Wheatstone Bridge - Derivation

Derivation

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

$\begin{align} I_3 - I_x + I_G &= 0 \\ I_1 - I_2 - I_G &= 0 \end{align}$

Then, Kirchhoff's second rule is used for finding the voltage in the loops ABD and BCD:

$\begin{align} (I_3 \cdot R_3) - (I_G \cdot R_G) - (I_1 \cdot R_1) &= 0 \\ (I_x \cdot R_x) - (I_2 \cdot R_2) + (I_G \cdot R_G) &= 0 \end{align}$

The bridge is balanced and, so the second set of equations can be rewritten as:

$\begin{align} I_3 \cdot R_3 &= I_1 \cdot R_1 \\ I_x \cdot R_x &= I_2 \cdot R_2 \end{align}$

Then, the equations are divided and rearranged, giving:

From the first rule, and . The desired value of is now known to be given as:

If all four resistor values and the supply voltage are known, and the resistance of the galvanometer is high enough that is negligible, the voltage across the bridge can be found by working out the voltage from each potential divider and subtracting one from the other. The equation for this is:

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

Read more about this topic: Wheatstone Bridge