Ok, so I know the point of the Thevenin is to make a simpler circuit so that that it results in the same current being sent to the other connected circuit. But why do we go through all the hassle of computing the Thevenin Voltage when we could choose any arbitary Voltage and a Resistor that would result in the same current?(adsbygoogle = window.adsbygoogle || []).push({});

For example: http://imgur.com/YOHCMr5

On the left circuit, the circuit is open on one end, so only the current goes through the R1 and R2 resistors. If you calculate the current, it is 2.67mA. The voltage drop across the R2 resistor is then 2.67V. Ok, so we use 2.67V as the Thevenin Voltage.

Now, we connect the open part of the circuit (still left hand figure), and calculate the current through R3 using KVL. The current through R3 is 1.6mA. If you recalculate it now, the current through R2 has changed: it is now 1.6mA, and the voltage drop across R2 is now 1.6V. But when the circuit was OPEN, the voltage drop across R2 was 2.67V <-- this was the Thevenin Voltage

So, why do we calculate Thevenin Voltage when the circuit is open (according to left diagram)? Why don't we use the voltage across R2 (1.6V) as Thevenin when everything is connected? You can still get the same current of 1.6mA by calculating the right resistor value.

If we use 2.67V as Thevenin Voltage, then to get 1.6mA current, Thevenin Resistance is 1.67kΩ.

If we use 1.6V as "Thevenin Voltage", then to get 1.6mA current, "Thevenin Resistance" is about 1kΩ.

As you can see, if the goal of Thevenin Equivalent circuit is to get the right resulting current of 1.6mA, why can't we just use ANY arbitrary voltage, and calculate the right resistance and call those Thevenin Voltage/Resistance?

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# Calculating Thevenin Equivalent Doesn't Seem Practical

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