thevenin's help

ok i have been given this as a part of my electonics assignment, ive spent ages pondering on it and have got nowhere. help would be greatly appreciated

Thévenin's theorem for linear electrical networks states that any combination of voltage sources, current sources, and resistors with two terminals is electrically equivalent to a single voltage source and a single series resistor. The single voltage source is the open circuit voltage (no load) at the two terminals. The single series resistor is the resistance looking back into the network from the two terminals with all the sources zeroed, i.e., voltage sources are shorted while current sources are opened.

So your circuit has four terminals: 0, 1, 2, 3. Which two will you choose to be the terminals of your Thévenin equivalent network?

I see it rather simply. The left and center resistors can be combined to a single series resistor equal to their parallel combination, excited by an AC source equal to the dividing ratio. Now you have a single source and a single resistor. So you have 6.67 Ohms in series with 33.33V source.

That is loaded by the 20 Ohms in series with the inductor. So you have, if you combine parts, 33.33 V in series with 26.67 Ohms and the inductor. Now with only a source, a resistor, and an inductor, you can compute the current.

what? How can two resistors in series be equal to their parallel equivalent?

Yes they can. Think of V1, R1 and R3 as a single voltage source with an equivalent output resistance given by the parallel connection of R1 and R3 AND you'll have to consider the reduced output voltage which is V1 divided by the voltage divider R1/R3.
If you do the math, it comes out the same.

Harald

Guess I will have to go back and re-read thevenin's theorem again. Because I am now lost on what I thought I knew. *shrugs*