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Thevenin's proof?

  1. May 2, 2006 #1
    Using thevenin equivalent is very good way to simply the circuit, because we know that the load can't "tell" the different between 2 circuits. But the thevenin's origin is really troublesome. I really don't know how to prove the thevenin's theorem or what's state of it. Kirchhoff Law has proof itself (electron into and out of a node must be equal). Node and Mesh current method are base on that proof too. But thevenin's theorem is different. It's really really annoy me!!! Please tell me how to prove it or which origin lead to thevenin's theorem...
     
  2. jcsd
  3. May 2, 2006 #2

    rbj

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    this is really an electrical engineering question. but the answer is pretty much what you said in your first sentence. put a voltage source of voltage [itex]V_T[/itex] in series with resistance [itex]R_T[/itex] in a black box and pull the two terminals out. now what is the "volt-amp" characteristic of that two terminal devices (the V-I characteristics fully describe the device, at least as far as the outside world is concerned)? if you define current coming out of the + terminal as positive (that is usually opposite of how we define it for loads), you will see that the V-I characteristic is

    [tex] V = V_T - R_T I [/tex]

    now do the same thing for a "norton source", an ideal current source [itex]I_N[/itex] in parallel with a resistance [itex]R_N[/itex] and pull the two terminals out. now what is the V-I characteristic?

    [tex] I = I_N - \frac{V}{R_N} [/tex]

    or

    [tex] V = R_N(I_N - I) = R_N I_N - R_N I [/tex]

    now you can easily see that the V-I characteristic of the two "different" 2 terminal devices are exactly the same if [itex] R_N = R_T [/itex] and [itex] V_T = R_N I_N = R_T I_N [/itex].
     
  4. May 2, 2006 #3

    robphy

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