Does Thevenin's Theorem cover the case where the voltage source is changing with time? As an example of such a circuit, I'd like to replace a sensor, that produces a voltage, with a Thevenin equivalent in a model. Presuming I have a model of the internal resistance of the sensor and a model of the voltage change with respect to stimulus change, am I allowed to say that by Thevenin's Theorem the equivalent circuit is a variable voltage source and a resistance in series?

Note that I am asking specifically if this is permitted under the theorem. I know that I can model my sensor this way, I'm just not sure if I can invoke Thevenin as justification.

  • \$\begingroup\$ Sure, Thevenin's Theorem applies to AC and DC circuits. \$\endgroup\$
    – Andy aka
    Dec 17 '19 at 14:36
  • \$\begingroup\$ An by AC we specifically do not mean just sinusoidal voltages, but any varying voltage? \$\endgroup\$ Dec 17 '19 at 14:37
  • \$\begingroup\$ Any varying voltage can be modeled by a summation of individual sine waves. \$\endgroup\$
    – Andy aka
    Dec 17 '19 at 14:38
  • 1
    \$\begingroup\$ @Swedgin: continuous functions can only approximate other continuous functions. Where there are discontinuities like jump discontinuities then there are many examples where you cannot approximate to arbitrary accuracy using continuous functions. One of the most famous is the Gibb's Phenomenon, which happens when you try to use sinusoids to approximate a square wave: mathworld.wolfram.com/GibbsPhenomenon.html You literally can't. \$\endgroup\$ Dec 17 '19 at 14:53
  • 1
    \$\begingroup\$ @MichaelStachowsky The ringing involved in the Gibbs phenomenon doesn't go to zero magnitude as you add more terms, sure, but it does go to zero time. An infinite sum of sinusoids can indeed be exactly a square wave. \$\endgroup\$
    – Hearth
    Dec 17 '19 at 15:22

If you put it as an answer I'll accept it

Thevenin's Theorem applies to AC and DC circuits.

Oh go on then I'll add some more stuff from hyperphysics: -

enter image description here

  • \$\begingroup\$ LoL +1 from me. \$\endgroup\$
    – G36
    Dec 17 '19 at 14:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.