Thevenin's Equivalent Help

Discussion in 'General Electronics Chat' started by sjgallagher2, Aug 13, 2013.

  1. sjgallagher2

    Thread Starter Member

    Feb 6, 2013
    111
    7
    For some reason I find immense difficulty in figuring out these circuits. I must be missing something somewhere :p Here's my issue. I have this circuit:
    [​IMG]
    And I need to find the Thevenins equivalent circuit. When I first got the problem I confidently threw some voltage divider equations around and got the thevenin voltage as 0.75V. But that's wrong. It's actually 0.5V. Why? Thats where I need help! If you could explain that to me I would be very thankful. Anyways, the next was the equivalent resistance, which I found by using parallel resistance formulae to be 50Ω. Wrong again, it's 67Ω. I've had difficulty with this problem before, and last time after an arduous time I finally solved it the right way, but I can't find those notes at the moment.
    If someone could please at least tell me what I did wrong I would be immensely thankful.
     
  2. sjgallagher2

    Thread Starter Member

    Feb 6, 2013
    111
    7
    To be more clear about what I did to find my answers--
    In solving the voltage:
    First I used a voltage divider equation on the 200ohm resistors. That resulted in 1/2(3V) = 1.5V. Then I put that voltage in the next voltage divider equation with the 100ohm resistors, which came out as 1/2(1.5) = 0.75. It seemed so simple, but I went wrong. Where?
    In solving the resistance:
    I found the equivalent resistance of the 200ohm resistors, which gave me 100ohms. Then I found the equivalent resistance of the 100ohm resistors, 50ohms. Putting the two together yielded my answer of 150ohms, where did I go wrong?
    Thanks
     
  3. sjgallagher2

    Thread Starter Member

    Feb 6, 2013
    111
    7
    Okay nevermind the resistance, I found what I had done wrong, I shouldn't have found both parallel resistances and added them, I should have found the first parallel equivalent, then added that to the 100ohm resistor, and then done another equivalent. That gets 67ohms! As for the voltage, I'm still working on it, all help is appreciated
     
  4. BobTPH

    Active Member

    Jun 5, 2013
    785
    114
    The two 100 Ohm resistors are in series making 200 Ohms.

    This is in parallel with a 200 Ohm going to ground, making 100 Ohms.

    So now you have a voltage divider with 200 Ohms on top and 100 Ohms on the bottom, for and output of 1V. The two 100 Ohm resistors divide this in half to get 0.5V between A and B.

    Bob
     
  5. sjgallagher2

    Thread Starter Member

    Feb 6, 2013
    111
    7
    So what you're saying I think is that you start with the original circuit:
    [​IMG]
    Then you simplify like this:
    [​IMG]
    and then you simplify again to get this:
    [​IMG]

    Well you lost me on the first step :p Where do the terminals go there? How come you can say the two are in series when the terminals are parallel to them? Aren't the terminals parallel to them?
     
  6. sjgallagher2

    Thread Starter Member

    Feb 6, 2013
    111
    7
    Oh I see, you were referencing the voltage divider! Problem solved now, thanks.
     
  7. PackratKing

    Well-Known Member

    Jul 13, 2008
    850
    215
    What parameter do you seek in the circuit that Ohms' Law won't give...
    IMHO Thevenin just adds unnecessary complication to arrive at an *equivalent *

    The easiest way to do this, is build it and measure it...
     
  8. vk6zgo

    Active Member

    Jul 21, 2012
    677
    85
    (1) Draw a dotted vertical line on your circuit,separating the two 100Ω resistors from the LHS pf the circuit.

    (2)Find the Thevenin equivalent of the circuit to the left of the line as
    follows:

    This circuit is a source with a voltage divider consisting of two 200Ω resistors connected across it.---what is the Thevenin equivalent voltage?

    The Thevenin equivalent resistance is obtained by replacing the 3V source with a short circuit----what is the equivalent resistance ?

    (3) Redraw the circuit with the LHS replaced by its Thevenin equivalent.
    Now find the Thevenin equivalent of the resulting simpler circuit.

    Sorry to be vague,but there is a fine line between helping you & doing the work for you.
    If you revise Thevenin analysis the steps I included will be spelled out in most texts.
     
  9. vk6zgo

    Active Member

    Jul 21, 2012
    677
    85
    Build it,measure it,& it will agree with Thevenin!
     
  10. WBahn

    Moderator

    Mar 31, 2012
    17,768
    4,804
    And therein lies your first mistake. You are picking equations and throwing them at a problem hoping something good will happen as a result. This is Analysis by Happening, and it seldom ends well.

    And therein lies your next mistake. Before you grab an equation and throw it confidently at a problem, you need to confidently establish that the said equation is actually valid for that problem. This comes from understand what the limitations on the equation are and this comes from understanding where the equation came from. The voltage divider equation is based on two series-connected resistors, meaning that whatever current is flowing in one resistor is the same current that is flowing in the other. Are the two 200Ω resistors in series? If not, then you can't throw the voltage divider equation at it and expect valid results.
     
  11. PackratKing

    Well-Known Member

    Jul 13, 2008
    850
    215
    Never said it wouldn't... :rolleyes:
    What is the point to computing redundant *equivalents* when Ohms' law will do nicely, with the same end result ??? :p

    Is it possible Thevenin couldn't hack Ohms' law ??? :D seems he was a wee bit of a putz
     
  12. WBahn

    Moderator

    Mar 31, 2012
    17,768
    4,804
    Thevenin and Norton equivalents are extremely power tools. Ever use a function generator? Did you take the schematic for the circuitry in the generator and add the schematic for your circuitry and then analyze the whole thing using just Ohm's law (and KCL/KVL)? Or did you set the function generator to a certain voltage and then use the fact that it has a 50Ω output impedance? If so, guess what? You used a Thevenin equivalent circuit for the entire function generator!
     
  13. PackratKing

    Well-Known Member

    Jul 13, 2008
    850
    215
    To be honest, my foray into electronics, is not so deep that I need to analyze circuits to that extent... things either work, or they do not. I am a repairman - not a designer per se.

    That kind of analysis is only necessary for those building or repairing the like of motherboards at the component level, which is WAY out of my league anyway...

    I have built several pieces of test equipment for my own use, -- a Hi-Pot analyzer for work on large motors and power tools, among other things... both an xenon-tube DC - 20K Hz analytic strobe for stop-motion study, and an LED version of same...
    Yes, I have used function gens, and built my own version of a *signal injector* ... which was not much more that a 555 tone generator I could trace signals through multi-stage stereo amplifiers with.

    After tech school for cameras on my GI Bill, circa 1978, my main claim to fame, is 30+ years work on high end Nikon / Canon/ Minolta / Leica 35mm SLR film cameras, Hasselblad and other similar larger format film units... I am getting into DSLR's as time allows.
     
Loading...