# Prove that in a *nonlinear* circuit, adding resistor between equipotential terminals draws no current

If a circuit is linear, then you can use Thevenin equivalence to prove that if a port of a circuit has an open-circuit voltage of 0 V (the two terminals are at the same potential), then the resistor will not draw any current.

What if the entire circuit is nonlinear? Intuitively I think the fact still holds but Thevenin equivalence can't prove it anymore. Is there a counterexample to this?

• I’m voting to close this question because it appears to be homework without a showing of attempts at solution. Commented Feb 13, 2021 at 23:10
• Hi @MathKeepsMeBusy, this is in fact not a homework question. I would be very surprised if any modern circuits classes ask questions similar to this at all since nonlinear circuit theory is not commonly taught in my experience. I certainly didn't find any reference to a question of this sort anywhere, which is why I asked it on this forum. Commented Feb 13, 2021 at 23:14
• OK. I have retracted my close vote. Commented Feb 13, 2021 at 23:16
• I'd like it if some teacher did, though. Get their students thinking. Commented Feb 13, 2021 at 23:23
• How can you prove that a ball won't start rolling along a level surface unless you push it? Commented Feb 14, 2021 at 0:25

Let's take a very simple circuit -

And non linear element has V-I characteristic

$$\begin{array}{ll} V=V_o\cos(I)&\mathrm{if}\,I<\frac{\pi}{2},\,\mathrm{else}\\ I=\frac{\pi}{2}&\mathrm{if}\,V<0. \end{array}$$

We can draw V-I characteristic of non linear element as -

Now introducing a load resistor ,we have new circuit something like -

And now we find out what will be current through load resistor will be for different values of load resistor -

Conclusion -without load resistor the potential between two terminals are same (equipotential ) but as soon as you introduce a new load resistor between two equipotential terminals , current will flow through resistor and hence in non linear circuits zero open circuit voltage doesn't ensure zero current if we introduce a resistor between equipotential terminals!

• Thanks a lot for this example- but shouldn't the V-I curve of the nonlinear component be flipped about the I axis? Since for the open-circuit condition to have 0 volts at the open port, the voltage across the device should be Vo, not -Vo, I think. Commented Feb 14, 2021 at 4:42
• @knzy yeah , whatever you wrote is absolutely correct ,it should be in 1st quadrant , I'll edit it ! But conclusion doesn't change . Commented Feb 14, 2021 at 4:56
• @user215805 I think that both operating points are valid, I think once the resistor is introduced it could technically be in either one of these operating points, but since it "starts" at (0, 0) like I mentioned in my other answer, it seems like it wouldn't go to the other operating point unless it were somehow forced to. But I think this definitely proves that a resistor introduced to a port of nonlinear circuit initially at (0,0) could potentially be in a different operating point. According to Thevenin, for a linear circuit it's impossible. Commented Feb 14, 2021 at 6:45
• So the load line for the resistor intersects the output VI curve of the circuit in two places. But none of the operating points between are viable for the resistor. So in order to move from the zero current operating point to the non zero current point, the current and voltage would need to change instantaneously upon application of the resistor. Normally we consider instantaneous changes impossible. Commented Feb 14, 2021 at 9:18
• But the voltage between the nodes isn't zero before you add the resistor, is it? Commented Feb 14, 2021 at 13:37

If the non-linear circuit has a steady-state voltage, any resistor to the same voltage can not inject any current and thus no matter what resistance it is, it will be perceived as an open circuit. But if some external load changes which might affect that voltage, then current will flow and affect the circuit.

Ridiculous Example:

A tiny coin cell at 3.05V with an internal resistance of 100 Ohms has a nonlinear ESR that changes with %SoC, voltage and temperature.. If you attach a 100A 3.05V power supply, there is no change to the battery voltage. However, an external load voltage would be affected by this connection.

## Practical interpretation.

Say you had a portable nonlinear circuit and wanted to find the average current load without disconnecting the batteries. Using a programmable power supply with 4 digit V,I,P displays (like the one I just got) Set the voltage lower than what you expect , connect and turn on. Vbat is display and I=0. Now set V to the same voltage

and notice current is still 0 mA.

