Wheatstone and metre bridge

An ammeter and voltmeter reading is only as good as the meter's own accuracy — a bent needle or a stiff spring quietly corrupts every measurement. The Wheatstone bridge sidesteps this problem entirely with a null-deflection method: adjust the circuit until a galvanometer reads exactly zero, and the unknown resistance drops out as a pure ratio of known resistances, with no dependence on the battery's EMF or the galvanometer's own resistance. The metre bridge is the everyday laboratory version of this idea, replacing two of the four resistors with two segments of a single uniform wire, so that a resistance value is found simply by locating a balance point with a metre scale. This article states the Wheatstone balance condition, builds the metre-bridge formula from it, and works through both a bridge problem and a metre-bridge problem, extending the numbers already used in the shorter version of this note.

The core insight

A Wheatstone bridge is balanced (zero galvanometer current) when $R_1/R_2 = R_3/R_4$ , and the metre bridge realises this with a uniform wire so that $X = P(100-l)/l$ locates an unknown resistance from a single length measurement.

The core idea

The four-arm network. A Wheatstone bridge has four resistors $R_1$ , $R_2$ , $R_3$ , $R_4$ forming a diamond of four arms between junctions A, B, C, D. A battery and key are connected across one diagonal (A to C) to drive current through the network; a galvanometer and key are connected across the other diagonal (B to D) to sense any current crossing between the two halves of the bridge.

The balance condition. As current from the battery splits and rejoins at A and C, the potentials at B and D generally differ, so current flows through the galvanometer. Tune one of the resistors and, at a particular setting, B and D reach the same potential — the galvanometer shows zero deflection, and the bridge is balanced. The CISCE syllabus explicitly waives the Kirchhoff's-law derivation of this condition; only the statement is required:

\frac{R_1}{R_2} = \frac{R_3}{R_4}

where: $R_1, R_2$ are the resistances of one pair of adjacent arms and $R_3, R_4$ of the other pair, matched to the actual circuit diagram. Because this ratio holds independently of the battery EMF and the galvanometer's own resistance, it is a genuinely precise way to find an unknown resistance — the third great advantage of null methods over direct meter readings (the others being immunity to meter zero-error and to contact resistance at the terminals).

Why a wire can replace two arms. For a uniform wire of resistivity $\rho$ , cross-section $A$ and length $\ell$ , resistance is $R = \rho\ell/A$ — so for two segments of the same wire, resistance is directly proportional to length. This single fact is what lets a metre bridge turn a length measurement into a resistance ratio.

The metre bridge apparatus. A one-metre uniform wire (manganin or constantan, chosen for a resistance per unit length that barely drifts with temperature) is stretched along a metre scale between two thick copper strips. A resistance box holding a known resistance $P$ sits in the gap on one side, the unknown resistance $X$ in the gap on the other side, both wired in series with the two ends of the wire — completing a Wheatstone bridge in which $P$ and $X$ are two arms and the two wire segments are the other two. A sliding contact called a jockey, wired through a galvanometer, touches the wire at any point to search for the null point.

Figure

Metre bridge circuit showing resistance box P and unknown X connected across a one metre wire with a sliding jockey and galvanometer

If the null point is found at length $l$ (measured from the end joined to $P$ ) and $(100-l)$ from the end joined to $X$ , the two segments act as the remaining two bridge arms, with resistance proportional to length, giving the metre-bridge formula derived below.

Derivation AThe metre-bridge formula X = P(100 minus l) over l

The goal is to turn a single length measurement, the balance point $l$ on a metre scale, into the value of an unknown resistance $X$ , by treating the two wire segments either side of the jockey as two arms of a Wheatstone bridge and combining that with the fact that a uniform wire's resistance is proportional to its length — no new physics beyond the balance condition already stated and $R = \rho\ell/A$ .

Step 1. Identify the four arms. The known resistance $P$ and unknown resistance $X$ form two arms in series with the wire; the wire segment of length $l$ (from the $P$ end to the jockey) forms the third arm, and the remaining segment of length $(100-l)$ forms the fourth. Label the wire-segment resistances $R_l$ and $R_{100-l}$ .

Step 2. Apply the Wheatstone balance condition to this arrangement. With $P$ paired against $R_l$ and $X$ paired against $R_{100-l}$ as adjacent arms, balance requires $\dfrac{P}{X} = \dfrac{R_l}{R_{100-l}}$ .

