Quick derivation at balance

At the point of balance, both the voltage and the current between the two midpoints (B and D) are zero. Therefore, ${\displaystyle I_{1}=I_{2}}$, ${\displaystyle I_{3}=I_{x}}$, ${\displaystyle V_{D}=V_{B}}$.

Because of ${\displaystyle V_{D}=V_{B}}$, then ${\displaystyle V_{DC}=V_{BC}}$ and ${\displaystyle V_{AD}=V_{AB}}$.

Dividing the last two equations by members and using the above currents equalities, then

${\displaystyle {\begin{aligned}{\frac {V_{DC}}{V_{AD}}}&={\frac {V_{BC}}{V_{AB}}}\\[4pt]\Rightarrow {\frac {I_{2}R_{2}}{I_{1}R_{1}}}&={\frac {I_{x}R_{x}}{I_{3}R_{3}}}\\[4pt]\Rightarrow R_{x}&={\frac {R_{2}}{R_{1}}}\cdot R_{3}\end{aligned}}}$

Full derivation using Kirchhoff's circuit laws

First, Kirchhoff's first law is used to find the currents in junctions B and D:

${\displaystyle {\begin{aligned}I_{3}-I_{x}+I_{G}&=0\\I_{1}-I_{2}-I_{G}&=0\end{aligned}}}$

Then, Kirchhoff's second law is used for finding the voltage in the loops ABDA and BCDB:

${\displaystyle {\begin{aligned}(I_{3}\cdot R_{3})-(I_{G}\cdot R_{G})-(I_{1}\cdot R_{1})&=0\\(I_{x}\cdot R_{x})-(I_{2}\cdot R_{2})+(I_{G}\cdot R_{G})&=0\end{aligned}}}$

When the bridge is balanced, then I_G = 0, so the second set of equations can be rewritten as:

${\displaystyle {\begin{aligned}I_{3}\cdot R_{3}&=I_{1}\cdot R_{1}\quad {\text{(1)}}\\I_{x}\cdot R_{x}&=I_{2}\cdot R_{2}\quad {\text{(2)}}\end{aligned}}}$

Then, equation (1) is divided by equation (2) and the resulting equation is rearranged, giving:

${\displaystyle R_{x}={{R_{2}\cdot I_{2}\cdot I_{3}\cdot R_{3}} \over {R_{1}\cdot I_{1}\cdot I_{x}}}}$

Due to I₃ = I_x and I₁ = I₂ being proportional from Kirchhoff's First Law, I₃I₂/I₁I_x cancels out of the above equation. The desired value of R_x is now known to be given as:

${\displaystyle R_{x}={{R_{3}\cdot R_{2}} \over {R_{1}}}}$

On the other hand, if the resistance of the galvanometer is high enough that I_G is negligible, it is possible to compute R_x from the three other resistor values and the supply voltage (V_S), or the supply voltage from all four resistor values. To do so, one has to work out the voltage from each potential divider and subtract one from the other. The equations for this are:

${\displaystyle {\begin{aligned}V_{G}&=\left({R_{2} \over {R_{1}+R_{2}}}-{R_{x} \over {R_{x}+R_{3}}}\right)V_{s}\\[6pt]R_{x}&={{R_{2}\cdot V_{s}-(R_{1}+R_{2})\cdot V_{G}} \over {R_{1}\cdot V_{s}+(R_{1}+R_{2})\cdot V_{G}}}R_{3}\end{aligned}}}$

where V_G is the voltage of node D relative to node B.