ARCHETYPE: Potentiometer Numericals

A voltmeter tells you a cell's voltage the way a leaky bucket tells you how much water a tap delivers: the act of measuring steals some of what it measures. Connect any voltmeter across a battery and a small current flows, the internal resistance takes its cut, and the reading lands below the true EMF — always, unavoidably. The potentiometer's answer is to measure by refusing to draw current at all: slide a contact along a long current-carrying wire until the wire's potential drop exactly cancels the cell's push, watch the galvanometer sit at zero, and read off a length. At that null, the cell is untouched, unloaded, honest — and the length is the EMF, in disguise. The principle, construction, and the reason a null beats a needle all live in the companion concept article, The Potentiometer. This article is the numericals gym built on top of it. ISC papers ask you to compare two EMFs from two balance lengths, to standardize the wire with a known cell and then measure an unknown, and — the crown jewel — to extract a cell's internal resistance from the shift in balance length when a shunt is thrown across it. CIE A-Level keeps the potentiometer alive as its favourite null method for comparing potential differences. Every one of these questions is the same three-symbol equation, $\varepsilon = k\,l$ , written twice and divided. Master the four setups here — comparison, standardization, internal resistance, and the diagnosis of a balance that refuses to appear — and the potentiometer becomes the most predictable question on the paper: a ratio, a subtraction, and marks.

Key takeaways

Everything runs on the potential gradient $k$ (volts per metre of wire): at balance against a cell on open circuit, $\varepsilon = k\,l$ .
Comparison of two EMFs needs no knowledge of $k$ : $\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{l_1}{l_2}$ — same wire, same current, $k$ cancels.
Standardization = using a known cell to measure $k$ ( $k = \varepsilon_s/l_s$ ), then reading any unknown as $\varepsilon = k\,l$ .
Internal resistance from two balances of the same cell: open circuit $l_1$ , then shunted by $R$ : $r = R\left(\dfrac{l_1}{l_2} - 1\right)$ .
Smaller gradient $k$ (longer wire or bigger series resistor) stretches every balance length — better resolution, same physics.
No balance anywhere on the wire means the driver is too weak ( $k L < \varepsilon$ ) or a polarity is reversed — a diagnosis question, not a calculation.
Key idea: every potentiometer numerical is $\varepsilon = k\,l$ written once per configuration; ratios kill $k$ , and the shunted-cell shift hands you the internal resistance.

The core insight

Because a balanced potentiometer draws zero current from the cell under test, each balance length is an honest sample of an EMF or terminal voltage via $\varepsilon = k\,l$ — so ratios of balance lengths equal ratios of voltages, and the drop from open-circuit length $l_1$ to shunted length $l_2$ measures internal resistance as $r = R\,(l_1/l_2 - 1)$ .

Where this lives today

The metre of wire is gone; the idea — measure by nulling, draw nothing from the source — never left. A modern digital multimeter on its voltage range presents an input resistance of ten megohms or more precisely to approximate the potentiometer's zero-current ideal, and precision "potentiometric" measurements in electrochemistry (pH meters, ion-selective electrodes) still insist on drawing femtoamps from cells whose chemistry would be wrecked by real current. In national metrology labs, the ghost of the standardization drill runs on: an unknown voltage is compared against a Josephson-junction array (the quantum descendant of the Clark cell) by a null detector — the same two-balance logic, executed in superconducting hardware. And the word itself survives in every volume knob: the three-terminal "potentiometer" component is Clark's wire bent into a circle, dividing potential by position exactly as $k\,l$ divided it along a board.

