From the figure to the right, the bridge current is represented as I5
Per Thévenin's theorem, finding the Thévenin equivalent circuit which is connected to the bridge load R5 and using the arbitrary current flow I5, we have:
Thevenin Source (Vth) is given by the formula:
V t h = ( R 2 R 1 + R 2 − R 4 R 3 + R 4 ) × U {\displaystyle V_{th}=\left({\frac {R_{2}}{R_{1}+R_{2}}}-{\frac {R_{4}}{R_{3}+R_{4}}}\right)\times U}
and the Thevenin resistance (Rth):
R t h = ( R 1 + R 3 ) × ( R 2 + R 4 ) R 1 + R 3 + R 2 + R 4 {\displaystyle R_{th}={\frac {(R_{1}+R_{3})\times (R_{2}+R_{4})}{R_{1}+R_{3}+R_{2}+R_{4}}}}
Therefore, the current flow (I5) through the bridge is given by Ohm's law:
I 5 = V t h R t h + R 5 {\displaystyle I_{5}={\frac {V_{th}}{R_{th}+R_{5}}}}
and the voltage (V5) across the load (R5) is given by the voltage divider formula:
V 5 = R 5 R t h + R 5 × V t h {\displaystyle V_{5}={\frac {R_{5}}{R_{th}+R_{5}}}\times V_{th}}
Bureau of Naval Personnel, Basic Electricity, p.114, Courier Dover Publications, 1970 ISBN 0-486-20973-3. /wiki/ISBN_(identifier) ↩
Clarence W. De Silva, Vibration monitoring, testing, and instrumentation, pp.2.43-2.49, CRC Press, 2007 ISBN 1-4200-5319-1 /wiki/ISBN_(identifier) ↩
All about circuits: AC bridge circuits http://www.allaboutcircuits.com/vol_2/chpt_12/5.html ↩