Metre Bridge and Potentiometer Experiments

In short

The metre bridge finds an unknown resistance by balancing a Wheatstone bridge on a one-metre uniform wire. The balance-length reading \ell (in cm from the left end) against a known S gives

\boxed{\;R \;=\; S\cdot\frac{\ell}{100-\ell}\;}\qquad\text{(unknown in left gap).}

To cancel the copper end-corrections, repeat with the two resistors swapped (balance now at \ell') and take the mean of the two values. The experiment is most accurate when S is chosen so the balance falls near 50 cm.

The potentiometer compares two EMFs by balancing each against the drop along a common uniform wire driven by a steady current. If \varepsilon_1 balances at \ell_1 and \varepsilon_2 balances at \ell_2,

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

To find the internal resistance r of a cell, record the open-circuit balance length \ell_1 and the loaded balance length \ell_2 with a shunt resistor R across the cell:

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

Both experiments are null methods: the answer comes from a length reading at the moment the galvanometer shows zero deflection, so the galvanometer's own calibration is irrelevant. A wooden metre bridge from a 1970s CBSE lab can therefore beat a ₹5000 digital multimeter — the design is immune to the error sources that plague direct-reading instruments.

Open the practical drawer of any government school in India and you will find the same two wooden boards: one a metre long with a shiny wire stretched taut above a brass scale, a jockey and a galvanometer at the side; the other longer — often four metres laid out in four parallel strips — with two plug-key gaps, a rheostat, and an accumulator cell. The first is a metre bridge. The second is a potentiometer. Between them these two instruments account for three of the six mandatory Class 12 practicals and at least one JEE Advanced lab question every year.

What both experiments have in common is the null method. Instead of measuring a current or a voltage, you slide a jockey along a wire until the galvanometer reads zero, and then you read a length off a scale. Length measurements on a metre rule are routinely accurate to 0.1%, far better than any galvanometer can report a current. And at the moment of zero deflection, the galvanometer's sensitivity, its zero-error, and its internal resistance are all irrelevant — the answer cannot be spoiled by the instrument that discovers it.

The physics of both instruments — the Wheatstone-bridge balance condition and the potential gradient along a uniform wire — is already covered in the Metre Bridge and Potentiometer article. This article is about the experiment itself: how to set it up, what to measure, what errors to watch for, and what to write in the observation table.

Measuring an unknown resistance with a metre bridge

What you have on the bench

A 100 cm uniform nichrome (or constantan) wire, about 0.3 mm thick, soldered between two copper end-strips labelled A (left) and C (right). Two small gaps in the copper — the left gap and the right gap — carry the resistance coils. A cell (typically a single Leclanché cell of about 1.5 V) with a plug-key drives current through the top rail. A galvanometer is connected from the copper midpoint B to a spring-loaded jockey that can touch the wire at any point.

The resistance box on the desk contains precisely-made coils of 1 Ω, 2 Ω, 5 Ω, 10 Ω, 20 Ω, 50 Ω, 100 Ω wired to plug-keys. Pulling out a plug switches that coil into the circuit; pushing it in short-circuits the coil. You build up any resistance from 1 Ω to 200 Ω in 1 Ω steps by choosing which plugs to pull.

The metre-bridge apparatus. Cell and key on the left drive current through the top rail via the unknown R, midpoint B, and the known S. A galvanometer from B to a sliding jockey J touches the wire along its length; slide J until the galvanometer reads zero, then read off the balance length $\ell$.

The procedure — step by step

You are about to find a coil of manganin wire whose nominal resistance is somewhere between 1 Ω and 10 Ω. Here is what you do.

Step 1. Connect the unknown R in the left gap and the resistance box S in the right gap. Tighten the terminal screws until the copper leads make firm contact — a loose screw adds contact resistance that shifts the balance.

Step 2. Pull a rough plug from the box — try S = 2\ \Omega. Close the key.

Step 3. Press the jockey gently near \ell = 20 cm and note the galvanometer deflection (say, to the left). Move the jockey to \ell = 80 cm and check the deflection (it should be to the right). If the deflection stays on the same side across the whole wire, the balance point is outside the wire — you need to change S.

Step 4. Once the two ends give opposite deflections, the balance point lies somewhere in between. Halve and re-halve: try 50, then 35, then 42, homing in on the point of zero deflection. Call this the balance length \ell.

Step 5. Lift the jockey, verify the balance once more by approaching from both sides, and record \ell to the nearest millimetre on the metre scale.

