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Thevenin's theorem

  1. Jul 13, 2014 #1
    In this part , m doubtful about finding equivalent resisitance and thevenin's resistance (Rth)..
    m giving a circuit in whi there we have to find current through 4 ohm resistancce connected acreoss terminals A& B.., on left side of A, B there is a battery of 8V and a resitance of 2 ohm in series with at & on right side of A,B there is a battery of 12 V and resitance 2 ohm in series with it..
    for finding out thevenin's resitance , the 2 resiitances both 2 ohm are taken as parallel... so Rth=1ohm
    but for finding out the total current in the circuit the 2 resitances are taken in series..!!
    this thng is making me reallt doubtful... I understand finding the equivalent resistance but why for thevenin resitance the resistances were taken as partallel.. its a short example but the are more complex ones..which m 'nt getting due to ths..plzz explain it..!!:confused:
     
  2. jcsd
  3. Jul 13, 2014 #2
    thevenin resistance and thevenin voltage are two different things. They don't have anything to do with each other until you combine them in the equivalent circuit at the end.

    also, post a schematic. Its very difficult to give specific advice without a schematic.
     
  4. Jul 13, 2014 #3
    this is the ckt diagram with 2 resistances of 2 ohm each and 2 batteries of 8V and 12V respectively and the 2 terminals A & B open ..now plzz clear my doubt..
     

    Attached Files:

  5. Jul 13, 2014 #4
    your Rth of 1 ohm is correct.

    this statement
    is technically correct, but I think you are solving it the hard way.

    To find the thevenin voltage, you just have to solve for the voltage at A recognizing B as the ground. There are many ways to do this, but Node Analysis is the best. Have you studied that yet? You could also use Superposition. (and really good for this particular problem). Lets use Superposition. We turn off the voltage sources and replace them with close circuits, then we turn them on one at a time and add up the results. Va is a simple voltage divider, so you can do it easily in your head.

    You would also not use Rth at all when calculating this voltage.
     
    Last edited: Jul 13, 2014
  6. Jul 13, 2014 #5
    I know the other ways to solve it out..but thts nt what m asking for.>!!! m talking about that particular thing...
    if we are finding out equivalent resistance then we r considering the 2 series...which actly is clear..but why the 2 r being taken parallel when we r finding thevenin resisitance..??? How its so..? is we see from the open terminals A , B even then 2 is in series with other 2 ohm..what's the difference is there while solving in the 2 cases..???
     
  7. Jul 13, 2014 #6
    Think of the thevenin resistance as the resistance an outside current going into the terminals would see.

    Rth : Turn off all sources and simplify the circuit until you get 1 resistor.
    Vth : Find the voltage drop at the specified point using any manner of linear techniques.
     
  8. Jul 13, 2014 #7
    m nt asking Vth.. jst Rth and Requvilent.. n in both cases we hv to turn off the sources n simplify bt why in one case resistance are in parallel while in other series??
     
  9. Jul 13, 2014 #8
    In one case, you are concerned about current flowing into and out of the circuit through the terminals. In the other, you are not.
     
  10. Jul 13, 2014 #9

    jim hardy

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    When beginning, it is important to follow the rules exactly.

    Here's how i was taught:

    1. Measure open circuit voltage.
    I think that'd be 10 volts? Halfway between the two supplies?

    2. Replace all sources with their internal impedances.
    That's zero for a voltage source, leaving two 2 ohm resistors in parallel.

    3. Measure (or calculate) impedance looking back into the circuit: result is Zthevenin


    With practice it'll become intuitive why the rules work, furthermore you will be unable to remember a time when it was the least bit mysterious.
     
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