Thevenin's with ground

Discussion in 'Homework Help' started by Hawkeye87, Oct 8, 2008.

  1. Hawkeye87

    Thread Starter Active Member

    Oct 7, 2008
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    ok, yet another Thevenin's theorem question. I looked through some of the forums and ebooks but couldn't find anything relating to ground. The questions says.....Using Norton's theorem, find the current through R1. Now i know that you have to make a Thevenin equivalent then use Norton's to find I1 but would you leave R1 out as the load and find the rest to be Rth? Or how would you go about this. I tried it a few times and i got a resistance of 5.52kΩ.
     
  2. guitarguy12387

    Active Member

    Apr 10, 2008
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    Do you have to use norton's thm? Seems like alot of extra work. I'd just use nodal analysis...
     
  3. hobbyist

    Distinguished Member

    Aug 10, 2008
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    For this particular question on the schem. there not refering anything to the current arrow or point (A). ?

    If not point (A) being in this question then my guess would be R1 taken out and then find the norton equiv. or use mesh analysis to solve for current through R1.

    That's my guess...
     
  4. Hawkeye87

    Thread Starter Active Member

    Oct 7, 2008
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    yes, unfortunately i do have to use the Norton Equivalent. I haven't learned node analysis i think. haha all this is confusing to me. Mind refreshing my mind on node analysis to see if i'm not confusing it with something else?
     
  5. Hawkeye87

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    Oct 7, 2008
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    They're actually using one figure to ask like 6 questions about it. Let me recalculate with R1 taken out and get back to you. Also, i think we're learning Mesh analysis in chapter 9 which we start on Fri. Right now we have to use Thevenin's and Norton's.
     
  6. Hawkeye87

    Thread Starter Active Member

    Oct 7, 2008
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    anybody get 13.24kΩ for Rth?
     
  7. guitarguy12387

    Active Member

    Apr 10, 2008
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    Nodal analysis = systematic application of KCL

    How do you use Thev/Norton without using some kind of analysis method (mesh/node)?
     
  8. Hawkeye87

    Thread Starter Active Member

    Oct 7, 2008
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    My teacher must refer to it as something else then. But i do know what KCL is. I would just think with this question it would be a bit extra work. Would it help if i say i'm in a beginners course on circuits? It's just the basic fundamentals of circuits.
     
  9. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Hi

    Always remember to put a short across the Voltage sources when calculating Rth.

    To help you out, if you short out the volt. source, what resistance in the circuit would not show up if you put a ohm meter across the input where R1 use to be at.

    Then to find the Vth. imagine putting a voltmeter across the leads where R1 use to be at, and always remember a voltage drop across a resistance only occurs when current is flowing through it. A voltmeter for analysis, will have no current flowing through it.
     
  10. Hawkeye87

    Thread Starter Active Member

    Oct 7, 2008
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    if i short the voltage source, i'm not sure which one would not show up because the source path is going through R4 and R2. Also, when calculating Rth, what viewpoint am i looking from? From R1 or from the voltage source?
     
  11. hobbyist

    Distinguished Member

    Aug 10, 2008
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    My fault,
    I looked at it wrong all resistances would be in the circuit.

    It's getting late now and I'm making mistakes here I need to really look at it and take my time with it.
    I'll try to get something by tomorrow.

    Maybe by then someone else could steer you in the right direction.
     
    Last edited: Oct 8, 2008
  12. Hawkeye87

    Thread Starter Active Member

    Oct 7, 2008
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    So what viewpoint am i looking at? From R1 or from the short?
     
  13. hitmen

    Active Member

    Sep 21, 2008
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    I simplified the circuit to the right after I. and then nodal analysis. the equations are very messy
     
  14. leonidus

    Member

    Sep 23, 2008
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    For finding eq. resistance u have to look thru the 2 pt.s towards the ckt across which the eq. resistance is asked. So,here from left.
    NORTON'S THEOREM(ANALYSIS):
    Short the 2 pt.s across which nrtn's ckt is asked & find current thru it.
    Replace all voltage & current sources by their internal resistances(voltage source by short ckt. & current src by open ckt.) & find eq. resistance across those 2 pt.s.
    Accordingly, here find the short circuit current thru R1 1st by shorting it.
    Then, short the voltage source. Now, replace the delta network formed by R2,R3,R4 by eq. star nertwork(formed by say, Ra,Rb,Rc). Ra is connected to R1,Rb to gnd thru short ckt & Rc to R5. Note that R1 is open circuited.
    Eq. resistance across R1=((((R7+R8)||R6)+R5+Rc)||Rb)+Ra

    U will now get it.[​IMG]
     
  15. Hawkeye87

    Thread Starter Active Member

    Oct 7, 2008
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    i get something like this..
    Rth = 13.24kΩ
    Vth = 33V
    In = 1.48 mA
    am i correct?
     
  16. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Number the nodes in the circuit, from left to right, 1,2,3,4,5.

    Node 1 is where R1 is connected.
    Node 2 is where the 48 volt source is connected.
    Node 4 is also designated A.

    If you remove R1 and calculate the Rth at node 1, I get 4458.252 ohms

    At that same point, with R1 still removed, I get Vth = 41.83 volts.

    It's possible to calculate Rth without using a Delta-Y transformation, but I don't see how to calculate Vth without it.
     
    Last edited: Oct 10, 2008
  17. hobbyist

    Distinguished Member

    Aug 10, 2008
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    I'm still not sure this is even right.

    on the last board R5 is in there in series with all that other stuff.
    Have to look back on board 2 to see what I mean.

    This is the only way I find to simplify it.
    from R1. input.
     
    Last edited: Oct 9, 2008
  18. Hawkeye87

    Thread Starter Active Member

    Oct 7, 2008
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    it looks like you copied the problem wrong.
     
  19. hobbyist

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    Aug 10, 2008
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    Maybe I did.
    I tried to rearange it so it was easier to see the series parrallel combinations, better,
    but maybe I rearranged to much.

    Sorry about that.
     
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