Step 6. Compute R = S\cdot\ell/(100-\ell).

Step 7. Now swap the unknown and the known — put R in the right gap and S in the left. Find the new balance length \ell'. Compute R' = S \cdot (100 - \ell')/\ell'.

Step 8. Take the mean of R and R'. This is your measurement of the unknown.

Step 9. Change S (say, to 5 Ω) and repeat from Step 3. Record a fresh R and R'. Three or four such rows give the observation table you submit.

Why press the jockey gently

The jockey is a sharp copper contact. Press it too hard onto the wire and three things go wrong.

You indent the wire, bending its cross-section and changing its resistance per cm locally. The wire is no longer uniform at that point — your next balance will read the wrong length.
You scratch the wire over time. The groove collects oxide, which raises the local resistance. Over the month of practicals, a heavily-used metre bridge develops bright bands at the favourite balance points where the wire is thinner than the average.
You heat the contact, because current flowing through a small contact patch develops a large local I^2R. A hot contact drifts in resistance during the measurement.

The discipline is: touch the jockey down, read the deflection, lift the jockey, slide to the next position, touch down again. Never drag the jockey along the wire while it is pressed down. On a well-kept metre bridge, the wire lasts decades. On a mistreated one, it develops a visible dip near 50 cm within a semester.

Why the balance should be near 50 cm — the sensitivity argument

The bridge formula is R = S\cdot\ell/(100-\ell). Differentiate with respect to \ell to see how the measured R responds to a small error d\ell in the balance length:

\frac{dR}{d\ell} \;=\; S\cdot\frac{(100-\ell)\cdot 1 - \ell\cdot(-1)}{(100-\ell)^2} \;=\; \frac{100\,S}{(100-\ell)^2}.

Why: quotient rule applied to \ell/(100-\ell), then the 100 in the numerator comes from combining the two terms. The denominator (100-\ell)^2 is what makes the slope blow up as \ell \to 100.

Divide by R = S\ell/(100-\ell) to get the relative error in R per unit length error:

\frac{1}{R}\cdot\frac{dR}{d\ell} \;=\; \frac{100\,S}{(100-\ell)^2}\cdot\frac{100-\ell}{S\,\ell} \;=\; \frac{100}{\ell(100-\ell)}.

Why: the factors of S cancel, and one power of (100-\ell) cancels. The result depends only on \ell — not on S or R.

The function \ell(100-\ell) is a downward-opening parabola, maximised at \ell = 50. At that point \ell(100-\ell) = 2500, and a 1 mm error in \ell gives a 0.1/2500 \times 100 = 0.004 = 0.4\% relative error in R.

At \ell = 90, \ell(100-\ell) = 900, and the same 1 mm error gives 0.1/900 \times 100 \approx 1.1\% — nearly three times worse. At \ell = 95 the error ratio is about 2\%, and the experiment becomes dominated by your ability to see the jockey.

This is why, after a first rough balance at \ell = 25 cm, you change S to shift the expected balance toward the middle. Choose S so that S \approx R — then R/S \approx 1 and the formula gives \ell \approx 50 cm.

Relative error in $R$ for a 1 mm uncertainty in the balance length. The error is smallest at $\ell = 50$ cm (0.4%) and triples by $\ell = 25$ or $\ell = 75$. Always choose $S$ so the balance lands near the middle of the wire.

Swapping and the end-correction

The theoretical balance formula assumes that the wire runs from exactly 0 cm to exactly 100 cm. Real metre bridges have small copper end-pieces at A and C; the current has to travel a few millimetres through these before it reaches the wire proper. These lengths contribute extra resistance — the end-corrections \alpha (at the A end) and \beta (at the C end). In the balance condition they appear as small extra lengths:

\frac{R}{S} \;=\; \frac{\ell + \alpha}{(100 - \ell) + \beta}.

These \alpha and \beta are not known — they are part of the instrument's systematic error. The trick to eliminate them is to do the same measurement with R and S swapped. When R sits in the right gap,

\frac{S}{R} \;=\; \frac{\ell' + \alpha}{(100 - \ell') + \beta}.

Take the mean of the two values of R you compute from the two balance lengths. To leading order in the small corrections, the \alpha and \beta errors are equal and opposite between the two readings, and they cancel in the mean. This is why every lab manual instructs you to do the experiment "both ways and average."

An even more careful protocol (used in IIT labs) determines \alpha and \beta separately by running the experiment with two known resistors in the gaps, solving for the corrections, and then applying them to the unknown. For CBSE Class 12 the mean-of-swapped-positions method is the standard.

