In this video, the tutor finds a negative Thevenin resistance for a circuit. How does this make sense? What is the physical meaning of this?

  • $\begingroup$ I have not heard of negative resistance in the context of an RLC circuit. Which was the video that you watched? $\endgroup$ – Farcher Sep 9 '16 at 4:52
  • $\begingroup$ youtube.com/watch?v=nE8zFgpvIwM $\endgroup$ – rathod dinesh Sep 9 '16 at 4:56
  • $\begingroup$ The text of your question has nothing to do with your actual question, so I rewrote it. $\endgroup$ – knzhou Sep 9 '16 at 6:16

Negative resistance is not something you see in normal circuits. However, it can show up in small-signal models of non-linear circuits.

A positive resistance means that if you increase the voltage, it increases the current. However, in some nonlinear circuits you can get into regions where, locally, a small increase in voltage can decrease the current that flows. If you do a small-signal model, some of those non-linear elements get replaced by resistors, and those virtual resistors may have a negative resistance in some cases.

I can't think of a scenario where I've seen it in a RLC circuit, but I've most certainly seen it in active circuits, where you can have amplifiers.


Thank you for the video reference which does not sem to be a RLC circuit.

However it is an example of the use of the Thevinen equivalent circuit for maximum power transfer when $R_{\text{Thevinen}} = R_{\text{Load}} $

The conversion of an actual circuit to a Thevinen equivalent is a mechanism for solving circuit problems and does not mean that the Thevinen equivalent circuit replaces the real circuit in all aspects.

For example when you have found the maximum power dissipated if the load resistor that is not necessarily equal to the power dissipated in the Thevinen equivalent circuit.

You only have to look at the circuit below and its Thevenin equivalent to see that they are not equivalent as far as internal power dissipation is concerned-.

enter image description here

With no load connected the power dissipation in the original circuit is $\frac 3 4 $ W and the power dissipated inside the Thevenin equivalent circuit is zero. The power dissipated in the load is zero.

You can then show that if a load resistor of 3 $\Omega$ is added to each of the circuits the power dissipated in the load resistor is the same in both cases $=\frac {3}{16}$ W whereas the power dissipated internally is not the same, $\frac {15}{16}$ W and $\frac {3}{16}$ W.

In your example, with a dependent voltage source, the resistance of $-4 \; \Omega$ is just a value which makes the calculations easier but it is not a real resistor.

In the video the lecturer makes the following transformation.

enter image description here

and notice that he has included a resistance of $+4 \;\Omega$ in the external circuit because he knows that on finding the Thevinen equivalent circuit a negative resistance of $-2 \;\Omega$ will appear.

To show that Thevinen actual works with an unrealisable $-2 \;\Omega$ resistor I suggest you try the following.

Get rid of the $+4 \;\Omega$ resistor in the load circuit and just have the $2 \; \Omega$ as the load.
Find the current through the external $+2 \; \Omega$ resistor using simple circuit analysis.
My guess it that you will find that it is infinite or indeterminate.

Now investigate the effect of having not having the $4 \; \Omega$ "load" resistor in the circuit both using the Thevinen equivalent circuit and simple circuit analysis and with $0 < R_L<2\; \Omega$
This should give an insight of how the circuit functions under different loads.

An example of a device which has a negative incremental/small signal/differential resistance is the tunnel diode.

enter image description here

Where if $v_1<V<v_2$ the reciprocal of the gradient of the graph $\frac{dV}{dI} = r_{\text{differential}}$ is negative.
This means that in that region if the voltage $V$ increases the current $I$ decreases.


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