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I have been told that I can take part of a complicated circuit and find the Thevenin Equivalent in order to simplify it. What I do not understand is how I can determine the voltage and current at a particular point without including the resistance from the component that I do not include.

If I have a load and I want to simplify the rest of the circuit not including the load and I treat the load as an open, won't the voltages and current be different once I include the load?

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  • \$\begingroup\$ The thevenin equivalent doesn't tell you just one voltage and one current. It gives you a relationship \$V = f(I)\$ that tells you how voltage depends on current. This plus the characteristics of whatever's connected to the thevenin network will give the actual operating point. \$\endgroup\$ – The Photon Jul 27 '17 at 21:28
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Here is the procedure for a simple voltage divider. It should have been part of your introduction to Thevenin equivalents. Start with a simple voltage divider and a voltage source. Ignore the load.

schematic

simulate this circuit – Schematic created using CircuitLab

With me so far? You've converted the original voltage source and two resistors to a new, equivalent source and a single equivalent resistor. Use the formulas you've been given.

Now add a load.

schematic

simulate this circuit

Notice anything? You now have a simple voltage divider made by the Thevenin equivalent resistor and the load resistor, driven by the Thevenin equivalent source. Neither the equivalent source nor the equivalent resistance are affected by the load, so you only have to calculate them once for any any particular set of values.

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A Thevenin source is independent of the actual load. Put another way, it's a way of describing what you get for any load.

Simplifying something to a Thevenin equivalent (assuming it can be at all) has nothing to do with the load. However, it can be useful pretend certain loads to help derive the Thevenin parameters. Keep in mind that a Thevenin source only has two parameter: The voltage source voltage and the series resistance.

Knowing the load voltage and current at any two different points allows you to compute the two Thevenin parameters. It is often easiest to use open and short as the two loads because the current is 0 in the first and the voltage is 0 in the second. However, any two different points will do.

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