Comparing two EMFs with a potentiometer

What the apparatus looks like

A four-metre uniform wire stretched taut along a wooden base, arranged in four parallel strips to save bench space. A driver cell (a 2 V lead-acid accumulator, or a Leclanché cell with a rheostat) drives a steady current along the whole wire. At the top of the board are two plug-keys: a main key connecting the driver, and a secondary key that selects which test cell (\varepsilon_1 or \varepsilon_2) is currently connected to the galvanometer. The galvanometer's other lead goes to a jockey that slides along the wire.

Potentiometer circuit for EMF comparison. The driver cell $\varepsilon_0$ maintains a steady current through the 400 cm wire via the rheostat Rh. A two-way key selects either $\varepsilon_1$ or $\varepsilon_2$ as the test cell, and the galvanometer leads to the jockey J. Positive terminal of each test cell points toward A (the high-potential end).

The procedure — step by step

You are to compare the EMFs of a fresh Leclanché cell (\varepsilon_1) and a slightly depleted dry cell (\varepsilon_2).

Step 1. Check that the driver cell's EMF and its terminal voltage across the wire (after the rheostat) are larger than the larger of \varepsilon_1 and \varepsilon_2. A 2 V accumulator driving a wire with potential gradient of ~0.4 V per metre is standard; this puts 1.6 V across the full wire, enough to balance any 1.5 V dry cell. If the gradient is too low, no balance exists anywhere along the wire.

Step 2. Connect the test cells with their positive terminals toward A (the high-potential end). This ensures that the wire's voltage drop and the test cell's EMF oppose each other in the galvanometer loop — the balance condition. If you reverse the polarity, the galvanometer deflects in the same direction no matter where the jockey sits, and no balance is found.

Step 3. Set the two-way key to \varepsilon_1. Close the driver key. Press the jockey at A — galvanometer deflects one way. Press at B — deflects the other way. Halve and re-halve to find the balance length \ell_1.

Step 4. Do not change the rheostat. Switch the two-way key to \varepsilon_2 and find the new balance length \ell_2.

Step 5. Compute \varepsilon_1/\varepsilon_2 = \ell_1/\ell_2.

Step 6. Repeat with the rheostat set to a different value (still larger than the test EMFs) and record fresh \ell_1, \ell_2. The ratio should be the same — if it drifts, something else is changing (contact resistance, driver cell temperature).

Why the rheostat must not move between readings

The balance condition \varepsilon = k\ell depends on the potential gradient k, which is set by the driver current through the rheostat. If you adjust the rheostat between the two balances, k changes. The ratio \ell_1/\ell_2 then no longer equals \varepsilon_1/\varepsilon_2 — you are comparing one cell at one gradient with another cell at another gradient.

This is the single most common error in the EMF-comparison experiment, and the examiner's favourite "spot the mistake" question. The rule in the lab book is written in boldface: between measurements in the same experiment, do not disturb the rheostat.

What if \varepsilon_1/\varepsilon_2 > \ell_1/\ell_2 by 2%?

Three possible reasons. First, contact resistance at the driver terminals has changed between the two measurements — clean the terminals. Second, the driver cell itself is sagging as the experiment runs (its EMF is dropping slightly as the accumulator discharges) — let it stabilise for a minute before each balance. Third, the jockey's copper contact oxidised between measurements — wipe the jockey tip and the wire with fine emery paper before the experiment starts.

Finding the internal resistance of a cell — the potentiometer method

The idea

A cell's terminal voltage V is less than its EMF \varepsilon when current flows through it, because of the internal Ir drop:

V \;=\; \varepsilon - Ir.

If you could measure the cell's EMF and its terminal voltage when a known current flows, you could solve for r. The potentiometer does both.

The procedure

Step 1. With the cell on open circuit (no external load), balance it on the potentiometer. Call this balance length \ell_1. Because no current flows through the cell at balance, you are reading the cell's true EMF: \varepsilon = k\ell_1.

Step 2. Now add an external shunt resistor R across the cell's terminals (say, R = 10\ \Omega). This draws a steady current I = \varepsilon/(R+r) from the cell. The cell's terminal voltage drops to V = \varepsilon - Ir = IR.

Step 3. Re-balance the potentiometer. The galvanometer is still at zero, so the galvanometer loop has no current — but the shunt loop carries current I. The balance length \ell_2 now reads the terminal voltage: V = k\ell_2.

Step 4. Compute r.

