In short

The meter bridge is a practical form of the Wheatstone bridge in which two of the four bridge arms are replaced by two segments of a single uniform resistance wire, stretched over a metre-long scale. Slide a jockey along the wire until the galvanometer reads zero. If the balance point is at length \ell (in cm, measured from the left end) and the known resistance is S in the right gap, the unknown R in the left gap is given by the balance condition

\boxed{\;R \;=\; S\cdot\frac{\ell}{100-\ell}\;}.

The potentiometer is a cell-in-parallel-with-a-uniform-wire device that measures a potential difference by comparing it to the drop along a known length of the wire — with no current drawn from the source at the moment of balance.

\boxed{\;\varepsilon \;=\; k\cdot \ell_\text{balance}\;}, \qquad \text{potential-gradient: } k \equiv \tfrac{V_\text{wire}}{L}.

To compare two EMFs \varepsilon_1 and \varepsilon_2, balance each one separately on the same wire and read off \ell_1 and \ell_2:

\frac{\varepsilon_1}{\varepsilon_2} \;=\; \frac{\ell_1}{\ell_2}.

To find a cell's internal resistance r, balance the cell's EMF at length \ell_1 (open-circuit), then connect an external shunt R across it and balance the reduced terminal voltage at \ell_2. Then

r \;=\; R\cdot\frac{\ell_1 - \ell_2}{\ell_2}.

Both instruments are "null" methods — they give an accurate answer at the moment of zero galvanometer deflection, so the galvanometer's own calibration is irrelevant. This is why a 1960s wooden potentiometer in a Delhi government-school physics lab can still give a more accurate EMF reading than a 2024 digital multimeter.

Open the side drawer of the physics lab in any CBSE-affiliated school and you find the same two instruments: a one-metre long wooden board with a shiny wire stretched taut along a metre-rule, two brass terminals at either end, a jockey (a spring-loaded pointer with a wire running to a galvanometer), and the small accompanying glass-topped box with its unknown resistance coils, standard cells, and a rheostat. This is the meter bridge. On the next shelf sits a similar board but with four uniform wires stitched end-to-end for a full four-metre length, a moveable jockey, and two "pots" (tap-off points) along the way. This is the potentiometer.

Neither instrument looks impressive. The wire is just nichrome or constantan of about 0.3 mm diameter. The jockey is a pencil-sized spring with a knob. There is no digital display, no LED, no USB port. In a country where every Class 12 student is handed a ₹150 digital multimeter that measures resistance, voltage, and current in parallel, why does the CBSE syllabus still demand that you learn these things — and why is the potentiometer experiment always worth at least one of the two practical-exam questions at JEE?

The answer is that both instruments are more accurate than a multimeter, in the specific sense that matters for physics. A digital multimeter reads potential differences by drawing a small current through its own internal resistance and measuring the voltage drop. That small current sags the voltage of whatever cell you're measuring by Ir — a real, non-zero correction if the cell is old or has high internal resistance. The potentiometer measures EMF by driving the galvanometer current to exactly zero at the moment of balance. Nothing is drawn from the cell under test. There is no Ir sag to correct for. The reading is the true EMF.

The meter bridge is the same idea applied to resistance: a Wheatstone bridge balanced by sliding a contact along a uniform wire, so that the reading is a pure length ratio — 25 cm against 75 cm — and length measurements on a metre-rule are routinely accurate to 0.1%, better than the tolerance of any laboratory resistor.

This article derives both instruments from Kirchhoff's laws, works out how to use them, and explains the rules of thumb the school lab manuals never tell you — why you press the jockey gently, why the balance point should ideally be near 50 cm, why the potentiometer driver cell must always have an EMF larger than any EMF you hope to measure.

The uniform wire — the common foundation

Both instruments use a single piece of uniform resistance wire: nichrome or constantan of constant cross-section, stretched along a metre rule. Let L be the total length of this wire (typically 100 cm for a meter bridge, 400 cm for a potentiometer), and R_\text{wire} its total resistance.

