Thevenin Theorem lab

Discussion in 'Homework Help' started by metelskiy, Nov 2, 2010.

  1. metelskiy

    Thread Starter Member

    Oct 22, 2010
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    I'm for about to do a lab about Thevenin theorem and i need to do some calculations before actually doing lab (I have just read Thevenin Theorem explanation here on Allaboutcircuit).
    Here i have a circuit:
    Lab 6.jpg
    And they ask to Calculate Thevenin voltage and resistance for the network to the left of points a-b using measured resistance values (R4 is the load). Show all work. First step i know i have to remove the load in that case would be R4 and so i got:
    Lab 6_1.png
    With given circuit, the voltage between the two points where the load resistor used to be attached needs to be determined. Need help with that.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    With a-b open, R1 & R3 in conjunction with the 12V source form a voltage divider. Vab will be the same as the voltage across R3 - no current is flowing in R2 (as denoted in the first circuit).

    You have two R3's in the second rendering.
     
    Last edited: Nov 2, 2010
  3. Georacer

    Moderator

    Nov 25, 2009
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    As for Rth, you need to nullify the source (make it a short circuit) and find the resistance between point A and Ground.
     
  4. JoeJester

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    Apr 26, 2005
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  5. metelskiy

    Thread Starter Member

    Oct 22, 2010
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    Thanks replies guys, i'm now clear on that circuit.
    But i have another one:
    28a.png
    Where they ask to determine the Thevenin equivalent as seen from terminals A and B. I have found Rth by shorting power source, now to find Vth i need to remove the RL (load resistor) and calculate Va-Vb=Vth but if RL will be removed, wouldn't voltage be the same at points A and B?
     
  6. JoeJester

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    Apr 26, 2005
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    What did you get for Rth?

    When you remove the load resistor, Vth should be easily found.
     
  7. metelskiy

    Thread Starter Member

    Oct 22, 2010
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    28a_Rth.png
    I shorted power source and found Rth between A and B by:
    Rth=R4+(R2||(R1+R3))= 72Ω

    The thing I'm confused about is in calculating Vth, if we remove RL, wouldn't Voltage at A and B be the same? What steps should i approach in order to find Vth? Thanks.
     
  8. JoeJester

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    Apr 26, 2005
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    The same as what?

    from the link I referenced earlier ...

    Your Rth is incorrect.

    Look at the attached and recalculate. Post your results.
     
    Last edited: Nov 3, 2010
  9. metelskiy

    Thread Starter Member

    Oct 22, 2010
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    JoeJester, I got clear on that problem i posted new one in post #5. Sorry probably confused you. :)
     
  10. Fraser_Integration

    Member

    Nov 28, 2009
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    no they wouldn't be the same voltage because there is still a current in that circuit going through R1 - R2 - R3 thus there would be a potential difference around R2 ( i.e. between a and b).
     
  11. JoeJester

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    Apr 26, 2005
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    Post your "new" original circuit ... with all the values.

    How do you know you got the other one correct? Was the answers present in your instructions from the professor?
     
  12. metelskiy

    Thread Starter Member

    Oct 22, 2010
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    Here is the original circuit in which i need to find Vth and Rth:
    28a.png
    I found Rth but shorting power source:
    28a_Rth.png
    And came out with 72Ω
    I need help to find Vth. Thanks.
     
  13. Fraser_Integration

    Member

    Nov 28, 2009
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    if i'm not mistaken, then Vth is simply the voltage at the top-centre node. As no current flows through R4, then A will be at the same voltage as the top-centre node.

    You find this by the voltage division of the R1 and the equivalent resistance of R2&R3.
     
  14. metelskiy

    Thread Starter Member

    Oct 22, 2010
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    Why wouldn't there be current flowing through R4? I believe in order to find Vth first we remove the RL than find Vth (Va-Vb) so instead of RL now there will be just a connection and R4 would be connected to R2&R3? Just my thoughts, please correct if I'm wrong.
     
  15. JoeJester

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    Apr 26, 2005
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    When you replace the load with a voltmeter ... the same potential appears at points A and B as that which is dropped across R2.

    There is no current flow through R3 until a load is connected. Yes, the meter is a load that is of sufficient high impedance that it doesn't affect the voltage reading.

    I am close to your 72 ohm Rth calculation ... My figure is higher because I choose to go with three decimal places.

    What did you compute for Vth? 1.39V is not the answer.
     
  16. metelskiy

    Thread Starter Member

    Oct 22, 2010
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    I used voltage devider formula to find V at junction of R2&R3, That I believe would be my Vth right? (47Ω/117Ω)*2.5V=1V
    Vth=1V and Rth=72Ω correct?
     
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