Deriving the internal-resistance formula

Divide the two expressions:

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

Why: the potential gradient k is the same in both measurements (driver current unchanged, rheostat untouched). It cancels in the ratio, so the ratio of EMF to terminal voltage equals the ratio of balance lengths.

Use V = IR and \varepsilon = I(R+r):

\frac{\varepsilon}{V} \;=\; \frac{I(R+r)}{IR} \;=\; \frac{R+r}{R} \;=\; 1 + \frac{r}{R}.

Why: both expressions describe the same current I through the shunt loop — I = \varepsilon/(R+r) sets the current, and V = IR is the drop across the shunt (which is what the potentiometer reads at the second balance).

Set the two ratios equal:

1 + \frac{r}{R} \;=\; \frac{\ell_1}{\ell_2} \quad\Longrightarrow\quad \frac{r}{R} \;=\; \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 final formula has only three measurable quantities on the right — the shunt R you chose, and the two balance lengths \ell_1 and \ell_2 you read. The cell's EMF itself has dropped out of the expression for r, which is remarkable: you find the internal resistance without ever writing down the EMF in volts. Length ratios do all the work.

Interactive: watch the balance shift as R changes

For a cell with $\varepsilon = 1.5$ V and internal resistance $r = 0.5\ \Omega$ on a wire with gradient $k = 0.005$ V/cm, the open-circuit balance stays at $\ell_1 = 300$ cm. As the shunt $R$ grows, the loaded balance $\ell_2$ climbs toward $\ell_1$ — a larger $R$ draws less current, so the $Ir$ drop shrinks. The readout recovers $r = R(\ell_1-\ell_2)/\ell_2$ to $\pm 0.01\ \Omega$ at every $R$.

Sensitivity, range, and limitations

Sensitivity — the smallest EMF change the potentiometer can detect

The galvanometer deflection at a given distance from balance is proportional to the voltage mismatch. If the mismatch is \Delta V, the galvanometer current is

I_G \;=\; \frac{\Delta V}{R_G + R_\text{wire seg} + r_\text{cell}},

where R_G is the galvanometer resistance and R_\text{wire seg} is the wire segment between jockey and A. The smallest detectable \Delta V is set by the galvanometer's minimum visible deflection — typically 1 mm on a lamp-and-scale arrangement, corresponding to about 10^{-7} A with a sensitive moving-coil instrument.

Converting back to \Delta V: if R_G + R_\text{wire seg} + r_\text{cell} \approx 100\ \Omega, then \Delta V_\text{min} \approx 10^{-7} \times 100 = 10^{-5} V. A potentiometer can resolve 10 μV — this is 100× better than a typical school digital voltmeter.

Range

Metre bridge: resistances from about 0.1\ \Omega (limited by contact resistance) to about 10^4\ \Omega (limited by galvanometer sensitivity — too little current flows to give a clean null).
Potentiometer: EMFs from about 10^{-3} V (below this, the galvanometer cannot find the null cleanly) to about 2 V (above this, the driver wire voltage cannot match). For higher voltages, use a voltage-divider attenuator first.

Limitations you should admit in the viva

Metre bridge fails for very low or very high R. Below 0.1 Ω, contact resistance at the terminal screws swamps the measurement — the fix is a Kelvin double bridge (not in CBSE syllabus). Above 10 kΩ, the bridge current is so small that no crisp null exists — use a digital multimeter in ohmmeter mode.
Metre bridge wire is not perfectly uniform. The manufacturer quotes a tolerance of \pm 0.2\% on the resistance per cm. A wire that has been pressed too hard develops local thin spots with higher resistance per cm. A \pm 0.2\% non-uniformity near the balance point translates directly into a \pm 0.2\% error in R.
Potentiometer requires a steady driver current. If the driver cell's EMF drifts during the experiment (dry cells do this as they age and as the ambient temperature changes), the potential gradient k drifts with it. You should record two pairs of (\ell_1, \ell_2) within a minute of each other and verify that the ratio is stable.
Both experiments assume a zero-resistance galvanometer at balance. At exactly zero deflection the galvanometer draws no current, so its internal resistance is irrelevant at that instant. But if you stop short of true null — say, the deflection is 2 mm instead of 0 — the reading is slightly off by a term proportional to R_G.

