# Thevenin Equivalent

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?

• 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. – The Photon Jul 27 '17 at 21:28

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.

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.