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Thevenin's Theorem question

  1. Jul 7, 2008 #1
    I have this question:

    A DC source has an open circuit voltage of 7.5 volts across its output terminals. the output voltage falls to 7.44 volts when a 600 ohm precision resistor is connected to the output. draw the TEC (Vt, Rt) of the source, showing all values. also show the calculations to obtain Vt and Rt.

    Any idea???
     
  2. jcsd
  3. Jul 7, 2008 #2

    berkeman

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    Staff: Mentor

    Thread moved from EE to Homework Help. Welcome to the PF, mechi. Homework/coursework questions need to be posted in this area of the PF, which is set aside for homework/coursework problems, and has special rules (see the Rules link at the top of the page).

    One of the rules is that you must show us your own work before we can offer tutorial help. We do not give out answers here on the PF.

    So tell us what you know about a real voltage source. It consists of an ideal voltage source and what? How does that help you to set up the equations to answer your original question?
     
  4. Jul 7, 2008 #3
    Thanks!
    Well... I'm just starting my course and I'm not that good at electrical stuff.
    The scheme might be like this:
    R = 600
    -----<><><>------ a
    |
    O V = 7.44v Voc= 7.5v
    |
    ------------------- b

    Being: O = V (voltage source) and <><><> = R (resistor), Voc (open circuit) = 7.5v
    It is an open circuit so a dn b are not connected.

    This is my understanding:
    - as it is an open circuit, Vab = Vt since I=0, and therefore Rt=0
    - for V=7.44 and R=600, using V=IR, I=V/R = 0.0124 A
    - Rt=Vab/Iab = 7.5/0.0124 = 604.8 ohms.

    Are these calculations and the above graph correct? or I'm really confuse?
    Thanks for your time!
     
  5. Jul 7, 2008 #4

    berkeman

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    Much better. Posting your work like that really helps you learn better, and helps us offer tutorial hints to guide you to figure out the answer yourself.

    So, my hint is that the resistor you've drawn should be the Rt resistance, not the 600 Ohm test load resistor. Re-label the diagram with the resistor shown as Rt, then connect the 600 Ohm test resistor to the terminals a & b on the right. Write the equation for the voltages and currents when the 600 Ohm resistor is connected, and see where that leaves you. Keep showing us your work as you go...
     
  6. Jul 7, 2008 #5
    So... here again...

    Rt
    ----<><>----a----
    | |
    O V=7.5 <> R = 600
    | <>
    | |
    -------------b-----

    - I=V/Rt = 7.5/Rt
    - Knowing that Vab = Vt = 7.44
    - Plug 1st formula into: I = V/(Rt+R) and getting Rt=75000 Ohms

    I'm going the right way???
     
  7. Jul 7, 2008 #6

    berkeman

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    Staff: Mentor

    I'm not really tracking what you are doing. When you connect the 600 Ohm resistor on the right side to a & b, you end up with a voltage divider between Rt (on the top leg of the circuit) and the 600 Ohm resistor on the right. Use the voltage divider equation to solve this problem...
     
  8. Jul 7, 2008 #7

    berkeman

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    Staff: Mentor

    I see one of the problems -- the forum software is taking your spaces out, which compresses everything in your figure to the left. You might use the forum tags for "code" to keep it from doing that, or just sketch it and post it as an attachment.
     
  9. Jul 7, 2008 #8
    I've attached the figure... hopefully will be understandable.

    Here is the equation:

    Vt=Vab=7.44
    Rx=600
    Vx=7.5

    Vx=(Vt*Rx)/(Rt+Rx)
    Therefore, Rt=-4.8 ohms
    Can be negative?
     

    Attached Files:

  10. Jul 8, 2008 #9

    berkeman

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    Staff: Mentor

    Nope, can't be negative in this problem. The voltage divider lowers the open circuit source voltage Vo = 7.5V a bit due to the voltage divider loading. Try the math again....
     
  11. Jul 10, 2008 #10
    See my attachment...

    << attachment with complete solution deleted by berkeman >>
     
    Last edited by a moderator: Jul 10, 2008
  12. Jul 10, 2008 #11
    No they arent!!!!!!! The open circuit and source voltages are the same!!!! In an open circuit there is no current, so there is no voltage drop across the Thev Resistor. However, Your numbers are correct in a way. You seem to be confused about what is meant by Thevenin
    resistance. It is the resistance of the "black box".
     
  13. Jul 10, 2008 #12

    berkeman

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    Staff: Mentor

    Welcome to the PF, justin. We do not allow posting of solutions to homework/coursework questions. We require the original poster (OP) to do the bulk of the work. We can offer tutorial help, hints, suggestions, etc., but no solutions should be given.

    We appreciate your help in providing tutorial help here.
     
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