. This lever, and the numerical families built on it, is drilled in ARCHETYPE: Potentiometer Numericals.

situation	consequence	reading
driver p.d. across wire < test emf	no balance point; one-way deflection along the whole wire	increase driver voltage or cut series resistance
test cell reversed at A	one-way deflection (the two sources now push together)	swap the test cell's terminals
balance found at $l$	$\varepsilon = kl$ exactly; zero current drawn	the true-emf reading
$k$ halved (series R doubled up)	every balance length doubles	finer precision, if the wire is long enough
wire non-uniform	$V \propto l$ fails; readings biased	why quality instruments use drawn alloy wire
driver current drifts mid-experiment	$k$ at measurement ≠ $k$ at standardization	re-standardize; suspect the rheostat or a tired driver cell
jockey between two null taps	deflection reverses across the null	bracket-and-bisect search strategy

Common trapA potentiometer is just a fancy voltmeter

It is the opposite of one, and the distinction is the most-tested sentence in this module. A voltmeter must pass current through its own coil to deflect, and that current, flowing through the cell's internal resistance, knocks the terminal voltage below the emf before the measurement even happens — an ideal voltmeter would need infinite resistance. The potentiometer at balance draws exactly zero current from the cell under test: zero current, zero internal drop, genuine emf — even with a mediocre galvanometer and dirty contacts, because null instruments only need to recognise zero. One caution keeps the claim honest: away from balance the potentiometer briefly draws current like anything else (why you tap, not press), and the driver circuit draws current continuously. The zero-current miracle applies to the test-cell branch, at the null, only.

Worked examples

Board level — can this potentiometer measure this cell? (ISC pattern)

A potentiometer wire is 10 m long with resistance 5.0 Ω. The driver cell has emf 2.0 V and negligible internal resistance, connected in series with a rheostat set to 15 Ω. (a) Find the current in the wire and the potential gradient. (b) Show that a cell of emf 1.5 V cannot be balanced at this setting. (c) The rheostat is reduced to zero. Find the new potential gradient and the balance length for the 1.5 V cell.

Step 1. Part (a): the driver loop has total resistance $15 + 5.0 = 20\ \Omega$ , so $I = 2.0/20 = 0.10\ \mathrm{A}$ . The p.d. across the wire is $V_{\text{wire}} = IR_{\text{wire}} = 0.10 \times 5.0 = 0.50\ \mathrm{V}$ , giving $k = 0.50/10 = \boxed{0.050\ \mathrm{V/m}}$ .

Step 2. Part (b): the largest voltage the wire can offer is at the far end: $kL = 0.050 \times 10 = 0.50\ \mathrm{V}$ . Since $1.5\ \mathrm{V} > 0.50\ \mathrm{V}$ , the null condition $\varepsilon = kl$ has no solution on the wire — the galvanometer deflects the same way at every point. The p.d. across the wire, not the driver's emf label, is what must exceed the test emf.

Step 3. Part (c): with the rheostat at zero, $I = 2.0/5.0 = 0.40\ \mathrm{A}$ , $V_{\text{wire}} = 0.40 \times 5.0 = 2.0\ \mathrm{V}$ , so $k = 2.0/10 = 0.20\ \mathrm{V/m}$ . Balance: $l = \varepsilon/k = 1.5/0.20 = \boxed{7.5\ \mathrm{m}}$ .

Step 4. The mark-earning sentence: "A balance point exists only if the p.d. across the whole wire exceeds the emf to be measured."

Competitive level — the gradient under real conditions (JEE-legacy / CIE pattern)

A driver cell of emf 4.0 V and internal resistance 1.0 Ω drives a potentiometer wire of length 8.0 m and resistance 15 Ω. (a) Find the potential gradient. (b) A cell of emf 1.02 V is balanced; find the balance length. (c) A 5.0 Ω resistor is now added in series with the driver. Explain, with the new balance length, why an experimenter might do this deliberately.

Step 1. Part (a): driver current $I = \dfrac{4.0}{1.0 + 15} = 0.25\ \mathrm{A}$ . Wire p.d.: $V_{\text{wire}} = 0.25 \times 15 = 3.75\ \mathrm{V}$ . Gradient: $k = 3.75/8.0 = \boxed{0.469\ \mathrm{V/m}}$ . Note the driver's internal resistance already stole $0.25\ \mathrm{V}$ — the wire never sees the full 4.0 V.