Step 3. Replace the wire resistances with lengths. Since the wire is uniform, $R_l = \rho l/A$ and $R_{100-l} = \rho(100-l)/A$ for the same resistivity $\rho$ and cross-section $A$ , so their ratio is simply $R_l/R_{100-l} = l/(100-l)$ — the $\rho$ and $A$ cancel entirely, which is exactly why a uniform wire can stand in for two calibrated resistors.

Step 4. Substitute and solve for $X$ . Combining Steps 2 and 3, $\dfrac{P}{X} = \dfrac{l}{100-l}$ , which rearranges to

X = P\,\frac{100-l}{l}

This is the working formula: measure $l$ with a metre scale, read $P$ off the resistance box, and $X$ follows directly — no ammeter or voltmeter accuracy enters the result at all.

Worked example

A metre-bridge experiment uses a known resistance $P = 15\ \Omega$ in the left gap and an unknown resistance $X$ in the right gap. The jockey finds the null point at $l = 60\ \text{cm}$ from the left end.

Step 1. Write the metre-bridge formula: $X = P\,\dfrac{100-l}{l}$ .

Step 2. Substitute $P = 15\ \Omega$ and $l = 60\ \text{cm}$ : $X = 15 \times \dfrac{100-60}{60} = 15 \times \dfrac{40}{60} = 15 \times \dfrac{2}{3}$ .

Step 3. Compute: $X = \mathbf{10\ \Omega}$ . The balance point at 60 cm sits comfortably inside the recommended 30–70 cm window, so the reading carries a small percentage error.

Competitive stretch

Suppose the student now swaps $P$ and $X$ between the two gaps and rebalances. By symmetry of the formula, the new balance length $l'$ satisfies $X = P\,l'/(100-l')$ (the roles of $l$ and $100-l$ swap with the resistors), so $10 = 15\,l'/(100-l')$ , giving $10(100-l') = 15 l'$ , $1000 = 25 l'$ , $l' = 40\ \text{cm}$ — exactly $100 - 60$ . This "end correction check": swapping $P$ and $X$ should flip the balance point across the 50 cm mark, and a real bridge shows a small, consistent shift from this ideal due to resistance in the end connections, which is why some labs report an average of the two readings.

Common traps

Attempting to derive the Wheatstone balance condition from Kirchhoff's laws. CISCE waives this derivation; state $R_1/R_2 = R_3/R_4$ and move on.
Measuring $l$ from the wrong end. By convention $l$ is measured from the end connected to the known resistance $P$ ; using $l$ from the $X$ end inverts the ratio and gives $X = Pl/(100-l)$ instead of the correct formula.
Letting the balance point drift near an end of the wire. A value of $P$ that pushes $l$ below about 10 cm or above 90 cm magnifies the percentage error in the length reading — choose $P$ so the null point falls near the middle, ideally between 30 cm and 70 cm.
Leaving the battery key closed too long. Sustained current heats the wire, changing its resistance per unit length mid-experiment and shifting the balance point away from its true value.

Ask the tutor: stuck on which end to measure $l$ from in your own metre-bridge diagram, or want another balance-point problem worked through step by step?

Why does the balance condition not depend on the battery's EMF or the galvanometer's resistance?

At balance, by definition, zero current flows through the galvanometer branch, so its resistance never enters any current or voltage relation used to derive the ratio. The battery's EMF sets how much current circulates around the outer loop, but the balance condition is a statement about equal potentials at B and D, which holds regardless of how large or small that circulating current is. This is exactly the appeal of a null method: it converts a resistance measurement into a ratio problem that no longer depends on the accuracy of the driving source or the detector.

What happens if I place the jockey exactly at the 50 cm mark and the galvanometer still deflects?

That simply means the balance point is not at 50 cm for your chosen values of $P$ and $X$ ; you slide the jockey along the wire, watching the galvanometer deflection shrink and change direction as you cross the true null point, and take the reading where the deflection is exactly zero. If you cannot find a null anywhere on the wire, the value of $P$ in the resistance box is badly mismatched to $X$ (usually too large or too small by more than a factor of about four), and you should pick a different $P$ so the balance point falls within the wire's length.

Could I use the metre bridge to find a resistance much larger than the wire's own resistance, like a few thousand ohms?

Yes, because the formula only ever involves a ratio and a known resistance box value $P$ , so as long as you can dial a comparable $P$ on the resistance box, the wire's own resistance never appears in the answer. What actually limits accuracy is not the size of $X$ but where the balance point lands: choose $P$ so that $l$ stays away from the two ends of the wire, whatever the absolute size of $X$ happens to be.

Wheatstone and metre bridge

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads