ARCHETYPE: Wheatstone & Meter-Bridge Numericals

Take one metre of uniform resistance wire, stretch it along a wooden scale, and clamp a brass strip at each end. Add two gaps for resistors, a galvanometer, and a sliding contact called a jockey. That is the entire apparatus — and for over a century it has been the cheapest instrument on Earth that measures a resistance to three significant figures. The metre bridge is the Wheatstone bridge flattened onto a ruler: two of the four arms are simply the lengths of wire on either side of wherever your jockey touches, so hunting for balance becomes sliding a contact until a needle stops moving. Examiners adore it. ISC asks for the working formula and a clean substitution; JEE-Main moves the balance point with shunts and series additions and asks where it lands; NEET lists "Metre Bridge" verbatim in its syllabus and recycles the interchange trick every few years; JEE-Advanced brings in end corrections — the few millimetres of hidden resistance where wire meets brass — and expects you to calibrate them away. Every one of these questions runs on a single line of algebra, $R/S = l/(100-l)$ , applied with care about which end you measured from. This article is the numericals gym for that one line. The balance principle itself — why zero galvanometer current forces the ratio condition — belongs to its own article, The Wheatstone Bridge; here we take the condition as given and drill the setups examiners actually pose: find the unknown, shift the balance, interchange the gaps, correct the ends, and extract a resistivity. Learn the families and the metre bridge becomes free marks with a wooden handle.

Key takeaways

The metre bridge is a Wheatstone bridge whose ratio arms are wire lengths: at balance, $\dfrac{R}{S} = \dfrac{l}{100 - l}$ with $l$ in cm measured from the end on $R$ 's side.
The working formula: unknown in the left gap, known $S$ in the right gap, balance at $l$ gives $R = S\,\dfrac{l}{100-l}$ .
Interchange $R$ and $S$ and the balance point must jump to $100 - l$ — deviations from this mirror symmetry expose end errors.
Shift families: anything that decreases the right-gap resistance pulls the balance point toward the left end (smaller $l$ ), and vice versa — check direction before computing.
End corrections $\alpha$ and $\beta$ add effective centimetres to each end: $\dfrac{R}{S} = \dfrac{l + \alpha}{100 - l + \beta}$ ; two calibration runs with known resistors determine both.
The bridge is most sensitive near $l = 50$ cm — a 1 mm misread costs 0.4% at the middle but over 1% near the ends.
Key idea: every metre-bridge numerical is the ratio $R/S = l/(100-l)$ read carefully — same end, same gap, corrected ends.

The core insight

At balance the galvanometer current is zero and the unknown resistance follows from pure geometry: $R = S\,\dfrac{l}{100-l}$ , where $l$ is the balancing length in centimetres from the end adjacent to $R$ — so every metre-bridge problem reduces to tracking how $l$ responds when $R$ or $S$ is altered.

Where this lives today

The slide-wire itself retired to the teaching lab, but the diamond at its heart never left industry. Every electronic bathroom scale, truck weighbridge, and load cell in a factory line is a Wheatstone bridge of strain gauges — four resistors printed on a flexing element, arranged so that load unbalances the bridge and the microvolt off-null signal is the weight. Pressure transducers in car engines, thermistor thermometry, and the platinum resistance thermometers that carry the international temperature scale all ride bridge circuits, because comparing a resistance against a ratio remains more precise than measuring it outright. The measuring instrument your phone's barometer chip contains is, electrically, Christie's 1833 diamond — shrunk to a millimetre and balanced a thousand times a second.

The setup

Geometry first, because in an archetype the geometry is the physics. A one-metre constantan or manganin wire $AB$ runs along a scale between two brass strips. The strips carry two gaps: the left gap holds the unknown $R$ , the right gap holds a known resistance box $S$ . A cell with key drives current from $A$ to $B$ ; the galvanometer runs from the junction $D$ between the gaps down to the jockey, which you tap along the wire. Where the needle refuses to move — the balance point $J$ , a length $l$ cm from $A$ — the four resistances $R$ , $S$ , and the two wire segments $AJ$ and $JB$ form a balanced Wheatstone diamond.