The setup

The stage is always the same, and drawing it correctly is the first mark. A driver cell (EMF $\varepsilon_d$ , sometimes with a series rheostat) pushes a steady current $I$ down a long uniform wire $AB$ — often several metres, folded on a board. That current makes the potential fall uniformly along the wire: the potential gradient

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

where: $V_{AB}$ — the potential drop across the wire itself (not the driver's EMF — the rheostat and the driver's internal resistance eat their share first); $L$ — the wire's length. Units: volts per metre, or per centimetre when balance lengths are quoted in cm. The gradient is the instrument's calibration constant, and every mistake in this archetype is ultimately a mistake about $k$ . The cell under test ( $\varepsilon_1$ ) connects between $A$ and a galvanometer that ends in a jockey, positive terminal to $A$ on both cells — the two positives fight. Tap along the wire: where the wire's drop from $A$ exactly equals $\varepsilon_1$ , no current flows through the galvanometer, and the balance condition is simply

\varepsilon_1 = k\,l_1 \qquad (2)

where: $l_1$ — the balance length from $A$ . The cell is on open circuit at this instant — zero current drawn — which is why equation (2) contains the true EMF and not some loaded terminal voltage. That single fact, proved and celebrated in The Potentiometer, powers every numerical below.

Figure

The potentiometer's two working setups — EMF comparison and internal resistance

The master solution pattern

Every potentiometer numerical is a bookkeeping exercise on equation (2). The drill:

Step 1. Establish the gradient's status. Is $k$ known (wire drop and length given)? To be found (standardization)? Or destined to cancel (comparison)? This decides the whole route.

Step 2. Write $\text{voltage} = k \times \text{length}$ once per configuration — one line per balance reading. Label each voltage honestly: true EMF if the test cell is on open circuit, terminal voltage if something is drawing current from it.

Step 3. Divide equations to kill $k$ (comparison, internal resistance), or compute $k$ first and substitute (standardization).

Step 4. Sanity-check with the lever: bigger EMF, longer balance; loaded cell, shorter balance (terminal voltage < EMF, always).

Comparing two EMFs

Balance cell 1, then swap in cell 2 by a two-way key without touching the driver circuit. Same current, same wire, same $k$ :

\frac{\varepsilon_1}{\varepsilon_2} = \frac{k\,l_1}{k\,l_2} = \frac{l_1}{l_2} \qquad (3)

where: the entire instrument has cancelled out of its own measurement. This is why the comparison form needs no standardization and why examiners can set it with shameless three-line brevity. If one EMF is known, the other follows immediately — that is standardization by stealth.

Internal resistance by the shunted cell

The cell under test balances twice. First with its shunt key open: no current drawn, the balance samples the true EMF, $\varepsilon = k\,l_1$ . Then the key closes a known resistance $R$ across the cell: the cell now drives a current $I = \varepsilon/(R + r)$ around its private loop, its terminal voltage sags to $V = \varepsilon - I r = \varepsilon R/(R+r)$ , and the potentiometer — still drawing nothing itself at null — dutifully reports the sag: $V = k\,l_2$ .

Derivation AInternal resistance from two balance lengths

We derive $r = R\,(l_1/l_2 - 1)$ , the workhorse of ISC internal-resistance experiments. The strategy: write the balance condition for the same cell in its two states — open circuit (true EMF) and shunted by a known $R$ (terminal voltage) — then divide, so the unknown gradient $k$ disappears and only measured lengths and the known $R$ remain. The subtlety to watch: at both balances the potentiometer draws zero current, so in the shunted state the only current through the cell is the one flowing around the cell– $R$ loop, and the ordinary EMF & Internal Resistance analysis of that loop applies untouched.

Step 1. Key open. The cell is on open circuit; balance samples the full EMF:

\varepsilon = k\,l_1

Step 2. Key closed. The cell drives $I = \dfrac{\varepsilon}{R + r}$ through its own loop, so its terminal voltage is

V = \varepsilon - I\,r = \varepsilon\,\frac{R}{R + r}

Why is the loop current undisturbed by the potentiometer? Because at null the galvanometer branch carries nothing — the potentiometer is a perfect, infinitely-polite voltmeter across the cell's terminals.