Cool.

Now raise the voltage 1% ( typical load regulation error) I had 3V from two AAA inside a 3 channel wireless digital thermometer) So I raised in 10 mV increments to verify battery charge raised current momentarily then dropped to read the external supply current where I assume the battery current was now 0. I could also estimate mAh and ESR of batteries by jogging voltage and computing dV/dI and C= I*dt/dV for the asymptotic decay time after raising the voltage. Now with my load current at 5 mA s.s., I could estimate if the 3 mos battery life met the spec of good batteries or not. I was getting only 2.4V and LCD was getting dim. I had just charged the alkaline primary batteries to 3V with a digital CC limit of 100 mA and then decided to perform this experiment, while watching TV and thinking about this question. (It's a Hanmatek 30V 3A) n.b. charging primary cells is possible but far less efficient than secondary cells. The only risk is corrosion leakage after a long time if using safe current limit.

You can simply resort to ohm's law: in a resistor, I = V/R. If V is zero, which it will be if the two ends of the resistor are at the same potential, then I must also be zero.

• The resistor doesn’t ‘know’ about the rest of the circuit; it only sees the voltage between its terminals. Similarly, the rest of the circuit can only ‘see’ that the resistance isn’t infinite by passing a current through it. Perhaps not a formal proof but quite good enough for me.
– Frog
Commented Feb 13, 2021 at 22:31
• Just because a circuit is non-linear doesn't mean we throw everything out the window and say the circuit can do whatever it wants, violate causality, and hack the internet to start a global thermonuclear war. I don't think proving a resistor with no potential across it has zero current is related to linearity. If you don't believe it for a linear circuit, then I don't think you should believe it for a non-linear circuit. And conversely. Commented Feb 14, 2021 at 6:16
• @mkeith Things get messy when you violate causality. I did that tomorrow and look what happened yesterday as a result! Commented Feb 14, 2021 at 6:19
• The question is which valid operating point the circuit will go to. My guess is that (as I pointed out in my answer to my own question below) since the "initial" operating point of the floating resistor matches the operating point of the (0V, 0A) port, that is the operating point that will manifest unless something else perturbs the circuit and forces it into a another operating point. Commented Feb 14, 2021 at 7:21
• @mkeith I already haven't deleted them. Commented Feb 14, 2021 at 21:39

Maybe this is a better example of a nonlinear circuit that shows no voltage across the R3 resistor.

• I would call that a linear, dynamic circuit. Where is the nonlinearity? Commented Feb 14, 2021 at 0:33
• steady state can apply to AC as well, but neither R nor C are nonlinear with respect to V/I for a sinusoidal dv/dt Commented Feb 14, 2021 at 0:42
• This is entirely linear, as far as I can tell. Commented Feb 14, 2021 at 2:33
• An example showing something doesnt happen in one situation doesnt prove there isnt some other situation where it doesnt happen.
– Matt
Commented Feb 14, 2021 at 2:35
• @Hearth the diodes are non-linear. Commented Feb 14, 2021 at 12:38

With no potential difference no current will flow in the resistor.

So you are essentially asking if can a circuit be disturbed without an extra current flowing. I think it should be clear that it cannot.

The only disturbance you're going to get from a resistor is Johnson (thermal) noise, but it could be argued that if the circuit is measuring Johnson noise between the terminals then they are not equipotential (else there would be no signal)

After a bit more thought I think this is the best I can come up with, similar to comments made by users in the other thread.

The nonlinear circuit's port of interest is, by the question's premise, at 0 V. Since it's also an open port, then the current between the terminals must also be 0 A.

The isolated resistor, before being attached to the port (and just floating in space somewhere), is clearly not conducting current, which by Ohm's law means its terminal voltage is 0 V.

So the circuit's port voltage/current are identical to the resistor's when it's floating in space, so the "operating point" of the resistor is immediately compatible with the port's when it gets attached, and so nothing changes and the resistor continues to draw no current and feels 0 V.