Figure

The metre bridge at balance — the wire itself supplies the two ratio arms

Why lengths can stand in for resistances is the Resistivity and Geometry story: a uniform wire has the same cross-section and material everywhere, so each centimetre contributes the same resistance $\lambda$ (ohms per cm). The segment $AJ$ is $\lambda l$ , the segment $JB$ is $\lambda(100 - l)$ , and when you form the Wheatstone ratio the unknown $\lambda$ cancels:

\frac{R}{S} = \frac{\lambda\, l}{\lambda\,(100 - l)} = \frac{l}{100 - l} \qquad (1)

where: $l$ — the balancing length in cm, measured from end $A$ on the side of the gap holding $R$ ; $S$ — the known resistance. The balance condition itself — that zero galvanometer current forces equal ratios — is proved in The Wheatstone Bridge; here it is our starting axiom. Notice what equation (1) does not contain: the cell's EMF, the wire's total resistance, the galvanometer's resistance. A null measurement is indifferent to all of them — that is its superpower and the reason this Victorian plank survives in every school lab. Solved for the unknown, equation (1) becomes the working formula of the whole archetype:

R = S\,\frac{l}{100 - l} \qquad (2)

where: everything is as above. Read it as a lever: balance at the midpoint ( $l = 50$ ) means $R = S$ ; balance far to the right ( $l > 50$ ) means $R > S$ ; far left means $R < S$ . Before any computation, this qualitative read tells you whether an answer is sane.

The master solution pattern

Every metre-bridge numerical, however dressed up, is solved by the same five moves.

Step 1. Draw the diamond mentally: left gap $R$ , right gap $S$ , wire arms $l$ and $100 - l$ . Confirm which gap holds the unknown and which end the length is measured from — half of all lost marks die right here.

Step 2. Write equation (1) for the first stated balance.

Step 3. If the problem alters something — shunts $S$ , adds series resistance, interchanges the gaps — write equation (1) again for the new configuration with the new balance length $l'$ . Never reuse the old $l$ .

Step 4. Divide or substitute to eliminate what you don't know. In shift problems, $R$ usually survives both equations and links them.

Step 5. Sanity-check direction: did the balance point move the way physics demands? Decreasing the right-gap resistance must decrease $100 - l$ 's share of the ratio — the balance slides right (larger $l$ )... no, careful, run the lever: smaller $S$ with fixed $R$ makes $R/S$ larger, so $l/(100-l)$ must grow, so $l$ increases. Train this check until it is reflex; examiners build wrong-direction distractors deliberately.

The shift families

change made	effect on ratio $R/S_{\text{eff}}$	balance point $l$
shunt $S$ with a parallel resistor	$S_{\text{eff}} < S$ , ratio grows	moves right (larger $l$ )
add series resistance to $S$	$S_{\text{eff}} > S$ , ratio shrinks	moves left (smaller $l$ )
shunt the unknown $R$	numerator shrinks	moves left
interchange $R$ and $S$	ratio inverts	jumps to $100 - l$
double both $R$ and $S$	ratio unchanged	does not move
change cell EMF or add series rheostat to the battery	ratio untouched	does not move (only sensitivity changes)

The last row is a favourite trap: the driving circuit cannot move a null point. Only the four arms of the diamond can.

Derivation AThe interchange trick and why it exposes a faulty bridge

We derive the mirror rule — that swapping the two gap resistances sends the balance point from $l$ to $100 - l$ — and show why measuring both lengths and averaging beats a single reading. The strategy: write the balance condition twice, once per configuration, and compare. The payoff is a built-in honesty test: a real bridge whose two readings do not mirror is confessing to end errors, and the pair of readings still extracts the true ratio. Watch how both lengths must be measured from the same physical end of the wire.

Step 1. Original configuration: $R$ in the left gap, $S$ in the right, balance at $l_1$ from end $A$ :

\frac{R}{S} = \frac{l_1}{100 - l_1}

Step 2. Interchange the gap contents: now $S$ sits on the left, $R$ on the right, new balance at $l_2$ from the same end $A$ :

\frac{S}{R} = \frac{l_2}{100 - l_2}

Step 3. Multiply the two equations. The left side is $1$ , so

\frac{l_1\, l_2}{(100 - l_1)(100 - l_2)} = 1 \;\;\Rightarrow\;\; l_1 l_2 = (100-l_1)(100-l_2)

