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Thevenin's Theorem


couch_potato

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Firstly you have written Rth as 8k in red, which is 10 times too much.

 

Edit I see you have .8 k sorry - a piece of advice never omit the leading zero before the decimal point. Rth = 0.8k

 

Secondly they say (correctly) exactly how the thevenin voltage is calculated in their text.

 

Remove the load (20k in this case)

 

Use simpler circuit laws, eg potential divider (in this case), KCL, KVL etc on what is left of the circuit to determine the open circuit voltage between the terminals of interest (A, B in this case).

 

Just follow that proceedure on a couple of examples to get the hang of it.

 

It seems a bit silly on such a simple example, try it on a more complicated one.

Edited by studiot
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For the benefit of whoever liked my previous post the thing about Thevenin is that it replaces the whole circuit to the left of the load between point A and B with a contant voltage generator in series with a resistor.

 

The purpose of the CV generator is to provide the voltage that is actually applied across AB by the missing (replaced) circuit.

The value of this voltage generator is called the Thevenin voltage.

The Thevenin voltage is not the voltage across both the CV generator and the series resistor, it is just the value of the CV generator.

 

The purpose of the resistor is to take account the loading effect of the load so in this case

 

Vth is the 8 volts that appears across AB without the load as given by the potential divider ie 4/5 of the 10 volts across the 4k resistor in series with a 1k resistor.

 

In the real real circuit the connection of the load 20k resistor on the right hand side of AB is in parallel with the 4k resistor so it reduce the the actual voltage across AB in the actual complete circuit below 8 volts.

Thus Rth is added to provide the Thevenin circuit with the 8 volts divided between Rth and the 20k load.

 

 

FYI Norton analysis works in exactly the same way except that a constant current generator with a parallel resistor is used.

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