Key property. Because the wire is uniform, the resistance of any segment of length \ell is proportional to \ell:

R(\ell) \;=\; R_\text{wire}\cdot\frac{\ell}{L}.

Why: resistance is \rho \ell/A, and on a uniform wire both \rho and A are the same everywhere. Doubling the length doubles the resistance; the piece from 0 to 30 cm has exactly 30% of the total resistance of the whole wire.

That one fact — resistance is proportional to length on a uniform wire — is what turns both instruments into pure length-reading devices. Every derivation in this article leans on it.

The meter bridge — a Wheatstone bridge on a wire

Construction

A one-metre uniform wire is stretched between two thick copper end-pieces labelled A and C (say). Between these two end-pieces sit two small gaps in the copper strip — the left gap (between A and a midpoint B) and the right gap (between B and C). The unknown resistance R goes in the left gap; a known resistance S (from a "resistance box") goes in the right gap. A cell (often a single Leclanché cell of about 1.5\ \text{V}) with a key drives current through the whole bridge. A galvanometer is connected from the midpoint B to a jockey — a spring-loaded contact that can be pressed onto any point along the wire.

Schematic of a meter bridgeA horizontal metre wire from A on the left to C on the right. Above the wire, two gaps contain resistor R (left, unknown) and resistor S (right, known), meeting at midpoint B. A cell drives current through the whole top rail. A galvanometer from B to a jockey touches point J on the wire at distance l from A. +−ε K R (unknown) B S (known) AC 050100 cm J G ℓ (balance length from A)100 − ℓ
Meter bridge schematic. A uniform resistance wire AC runs along a metre scale. The unknown resistor $R$ sits in the left gap, known $S$ in the right gap, joined at the midpoint B. A galvanometer connects B to the jockey J. Slide J until the galvanometer reads zero; then the balance length $\ell$ from A determines $R$ via the ratio condition.

The circuit is exactly a Wheatstone bridge with four arms:

The galvanometer is the bridge's cross-arm.

Deriving the balance condition

You established in the Wheatstone bridge chapter that a Wheatstone bridge balances — zero current through the cross-arm galvanometer — when the products of opposite arms are equal, or equivalently, when the ratios of adjacent arms match:

\frac{R}{S} \;=\; \frac{R_\text{AJ}}{R_\text{JC}}.

For the meter bridge, substitute the segment resistances:

\frac{R}{S} \;=\; \frac{r\ell}{r(100 - \ell)} \;=\; \frac{\ell}{100 - \ell}.

Why: the wire is uniform, so both bottom-arm resistances are the same r per cm times their respective lengths. That constant r cancels, and the bridge condition becomes a pure ratio of lengths. You never need to know the resistance per cm of the wire itself.

Solving for R:

\boxed{\;R \;=\; S\cdot\frac{\ell}{100 - \ell}\;}.

How to use the meter bridge in practice

  1. Close the key to send current through the bridge.
  2. Slide the jockey along the wire while gently pressing it down. Watch the galvanometer.
  3. When the deflection is small and changes sign as you cross a specific point on the wire, you've found the balance length \ell. Read it off the metre scale.
  4. Compute R = S \cdot \ell/(100 - \ell).
  5. Repeat with R and S swapped — put the unknown in the right gap and the known in the left. The balance is now at 100 - \ell (check this — the circuit is symmetric). Take the mean of the two readings to cancel any asymmetry in the end-strip copper.

Why a balance near 50 cm is ideal — the end-correction

The theoretical balance condition assumes that the wire ends at exactly 0 cm and 100 cm — that is, that the copper end-pieces are perfect connectors of zero length. In reality, the copper strips at A and C contribute small end-corrections, typically a few millimetres each. If the balance point is at \ell = 12 cm, any error of 1 mm in the end-correction is a ~1% relative error on the short 12 cm side. But if \ell = 50 cm, the same 1 mm end-correction is only a 0.2% relative error. The sensitivity is maximum near the centre of the wire.