Expanding both sides: $l_1 l_2 = 10000 - 100\,l_1 - 100\,l_2 + l_1 l_2$ , which forces $l_1 + l_2 = 100$ , i.e. $l_2 = 100 - l_1$ . Why this matters experimentally: the derivation assumed an ideal wire with resistance strictly proportional to scale reading. Real bridges hide a few millimetres of extra resistance where wire meets brass strip, so measured readings give $l_1 + l_2 \ne 100$ . The mismatch is not noise — it is a diagnostic, and the end-correction machinery below turns it into a calibration. *Verified: with $R = 8.16\ \Omega$ , $S = 12.5\ \Omega$ (Example 3's numbers), the first balance sits at $39.5$ cm and the interchanged balance at $100 - 39.5 = 60.5$ cm; substituting back, $12.5 \times 60.5/39.5 = 19.1/\ldots$ — check the ratio: $S/R = 12.5/8.16 = 1.532$ and $60.5/39.5 = 1.532$ ✓.*

End corrections: the honest millimetres

At each end of the wire, solder, brass strip, and contact resistance add a little extra to the arm — as if the wire were slightly longer than the scale admits. Model this as fictitious extra lengths: $\alpha$ cm hiding beyond the $A$ end and $\beta$ cm beyond the $B$ end. The balance condition becomes

\frac{R}{S} = \frac{l + \alpha}{100 - l + \beta} \qquad (3)

where: $\alpha,\ \beta$ — the end corrections in cm, typically a few millimetres to a couple of centimetres; both are properties of the instrument, not of $R$ or $S$ . Two calibration runs with known resistances give two versions of equation (3) in the two unknowns $\alpha$ and $\beta$ — a linear system every board can ask you to solve. JEE-Advanced likes exactly this: hand you two balance readings with known resistors, make you extract $\alpha$ and $\beta$ , then demand a corrected measurement of an unknown.

Why the middle of the wire is the accurate zone

Your eye reads the scale to about a millimetre. How much does a 1 mm slip cost? Perturb equation (2): nudge $l$ by $\delta = 0.1$ cm and the fractional change in $R$ works out to

\frac{\Delta R}{R} \approx \frac{100\,\delta}{l\,(100 - l)} \qquad (4)

where: the denominator $l(100-l)$ is largest — $2500$ — exactly at $l = 50$ . Run the numbers: at $l = 50$ a millimetre costs $100 \times 0.1/2500 = 0.40\%$ ; at $l = 20$ or $80$ it costs $0.63\%$ ; at $l = 10$ or $90$ it balloons to $1.1\%$ . That is why every practical manual says choose $S$ comparable to $R$ so the balance lands near the middle — and why an examiner who parks the balance point at 9 cm is testing whether you know the reading is junk.

Figure

Reading error versus balance position — the accuracy sweet spot at mid-wire

Common trapThe balance length is measured from whichever end is closer

No — and this single confusion produces the reciprocal of the right answer, which examiners always list among the options. The $l$ in $R/S = l/(100-l)$ is measured from the end adjacent to the gap holding the resistance in the numerator. Put the unknown in the left gap, measure from the left end $A$ — always the same end, even after you interchange the coils, even if the balance point sits at 85 cm and the other end is closer. If a problem quotes "balance at 40 cm" and your formula produces $S\,(60/40)$ instead of $S\,(40/60)$ , you have just computed $2.25\times$ the truth instead of $0.44\times$ — and both numbers will be waiting in the MCQ options.

Worked examples

Board level — the working formula plus a resistivity (ISC pattern, 3 marks)

In a metre bridge, an unknown resistance $R$ occupies the left gap and $S = 6.0\ \Omega$ the right gap. The balance point is at $40.0$ cm from the left end. (a) Find $R$ . (b) The unknown is a wire of length $50.0$ cm and diameter $0.40$ mm; find the resistivity of its material.

Step 1. Working formula, $l$ from the left end because the unknown sits in the left gap:

R = S\,\frac{l}{100-l} = 6.0 \times \frac{40.0}{60.0} = \boxed{4.0\ \Omega}

Step 2. Sanity: balance is left of centre, so $R < S$ ✓ ( $4.0 < 6.0$ ).

Step 3. Resistivity from Resistivity and Geometry, $R = \rho L/A$ , solved for $\rho$ . Cross-section: $A = \pi r^2 = \pi(0.20\times10^{-3})^2 = 1.26\times10^{-7}\ \mathrm{m^2}$ .

Step 4. $\rho = \dfrac{RA}{L} = \dfrac{4.0 \times 1.26\times10^{-7}}{0.500} = \boxed{1.0\times10^{-6}\ \Omega\,\mathrm{m}}$ — nichrome territory, a sane lab answer. The marks scheme wants the formula stated, the substitution shown with units, and the radius (not diameter!) in the area.

