# computing Impedance related with Voltage Vx

I have a circuit like:

How do you get Thevenin ( $V_x = I_x * R$) circuit?

I have these formulas, but I am a little stucked...

$\ I_0 + Ix = (\frac{V\pi}{r\pi}) + gmV\pi$ --- 1
$VR_1 - VR_2 = Vx$ --- 2
$\ I_0 = (\frac{VR_2}{R_2})$ --- 3

How do you get $V_{R_1}$?

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Please phrase your question more clearly and provide more information about the problem. You are looking for the thevenin equivalent circuit for that entire thing? Or just for the input source (looks like input source might be Vx & R1 & R4 & R3, is that right?) This looks a lot like a hybrid pi BJT small signal model with some other resistor stuff around it... is that right? –  Adam P Apr 9 '11 at 16:58
sounds like a homework question? –  vicatcu Apr 10 '11 at 3:58
maybe exam question –  cMinor Apr 12 '11 at 22:03
@vicatcu, we have agreed in the past to handle homework problems. Now, it is important that no one gives the solution, but just explanation on how to approach the problem and how circuits are approached. –  Kortuk May 15 '11 at 4:04
what's the deal with using the 'earth ground' symbol three times across the bottom, and then using a 'chassis ground' looking thing in the upper right? –  JustJeff Jun 12 '11 at 1:59

I taught this at first year level for a few years, and here's the method I found students had the most success with:

1. Identify a part of the circuit where two passive circuit elements can be combined.
2. Combine those elements, calculating the new value.
3. Redraw the circuit. If you're learning how to do this, don't leave that step out! It's valuable for many reasons, not least that it helps the marker follow your reasoning.
4. Repeat from 1. until you have simplified the circuit as much as possible.
5. Perform a source transformation, where you convert a voltage source in series with a resistor into a current source in parallel with a resistor (or vice versa).
6. Redraw the circuit with the new values.
7. Repeat from 1. again.

Eventually you'll get down to one voltage source and a resistor in series with it - the Thevenin equivalent circuit. If you end up with a current source in parallel with a resistor you've got the Norton equivalent - perform a source transform to swap between them.

It also helps to draw a circuit diagram that is equivalent to the one you're given, but arranged to make the problem easier. I've had a stab at that in this image:

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$R1\|R2\|R3 = \frac{1}{\frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3}}$