Practical consequence. Choose the known resistance S in the right gap so that you expect \ell \approx 50 cm — i.e., choose S \approx R. This gives you the smallest end-correction error and the largest galvanometer sensitivity at balance. If your first balance comes out at 15 cm, change S to a smaller value (about one-fifth of the original) and rebalance.

Interactive: meter bridge balance length vs unknown RPlot of unknown resistance R versus balance length ell for a fixed known resistor S = 5 ohm. The curve R = S ell / (100 - ell) rises sharply near ell = 100 and is near zero near ell = 0. A draggable marker along the length axis shows the balance length and resulting R live. balance length ℓ (cm)R (Ω) with S = 5 Ω 1535557595 5101520 R = S = 5 Ω at ℓ = 50 drag the red point along ℓ
Drag the balance length $\ell$ from 5 cm to 95 cm with known $S = 5\ \Omega$. The curve $R = S\ell/(100-\ell)$ shows the unknown resistance you'd compute at each balance point. Notice the steep rise beyond $\ell = 70$ cm — a 1 mm error in length at $\ell = 90$ cm costs you a 10% error in $R$; at $\ell = 50$ cm the same 1 mm error costs less than 1%. This is why you always try to balance near the middle.

The potentiometer — measuring EMF without drawing current

The core idea

A cell under test is connected so that its EMF opposes part of the voltage drop along a uniformly-dropping wire. Slide the jockey along the wire until the two voltages exactly cancel — the galvanometer reads zero. At that moment, no current flows from the cell under test, so no Ir sag. The wire segment from one end to the jockey must have a voltage drop equal to the cell's EMF. Because the wire is uniform, that wire-segment voltage drop is proportional to the balance length:

V(\ell) \;=\; k\,\ell, \qquad k \equiv \text{potential gradient (volts per cm)}.

So the balance length itself is the measurement: \varepsilon = k\,\ell_\text{balance}.

Construction

Schematic of a potentiometerA driver cell ε0 with a rheostat Rh drives a steady current through a long uniform potentiometer wire AB of length L. A test cell ε1 is connected with its positive terminal facing the positive end of the wire, through a galvanometer and a jockey J that can touch any point along the wire. At balance, no current flows through the galvanometer. ε₀ K Rh A (+)B (−) 0ℓL G ε₁ J V(ℓ) = kℓ
Potentiometer schematic. A steady driver current from cell $\varepsilon_0$ (regulated by rheostat Rh) flows through a uniform wire from A (high potential) to B (low potential). The test cell $\varepsilon_1$ is connected with its $+$ end toward A through a galvanometer and a jockey J. At balance, no current flows through the galvanometer, and the wire voltage drop from A to J equals $\varepsilon_1$.

Let \varepsilon_0 be the driver cell's EMF, R_\text{wire} the total resistance of the potentiometer wire, R_\text{rest} the combined resistance of the rheostat and driver internal resistance, and L the wire's length. The steady current through the wire is

I \;=\; \frac{\varepsilon_0}{R_\text{wire} + R_\text{rest}}.

The voltage drop across the full wire is V_\text{wire} = I R_\text{wire}, and the potential gradient along the wire is

k \;=\; \frac{V_\text{wire}}{L}.

At any point a distance \ell from A, the potential is V_A - k\ell.

Deriving the balance condition

At balance, the galvanometer reads zero. Apply Kirchhoff's voltage law to the loop that contains the test cell \varepsilon_1, the galvanometer (carrying zero current, so no IR_G drop), the jockey J, and the wire segment from A to J.

Walk around this small loop starting at A, going through the wire to J, through the jockey and galvanometer, through the test cell \varepsilon_1, back to A.

Along the wire from A to J: voltage drops by k\ell. Contribution: -k\ell.

Through the galvanometer (zero current): no voltage drop. Contribution: 0.

Through the test cell from - to +: voltage rises by \varepsilon_1. Contribution: +\varepsilon_1.

