# Thevenin Equivalent Circuits

1. May 13, 2016

### Mjmuk

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Hiya, this is my first time posting so apologies for any errors in formatting.
I am confused as to whether when Thevenising a portion of a circuit is it to simplify BUT provide the same output characteristics, meaning the workings of the Thevenin part of the circuit may behave completely differently to the original circuit?

I have an example which may explain this better (hopefully it attached, if not here's a link http://imgur.com/VGa0TI5 ) by Thevenising the left hand side initially I assumed it would make the question easier but the answers turn out to be different when concerning the resistor in particular, is this because the Inductor 'sees' the same characteristics as before but the resistor behaves differently?

Many Thanks

Last edited by a moderator: May 13, 2016
2. May 13, 2016

### phinds

If the "Thevenin equivalent" is not exactly equivalent as far as the rest of the circuit is concerned, then it is not equivalent, so not a Thevenin equivalent. How it works internally is irrelevant as long as the rest of the circuit can't tell the difference.

As far as I am aware, Thevenin equivalents are for resistive circuits, not reactive circuits.

3. May 13, 2016

### Staff: Mentor

There were 3 files attached. They seemed to be identical so I have deleted two.

4. May 13, 2016

### cnh1995

I believe Thevenin equivalents are also for reactive circuits.(e.g. In case of ac circuits, maximum power is transferred to the load if the load impedance is complex conjugate of the Thevenin impedance viewed from the load terminals.)
The circuit shown in the OP can be reduced to its Thevenin equivalent.

5. May 13, 2016

### phinds

OK, it's been a very long time since I took EE and if I ever knew that I had forgotten it. Thanks.

6. May 13, 2016

### cnh1995

The Norton source can be replaced by its Thevenin equivalent. A source transformation will do. To find the time constant, effective R and L need to be in series.