Step 2. Part (b): $l = \varepsilon/k = 1.02/0.46875 = \boxed{2.18\ \mathrm{m}}$ (keep the unrounded $k$ in the division; round once).

Step 3. Part (c): $I' = \dfrac{4.0}{1.0 + 15 + 5.0} = 0.190\ \mathrm{A}$ ; wire p.d. $= 2.86\ \mathrm{V}$ ; $k' = 2.86/8.0 = 0.357\ \mathrm{V/m}$ ; balance $l' = 1.02/0.35714 = \boxed{2.86\ \mathrm{m}}$ .

Step 4. The point: the balance length grew from 2.18 m to 2.86 m for the same cell, so a ±1 mm reading error shrank from $0.046\%$ to $0.035\%$ of the length — finer precision, with the balance still safely on the wire ( $2.86\ \mathrm{V} > 1.02\ \mathrm{V}$ ). Reducing the gradient stretches the measurement; the limit is running out of wire. Timing note: about 3 minutes. The whole question is one chain — loop current, wire p.d., gradient, divide — done twice. The classic slip is using the driver emf 4.0 V as the wire p.d.; the internal resistance and any series resistor must take their cut first. Practise the chain until it runs in one breath: it is the opening move of every potentiometer numerical in the gym module.

Exam traps

The rationalization footnote first. CBSE deleted the potentiometer in the 2023-24 rationalization — it carries no CB tag, and JEE-Main and NEET dropped it in their 2024 alignments too. It remains squarely examinable in ISC (named practical and theory topic) and CIE A-Level 9702 (the canonical null method for comparing p.d.s). CBSE students: treat this as enrichment and voltmeter-error insurance; ISC and A-Level: live ammunition.

Driver emf vs wire p.d. The balance condition needs the p.d. across the wire to exceed the test emf. "Driver emf 2.0 V > 1.5 V, so balance exists" fails whenever a rheostat or internal resistance eats the margin — Worked Example 1's wire carried only 0.50 V.

"The potentiometer draws no current." Half-true, and examiners punish the sloppy half. The test-cell branch carries zero current at balance only; the driver circuit draws current the whole time — that current creates the gradient.

Units of k. Boards mix V/m and V/cm freely: $0.050\ \mathrm{V/m} = 5.0 \times 10^{-4}\ \mathrm{V/cm}$ . Carry the unit through equation (3) or eat a silent factor of 100.

One-way deflection diagnosis. Exactly two standard causes: wire p.d. smaller than the test emf, or reversed test-cell polarity. State the symptom (same direction at both ends) and both causes for full marks.

Uniform and steady. Any "state the principle" answer must include both conditions: uniform wire and steady driver current. $V \propto l$ is a property of uniform wires carrying constant current, not of wires.

Constants and conventions. No $g$ here. The Weston standard-cell value is $1.0186\ \mathrm{V}$ (ISC papers supply it). A-Level phrases the instrument as "a null method of comparing p.d.s" — same physics, jockey and all.

Origins · 1 minJohann Christian Poggendorff and how this was discovered

Johann Christian Poggendorff

1796–1877 · Berlin

Poggendorff spent his life measuring things other physicists were content to estimate. An apothecary's apprentice turned physicist, he built one of the century's most sensitive galvanometer readouts and, in 1841, published the compensation method: balance an unknown emf against the calibrated drop along a current-carrying wire, detect the null, and read the answer with zero current drawn. The idea's genius is what it removes — at balance, the errors from internal resistance, contact resistance and galvanometer calibration all multiply zero current and vanish together. For the next century, potentiometric measurement was the gold standard of electrical laboratories. Poggendorff also edited the Annalen der Physik for 52 years, shepherding much of nineteenth-century physics into print — a man who spent his days balancing other people's claims against a known standard, in every sense.

History

1841

— Poggendorff publishes the compensation method: emf measured by balancing against a wire's potential drop, at zero drawn current

1873

— Latimer Clark's standard cell provides a stable reference emf, turning compensation into absolute voltage measurement

1893

— The Chicago International Electrical Congress defines the legal volt through standard-cell potentiometry

1908

— The Weston cell (1.0186 V) becomes the international emf standard, anchoring potentiometer practice for six decades

1990

— Josephson-junction arrays replace standard cells as the volt's definition; the null-comparison philosophy survives the slide-wire

Going further

Texts. ISC Physics Class 12, Current Electricity unit (potentiometer theory + the two named experiments) · CIE 9702 syllabus §7.3 (null methods, comparing p.d.s) · NCERT Class 12 Ch. 3 (pre-2023 editions — the deleted section remains the clearest short treatment in print) · H.C. Verma, Concepts of Physics Vol. 2, Ch. 33.