Worked examples

Example 1: Metre bridge — unknown manganin coil

A Class 12 student at a Kolkata ISC school sets up a metre bridge to measure the resistance of a manganin coil. With S = 3\ \Omega in the right gap and the coil in the left, the balance point is at \ell = 42.5 cm. With S and the coil swapped (coil in right gap, 3\ \Omega in left), the new balance is at \ell' = 57.3 cm. Find the resistance of the coil, estimate the end-correction error, and judge whether the student should have chosen a different S.

The same manganin coil measured with its position swapped. Balance shifts from 42.5 cm to 57.3 cm — the two readings bracket 50 cm, which is the mark of a well-chosen $S$.

Step 1. Coil in left gap, \ell = 42.5 cm.

R_1 \;=\; S\cdot\frac{\ell}{100-\ell} \;=\; 3 \times \frac{42.5}{57.5} \;=\; 3 \times 0.7391 \;=\; 2.217\ \Omega.

Why: the coil is in the left gap, so the standard metre bridge formula R = S\ell/(100-\ell) applies directly. Keep four significant figures for the intermediate calculation — you will average two values in a moment and rounding early propagates error.

Step 2. Coil in right gap, \ell' = 57.3 cm.

With positions swapped, the coil is now on the right; the left gap holds S = 3\ \Omega. The balance formula (\text{left})/(\text{right}) = \ell'/(100-\ell') becomes 3/R_2 = 57.3/42.7, so

R_2 \;=\; 3 \times \frac{100 - \ell'}{\ell'} \;=\; 3 \times \frac{42.7}{57.3} \;=\; 3 \times 0.7452 \;=\; 2.236\ \Omega.

Why: after swapping, the unknown sits on the right. The bridge condition is still (\text{left resistance})/(\text{right resistance}) = \ell'/(100-\ell'), but now the "left resistance" is S and the "right resistance" is R. Rearranging gives R = S(100-\ell')/\ell'.

Step 3. Take the mean.

R \;=\; \frac{R_1 + R_2}{2} \;=\; \frac{2.217 + 2.236}{2} \;=\; 2.227\ \Omega \;\approx\; 2.23\ \Omega.

Step 4. Estimate the end-correction.

The spread R_2 - R_1 = 0.019\ \Omega is a measure of the end-correction bias. Half this spread, \pm 0.010\ \Omega or about \pm 0.4\% of R, is the systematic error remaining before averaging. After averaging, the residual is much smaller — perhaps \pm 0.1\% or 0.002 Ω.

Why: if the two readings were perfectly consistent, they would give the same R exactly. Their difference measures the asymmetry between the two end-corrections at A and C. The mean-of-swapped-positions trick cancels the leading-order term; the residual (second-order) is roughly the square of the fractional end-correction times R itself.

Step 5. Choice of S.

The balance lengths 42.5 and 57.3 both sit comfortably in the 40–60 cm range, where sensitivity is near its best. The student's choice of S = 3\ \Omega was good: a smaller S would have pushed the first balance toward 55 cm and the second toward 45 cm — still fine; a larger S (say 10 Ω) would have given balances near 18 cm and 82 cm, where the relative error per mm is about twice as large.

Result. R = (2.23 \pm 0.01)\ \Omega.

What this shows. The metre bridge experiment's three error-reducing habits — choose S to centre the balance, press the jockey gently, and average over swapped positions — each contribute a factor of two or more to the precision. Without them, a CBSE practical lab returns answers good to about 2%. With them, the same student returns answers good to 0.1% — good enough to distinguish a manganin coil from a constantan coil of the same nominal resistance.

Example 2: Potentiometer — comparing two dry cells

A student at a Chennai government school uses a 400 cm potentiometer wire driven by a 2 V accumulator. A fresh dry cell (call it \varepsilon_1) balances at \ell_1 = 292 cm. A second dry cell from the same carton, left in a desk drawer for six months, balances at \ell_2 = 276 cm. If the fresh cell's EMF is known to be 1.460 V, find (a) the potential gradient k, (b) the EMF of the older cell, and (c) how much EMF the older cell has lost in storage.

The older cell balances at a shorter length — its EMF has dropped, so a shorter wire segment balances it.

Step 1. Find the potential gradient k from the known fresh cell.

k \;=\; \frac{\varepsilon_1}{\ell_1} \;=\; \frac{1.460}{292} \;=\; 0.005000\ \text{V/cm} \;=\; 0.5\ \text{V/m}.

Why: the fresh cell is the reference (the "standard" in this experiment). Its EMF and its balance length together pin down the one unknown of the potentiometer — the potential gradient. Once k is known, every later balance length converts straight to volts.

Step 2. Find the EMF of the older cell.

