[Circuit theory] Current does not branch in these problems? [Visual examples]

Discussion in 'Homework Help' started by ZeroTorrent, Mar 29, 2012.

  1. ZeroTorrent

    Thread Starter New Member

    Mar 28, 2012
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    Hello again!

    I'm always getting stuck on problems where I have several current and/or voltage sources in various combinations. In both examples i'm supposed to create a Thevinin circuit but I can manage R_TH myself.

    Example 1 shows how I normally assume currents go and attempting to calculate this will be wrong every time except when I assume that the two loops does not interact with eachother.

    Example 2 is pretty much the same deal. If I assume that the entire circuit is made up by 3 closed circuits I get the right answer.

    Can someone please explain why the current I does not stray from the closed loops despite it being connected to the other loops?
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,959
    1,097
    In the first Example 1 all you components are connect to the same node.

    [​IMG]

    I hope that now you see why you can treat this circuit separately.


    As for Example 2 the answer is not so intuitive.
    But I think that "close loop" is the right answer.

    Current will flow only in closed loop circuit. Can you see in the diagram any closed path (closed loop) for current to flow between E1 and E2? Or between E1 or E2 and current source?
    First Kirchhoff's law says that current leaving the "+" of a voltage source must be equal to the current that is flowing into "-" terminal.

    [​IMG]
     
    Last edited: Mar 29, 2012
    ZeroTorrent likes this.
  3. ZeroTorrent

    Thread Starter New Member

    Mar 28, 2012
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    Very nice picture that last one :) It's very clear what happends and it adds some intuition!

    Thank you!
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,760
    4,800
    [QUOTE}Can someone please explain why the current I does not stray from the closed loops despite it being connected to the other loops? [/QUOTE]

    Because in order for current to stray out of a loop, it has to have some way to get back eventually.

    In your first diagram, is the red current supposed to be from the voltage source and the blue current supposed to be from the current source? If so, then remember that current has to flow in a closed circuit. With only one conductor connecting the left and right halves, there is no way for current that comes from either side to get back. Hence, no current is flowing in that wire. As soon as you put a load across the terminals, that will change, but for calculating the open circuit voltage, you only have one point of connection and you have to have at least two to make a circuit.

    In the second diagram, consider the slightly modified diagram attached where all I have done is rearrange the wires a bit without changing the connections at all. Can you see how the observations regarding the first diagram come into play? Consider any of the three loops and you will see that there is no way for current to leave the loop into the rest of the circuit and ever return. Therefore, the current in the little corner wires will be zero. This allows you to write down the open circuit voltage by inspection.

    Frequently, redrawing the circuit so that it looks more like you are used to seeing things can really help. In this case, when I first looked at the second diagram, I thought, wow, quite a few loops to wade through. But with just a moment's consideration, you can see how the loops can't interact (until a load is added).

    Consider the more heavily redrawn version in Example 3. Here you should be able to convince yourself very quickly that the open circuit voltage is:

    <br />
V_{oc}=-R_4\frac{E_2}{R_4+R_5}+I*R_1-R_3\frac{E_1}{R_2+R_3}<br />
     
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