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Homework Help: Trying to find the short circuit current

  1. May 27, 2010 #1
    1. The problem statement, all variables and given/known data
    i have a circuit like this

    https://www.physicsforums.com/attachment.php?attachmentid=26017&stc=1&d=1274949363

    and i want to reduce it to thevenin's equivalent as seen from the terminals of R5

    2. Relevant equations

    IF i short circuit the resistor R5 in order to find the short circuit current at its terminals,then
    will this attempt also short circuit the 0.5A current source ?
    And then can i remove it ?


    3. The attempt at a solution
     

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  3. May 27, 2010 #2

    vela

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    It will short the current source, but you can't remove it.
     
  4. May 27, 2010 #3
    Can you please explain why i can't remove it?

    I already know that if the voltage drop or rise across it terminals is 0V then there is no current flow.

    Am i right?

    So i will have to include the sourse in the equations?
     
  5. May 27, 2010 #4

    vela

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    The current source will have a 0.5-A current flowing through it regardless of the voltage drop across its terminals. That's what a current source does by definition. If you want to find the short-circuit current, you'll have to include the current source in your calculations.

    If you're trying to find the equivalent resistance of the circuit, you can set all the sources to zero -- that means removing all the current sources and shorting all the voltage sources -- and find the resistance of the remaining circuit.
     
  6. May 27, 2010 #5
    OK thanks a lot for replying.
    but If it is a resistor instead of a current source what should i do?
     
  7. May 27, 2010 #6

    vela

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    What do you mean?
     
  8. May 27, 2010 #7
    if i replace the current source with a resistor and then make a short circuit at it's terminals then should i remove the resistor?
     
  9. May 27, 2010 #8

    vela

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    Why would you replace the current source with a resistor?
     
  10. May 27, 2010 #9
    Let me see. If I short a current source, then it inputs as much current into a node as it removes, and I can disregard it.
     
  11. May 27, 2010 #10

    vela

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    Not if you're looking for the current flowing through the node.
     
  12. May 27, 2010 #11
    just an example because i am a little confused.

    So if i short circuit a current source it continues to deliver current to the circuit ,but if i short circuit a resistor then no current flows throw it's terminals right?
     
  13. May 27, 2010 #12

    vela

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    Right. If you short a resistor, the voltage across the resistor will be zero, so according to Ohm's law, the current through it will be zero.


    Let's go back to your original question about shorting/removing a current source. I thought of a simple example of showing if you remove the current source, you'll get the wrong answer: Consider a current source I in parallel with a resistor R, and say you want to find its Thevenin equivalent (which, as you should know, is a voltage source V=IR in series with a resistor R). The Thevenin resistance is the ratio of the open-circuit voltage Voc and short-circuit current Isc. The open-circuit voltage will just be Voc=IR. Now, if you say you can remove the current source because you're shorting it out, you'd then be left with just the resistor R connected to the short. The short-circuit current would therefore be Isc=0, and the Thevenin resistance would be infinite. Clearly, that's not correct.

    The correct analysis is: When you short the nodes, you can remove the resistor, and all the current from the source will flow through the short, i.e. Isc=I. The Thevenin resistance will therefore be Voc/Isc=IR/I=R. The Thevenin equivalent will consist of a voltage source Voc=IR in series with resistance R, as you'd expect.
     
  14. May 28, 2010 #13

    The Electrician

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    One way to solve this is to remove the I2 source, solve the network for the short circuit output current, and then add 1/2 amp to that result.
     
  15. May 28, 2010 #14
    Thanks a lot .I think i understood your point ,i will just make some more examples to make it clearer.
     
  16. May 28, 2010 #15
    OK, that makes sense, though current doesn't flow through a node, but in and out of a node to curcuit elements. And we are really taking the limit as R5--> its infinitesimal. We are looking for the current between two nodes as R5 becomes small.
     
    Last edited: May 28, 2010
  17. May 28, 2010 #16

    The Electrician

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    I'm sure by now you have learned how to do a mesh (or loop) analysis. Look at your circuit--there are 4 obvious meshes; one of them includes R5. If you were to perform a loop analysis, you could calculate a current in a loop that includes R5. You could then replace R5 in the schematic with a straight line, representing a wire. That same loop would still be there, and the current in the loop would be the current in the wire (short) that replaced R5.
     
  18. May 28, 2010 #17

    The Electrician

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    You don't even need to short the R5 terminals to get the Thevenin eqivalent.

    Remove R5 and caculate the open circuit voltage at those terminals.

    Short the voltage source, open the current sources and calculate the equivalent resistance seen from the R5 terminals, and you now have the Thevenin equivalent.
     
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