\varepsilon_2 \;=\; k\ell_2 \;=\; 0.005000 \times 276 \;=\; 1.380\ \text{V}.

Why: with k fixed, the balance length is a direct voltage reading. Any test cell balanced on the same wire under the same driver conditions reads at a length equal to its EMF divided by k.

Step 3. Confirm via the length ratio (sanity check).

\frac{\varepsilon_2}{\varepsilon_1} \;=\; \frac{\ell_2}{\ell_1} \;=\; \frac{276}{292} \;=\; 0.9452 \quad\Longrightarrow\quad \varepsilon_2 \;=\; 0.9452 \times 1.460 \;=\; 1.380\ \text{V}.

Why: the ratio form does not need k explicitly. It must agree with the direct calculation in Step 2, and it does to four significant figures — the consistency check is satisfied.

Step 4. EMF lost in storage.

\Delta\varepsilon \;=\; \varepsilon_1 - \varepsilon_2 \;=\; 1.460 - 1.380 \;=\; 0.080\ \text{V},

a loss of 0.080/1.460 \approx 5.5\% over six months. For a zinc-carbon dry cell stored at Chennai room temperature (~30 °C), this is about the expected shelf-life loss — the zinc anode slowly oxidises, reducing the open-circuit voltage.

Result. The potential gradient is 0.005 V/cm; the old cell has EMF 1.380 V, having lost about 5.5% of its original EMF to storage.

What this shows. The potentiometer reads both EMFs as pure length measurements and gives a difference of 16 cm that translates to 80 mV. A digital multimeter (input impedance 10 MΩ) would have given the same difference as 80 mV — but it would have drawn ~100 nA from each cell, an Ir drop of 100 μV for a 1 Ω internal resistance, which matters if you want the third decimal. The potentiometer draws exactly zero current at balance, which is why the Bureau of Indian Standards still uses a potentiometer to calibrate laboratory standard cells, not a digital voltmeter.

Example 3: Potentiometer — internal resistance of a Leclanché cell

Using a 400 cm potentiometer with potential gradient k = 0.005 V/cm, a Leclanché cell's open-circuit balance is \ell_1 = 288.0 cm. With a shunt resistor R = 8.0\ \Omega across the cell, the balance shifts to \ell_2 = 272.0 cm. Find the cell's EMF, internal resistance, and the current drawn from it when the shunt is connected.

Step 1. EMF from the open-circuit balance.

\varepsilon \;=\; k\ell_1 \;=\; 0.005 \times 288.0 \;=\; 1.440\ \text{V}.

Why: the open-circuit balance length converts to volts via the potential gradient. No current flows through the cell at balance, so there is no Ir sag — the reading is the true EMF.

Step 2. Internal resistance from the shift formula.

r \;=\; R\cdot\frac{\ell_1 - \ell_2}{\ell_2} \;=\; 8.0 \times \frac{288.0 - 272.0}{272.0} \;=\; 8.0 \times \frac{16.0}{272.0} \;=\; 8.0 \times 0.05882 \;=\; 0.471\ \Omega.

Why: the shift formula needs only the two balance lengths and the shunt resistance — the EMF itself drops out. A 16 cm shortening on a 272 cm loaded length corresponds to r/R = 16/272 \approx 1/17, so r is about one-seventeenth of the 8 Ω shunt, i.e. about 0.47 Ω.

Step 3. Current through the shunt loop.

I \;=\; \frac{\varepsilon}{R + r} \;=\; \frac{1.440}{8.0 + 0.471} \;=\; \frac{1.440}{8.471} \;=\; 0.170\ \text{A}.

Why: Ohm's law for the complete shunt loop — the EMF drives a current through the series combination of the external shunt R and the cell's internal resistance r.

Step 4. Consistency check via terminal voltage.

Terminal voltage: V = IR = 0.170 \times 8.0 = 1.360\ \text{V}. Wire voltage at \ell_2: k\ell_2 = 0.005 \times 272.0 = 1.360\ \text{V}. Match. ✓

Ir drop: 0.170 \times 0.471 = 0.080\ \text{V}. \varepsilon - V = 1.440 - 1.360 = 0.080\ \text{V}. Match. ✓

Result. EMF = 1.44 V, internal resistance = 0.47\ \Omega, shunt current = 170 mA.

What this shows. The length-only formula r = R(\ell_1-\ell_2)/\ell_2 gives an internal resistance accurate to about 2% on a 0.5 Ω target — considerably better than a direct voltmeter measurement, where a 10 mV error in reading the terminal voltage propagates to a 15% error in r. This accuracy is why every JEE Advanced lab-based problem on cell characteristics uses the potentiometer method.

