The Wheatstone Bridge circuit
consists of two potential dividers connected between the supply and ground. One divider is
formed from R_{1} and R_{2}, the other from R_{3} and R_{4}.
The voltage at the point between R_{1} and R_{2}, and that between R_{3} and R_{4}
will vary according to the resistor value. V_{bridge} is the voltage between these
points, as shown in the diagram below:

When the voltages are equal V_{bridge} will
be zero, at which point the bridge is said to be balanced. At this point the value of R_{4}
can be found from:

R_{4} = (R_{3} * R_{2}) / R_{1}

A typical application for a Wheatstone Bridge is for
measuring resistance; if R_{1} is variable and is adjusted it until V_{bridge}
= 0 the value of R_{4} can then be found from the above equation.

If all four resistor values (R_{1} to R_{4})
and the supply voltage (V) are known the voltage across the bridge (V_{bridge})
can be found by working out the voltage from each potential divider and subtracting one
from the other. The equation for this is:

V_{bridge} = (R_{4} / (R_{3} + R_{4})
* V) - (R_{2} / (R_{1} + R_{2}) * V)

This can be simplified to:

V_{bridge} = ((R_{4} / (R_{3} + R_{4}))
- (R_{2} / (R_{1} + R_{2}))) * V

If the bridge is balanced the equivalent resistance of the circuit between V and GND is:

R_{1} + R_{2} in parallel with R_{3} + R_{4}

R_{E} = ((R_{1} + R_{2}) * (R_{3}
+ R_{4})) / (R_{1} + R_{2} + R_{3} + R_{4})