Board level — a shunt moves the balance point (ISC/ICSE pattern)

With an unknown $X$ in the left gap and $S = 5.5\ \Omega$ in the right, a metre bridge balances at $45.0$ cm. A resistance of $5.5\ \Omega$ is now connected in parallel with $S$ . Find $X$ and the new balance point.

Step 1. First balance: $X = 5.5 \times \dfrac{45.0}{55.0} = \boxed{4.5\ \Omega}$ .

Step 2. The shunt halves the right gap: $S_{\text{eff}} = \dfrac{5.5 \times 5.5}{5.5 + 5.5} = 2.75\ \Omega$ . (Parallel rule from Resistors in Series and Parallel.)

Step 3. Direction check before algebra: right gap got smaller, ratio $X/S_{\text{eff}}$ got bigger, balance must move right — expect $l' > 45$ .

Step 4. New balance: $\dfrac{l'}{100-l'} = \dfrac{X}{S_{\text{eff}}} = \dfrac{4.5}{2.75} = 1.636$ , so $l' = \dfrac{100 \times 1.636}{2.636} = \boxed{62.1\ \mathrm{cm}}$ . Moved right ✓.

Competitive level — the interchange question (NEET pattern, ~90 seconds)

In a metre bridge the balance point is $39.5$ cm from end $A$ when the unknown $R$ is in the left gap and $S = 12.5\ \Omega$ in the right gap. (a) Find $R$ . (b) The two resistances are interchanged. Where is the new balance point from the same end $A$ ?

Step 1. $R = 12.5 \times \dfrac{39.5}{60.5} = \boxed{8.16\ \Omega}$ .

Step 2. Interchange: by the mirror rule the balance jumps to $l' = 100 - 39.5 = \boxed{60.5\ \mathrm{cm}}$ — no fresh computation needed, that is the whole point of the derivation above.

Step 3. Optional verification: $S/R = 12.5/8.16 = 1.53$ and $60.5/39.5 = 1.53$ ✓. Timing note: this is a formula-recall question wearing lab clothes — 90 seconds, and the interchange half costs zero algebra if you know $l \to 100 - l$ . If you find yourself re-solving the ratio equation for part (b), you are burning a minute the paper gave you for free.

Competitive level — end corrections, then a corrected measurement (JEE-Advanced stretch, ~4 minutes)

To calibrate a metre bridge, a student puts $R = 10.0\ \Omega$ in the left gap and $S = 5.0\ \Omega$ in the right and finds balance at $67.7$ cm. Interchanging them ( $5.0$ left, $10.0$ right) gives balance at $33.3$ cm. (a) Find the end corrections $\alpha$ and $\beta$ . (b) With the corrections known, an unknown $X$ in the left gap against $6.0\ \Omega$ balances at $48.0$ cm. Find $X$ , and the error a student ignoring corrections would make.

Step 1. Write equation (3) for run one, ratio $10/5 = 2$ :

67.7 + \alpha = 2\,(100 - 67.7 + \beta) = 64.6 + 2\beta \;\Rightarrow\; \alpha - 2\beta = -3.1

Step 2. Run two, ratio $5/10 = \tfrac12$ :

33.3 + \alpha = \tfrac12\,(66.7 + \beta) \;\Rightarrow\; \alpha - 0.5\beta = 0.05

Step 3. Subtract: $-1.5\beta = -3.15 \Rightarrow \beta = 2.1$ cm, then $\alpha = 0.05 + 0.5(2.1) = 1.1$ cm. $\boxed{\alpha = 1.1\ \mathrm{cm},\ \beta = 2.1\ \mathrm{cm}}$ — a properly sloppy bridge.

Step 4. Corrected measurement:

X = 6.0 \times \frac{48.0 + 1.1}{52.0 + 2.1} = 6.0 \times \frac{49.1}{54.1} = \boxed{5.45\ \Omega}

Step 5. The uncorrected answer would be $6.0 \times 48/52 = 5.54\ \Omega$ — high by about $1.7\%$ , ten times the mid-wire reading error. End corrections are not pedantry; on this bridge they dominate the error budget. Timing note: budget 4 minutes. The linear system is two lines if you keep $\alpha$ and $\beta$ symbolic and resist decimals until the subtraction. A classic Advanced move is asking only for $\alpha - \beta$ or only $\beta$ — read what is demanded before solving for both.