Provenance. Authored fresh for the HS E&M track (rationalized-NCERT module kept for ISC/legacy-JEE/CIE demand). A verifier should recompute: Example 1 ( $I = 0.10\ \mathrm{A}$ , $k = 0.050\ \mathrm{V/m}$ , no balance for 1.5 V; then $k = 0.20\ \mathrm{V/m}$ , $l = 7.5\ \mathrm{m}$ ); Example 2 ( $k = 0.46875\ \mathrm{V/m}$ , $l = 2.18\ \mathrm{m}$ ; with 5 Ω added, $k' = 0.357\ \mathrm{V/m}$ , $l' = 2.86\ \mathrm{m}$ ); the voltmeter comparison ( $1.5 \times 9/9.9 = 1.36\ \mathrm{V}$ ).

Prerequisites. EMF, Terminal Voltage & Internal Resistance · Potential Dividers. Leads to. ARCHETYPE: Potentiometer Numericals · ARCHETYPE: Wheatstone & Meter-Bridge Numericals · EMF and Internal Resistance (UG deep dive).

If the galvanometer isn't calibrated, how can the measurement be accurate?

Because the galvanometer is never asked to measure anything — only to recognise zero, which needs sensitivity, not calibration. Off balance it deflects some amount (who cares how much); at balance it sits at zero, the one reading every galvanometer gets right. The actual measuring is done by the scale under the wire and the standardization that fixed $k$ . This is the deep pattern of null methods: convert "measure a value" into "detect an equality", and the detector only needs sensitivity near zero. The Wheatstone Bridge plays the identical trick with resistances.

Why must both positive terminals connect at A?

Because the wire's drop and the cell's emf must oppose each other around the test loop for a stalemate to exist. With positives together at A, the loop gains potential through the wire segment and loses it through the cell — the two can cancel at the right length. Reverse the cell and both sources push the galvanometer current the same way everywhere: the deflection never crosses zero, only weakens toward one end. Same-direction deflection at both extremes is precisely the exam symptom of a reversed cell (or an underpowered driver).

Is the potentiometer just a potential divider with a meter attached?

They share the same hardware idea — a resistive track with a movable tap — but run it for opposite purposes. A potential divider outputs a chosen fraction of a supply to feed a load, and the load's current actually flows through the track (the divider formula and its loading headaches live in the Potential Dividers module). A measurement potentiometer compares: the tap feeds a galvanometer whose job is to carry no current at the end, so the track is never loaded by the thing being measured. Divider = adjustable source; potentiometer = null comparator. The volume-knob component called a "pot" is named for the hardware, used as a divider.

Why is a longer balance length more precise?

Your reading uncertainty is roughly fixed in millimetres — the scale's least count plus judgement about where the null sits. What matters is that uncertainty as a fraction of the length: ±1 mm on 2.0 m is 0.05%, on 8.0 m it is 0.0125%. Since $\varepsilon = kl$ , the fractional error in $l$ passes straight into $\varepsilon$ . So experimenters lower the gradient until the balance sits near the far end of the scale. The gym module drills the numerical versions — including the trap where lowering the gradient too far pushes the balance off the wire entirely.

Does the potentiometer work for measuring terminal voltage, or only emf?

Both — it measures whatever p.d. is presented to its terminals, drawing nothing. A cell alone gives its emf. The same cell driving an external resistor gives its terminal voltage — smaller by the internal drop the load current causes (EMF, Terminal Voltage & Internal Resistance tells that story). Comparing the two readings is the classic method for extracting internal resistance, drilled in ARCHETYPE: Potentiometer Numericals.

What is a standard cell, and why standardize at all?

A standard cell is built for one virtue: an emf precisely known and boringly stable (the Weston cell holds 1.0186 V to parts in a hundred thousand, if never asked to deliver real current). Standardizing means balancing this known emf first: its balance length gives $k = \varepsilon_{\text{std}}/l_{\text{std}}$ directly, calibrating the instrument against reality rather than the driver cell's optimistic label — converting the systematic unknowns (true driver emf, exact wire resistance) into one measured number. Repeat the standardization after the experiment: if $k$ drifted, the driver current wandered and the run is suspect.