Thevenin's Theorem Help

Discussion in 'Homework Help' started by Denton91, Dec 16, 2014.

  1. Denton91

    Thread Starter New Member

    Dec 16, 2014
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    Hello

    I have been given the following circuit, asking, using Thevenin's Theorem, to calculate the current in the 3.2 ohm resistor. Now, I'm familiar with Thevenin's Theorem, and replacing voltage sources with short circuits and current sources with open circuits, but truth is, the layout of this circuit is confusing me...hope someone can help me in the right direction.
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    1) Don't make people download a nearly 2MB file when you can almost certainly reduce it to about 1% to 2% of that size by simply resizing it to something like 400 pixels wide.

    2) You need to show YOUR best attempt to solve YOUR homework problem. We can't help you in the right direction until we see what direction you are headed in first.
     
  3. Denton91

    Thread Starter New Member

    Dec 16, 2014
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    Sorry, on my phone it's telling me the file is only 0.6mb, I'll have to sort it out when I get home unfortunately...and I'm trying to find the easiest way to put my workings up on here at the minute, without it sounding long winded.
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Last edited: Dec 16, 2014
  5. crutschow

    Expert

    Mar 14, 2008
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    For example if you short the voltage sources, then you can combine the series and parallel resistors into two resistors to calculate the voltages and currents from the current source.
    Then do a similar thing when you open the current source for the voltage sources.
     
  6. Denton91

    Thread Starter New Member

    Dec 16, 2014
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    That's how I had started originally, by removing the 3.2 resistor to make an AB platform, but i just didn't know whether to then to start working with the current or the voltage...i'll have a go suggesting your help, and feedback with my results. Thanks
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    Pick whichever you want. You need two pieces of information out of three possible:

    1) The voltage across the terminals with the load replaced by an open circuit.

    2) The current through the terminals with the load replaced by a short circuit.

    3) The resistance seen between the terminals with the load removed and all of the sources set to zero.

    Given any two, you can calculate the third. If you find (1) and (3) then you are done since you don't need (2) except to find (1) or (3) from the other.

    In each case, redraw the circuit as it exists for that case and then perform the analysis. At that point it's just a circuit like any other circuit you've worked with. Any technique that you've learned can be applied.
     
  8. Denton91

    Thread Starter New Member

    Dec 16, 2014
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    Thank you all for your help, I'm going to give it a real good go after dinner...sorry for the size of the picture, and not immediately putting my original workings, its hard work with the state my phone is in.
     
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