Step 3. The new balance reads that terminal voltage:

\varepsilon\,\frac{R}{R+r} = k\,l_2

Step 4. Divide Step 1 by Step 3 — $k$ and $\varepsilon$ both cancel:

\frac{R + r}{R} = \frac{l_1}{l_2} \;\;\Rightarrow\;\; r = R\left(\frac{l_1}{l_2} - 1\right)

Why the formula is trustworthy: it contains only a resistance box value and a ratio of lengths — no meter readings, no gradient, no driver details. Every hard-to-know quantity has been divided away. *Verified: limits behave — a perfect cell ( $r = 0$ ) gives $l_2 = l_1$ (no sag); $R \to \infty$ (barely loaded) also gives $l_2 \to l_1$ ; halving $l_2$ at fixed $R$ means $r = R$ . Numbers from Example 2: $R = 9.5\ \Omega$ , $l_1 = 76.3$ cm, $l_2 = 64.8$ cm give $r = 9.5(76.3/64.8 - 1) = 1.69\ \Omega$ ✓.*

The gradient as a design variable

change to the driver circuit	effect on $k$	effect on balance lengths
increase series rheostat resistance	$k$ drops	all lengths stretch (finer resolution)
longer wire, same wire drop	$k$ drops	lengths stretch
driver cell runs down	$k$ drops slowly	lengths creep longer during the session
$k\,L < \varepsilon_{\text{test}}$	—	no balance anywhere: null falls off the wire
test-cell positive to the wrong terminal	—	galvanometer deflects one way at both ends

Figure

The gradient lever — one cell, two driver settings, and the no-balance failure

The last two rows are the diagnosis family — questions with no formula, only understanding. One-way deflection at both ends of the wire means the two cells are aiding instead of opposing (polarity error) or the wire's total drop cannot reach the test EMF (driver too weak). Both appear regularly as reasoning questions on ISC papers and as "suggest why no balance point is found" on CIE practical-skills papers.

Worked examples

Board level — comparing two EMFs (ISC pattern, 3 marks)

In a potentiometer experiment, cell 1 balances at $l_1 = 350$ cm and cell 2 at $l_2 = 280$ cm with the same driver setting. (a) Find $\varepsilon_1/\varepsilon_2$ . (b) If cell 1 is a standardized cell of EMF $1.50$ V, find $\varepsilon_2$ .

Step 1. Same wire, same current, same $k$ — equation (3) applies:

\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2} = \frac{350}{280} = \boxed{1.25}

Step 2. With $\varepsilon_1 = 1.50$ V known:

\varepsilon_2 = 1.50 \times \frac{280}{350} = \boxed{1.20\ \mathrm{V}}

Step 3. Sanity: the shorter balance belongs to the smaller EMF ✓. Note what was never needed: the driver EMF, the wire's resistance, the gradient. Ratios are the potentiometer's native language.

Board level — internal resistance of a cell (ISC pattern, NCERT-classic numbers)

A cell balances at $76.3$ cm on open circuit. When a resistance of $9.5\ \Omega$ is connected across the cell, the balance point shifts to $64.8$ cm. Find the internal resistance of the cell.

Step 1. The shift is the physics: loading the cell sagged its terminal voltage from $\varepsilon$ to $V$ , and the potentiometer faithfully reported the sag as a shorter length. Expect $r > 0$ .

Step 2. Apply the two-balance formula:

r = R\left(\frac{l_1}{l_2} - 1\right) = 9.5\left(\frac{76.3}{64.8} - 1\right)

Step 3. Compute the ratio first, then subtract: $76.3/64.8 = 1.1775$ , so $r = 9.5 \times 0.1775 = \boxed{1.7\ \Omega}$ (two significant figures, as the data deserve).

Step 4. Cross-check by reconstructing: with $\varepsilon = k\,l_1$ and $V = k\,l_2$ , the sag fraction is $V/\varepsilon = 64.8/76.3 = 0.849$ , and $R/(R+r) = 9.5/11.2 = 0.848$ ✓.

Competitive level — designing the gradient, then reading a cell (A-Level style, ~3 minutes)

A potentiometer wire is $4.00$ m long with resistance $8.0\ \Omega$ . The driver is a $3.0$ V cell of internal resistance $0.50\ \Omega$ in series with a rheostat set to $R_s$ . (a) Find $R_s$ so the potential gradient is $0.50$ V/m. (b) Where does a $1.50$ V cell then balance? (c) Explain whether a $2.5$ V cell can be measured at this setting.

