Connect with us

Stumped by a Thevenin Problem

Discussion in 'General Electronics' started by Mrs. Kerchief, Nov 4, 2003.

  1. I have to get a better understanding of Thevenin's theorem and have
    been unable even with the help of a tutor to come up with the Rth
    specified by my textbook for the following circuit:

    Vs--50V

    R1=22 Ohms || R2=22 Ohms, leading at the first node to R3=47 Ohms,
    then to the second node R4=33 Ohms. I've just started to learn how to
    Thevenize a circuit, and looking back from the open terminals, the
    circuit looks like this:


    ___22___
    _____| |______*______________A
    +| |__22___| |
    50| |
    -| 47
    | |
    |_______33__________|______________B

    I apparently got the right Vth by using the voltage divider formula
    (47/91) x 50V. But when I tried to get the "right" (according to the
    textbook) Rth of 22.7, there was no way I could get that. Looking
    back from the open terminals, I put 11 Ohms (the first two resistors
    equivalent resistance) in parallel with 47, then that in parallel with
    33. I continually came up with an answer of approximately 7.01, not
    22.7.

    If anyone could tell me what I did wrong, I would appreciate greatly
    hearing from you here on this group. Thanks greatly.
     
  2. Ken Taylor

    Ken Taylor Guest

    Do you need to parallel up any resistances - look more closely at the
    circuit.

    Cheers.

    Ken
     
  3. You s/b putting the 47 in parallel w/ the series combo of 11(22||22)in
    series w/ the 33. ie. 44||33
     
  4. JeffM

    JeffM Guest

    To do the resistance analysis,
    replace each voltace source with its internal resistance (zero ohms)
    and each current source with its internal resistance (open circuit).
    Then you get 44 || 47.
     
  5. jeffm wrote
    You added the equivalent resistance of the 22 resistors to the 33
    resistor and put that in parallel with 47, but I don't understand.
    The textbook arrives at Vth=25.8 (which could be arrived at only if
    you use RT of 91), so why does RT "change?"

    According to the book ("Principles of Electric Circuits, Conventional
    Current Version, Floyd), "The Thevenin equivalent resistance is the
    total resistance appearing between two terminals in a given circuit
    with all sources replaced by their internal resistances." Heck, maybe
    "internal resistance" is somehow different from plain "resistance."
    I'm a newbie, what do I know.

    The author's first schematic explaining Thevenin Resistance is:
    ________R1_______._________R3_________.A
    | |
    | |
    | |
    (Vs replaced by short)
    |
    | R2 |
    | |
    | |
    |________________._____________________.B

    "Rth=R3 + R1 || R2."
    -------------------------------------------------------------

    In the problem I asked help with, there's no resistor where R3 is in
    the above circuit. But if R1 and R2 are to be considered in parallel,
    what I don't understand is why R4 in the original problem (at the
    antipodes of the circuit) shouldn't also be considered in parallel.

    If anyone would be kind enough to offer a little more disquisition on
    the logic/procedure for computing Rth, I would be as grateful as I am
    for the answers you already provided. Thanks to everybody.
     
  6. JeffM

    JeffM Guest

    I don't understand is why R4...shouldn't also be considered in parallel
    The question is "Why should it?".

    You're making it harder than it is by making up new rules.

    There are only 2 steps needed here to solve for R equiv:
    1) Replace each voltage source with a short.
    2) Solve for the resistance between the specified points.
     
  7. Fraze

    Fraze Guest

    Everyone is telling you the right answer, but perhaps you just need to
    see the circuit drawn a little different. First off, replace the 50V
    supply with a short. Then move R1 and R2 down the line to the left
    side.

    ___________________*______________A
    | |
    ___|___ |
    | | |
    | | |
    22 22 47
    | | |
    | | |
    |______| |
    | |
    Now, I think we can agree that R1 and R2 are in Parallel. Lets go
    ahead and make an R equivalent instead of the R1 and R2.

    _________________*______________A
    | |
    | |
    | |
    | |
    11 47
    | |
    | |
    | |
    | |

    Now lets move R4, the 33-ohm resistor down the line to the left.

    _________________*______________A
    | |
    | |
    | |
    | |
    11 47
    | |
    | |
    33 |
    | |
    Now, our Req of 11 ohms and R4, 33 ohms are obviously in series, so
    lets add them and make another Req of 44 ohms.

    _________________*______________A
    | |
    | |
    | |
    | |
    44 47
    | |
    | |
    | |
    | |
    From here it should be evident that the new Req and R3(47ohms) are in
    parallel. This combination would equal 22.7ohms and there is your
    answer.

    When deriving Vth the circuit would look a little different.
    (R1||R2)+R3+R4=Rt You apparently did this right and came up the
    correct Vth of 25.2688V. When the power supply is a short and you are
    looking in from terminals A & B, the circuit would look like the above
    figure.

    Hope this helps.
    Fraze
     
  8. Fraze wrote
    Yeah! Now I finally saw it. THanks, Fraze. I never thought I'd
    spend so much time on a simple problem.
     
Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-