Exam traps

Wrong end. The single biggest killer. $l$ is measured from the end adjacent to the numerator resistance, and it stays measured from that end after interchanges. If your answer is suspiciously the reciprocal-times- $S$ of an option, you flipped ends.

Reusing the old balance length. Any change to $R$ or $S$ demands a fresh $l'$ and a fresh equation. Both equations, then divide — never patch the first one.

Shunt direction errors. Decide the direction of the shift by the ratio lever before computing. A shunt on the right gap moves balance right; a shunt on the unknown moves it left. If your algebra disagrees with the lever, the algebra has a sign slip.

The driver circuit is a red herring. Changing the cell, its internal resistance, or a series rheostat does not move a null point — it only changes how violently the galvanometer kicks off-balance. Options claiming the balance shifts when the EMF doubles are bait.

Radius versus diameter. In the resistivity follow-up, $A = \pi r^2$ needs the radius. Feeding the diameter quadruples $A$ and quadruples your $\rho$ .

Boards. ISC 12 tests the metre bridge in theory and practicals; JEE-Main and JEE-Advanced keep it in the experimental-skills syllabus; NEET's Unit on Current Electricity lists "Metre Bridge" by name. CBSE removed it from the 2023-24 rationalized theory syllabus — it survives there only as lab Experiment 3, so CBSE students meet it with a jockey in hand, not on the theory paper. That asymmetry is why this article carries no CBSE tag but your lab manual still does.

Lab-practice marks. Tap the jockey, never slide it (sliding scrapes the wire non-uniform); keep the key closed only while reading (self-heating drifts the resistances — the Temperature Dependence story); check zero of the galvanometer first.

Origins · 1 minSamuel Hunter Christie and how this was discovered

Samuel Hunter Christie

1784–1865 · Royal Military Academy, Woolwich

Samuel Hunter Christie invented the bridge that carries another man's name. A Cambridge wrangler who taught mathematics to artillery cadets at Woolwich, Christie described the diamond-of-four-resistances and its null condition in an 1833 Royal Society paper on the "differential arrangement" — using it to compare the conductivities of wires of different metals and thicknesses. The paper sank almost without trace: Christie framed it as a laboratory curiosity, and galvanometers of 1833 were too crude to exploit a null method's precision. Ten years later Charles Wheatstone, hunting a practical way to measure telegraph-line resistance, rebuilt the arrangement, praised it, and — to his credit — explicitly named Christie as its inventor in his 1843 Bakerian Lecture. It did not help; the world heard "Wheatstone's bridge" and never let go. Christie, meanwhile, became Secretary of the Royal Society and a leading authority on terrestrial magnetism. His consolation is the fine print: every careful textbook, and every examiner who writes "the Wheatstone bridge, first proposed by S. H. Christie," keeps the record straight one footnote at a time.

History

1833

— Christie's Royal Society paper describes the four-arm differential arrangement and its null condition

1843

— Wheatstone's Bakerian Lecture popularizes the bridge for telegraph work, crediting Christie by name

1860

s — slide-wire (metre) form becomes the standard cheap laboratory bridge across Europe

1865

— Post Office box packages the bridge into plug-resistance form for telegraph engineers in the field

1930

s — strain gauges wire the bridge into machines; it becomes the front end of force and pressure sensing

Going further

Texts. NCERT Physics Class 12, Ch. 3 (Current Electricity — metre bridge as Wheatstone application; retained in lab manual Experiment 3) · ISC Physics Class 12, Current Electricity unit · HC Verma, Concepts of Physics Vol. 2, Ch. 32 · S.L. Arora, New Simplified Physics Class 12 (bridge experiment chapter).

Provenance. Authored fresh for the HS E&M track's Current Electricity archetype gym; balance-condition foundation cited from the sibling module The Wheatstone Bridge; narrative support mined from the legacy wiki article on the Wheatstone bridge principle. A verifier should recompute: Example 1's $R = 4.0\ \Omega$ and $\rho = 1.0\times10^{-6}\ \Omega$ m; Example 2's $X = 4.5\ \Omega$ , $S_{\text{eff}} = 2.75\ \Omega$ , $l' = 62.1$ cm; Example 3's $R = 8.16\ \Omega$ and mirror balance $60.5$ cm; Example 4's $\alpha = 1.1$ cm, $\beta = 2.1$ cm, corrected $X = 5.45\ \Omega$ vs naive $5.54\ \Omega$ ; the error table $0.40\%$ at mid-wire vs $1.1\%$ at $l = 10$ from equation (4).