Step 1. Demand the wire drop: $V_{\text{wire}} = k\,L = 0.50 \times 4.00 = 2.0$ V.

Step 2. The wire current follows from the wire's own resistance: $I = V_{\text{wire}}/R_{\text{wire}} = 2.0/8.0 = 0.25$ A.

Step 3. The driver loop must push that current through everything: $\varepsilon_d = I\,(R_s + R_{\text{wire}} + r_d)$ , so

R_s = \frac{3.0}{0.25} - 8.0 - 0.50 = 12.0 - 8.5 = \boxed{3.5\ \Omega}

Step 4. Balance of the $1.50$ V cell: $l = \varepsilon/k = 1.50/0.50 = \boxed{3.00\ \mathrm{m}}$ — comfortably on the 4 m wire.

Step 5. The wire's total drop is only $2.0$ V. A $2.5$ V cell needs $l = 5.0$ m of this gradient — off the end of the wire. No balance exists; the galvanometer deflects the same way even at $B$ . Fix: reduce $R_s$ to raise $k$ (coarser but reaching higher), or accept the diagnosis as the answer. Timing note: ~3 minutes, and part (a) is the whole test — the classic error is setting the full $3.0$ V across the wire and "finding" $R_s$ from nothing. The driver's EMF divides between rheostat, wire, and internal resistance; only the wire's share defines $k$ . Parts (b) and (c) are ten-second reads once $k$ is trusted.

Competitive level — potentiometer versus voltmeter, as a numerical (ISC reasoning-plus-numbers, ~3 minutes)

A cell has EMF $1.500$ V and internal resistance $2.0\ \Omega$ . (a) A voltmeter of resistance $200\ \Omega$ is connected across the cell. What does it read? (b) The same cell is balanced on a potentiometer with gradient $k = 0.500$ V/m. Find the balance length and the reading error of each method. (c) With lengths readable to $\pm 2$ mm, what voltage resolution does the potentiometer achieve?

Step 1. The voltmeter loads the cell: current $I = \dfrac{1.500}{200 + 2.0} = 7.43$ mA, and the reading is the terminal voltage

V = \varepsilon\,\frac{R_V}{R_V + r} = 1.500 \times \frac{200}{202} = \boxed{1.485\ \mathrm{V}}

— low by $15$ mV, a systematic error of $\approx 1.0\%$ that no care in reading can remove.

Step 2. The potentiometer at null draws zero current: it balances the true EMF at

l = \frac{\varepsilon}{k} = \frac{1.500}{0.500} = \boxed{3.000\ \mathrm{m}}

with zero loading error, by construction.

Step 3. Resolution: $\pm 2$ mm of length at $0.500$ V/m is $\pm 0.500 \times 0.002 = \pm 1$ mV. The voltmeter's systematic $15$ mV error is fifteen times the potentiometer's whole reading uncertainty.

Step 4. The exam sentence worth both marks: the voltmeter's error is systematic and grows with $r/R_V$ ; the potentiometer's null method draws no current at balance, so it reads the true EMF, limited only by length resolution. *Timing note: ~3 minutes. Notice the anatomy — part (a) is the EMF & Internal Resistance loop analysis, part (b) is one division, part (c) is one multiplication. The physics content is knowing which voltage each instrument reports. This comparison reappears from the other side in Real Meters — Loading Effects & Reading Errors, next in the gym.*

Exam traps

Terminal voltage masquerading as EMF. The moment anything draws current from the test cell — a closed shunt key, a voltmeter left connected — the balance length reads terminal voltage, not EMF. The open-circuit reading, and only that one, is $\varepsilon$ .

Gradient from the wrong voltage. $k$ is the drop across the wire divided by the wire's length — never the driver's EMF divided by the length. The rheostat and the driver's internal resistance take their cut first (Example 3's whole point).