KVL: -k\ell + 0 + \varepsilon_1 = 0, so

\boxed{\;\varepsilon_1 \;=\; k\ell.\;}

Why: at balance, zero galvanometer current means the test cell is effectively disconnected from the rest of the circuit. Its EMF must exactly balance the wire drop from A to J. No current flows through \varepsilon_1, so there is no Ir sag — you are reading the true EMF, not the terminal voltage.

Comparing two EMFs

Balance the first test cell \varepsilon_1 at length \ell_1, then swap in the second cell \varepsilon_2 and balance at length \ell_2. Since k is the same for both measurements (driver current held constant):

\varepsilon_1 \;=\; k\ell_1, \qquad \varepsilon_2 \;=\; k\ell_2.

Divide:

\boxed{\;\frac{\varepsilon_1}{\varepsilon_2} \;=\; \frac{\ell_1}{\ell_2}.\;}

Why: the potential gradient k cancels — you don't need to know it. You measure two lengths on the same wire under the same driver conditions and take their ratio. This is why the potentiometer is a comparison instrument — it converts an unknown EMF into a known length ratio against a reference.

To actually find \varepsilon_1, you'd use a standard cell (a Weston cadmium cell, say, with EMF exactly 1.01860 V at 20 °C) as \varepsilon_2, balance both, and compute \varepsilon_1 = \varepsilon_2 \cdot \ell_1/\ell_2. Accuracy limited by the length measurement.

Finding internal resistance of a cell

Here is the clever bit. Connect the test cell \varepsilon to the potentiometer with no external load — balance at length \ell_1. Because no current flows through the cell at balance, the reading k\ell_1 is the EMF \varepsilon itself:

\varepsilon \;=\; k\ell_1.

Now add an external shunt resistor R across the test cell (so that current I = \varepsilon/(R + r) flows through it from its own branch, even when the galvanometer still reads zero at balance). At this new balance, the potentiometer reads the cell's terminal voltage V = \varepsilon - Ir — the drop across the shunt, which is IR. Let the new balance length be \ell_2:

V \;=\; IR \;=\; k\ell_2.

Divide:

\frac{\varepsilon}{V} \;=\; \frac{\ell_1}{\ell_2}.

Solve for r using V = \varepsilon - Ir and V = IR:

\frac{\varepsilon}{V} \;=\; 1 + \frac{r}{R} \quad\Longrightarrow\quad \frac{r}{R} \;=\; \frac{\varepsilon}{V} - 1 \;=\; \frac{\ell_1}{\ell_2} - 1 \;=\; \frac{\ell_1 - \ell_2}{\ell_2}.
\boxed{\;r \;=\; R\cdot\frac{\ell_1 - \ell_2}{\ell_2}.\;}

Why: the first balance reads the true EMF (no current through the cell). The second balance reads the terminal voltage when the cell is loaded. Their ratio gives you \varepsilon/V = (R+r)/R, and a line of algebra isolates r. All four quantities on the right-hand side — R, \ell_1, \ell_2 — are directly measured.

Why the driver EMF must exceed any test EMF

Essential subtlety. The total voltage drop along the full potentiometer wire is V_\text{wire} = kL. If the test cell's EMF \varepsilon_1 is larger than this, the wire can never produce a matching drop anywhere along its length — the galvanometer never shows zero deflection, no balance exists. The driver cell must therefore have EMF and terminal voltage such that kL > \varepsilon_\text{max test}.

In practice this is enforced by choosing a driver cell (an accumulator or a Leclanché cell) of EMF at least \sim 2\ \text{V} and adjusting the rheostat so that the driver terminal voltage across the wire is just over 2 V — comfortably above the 1.5 V of typical dry-cell test samples. If you ever try a potentiometer experiment and can't find a balance point, this is usually the reason.

