Wheatstone Bridge

Many physical quantities besides a wire's own material and geometry can shift the value of a resistance -- temperature, mechanical strain, humidity, the level of a liquid, and more all change how much a given resistor resists current. Because of this, measuring one of THOSE quantities accurately can often be turned into the (easier) problem of measuring a resistance accurately, provided the resistance-measuring method itself is precise enough. Depending on the expected size of the resistance -- anywhere from a few milliohms up to hundreds of megohms -- different measurement methods suit different ranges; the Wheatstone bridge, in particular, is the standard method for resistances from roughly a few tens of ohms up to a few hundred ohms.

The Wheatstone bridge circuit was originally developed by Charles Wheatstone (1802-1875) to measure unknown resistances, and it is also used to calibrate other measuring instruments such as voltmeters and ammeters. Four resistances -- call them P, Q, R and S -- are wired to form a quadrilateral (a four-sided closed loop) ABCD. A battery of emf $\varepsilon$, in series with a key, is connected across ONE diagonal of this quadrilateral (A to C), with point A held at the higher potential; a sensitive galvanometer of internal resistance G is connected across the OTHER diagonal (B to D).

When the key is closed, a total current I enters the bridge at A and splits into two parts: $I_1$ through arm P (towards B) and $I_2$ through arm S (towards D), so that

At B, $I_1$ splits further: a current $I_g$ flows through the galvanometer branch (from B to D), while the rest, $(I_1 - I_g)$, continues on through arm Q. Meanwhile at D, the current arriving from S combines with the galvanometer current, so that $(I_2 + I_g)$ flows on through arm R; both paths then recombine at C, on their way back to the battery.

Applying Kirchhoff's voltage law (clockwise) to loop A-B-D-A gives

and applying it (clockwise) to loop B-C-D-B gives

A special, very useful case arises when the galvanometer carries NO current at all, $I_g = 0$; the bridge is then said to be BALANCED, and this balance can always be reached by suitably adjusting P, Q, R and S. Substituting $I_g=0$ into Eqs. (9.4) and (9.5) gives, respectively,

Dividing Eq. (9.6) by Eq. (9.7) eliminates both branch currents entirely, leaving the famous balance (or bridge) condition, purely in terms of the four resistances:

$\frac{P}{Q} = \frac{S}{R} \qquad \text{--- (9.8)}$ …