Ratio upside down. In $r = R(l_1/l_2 - 1)$ , $l_1$ is the open-circuit (longer) length. Inverting the ratio produces a negative $r$ — instantly recognizable nonsense; treat it as a built-in alarm, fix the ratio, move on.

Units in $k$ . Volts per metre versus volts per centimetre. A balance quoted at $350$ cm with $k$ in V/m needs the conversion before multiplying — the classic ISC half-mark leak.

Touching the rheostat between readings. Any driver change reprices every length: comparisons and standardizations are valid only within one driver setting. If a question says the rheostat was adjusted between balances, the lengths cannot be ratioed — that is the answer.

No-balance diagnosis. One-way deflection at both ends = reversed test-cell polarity or $kL < \varepsilon$ . Say which and why; a formula is not being asked for.

Boards. ISC 12 carries the potentiometer in full (comparison + internal resistance experiments and their numericals). CIE A-Level (9702) keeps it as the standard null method for comparing potential differences in its DC-circuits practical repertoire. CBSE removed the potentiometer in the 2023-24 rationalization and JEE-Main/NEET followed — for those exams this is legacy material, which is exactly why it remains fair game (and beloved) on ISC and CIE papers.

Origins · 1 minLatimer Clark and how this was discovered

Latimer Clark

1822–1898 · Telegraph engineering, London

Latimer Clark spent his career making electricity measurable. Chief engineer of the Electric Telegraph Company and later a partner building submarine cables across the world's oceans, he learned early that a telegraph network is unmanageable if no two stations agree on what a volt is. With Charles Bright in 1861 he proposed to the British Association that electrical units be named after great experimenters — ohm, volt, farad — the very words on every exam paper since. But names need embodiments: in 1873 he introduced the Clark standard cell, a zinc–mercury chemistry so reproducible that any laboratory could build one and hold $1.4328$ volts at $15\,^\circ$ C. For a generation, "one volt" meant a stated fraction of a Clark cell — and the potentiometer was the instrument that transferred that standard onto working circuits, comparing any unknown EMF to the Clark cell by the ratio of two balance lengths, no current drawn from either. The 1893 Chicago International Electrical Congress wrote the Clark cell into the legal definition of the volt. Every standardization numerical in this article — balance the known, price the wire, read the unknown — is Clark's metrological legacy in miniature.

History

1841

— Poggendorff's compensation method shows an EMF can be measured by opposing it until no current flows

1861

— Clark and Bright propose naming units after experimenters: the ohm, volt, and farad enter the language

1873

— Clark introduces his standard cell; balance lengths can now be priced in reproducible volts

1893

— Chicago congress defines the legal volt via the Clark cell; potentiometers become the transfer instruments of world metrology

1908

— the flatter-tempered Weston cell supersedes the Clark; slide-wire potentiometry rules precision measurement for another half-century

1990

— the Josephson effect redefines the practical volt; the null method survives inside every self-balancing digital instrument

Going further

Texts. ISC Physics Class 12 (Nootan/Nageen), Current Electricity — potentiometer experiments · NCERT Physics Class 12 Ch. 3 (legacy editions; retained in many lab manuals) · CIE 9702 syllabus, DC circuits and practical skills (null methods) · S.L. Arora, New Simplified Physics, potentiometer numericals bank.

Provenance. Authored fresh for the HS E&M track's Current Electricity archetype gym; principle and construction cited from the lane-mate concept module The Potentiometer; loop analysis of the shunted cell cited from the sibling module EMF, Terminal Voltage & Internal Resistance. A verifier should recompute: Example 1's ratio $1.25$ and $\varepsilon_2 = 1.20$ V; Example 2's $r = 9.5(76.3/64.8 - 1) = 1.69 \approx 1.7\ \Omega$ and the sag cross-check $0.849$ vs $0.848$ ; Example 3's $I = 0.25$ A, $R_s = 3.5\ \Omega$ , balance $3.00$ m, and the no-balance verdict for $2.5$ V; Example 4's voltmeter reading $1.485$ V ( $1.0\%$ low), balance $3.000$ m, resolution $\pm 1$ mV.

