Wheatstone bridge balance ($R_x=R_3{R}_2/R_1$)

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:Setting this difference to zero and cross-multiplying givesThe 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.^[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 $R_3=1.500\mathrm{k\Omega }$. ThenFor $V_s=3.000\mathrm{V}$, 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.