Common confusions

"The metre bridge needs to know the wire's resistance per cm." No — only that the wire is uniform. The per-cm resistance r cancels in the balance ratio R/S = \ell/(100-\ell). Uniformity is the only property of the wire that enters the answer.
"Pressing the jockey harder gives a more accurate reading." The opposite — firm contact adds more than gentle contact, and the extra pressure indents the wire, ruining its uniformity for future measurements. Practise the light touch: just enough pressure for electrical contact, no more.
"The potentiometer galvanometer must be very sensitive." Only near balance. At exact balance no current flows, so the galvanometer's sensitivity does not affect the result. What sensitivity does determine is how precisely you can identify the null — a sluggish galvanometer might let you balance within \pm 2 mm, while a tangent galvanometer on a lamp-and-scale reads to \pm 0.2 mm.
"You should adjust the rheostat to find a balance." Never between two balance measurements in the same experiment. The rheostat sets the potential gradient k, and changing k changes the conversion between balance length and voltage. Adjust the rheostat once at the start to make sure the driver wire voltage exceeds the largest test EMF, then leave it alone.
"The driver cell's EMF shows up in the final answer." For EMF comparison it does not — it cancels in the ratio \ell_1/\ell_2. For the direct EMF calculation via \varepsilon = k\ell, it sets the gradient through k = V_\text{wire}/L, but you calibrate k with a known standard cell rather than computing it from the driver cell. The driver cell contributes to the experiment only through the stability of its current.
"You can use the metre bridge to measure very small resistances, like a short piece of copper wire." Below 0.1 Ω the terminal contact resistance (where the coils plug into the copper strips) swamps the measurement. A 10 mΩ contact resistance against a 50 mΩ unknown is a 20% error before you even start. For sub-ohm resistances, a Kelvin double bridge is the standard — beyond CBSE, but a JEE Advanced favourite.
"The potentiometer is always more accurate than a digital voltmeter." Not necessarily. A modern 6½-digit laboratory DMM has input impedance > 10 GΩ and reads to 10 μV, comparable to a good potentiometer. What the potentiometer does reliably, at almost no cost, is calibrate cheaper voltmeters — the National Physical Laboratory in Delhi uses a Lindeck potentiometer as a secondary voltage standard for exactly this reason.

If you came here to perform the CBSE or ISC practical and write the observation table, you have what you need. What follows is for readers interested in the precision limits of each instrument, the propagation of measurement errors, and the JEE Advanced-level optimisations that separate a good reading from a great one.

Error propagation in the metre bridge

For the metre bridge formula R = S\ell/(100-\ell), take the logarithmic derivative:

\frac{dR}{R} \;=\; \frac{dS}{S} + \frac{d\ell}{\ell} - \frac{d(100-\ell)}{100-\ell} \;=\; \frac{dS}{S} + \frac{100\,d\ell}{\ell(100-\ell)}.

Why: the logarithmic derivative turns products and quotients into sums — a standard error-propagation trick. The d(100-\ell) = -d\ell gives an extra minus that flips the third term to a plus, and combining the second and third terms with a common denominator gives the 100 in the numerator.

Typical values: dS/S \approx 0.2\% (resistance box tolerance), d\ell = 1 mm = 0.1 cm, \ell = 50 cm. Plug in:

\frac{dR}{R} \;=\; 0.002 + \frac{100 \times 0.1}{50 \times 50} \;=\; 0.002 + 0.004 \;=\; 0.006 \;=\; 0.6\%.

So a single balance at 50 cm gives R to about 0.6%. After swapping and averaging, the 0.2% end-correction is removed, leaving only the statistical error. With five independent readings, the error falls to 0.006/\sqrt{5} \approx 0.27\%. A careful hour of bench work gets you R to about \pm 0.3\%.

At \ell = 25 cm: dR/R = 0.002 + 100 \times 0.1/(25 \times 75) = 0.002 + 0.0053 = 0.73\%. The length-error term grows as \ell moves away from 50 cm — the source of the "balance near the middle" rule.

Finding end-corrections directly

With a single pair of measurements the end-corrections \alpha and \beta cannot be separated — they only appear in the combination (\ell+\alpha)/((100-\ell)+\beta). But if you do the experiment with two different known-resistor ratios (say, first a known 1 Ω and 2 Ω, then a known 2 Ω and 3 Ω), you get two equations in \alpha and \beta and can solve for each separately. This is the Carey Foster bridge method — a refinement of the metre bridge that reports \alpha and \beta at about \pm 1 mm each. Once known, these end-corrections improve any subsequent measurement of an unknown resistance on the same bridge.