Prerequisites. The Potentiometer (principle, construction, why null beats needle) · EMF, Terminal Voltage & Internal Resistance ( $V = \varepsilon - Ir$ ). Leads to. ARCHETYPE: Real Meters — Loading Effects & Reading Errors (what loading does when you don't null) · Potential Dividers (the loaded-divider mathematics from the other side).

Why must both positive terminals connect to the same end A?

Because the null exists only if the two sources oppose each other around the galvanometer loop. Positive-to-positive at $A$ means the wire's potential drop pushes current one way through the galvanometer while the test cell pushes the other way; somewhere along the wire the two pushes cancel — that point is your balance. Connect the test cell backwards and the pushes add: current flows the same direction through the galvanometer no matter where you tap, the deflection never reverses, and no null exists anywhere. That one-way deflection is the fingerprint examiners want named in diagnosis questions.

Zero current at balance — so why does the jockey position matter at all?

Zero current is the destination, not the journey. Off balance, the mismatch between $k\,l$ and the cell's EMF drives a current through the galvanometer whose direction tells you which way to move. The measurement is a bisection hunt: deflection left at 240 cm, right at 260 cm, so the null hides between. At the null itself the cell is momentarily on open circuit — which is precisely when its terminal voltage equals its EMF, and why the length you record prices the true $\varepsilon$ rather than some loaded, sagging voltage.

Where does r = R(l1/l2 − 1) actually come from — can I avoid memorizing it?

You can rebuild it in three lines, and should at least once. Open key: $\varepsilon = k l_1$ . Closed key: the cell drives current only around its private loop with $R$ , so its terminal voltage is $\varepsilon R/(R+r)$ , and the balance reports it: $\varepsilon R/(R+r) = k l_2$ . Divide the two equations — $k$ and $\varepsilon$ both vanish — leaving $(R+r)/R = l_1/l_2$ , i.e. $r = R(l_1/l_2 - 1)$ . The derivation is the memory aid: longer-over-shorter, minus one, times the shunt.

Does the potentiometer really draw no current, ever?

At exact balance, from the test cell — yes, exactly zero: that is what the galvanometer's null certifies. But two honest caveats. While you hunt for balance, small currents do flow through the test cell (this is why you tap briefly rather than hold). And the driver circuit carries a steady working current the whole time — the wire is warm with it; zero-current applies to the measured branch only. In real experiments the cell's chemistry recovers between taps, and the recorded balance is taken as a clean open-circuit sample.

Why is a longer potentiometer wire better if the physics only needs a ratio?

Resolution. Your eye and jockey resolve about a millimetre of wire regardless of length; the question is how many volts that millimetre is worth. A ten-metre wire carrying a 2 V drop has $k = 0.2$ V/m, so a millimetre is $0.2$ mV — while a one-metre wire at the same drop prices the same millimetre at ten times more. Stretching the wire (or lowering $k$ with the rheostat) spreads the volts over more centimetres, sharpening every reading. The limit: $k L$ must still exceed the largest EMF you plan to balance, or the null falls off the end.

What actually goes wrong if the driver cell runs down during the experiment?

The gradient $k$ sags slowly, so every balance length creeps longer as the session proceeds — and a comparison between a length measured early and one measured late is quietly poisoned. This is why careful procedure standardizes at the start and re-checks at the end: balance the standard cell, do the measurements, balance the standard again. If the two standardizations disagree beyond reading error, the run is discarded. A-Level practical papers love asking why the standard cell is re-balanced last; this drift is the answer.

Is the lab potentiometer related to the volume-knob potentiometer in electronics?

Directly — same word, same principle, different packaging. The knob component is a resistive track with a sliding wiper: the wiper picks off a fraction of the voltage across the track, exactly as the jockey picks off $k\,l$ volts of the wire's drop. The lab instrument adds a galvanometer and a standard so the picked-off voltage can be measured rather than merely used. In audio gear the wiper feeds an amplifier of enormous input resistance — the same draw-no-current ideal — so the divider ratio, like the balance condition, stays honest under load.