I'm trying to understand the justification for turning off independent sources when calculating the equivalent resistance in Thévenin's theorem.

This answer explains has a good explanation but it assumes that resistance is the derivative of the voltage with respect to the current, dV/dI. I'm trying to relate this to Ohm's law.

For a resistor, the relationship is linear so dV/dI = V/I = R.

But for an ideal voltage source, dV/dI = 0 (short circuit) while the ratio V/I has some constant value depending on the current passing through it.

Similarly for an ideal current source, dV/dI = Infinity (open circuit) while the ratio V/I has some constant value depending on the voltage applied accross it.

So my question is does this mean that Ohm's law has a general form which is R = dV/dI and for the resistor (linear case) this is also equal to R = V/I? Or am I missing something?

Thanks in advance.

  • \$\begingroup\$ @DKNguyen the question is about dV/dI not dV/dt \$\endgroup\$ Commented Aug 26, 2020 at 14:15
  • \$\begingroup\$ Also AC analysis often works with dV/dI at a certain operating point. \$\endgroup\$ Commented Aug 26, 2020 at 14:16
  • \$\begingroup\$ @user253751 Oh, I see. Let me re-read. \$\endgroup\$
    – DKNguyen
    Commented Aug 26, 2020 at 14:16
  • 2
    \$\begingroup\$ Hmmm I still think it's not generally true since nonlinear resistors exist, and the resistance at any one point is the ratio, but not the slope. V/I at that point on the graph, not how quickly V is changing relative to I on that graph, or else I^2R does not equal V*I. \$\endgroup\$
    – DKNguyen
    Commented Aug 26, 2020 at 14:17
  • \$\begingroup\$ @DKNguyen So does this mean that the justification in the answer linked is incorrect? \$\endgroup\$ Commented Aug 26, 2020 at 14:24

3 Answers 3


Ohm's Law is the statement that \$V = I\cdot R\$. Two-terminal devices for which this relation holds are devices that we can model as a resistor.

On the other hand, the distinction between R = V/I and R = dV/dI is one of small-signal versus large-signal model, and is very important to the analysis of analog, linear circuitry such as amplifiers driven with small signals or in feedback.

Consider the following circuit:


simulate this circuit – Schematic created using CircuitLab

This is an inverting, common-source amplifier. I've neglected biasing details, but suppose that it is biased at a suitable operating point where it acts like an amplifier. We can model the drain current of the MOSFET as

$$I_d = k(V_{in} - V_{th})^2(1 + \lambda\cdot V_{ds})$$

where k and lambda are a consequence of the transistor's construction. For a small signal excitation around our operating point we can take a first-order Taylor expansion and model the drain current instead as

$$ I_d = I_0 + g_m\cdot(V_{in} - V_{in,0}) + g_o\cdot(V_{ds} - V_{ds,0}) $$

corresponding to the following small-signal model:


simulate this circuit

This model now corresponds to the small dI/dV derivatives that you ask about, and is valid for small excitations about the operating point. As you can see, the model draws resistors and linear dependent sources, even though they are non-physical--the MOSFET does not contain a parallel resistor that can be modeled by Ohm's law in the large-signal regime. Because it is linear, we use superposition to combine it with the constant voltages and currents seen at the operating point.

The take-away here is that we have all of our mathematical tools for linear circuits available when we deal with these small-signal models. Even though our MOSFET was hopelessly non-linear, we linearized it and can now do analyses such as finding Thévenin equivalents. We now deal with "resistors" that model the small-signal resistances dV/dI instead.

Looking at the transfer characteristic of our circuit and of the small-signal model, you can see that the name makes sense: it's a good model for small excitations about the operating point, and a poor model for larger excitations. The black trace is the real response, while the green dashed line is that predicted by the small-signal model.

enter image description here


For a pure resistor you get this relationship between voltage and current: -

enter image description here

The gradient is the resistance hence \$\dfrac{V}{I} = \dfrac{dV}{dI}\$

If the resistor were non-linear it would be a different story.


So my question is does this mean that Ohm's law has a general form which is R = dV/dI and for the resistor (linear case) this is also equal to R = V/I? Or am I missing something?

Ohm's law applies to resistors (and devices behaving like resistors, i.e. current being proportional to voltage) only.

So, no, there's no "more general" Ohm's law. We just call such things "differential resistance" or so, but it usually goes with the requirement of stating where exactly we're working at. nanofarad's answer illustrates that.


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