= Wheatstone bridge balance
{c}
{title2=$R_x=R_3R_2/R_1$}
First disconnect the detector. Each side is a <voltage divider>. <Ohm's law> and <Kirchhoff's voltage law> give the open-circuit intermediate <electric potentials> and their difference:
$$
V_B=V_s\frac{R_2}{R_1+R_2},\qquad
V_C=V_s\frac{R_x}{R_3+R_x},\qquad
V_{\mathrm{oc}}=V_B-V_C
=V_s\left(\frac{R_2}{R_1+R_2}-\frac{R_x}{R_3+R_x}\right).
$$
Setting this difference to zero and cross-multiplying gives
$$
R_2R_3=R_1R_x,\qquad
\frac{R_1}{R_2}=\frac{R_3}{R_x},\qquad
R_x=R_3\frac{R_2}{R_1}.
$$
The balance compares ratios, not necessarily equal arm values or equal branch <electric currents>. At balance the detector carries no <electric current>, so <Kirchhoff's current law> again makes the <electric current> constant along each two-arm path.<Wheatstone bridge by PHYWE (P2410200)>[PHYWE, Wheatstone bridge, P2410200, p. 4, Theory (1/3)–(2/3), equations (1)–(4): divider potentials and balance]{ref}
For a worked <null measurement>, suppose $R_1=1.000\,\mathrm{k\Omega}$ and $R_2=2.000\,\mathrm{k\Omega}$, and the detector null occurs at adjustable $R_3=1.500\,\mathrm{k\Omega}$. Then
$$
R_x=(1.500\,\mathrm{k\Omega})\frac{2.000\,\mathrm{k\Omega}}{1.000\,\mathrm{k\Omega}}
=3.000\,\mathrm{k\Omega}.
$$
For $V_s=3.000\,\mathrm V$, both intermediate <electric potentials> are $2.000\,\mathrm V$. The left branch carries $1.000\,\mathrm{mA}$ and the right branch $0.6667\,\mathrm{mA}$: the zero detector reading does not mean that the unknown carries no <electric current>. These are stipulated ideal values illustrating the calculation, not experimental observations.
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