Sensitivity of a potentiometer: galvanometer deflection per mm off balance

When the jockey is at length \ell + \delta with true balance at \ell, the voltage mismatch across the galvanometer loop is \Delta V = k\delta (using the potential gradient). The current through the galvanometer is then

I_G \;=\; \frac{k\delta}{R_G + R_\text{wire seg}(0 \to \ell+\delta) + r_\text{cell}}.

For a typical setup: k = 0.005 V/cm, R_G = 50\ \Omega, R_\text{wire seg} \approx 30\ \Omega (a \ell=300 cm segment of a 10 Ω/m wire), r_\text{cell} = 0.5\ \Omega. Then a \delta = 1 mm off-balance gives

I_G \;=\; \frac{0.005 \times 0.1}{50 + 30 + 0.5} \;=\; \frac{5 \times 10^{-4}}{80.5} \;\approx\; 6.2\ \mu\text{A}.

If the galvanometer reads 1 mm per μA, the 1 mm off-balance shows up as a 6 mm deflection — easily visible. The potentiometer is sensitive enough to locate balance to within 1 mm for ordinary dry cells.

To increase sensitivity (for smaller EMFs), reduce R_G + R_\text{wire seg}: use a galvanometer of lower internal resistance, and choose a potentiometer wire of higher total resistance so the per-cm resistance k/V is larger. A factor-of-two higher k doubles the deflection per mm, but also halves the range of EMFs you can match — a trade-off the CBSE lab manual does not mention but the JEE Advanced examiner sometimes tests.

Why the potentiometer wire is usually made of nichrome or constantan

Both alloys have high resistivity (~1 μΩ·m — about 60× copper) so a 1 m length has a measurable few-ohm resistance. Both have low temperature coefficients of resistance (~0.01% per °C for constantan) so the wire's resistance, and hence the potential gradient, barely drift as the driver current warms the wire by a few degrees during the experiment. A potentiometer wire made of copper would drift by ~0.4% per °C and give useless measurements after ten minutes of current flow.

Kelvin's double bridge — beyond the metre bridge

For sub-ohm resistances the contact and lead resistances (few mΩ to tens of mΩ) are a significant fraction of the unknown. The Kelvin double bridge is a metre-bridge variant with two galvanometer connections — an inner pair that measures only the resistance of the coil itself, excluding the leads. The balance condition becomes

R_x \;=\; S \cdot \frac{\ell}{100 - \ell} + \text{(small correction involving lead resistances)}.

The correction term is determined by the auxiliary galvanometer connection and vanishes when the two connections satisfy a second balance condition simultaneously. Kelvin bridges routinely measure contact resistances of a few microohms — good enough to find the resistance of a metre of copper wire, or of a switch contact, where a metre bridge would give noise.

Historical footnote — why the potentiometer outlives the analogue voltmeter

An analogue voltmeter is a galvanometer plus a large series resistor, calibrated in volts. Its reading depends on the galvanometer's sensitivity, which drifts with temperature, magnet aging, and spring fatigue. Every year or two the voltmeter must be re-calibrated against a standard — and the standard is nearly always a potentiometer balanced against a saturated Weston cadmium cell. The potentiometer's reading, being a length ratio, is absolute: it cannot drift without the metre rule itself distorting, and metre rules distort only with temperature in utterly predictable ways. This is why every secondary-standards laboratory in India — the National Physical Laboratory in Delhi, the Indian Standards Institute labs in Bengaluru — still has a Lindeck potentiometer on a vibration-isolated bench somewhere, even alongside rooms full of digital calibration equipment. The digital instruments are cross-checked against the potentiometer, not the other way around.

Where this leads next

Metre Bridge and Potentiometer — the underlying physics: Wheatstone-bridge balance condition and the potential-gradient derivation.
Wheatstone Bridge — the four-arm circuit that the metre bridge implements with a sliding contact.
Potentiometer and Bridge-Based Measurements — advanced bridge techniques including AC bridges (Anderson, Maxwell, de Sauty) for inductors and capacitors.
EMF and Internal Resistance — the \varepsilon-vs-V distinction that makes the potentiometer indispensable.
Errors in Measurement — the general framework for the end-correction averaging and error-propagation arguments used in this article.