Worked examples

Example 1: CBSE Class 12 practical — meter bridge

A student in a Delhi government school sets up a meter bridge to find the unknown resistance of a coil of manganin wire. The known resistance in the right gap is S = 4\ \Omega. The balance point is at \ell = 36.0 cm from the left end A. With the positions swapped (unknown in right gap, known in left), the new balance is at \ell' = 63.8 cm from A. Find the unknown resistance and estimate the effect of the end-correction.

Meter bridge with balance point at 36 cmA one-metre wire from 0 to 100 cm with a red triangle at 36 cm marking the balance point. Above the wire, the unknown R sits in the left gap and the known S = 4 ohm in the right gap.AC 03650100 cm R (?) S = 4 Ω J (balance)
Balance point at 36 cm with the unknown R in the left gap and $S = 4\ \Omega$ in the right.

Step 1. Apply the meter bridge formula.

R \;=\; S\cdot\frac{\ell}{100 - \ell} \;=\; 4 \times \frac{36.0}{100 - 36.0} \;=\; 4 \times \frac{36.0}{64.0} \;=\; 4 \times 0.5625 \;=\; 2.25\ \Omega.

Why: direct substitution of the balance condition. No intermediate quantities needed — the unknown resistance is fixed entirely by the length ratio and the known S.

Step 2. Repeat with positions swapped.

With R in the right gap and S = 4\ \Omega in the left gap, the balance formula flips: the left-gap resistance (now S) equals the right-gap resistance (now R) times \ell'/(100 - \ell'):

S \;=\; R \cdot \frac{\ell'}{100 - \ell'} \quad\Longrightarrow\quad R \;=\; S \cdot \frac{100 - \ell'}{\ell'} \;=\; 4 \times \frac{100 - 63.8}{63.8} \;=\; 4 \times \frac{36.2}{63.8} \;=\; 4 \times 0.5674 \;=\; 2.27\ \Omega.

Why: after swapping positions, the gap that originally held the unknown is now on the right. The balance equation is the same Wheatstone structure, just with the unknown and known swapped. The second reading gives an independent check.

Step 3. Take the mean to cancel end-corrections.

R_\text{best} \;=\; \frac{2.25 + 2.27}{2} \;=\; 2.26\ \Omega.

Why: the two measurements are affected by the small end-corrections in opposite ways — if the copper strip at the left contributes an effective extra few millimetres, the first balance over-estimates the left-arm length and the second balance over-estimates the right-arm length. Averaging cancels this systematic error to leading order. This is the reason the lab manual always insists on doing the experiment "both ways" and taking the mean.

Result. The unknown resistance is R \approx 2.26\ \Omega after end-correction.

What this shows. The meter bridge's accuracy comes from the null-method plus the length-ratio principle plus the mean-of-swapped-positions trick. All three are needed. If you did only the first measurement, end-corrections would give you about 2.25 Ω with a systematic error of perhaps 1%. With both readings you get about 2.26 Ω with an error closer to 0.1% — about the precision of the metre rule itself. This is how a ₹200 wooden board beats a ₹5000 digital ohmmeter in a CBSE lab.

Example 2: Finding the internal resistance of a Leclanché cell using a potentiometer

In a JEE Advanced lab problem, a potentiometer of length L = 400 cm is driven by a 4 V accumulator with the rheostat set so that the potential gradient along the wire is k = 0.005 V/cm. A Leclanché cell is connected as the test cell. With no external load across it, the balance length is \ell_1 = 286.0 cm. A shunt resistor R = 10.0\ \Omega is then placed across the Leclanché cell, and the balance length shifts to \ell_2 = 275.0 cm. Find the cell's EMF and internal resistance.

Potentiometer balance at two lengthsA long potentiometer wire showing two balance points: one at 286 cm (open circuit, EMF) and one at 275 cm (loaded, terminal voltage). The difference between the two balance positions corresponds to the IR drop across the internal resistance.AB 0100200300400 cm ℓ₁ = 286 (open) ℓ₂ = 275 (loaded)
Balance shifts from $\ell_1 = 286$ cm (cell on open circuit) to $\ell_2 = 275$ cm (cell loaded with a 10 Ω shunt). The 11 cm shortening reflects the voltage sag $Ir$ across the internal resistance when current flows.

