Thévenin's theorem

by ECET undergraduate student Justin Flick.


The application of Thévenin's theorem enables multiple voltage and/or current sources and resistors to be represented by a single voltge source and resistor. This greatly simplifies analysis and troubleshooting a circuit when changes are made in areas other than the power supply.


One set of rules to 'Thévenize' a circuit is as follows.

  • Find the unloaded voltage output of the source using voltage divider rules.
  • Find the total resistance of the circuit across the nodes 'connected to the outside'. This can be done by replacing all voltage sources with a short all current sources with an open circuit.


Example:


A 15Vdc source is connected to a 4 resistor series parallel circuit with output nodes A and B.

Thevenin 001.PNG

Step 1. Use voltage divider rules to find the unloaded voltage at the output.

$ V_{out}=15V*\frac{5k\Omega}{(2k\Omega+3k\Omega)}=7.5V $

$ R_{total} = 8k\Omega + \frac{1}{(2k\Omega+3k\Omega)^{-1}+ 5k\Omega^{-1}} = 10.5k\Omega $


Verification:

Testing a Thevenin power supply under load should reveal that it is equivilent to the original. placing a 13kohm load in parallel with the output will load both circuits similarly. Thevenin 002.PNG

$ 15V_{Ein}-\frac{15V_{Ein}*5k\Omega_{load}}{((2k\Omega_{R3}+3k\Omega_{R4})^{-1}+(8k\Omega_{R2}+13k\Omega_{Rload})^{-1})^{-1}+5) }*\frac{13k\Omega_{R1}}{(8k\Omega_{R2}+13k\Omega_{Rload})}=4.148V $

$ 7.5V*\frac{13k\Omega}{(10.5k\Omega+13k\Omega)}=4.148V $


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