Editing Wheatstone bridge (section)

== Derivation ==
[[File:Wheatstonebridge current.svg|thumb|300px|Directions of currents arbitrarily assigned]]

=== Quick derivation at balance ===
At the point of balance, both the [[voltage]] and the [[Electric current|current]] between the two midpoints (B and D) are zero. Therefore, {{math|1=''I''<sub>1</sub> = ''I''<sub>2</sub>}}, {{math|1=''I''<sub>3</sub> = ''I''<sub>''x''</sub>}}, {{math|1=''V''<sub>D</sub> = ''V''<sub>B</sub>}}.

Because of {{math|1=''V''<sub>D</sub> = ''V''<sub>B</sub>}}, then {{math|1=''V''<sub>DC</sub> = ''V''<sub>BC</sub>}} and {{math|1=''V''<sub>AD</sub> = ''V''<sub>AB</sub>}}.

Dividing the last two equations by members and using the above currents equalities, then
: <math>\begin{align}
  \frac{V_{DC}}{V_{AD}}&=\frac{V_{BC}}{V_{AB}} \\[4pt]
 \Rightarrow \frac{I_2R_2}{I_1R_1} &= \frac{I_xR_x}{I_3R_3}\\[4pt]
  \Rightarrow R_x &= \frac{R_2}{R_1} \cdot R_3
\end{align}</math>

=== Alternative Derivation at Balance using Voltage Divider Expressions ===
ADC and ABC form two [[voltage divider]]s, with <math> V_G </math> equal to the difference in output voltages. Thus
: <math>\begin{align}
V_{DC} &= V_{BC} \\
I_2 R_2 &= I_x R_x \\
V_{AC} \frac{R_2}{R_1 + R_2} &= V_{AC} \frac{R_x}{R_3 + R_x} \\
\frac{R_2}{R_1 + R_2} &= \frac{R_x}{R_3 + R_x} \\
\frac{R_1 + R_2}{R_2} &= \frac{R_3 + R_x}{R_x} \\
1 + \frac{R_1}{R_2} &= 1 + \frac{R_3}{R_x} \\
\frac{R_1}{R_2} &= \frac{R_3}{R_x} \\
\end{align}</math>

=== Full derivation using Kirchhoff's circuit laws ===
First, [[Kirchoff's first law|Kirchhoff's first law]] is used to find the currents in junctions B and D:
: <math>\begin{align}
  I_3 - I_x + I_G &= 0 \\
  I_1 - I_2 - I_G &= 0
\end{align}</math>

Then, [[Kirchhoff's circuit laws#Kirchhoff's voltage law (KVL)|Kirchhoff's second law]] is used for finding the voltage in the loops ABDA and BCDB:
: <math>\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}</math>

When the bridge is balanced, then {{math|''I''<sub>''G''</sub> {{=}} 0}}, so the second set of equations can be rewritten as:
: <math>\begin{align}
  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{align}</math>

Then, equation (1) is divided by equation (2) and the resulting equation is rearranged, giving:
: <math>R_x = {{R_2 \cdot I_2 \cdot I_3 \cdot R_3}\over{R_1 \cdot I_1 \cdot I_x}}</math>

Due to {{math|1=''I''<sub>3</sub> = ''I''<sub>''x''</sub>}} and {{math|1=''I''<sub>1</sub> = ''I''<sub>2</sub>}} being proportional from Kirchhoff's First Law,  {{math|''I''<sub>3</sub>''I''<sub>2</sub>/''I''<sub>1</sub>''I''<sub>x</sub>}} cancels out of the above equation.  The desired value of {{math|''R''<sub>''x''</sub>}} is now known to be given as:
: <math>R_x = {{R_3 \cdot R_2}\over{R_1}}</math>

On the other hand, if the resistance of the galvanometer is high enough that {{math|''I''<sub>''G''</sub>}} is negligible, it is possible to compute {{math|''R''<sub>''x''</sub>}} from the three other resistor values and the supply voltage ({{math|''V''<sub>''S''</sub>}}), 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:
: <math>
\begin{align}
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{align}
</math>
where {{math|''V''<sub>''G''</sub>}} is the voltage of node D relative to node B.