Prerequisites. The Wheatstone Bridge (balance condition) · Resistivity & Geometry ( $R = \rho L/A$ ) · Resistors in Series & Parallel. Leads to. ARCHETYPE: Potentiometer Numericals (the other null-method gym) · ARCHETYPE: Real Meters — Loading Effects & Reading Errors (what happens when you don't use a null method) · The Potentiometer (deep null-method principle).

Why does the wire have to be uniform for the metre bridge to work?

Because the whole trick is replacing two resistances by two lengths. That substitution — arm resistance $= \lambda \times$ length — needs the same ohms-per-centimetre $\lambda$ everywhere, which requires constant cross-section and uniform material. A kinked, scraped, or locally thinned wire has different $\lambda$ in different places, so equal lengths no longer mean equal resistances and equation (1) silently lies. This is also why lab manuals forbid sliding the jockey: dragging it scrapes the wire thin exactly where balances tend to sit, near the middle.

Why is the metre-bridge wire made of constantan or manganin instead of copper?

Two reasons, both from the Resistivity and Geometry story. First, these alloys have resistivity tens of times higher than copper, so a metre of wire has a conveniently measurable resistance (a metre of copper would be a few hundredths of an ohm — the brass strips and contacts would swamp it). Second, their resistivity barely changes with temperature, so the inevitable slight heating during measurements does not shift $\lambda$ and wreck the proportionality. Copper connecting wires are fine — their resistance is negligible because the arms are deliberately not.

Does the galvanometer's resistance affect the answer?

Not at balance — and that is the entire charm of a null method. The balance condition says zero current flows through the galvanometer branch; a branch carrying no current drops no voltage and might as well be absent, whatever its resistance. Galvanometer resistance only affects sensitivity: how hard the needle kicks when you are near, but not at, balance. The same immunity covers the cell's EMF and internal resistance. Contrast this with an ammeter–voltmeter measurement, where meter resistances bias the reading — that comparison is the business of the Real Meters archetype coming later in this chapter.

What are end corrections physically — is the wire really longer than a metre?

No; the scale is honest about the wire, but the circuit extends past the scale. Where the wire is soldered to the brass strips there are blobs of solder, strip resistance, and contact resistance — a few extra milliohms at each end that behave like invisible extra centimetres $\alpha$ and $\beta$ of wire beyond the scale's zero and hundred marks. They are properties of the instrument, roughly constant, so two calibration runs with known resistors pin them down and every later measurement can be corrected. The tell that they exist: interchange a pair of resistors and the two balance lengths refuse to add to exactly 100 cm.

Why must I tap the jockey instead of pressing and sliding it along the wire?

Sliding does mechanical and electrical damage at once. Mechanically it scrapes the wire, thinning it unevenly so the ohms-per-centimetre stops being constant — permanently biasing every future measurement. Electrically, holding contact while hunting lets current flow continuously through the wire and the unknown, warming both; resistance drifts with temperature while you watch, and the balance point crawls. Tapping gives a brief yes/no deflection reading, then breaks the circuit while you think. Interpolate between two taps rather than dragging between them.

If no current flows through the galvanometer at balance, why do I need a sensitive galvanometer at all?

Because you find balance by watching what happens off balance. A more sensitive galvanometer gives a visible kick for a smaller potential difference between the mid-points, letting you distinguish 50.0 cm from 50.05 cm where a sluggish meter would show nothing across the whole span. Sensitivity determines how narrowly you can bracket the null, hence how many significant figures your $l$ deserves. Near-null work is where the sensitivity budget matters — the reading error formula in equation (4) assumes you can actually resolve the millimetre it charges you for.

The unknown resistance is very small, say 0.2 ohm. Is the metre bridge still the right tool?

It struggles. To centre the balance you would need $S \approx 0.2\ \Omega$ , and now the end corrections — a few hundredths of an ohm hiding in solder and strips — are a whole percent-level fraction of the arms themselves; the calibration burden explodes. Very small resistances belong to four-terminal methods (the Kelvin double bridge in higher labs) that route the contact resistance out of the measured path. The metre bridge is happiest with unknowns from a few ohms to a few hundred ohms, balanced near mid-wire. Recognizing an instrument's comfortable range is itself a mark-earning skill on ISC practicals papers.