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Thévenin equivalent with dependent source

  1. Feb 26, 2014 #1

    Maylis

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    Gold Member

    Hello,

    I had read in my textbook that when a dependent source is in a circuit, and you wish to find the thévenin resistance, then the method to solve for the thevenin resistance is to turn off the independent sources and add an external voltage, then find the external current.

    The external voltage divided by the external current will give the thévenin resistance.

    However, when the textbook actually does the problem, the example does not even follow the method described. It just adds a short circuit across the terminal and treats it no differently than a problem would have been if there was only independent sources.

    I am now confused because they don't follow the method they said works, and I am trying to do it using the method described and cannot get the answer. Also, I know how to convert from thevenin to norton, so just assume I was looking for the thévenin first, and will convert to norton later.

    In fact, I will quote the paragraph that is hypocritical

    ''The equivalent-resistance method described previously [independent sources] does not apply to circuits containing dependent sources. Hence, and alternative variation is called for. Independent sources again are deactivated (but dependent sources are left alone) and an external voltage Vex is introduced to excite the citcuit. After analyzing the circuit to determine the current Iex, Rth is found by applying

    Rth = Vex/Iex''

    This paragraph is a blatant lie as far as I can tell, since the example posted is done using a previous method.
     

    Attached Files:

  2. jcsd
  3. Feb 27, 2014 #2

    NascentOxygen

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    Staff: Mentor

    Could someone confirm that Maylis's pdf file is a valid readable pdf file?
     
  4. Feb 27, 2014 #3

    gneill

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    It opened fine for me.

    Here's a snapshot of the circuit in question taken from the pdf:

    attachment.php?attachmentid=67079&stc=1&d=1393501247.gif
     

    Attached Files:

  5. Mar 1, 2014 #4

    NascentOxygen

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    Staff: Mentor

     
  6. Mar 1, 2014 #5

    NascentOxygen

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    Staff: Mentor

    Yes, you can do that. But you don't have to turn off internal sources, you can let them be and still add an external source, if you wish. They are linear circuits, so their behaviour doesn't change with drive level. I'd prefer that it be said that impedance is calculated as

    R = ΔV / ΔI

    rather than R = V/I so then we'd have just one equation that accommodates all circumstances.

    The variation at which you say you have taken umbrage carries nothing untoward. They shorted the ouput terminal to ground. That is nothing more than connecting a voltage source (of 0 volts) to the output terminals, then measuring the (change of ) current. Then using R = ΔV / ΔI

    All perfectly legal. :smile:

    What you can't do is short out voltage sources and rub out current sources and then try to simplify the internal resistor network. You can't do that to a circuit that contains dependent sources.
     
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