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Most physics text books say that a power source can be modelled as an EMF with a internal resistance. This is also know as Thevenin's theorem or Norton's theorem. However I have read in some sources that this is not always the case.

When does this model not apply/break down and for what reasons?

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Is electronics.stackexchange.com a better home for this question? –  Qmechanic Apr 25 '13 at 20:51
@Qmechanic, the presence of the thevenin tag suggests that it might get a better audience over there (and may even be a duplicate over there). But I also think it's on-topic here. –  Colin McFaul Apr 25 '13 at 21:43

4 Answers 4

There are actually two slightly different versions of Thevenin's theorem. I think what you are describing is the weaker of the two: you can replace any circuit with a single voltage/current source and a single resistor. That version holds for any two-terminal network made up only of voltage/current sources and ohmic resistors. It fails as soon as you add non-ohmic components to the circuit. The first non-ohmic components you'll likely encounter are capacitors and inductors. If all you have are resistors, capacitors, inductors, and voltage/current sources, then you can use the stronger version of Thevinin's theorem. In this case, you can still replace the network with a singe voltage/current source and a single impedance; as long as you use the complex impedance to describe the capacitors and the inductors in the network, then this works exactly as in the case with just resistors.

That stronger version holds for any circuit at all, regardless of how complex. Diodes, transistors, power supplies, voltmeters, computers, etc, all have an equivalent Thevenin circuit. Actually calculating the Thevinin impedance and voltage are horribly complicated, so in practice you almost always just measure them directly.

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Thanks for the answer. What I was trying to ask was what features of a power supply prevent this model from being used. For example can all batteries reduced to such a model? –  Josh Apr 26 '13 at 17:35
All batteries can be reduced to this model. A bench-top power supply can, as well. What makes it difficult to calculate the equivalent impedance for a power supply is the diodes and transistors that make it up. –  Colin McFaul Apr 26 '13 at 19:35
@ColinMcFaul: Thevenin's theorem only applies to linear networks. They can be time-varying (so the concept of impedance no longer applies) but they must be linear. (On the other hand, the load can be non-linear.) –  Art Brown May 25 '13 at 22:33

Since you mention batteries in a comment:

The voltage source + resistance model works well in many circumstances. At the extremes:

  1. high frequency / fast rise time waveforms: There will always be some inductance, from wiring if nothing else, which appears as an inductance, an easy add to the model.
  2. low frequency / long durations: As the battery discharges (or charges), its terminal voltage falls (rises). The simplest model substitutes a (large) capacitor for the fixed voltage source. More accurate models add a variety of distributed resistance and capacitive elements to simulate the electrochemistry inside the cell. More accurate yet is to make these elements non-linear (with capacitance a function of the voltage). People measure and study these characteristics intensively these days.
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When does this model not apply/break down and for what reasons?

Thevenin's theorem assumes linearity. So, if your circuit is non-linear, the theorem doesn't apply. However, one can linearize the circuit about an operating point and find a small-signal Thevenin equivalent circuit.

To find the Thevenin equivalent of a battery, you would (conceptually, at least) measure the open circuit voltage and then the short circuit current (not advisable in practice). The internal resistance is then just the ratio:

$R_{TH} = \dfrac{V_{OC}}{I_{SC}}$

Then, the voltage across the battery is:

$V_B = V_{OC} - I_B \, R_{TH}$

However, this model only approximates a real battery.

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It is doubtful if Thevenin's eqv also supports transients/dynamics. Suppose the ïnternals of a battery include a resistance as is usual and also an inductance in series with it. Concept of ïmpedance does not hold as supply is DC, and the same cannot also be converted to an eqv Norton's. If the supply is AC and the "source" internals include a series resistance + an inductance ALONG with a capacitance across the output terminals of this "practical AC source" will the Thevenin's eqv hold for transients? No.

Test by shorting the outputs. The original with a capacitive termination will give an infinite current at t= 0+, while the eqv Zth, if inductive, will give Zero current at t = 0+.

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