Wheatstone Bridge & Metre Bridge

PHYMeasuring Instruments

The Wheatstone bridge measures an unknown resistance by balancing four arms so that no current flows through the galvanometer; the metre bridge is its practical slide-wire version. The central balance condition $P/Q=R/S$ lets you find any one arm from the other three.

ISC asks for the unknown arm in a balanced bridge and for the unknown resistance from a metre-bridge balance length.

Wheatstone balance condition

\dfrac{P}{Q} = \dfrac{R}{S}

P, Q, R, S

the four arm resistances (

\Omega

). At balance the galvanometer current is zero; the unknown follows, e.g.

S = QR/P

Derivation

At balance no current flows through the galvanometer, so the series branch $P$ – $Q$ carries a single current $I_1$ throughout, and the series branch $R$ – $S$ carries a single current $I_2$ throughout. Balance also means the galvanometer's two ends ( $B$ and $D$ ) are at the same potential, so the drop from the battery's $+$ terminal to that common potential is equal along both branches:
$I_1 P = I_2 R$
and, by the same argument, the drop from that common potential to the battery's $-$ terminal is equal along both branches:
$I_1 Q = I_2 S$
Dividing the first relation by the second eliminates the unknown currents:
$\dfrac{I_1 P}{I_1 Q} = \dfrac{I_2 R}{I_2 S} \;\Rightarrow\; \dfrac{P}{Q} = \dfrac{R}{S}$

Metre bridge balance

\dfrac{R}{S} = \dfrac{l}{100 - l}

R

unknown in the left gap,

S

known in the right gap,

l

balance length (cm) from the left end. The wire's resistance per length cancels.

Unknown from metre bridge

R = S \, \dfrac{l}{100 - l}

l

measured from the left (the side of the unknown

R

). If

S

is in the right gap, the unknown in the left gap uses the ratio

l/(100-l)

At balance no current flows through the galvanometer, so the bridge is independent of the galvanometer resistance and the supply voltage.
The metre bridge applies the Wheatstone condition with the two wire segments acting as two arms whose resistances are proportional to their lengths.
Balance length is measured from the end nearest the resistor placed in the corresponding gap — keep track of which arm is which.
Sensitivity is best when the balance point is near the middle of the wire, so choose the known resistance comparable to the unknown.
End corrections (contact resistances at the ends of the wire) shift the balance point and are reduced by interchanging gaps and averaging.
The bridge measures a ratio, so a uniform wire's resistance per unit length cancels out of the final result.
A balanced bridge can be redrawn with the galvanometer branch removed, simplifying any further analysis.

Where the marks go

Misreading the balance condition as $P/Q=S/R$ instead of $P/Q=R/S$ , inverting the unknown.
Using $l/100$ instead of $l/(100-l)$ for the length ratio in the metre bridge.
Measuring the balance length from the wrong end, swapping which gap holds the known resistance.
Assuming the supply voltage or galvanometer resistance affects the balance point — at balance neither does.

How the board asks it

Numericalthe metre bridge balance length
In a metre bridge, the balance point is obtained at $40\,cm$ from the left end when an unknown resistance $X$ is in the left gap and a known resistance of $6\,\Omega$ is in the right gap. Calculate the value of $X$ .
Numericalunknown arm from the balance condition
In a balanced Wheatstone bridge the resistances in three arms are $P = 10\,\Omega$ , $Q = 15\,\Omega$ and $R = 6\,\Omega$ . Find the resistance $S$ in the fourth arm at balance.
Diagram / graphthe balanced bridge layout
Draw a labelled circuit diagram of a Wheatstone bridge and state the condition under which it is said to be balanced.
Derive / provethe wheatstone balance condition
Using Kirchhoff's laws, derive the balance condition $\frac{P}{Q} = \frac{R}{S}$ for a Wheatstone bridge network.
Give reasonsindependence from galvanometer and supply
Explain why the balance point of a Wheatstone bridge is independent of the galvanometer resistance and of the emf of the cell.
Applicationsensitivity and end corrections
In a metre bridge experiment, explain why the known resistance is chosen so that the balance point lies near the middle of the wire, and state how end corrections can be minimised.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.