Step 1. Read the EMF from the open-circuit balance.

\varepsilon \;=\; k\,\ell_1 \;=\; 0.005 \times 286.0 \;=\; 1.430\ \text{V}.

Why: at the open-circuit balance, no current is drawn from the Leclanché cell (galvanometer has zero current and the shunt is not yet connected). The cell's own Ir drop is zero, so the wire voltage k\ell_1 equals the cell's true EMF.

Step 2. Find the internal resistance using the shift formula.

r \;=\; R\cdot\frac{\ell_1 - \ell_2}{\ell_2} \;=\; 10.0 \times \frac{286.0 - 275.0}{275.0} \;=\; 10.0 \times \frac{11.0}{275.0} \;=\; 10.0 \times 0.0400 \;=\; 0.400\ \Omega.

Why: with the 10 Ω shunt connected, a current I = \varepsilon/(R+r) flows through the loop cell + shunt (not through the galvanometer at balance). The terminal voltage — what the potentiometer wire matches at the new balance length — is V = IR = \varepsilon R/(R+r). The ratio \varepsilon/V = (R+r)/R equals \ell_1/\ell_2, which rearranges to the boxed formula.

Step 3. Sanity-check using the current through the shunt.

Current when loaded: I = \varepsilon/(R + r) = 1.430/(10.0 + 0.400) = 1.430/10.400 = 0.1375 A.

Terminal voltage: V = IR = 0.1375 \times 10.0 = 1.375 V.

Wire voltage at \ell_2: k\ell_2 = 0.005 \times 275.0 = 1.375 V. ✓

Why: the terminal voltage computed from the full Kirchhoff analysis of the shunt circuit equals exactly the wire voltage at the second balance. That self-consistency is the signature of a properly set-up potentiometer.

Step 4. Verify the voltage loss.

Ir drop: 0.1375 \times 0.400 = 0.0550 V.

EMF − V: 1.430 - 1.375 = 0.055 V. ✓ Matches.

Result. The Leclanché cell has EMF \varepsilon = 1.430 V and internal resistance r = 0.400\ \Omega.

What this shows. This is precisely the technique used in every JEE Advanced lab-based question on cell characteristics. The trick is the null method — both balance points are moments of zero galvanometer current, so the galvanometer's own calibration, its internal resistance, and any slight nonlinearity are all irrelevant. The answer depends only on two length readings and one known resistor. This is also why old Class 12 lab books from the 1960s (with brass-and-polished-wood potentiometers) can give answers more accurate than a modern handheld multimeter — the instrument's design exploits the length-ratio principle, which is immune to the common error sources of voltmeters.

Common confusions

If you came here to set up and analyse a meter bridge or potentiometer for a CBSE practical or JEE Advanced question, you already have the tools. What follows is the deeper view: why null methods are fundamentally more accurate than deflection methods, the sensitivity of the balance condition, and two generalisations (the Carey-Foster bridge and the AC potentiometer).

Why null methods beat deflection methods — the Wheatstone argument

A deflection-based ohmmeter measures resistance by passing a known current through the unknown and reading the galvanometer deflection. The accuracy depends on:

  1. The stability of the reference current (must be constant).
  2. The linearity of the galvanometer (deflection proportional to current).
  3. The accuracy of the galvanometer's calibration.
  4. The temperature coefficient of the galvanometer's springs.
  5. Zero-error of the galvanometer.
  6. Non-uniformity of the magnetic field the coil sees.

All six must be accurate for the reading to be trustworthy.

A null method eliminates every single one of these. At balance, the galvanometer reads zero. It doesn't matter if its scale is nonlinear (you're not using the scale). It doesn't matter if it has a zero-error (you're reading zero). It doesn't matter if it is sensitive or insensitive (as long as it reads zero when zero current flows). The only thing that matters is that the galvanometer correctly identifies the direction of current — which way it deflects on either side of balance — and that information is qualitative, not quantitative.

The accuracy of the null method is thus transferred from the galvanometer to the standards — the known resistance S, the known length \ell, the known standard cell. Those can be manufactured to much tighter tolerances than any galvanometer.

Sensitivity of the balance condition

How precisely can you locate the balance point? The sensitivity of the meter bridge is defined as the smallest detectable deflection of the galvanometer per unit displacement of the jockey.

Let \Delta I_g be the galvanometer current for a jockey displacement \Delta\ell from balance. A careful Kirchhoff analysis (done in the Wheatstone-bridge article) gives

\frac{dI_g}{d\ell} \;\propto\; \frac{\varepsilon \cdot r\,\Delta R}{\text{(sum of arm resistances)}^2 \cdot R_G},

where R_G is the galvanometer resistance and \Delta R is the imbalance. The sensitivity is maximised when:

  1. The four arms are equal — i.e., balance near \ell = 50 cm with R \approx S \approx r\ell/2. (This is also why the end-correction argument pointed at 50 cm.)
  2. The galvanometer resistance matches a specific combination of the arm resistances (a formal result from Wheatstone analysis).
  3. The driver EMF is as large as safely tolerable.

The practical implication: always tune S to put the balance near 50 cm, and always use a driver cell that gives as much current as the wire and galvanometer can handle without overheating or saturating.

Carey-Foster bridge — eliminating end-corrections completely

The meter bridge has a systematic error from the end-correction: a small but non-zero resistance in the copper terminals at either end, which effectively shifts the "true" 0 cm and 100 cm points by a few millimetres. Taking the mean of two readings (with positions swapped) cancels this to leading order but not to higher order.

The Carey-Foster bridge is a modification that eliminates the end-correction entirely. It uses four equal arms with the unknown and a known standard resistor in the bottom two arms (along the wire), and two equal fixed resistors in the top two arms. Balance the bridge once with the unknown on the left; then interchange the unknown and the known, and balance again. The difference (\ell_1 - \ell_2) in balance lengths is now directly the difference (R - S) divided by the resistance per cm — entirely free of end-corrections.

Carey-Foster measurements routinely reach 10^{-4} relative accuracy in length, which is better than most commercial resistance standards. This is the circuit used in India's National Physical Laboratory (NPL Delhi) for calibrating low-resistance standards.

Beyond DC — the AC potentiometer

Everything in this article assumed steady DC. Can you build an AC potentiometer to measure AC voltages? Yes, but with two complications:

  1. The "standard cell" becomes a standard AC source — typically a 50 Hz mains-derived reference or a quartz oscillator.
  2. The galvanometer becomes a phase-sensitive detector — you need zero in-phase component AND zero quadrature component for a null, so you need a two-dimensional balance (magnitude and phase).

The AC potentiometer is the ancestor of the lock-in amplifier used in every physics research lab today. The lock-in is a DC null method in the rotating frame of a reference oscillator — exactly a potentiometer, but with the "wire" replaced by an internal multiplier and the "jockey" replaced by a tuneable phase shifter.

The potentiometer is a voltage source in two opposing forms

A subtle but beautiful point: a potentiometer wire with a sliding tap creates a continuously adjustable voltage divider — pull off a tap anywhere and you get a voltage k\ell relative to one end. The instrument used for measurement is this device running in comparator mode (balance against an external cell). But the same device in divider mode is what a volume-control knob on a stereo is — or, in higher-power versions, what a variable-speed drive uses to control a motor. Rotate the knob; the tap slides along a helical wire; the voltage to the speaker decreases. Same physics, different use case.

In circuit diagrams the two usages share a symbol (the resistor with a sliding arrow). Whenever you see that symbol, mentally ask: is this a divider being used to adjust a downstream voltage, or a comparator being used to null against a reference? Both roles exist in the potentiometer chapter's instrument.

Where this leads next