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Thevenin Equivalent of a Circuit

  1. Aug 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the Thévenin equivalent of the circuit shown in FIGURE 1.

    2. Relevant equations

    Any two-terminal network can be replaced by an equivalent circuit consisting of a voltage source and a series resistance equal to the internal resistance seen looking into the two terminals.


    3. The attempt at a solution

    Figures 2 & 3

    I'm unable to figure out how to proceed from here or if what i've done so far is correct. Can't seem to find any examples anywhere of anything similar . Any help or useful sites would be much appreciated.
     

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  2. jcsd
  3. Aug 17, 2013 #2

    vela

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    Your figure 2 looks fine. You simply set the voltage source to 0. I'm not sure how you got to figure 3 though.

    It would probably help if you redrew the circuit by taking the two resistors at the top and placing them so that they're obviously in parallel with some of the other resistors. (Individually, that is. They're clearly not in parallel to each other.)
     
  4. Aug 18, 2013 #3
    Thanks for getting back so quick. I've amended the figure as recommended, please see attached.

    I know that for resistors connected in parallel, the supply current is equal to to the sum of the currents through each resistor, and that when resistors are connected in series, the total resistance of a number of resistors is equal to the sum of all the individual resistances. Is there any advice for how I can go about simplifying the network?
     

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  5. Aug 18, 2013 #4

    vela

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    That's not correct. You removed the wire on the left.

    What two nodes is the upper left resistor connected to? What other resistor is connected to the same two nodes? Those two resistors are connected in parallel. You can redraw the circuit to make that fact more obvious. Do the same analysis for the upper right resistor.
     
  6. Aug 18, 2013 #5
    Ok thanks,

    So a node is, any point on a circuit where three or more circuit elements meet. I've tried to colour the diagram to show this. If this is right then the resistor which has the same two nodes as the top left resistor is the bottom left resistor. At the moment these are connected in series or am I being nieve?
     

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  7. Aug 18, 2013 #6

    vela

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    Because they're connected to the same two nodes, they're in parallel.

    They're not in series. If they were in series, the current that flows through one has to flow through the other. That's clearly not the case. Series means they're connected in a line, with nothing branching off in the middle.
     
  8. Aug 18, 2013 #7
    Thanks,

    So for the left hand side would the attached be acceptable? With the addition of the load on the right side of the circuit i'm really struggling to see how something similar could be achieved here?

    Thanks for your patience here, i'm really struggling to get a grasp of what's going on.
     

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  9. Aug 18, 2013 #8

    vela

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    Yes, that's fine. In fact, you could bring it all the way down so the two lefthand resistors are actually drawn parallel to each other.

    For the resistor on the right, have the upper wire loop down on the right side of the circuit instead of going around to the left. As long as that end of the resistor is connected to the same node as before, it doesn't matter how you draw the wire.
     
  10. Aug 18, 2013 #9
    Right so the circuit looks like the attached, thanks. I'll have a go at simplifying.
     

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