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Thevenin's equivalent circuit

  1. Mar 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the voltage and the current through R5

    http://i.imgur.com/bL9AFVo.jpg?1

    2. Relevant equations

    Thevenin's theorem.

    3. The attempt at a solution

    http://i.imgur.com/Lz9k8l2.jpg?1

    * To find Vth:
    1- Open circuit R5.
    2- Find the voltage at A and B by determining the current through R1, R2, R3 and R4 (Their Req=(R1+R4)//(R2+R3)=368.4 ohms
    3- The current is equal to 0.24 A, hence the Vab=Vth=30V

    http://i.imgur.com/Lz9k8l2.jpg?1

    * To find Rth (=Rab):
    1- Open circuit R5 and short circuit the voltage source.
    2- (R1+R4)//(R2+R3)=368.4 ohms

    My problem is here, does the wire (caused by short circuiting the source) cause short on (R1+R4)?
     
    Last edited: Mar 13, 2014
  2. jcsd
  3. Mar 13, 2014 #2

    phinds

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    You need to make your figures just a little more fuzzy ... I can almost read these.
     
  4. Mar 13, 2014 #3

    gneill

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    Staff: Mentor

    Shorting the battery joins the top rail to the bottom rail making them all one node. That means R1 and R4 each have one lead that shares that one node... what about their other leads?
     
  5. Mar 13, 2014 #4
    Edited.

    Sorry.
     
  6. Mar 13, 2014 #5
    Their other leads are connecting together with B, does that make them in series?
     
  7. Mar 13, 2014 #6

    gneill

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    Let's see... two components whose leads share two nodes... nope not series... what's the other choice? :smile:
     
  8. Mar 13, 2014 #7
    I see, do I have to consider B as an element?
    I know that when two elements shares two extraordinary nodes (A node connects more than two elements), they're are in parallel; and when two elements shares one ordinary node (A node connects only two elements), they're in series.
     
  9. Mar 13, 2014 #8

    gneill

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    Staff: Mentor

    Not if the components in question are in parallel; It doesn't matter what else connects to the nodes that they share. Besides, in this case you're looking at the case where the terminals A and B are open.

    Not sure about the terminology "extraordinary" and "ordinary" nodes. Must be something new. I was taught about "essential" and "non-essential" nodes, a non-essential node being a node embedded in a serially connected branch.
     
  10. Mar 13, 2014 #9
    So this problem is similar to this one?
    I181UpO.jpg
     
  11. Mar 13, 2014 #10

    gneill

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    Very similar indeed.
     
  12. Mar 13, 2014 #11
    So my Rth would be: (R1//R4)+(R2//R3)=405.7 ohms.

    Can I assume - in similar cases - that there is an "imaginary" voltage source to compute the total resistance?
     
  13. Mar 13, 2014 #12

    gneill

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    You can if you like, but if the resistors can be reduced by the usual parallel/serial combination methods it's not needed.
     
  14. Mar 13, 2014 #13
    Thank you for your quick responds :)

    Would you explain the difference in these two cases? (Where I am asked to evaluate the equivalent between a and b)

    http://i.imgur.com/eHYrn47.jpg?1
     
  15. Mar 13, 2014 #14

    gneill

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    Looks like you've spotted the difference already. The second one has a direct wire running from a to b so its equivalent resistance is zero (short circuit as you state). You've made a calculation error for the equivalent resistance of the first case; the method is right but your result is not.
     
  16. Mar 14, 2014 #15
    Yes, it should 10 not 5.

    But for this kind of problems, how should I start? where terminals a and b are not on the same side.
    http://i.imgur.com/rs6IAmJ.jpg?1
     
  17. Mar 14, 2014 #16

    gneill

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    All of a given node is the same place electrically, so you can slide connections to a given node to anywhere on the node (the contiguous conducting path) and not change the circuit. So nestle the a and b terminal up to either side of the 5 Ohm resistor. Ta da! Same side.

    attachment.php?attachmentid=67616&stc=1&d=1394800812.gif

    If the terminals in question are truly separated by a lot of circuitry though, look to simplifying other portions of the circuit as you can. At times there may be more intuition than canned rules involved, and experience is the best guide. There's no one standard method that applies to all cases.
     

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  18. Mar 14, 2014 #17
    Okay, so if I assumed a voltage source between a and b, then there will be an essential between 8, 20 and 30. How could 8 and 20 be in series?
    http://i.imgur.com/ICCTPwn.jpg?1
     
  19. Mar 14, 2014 #18

    gneill

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    The 8 and 20 Ohm resistors are not in series due to the presence of the 30 Ohm's connection at their junction.

    What's the relationship between the 20 and 30 Ohm resistors?
     
  20. Mar 14, 2014 #19
    They're in parallel (c and b are common)
     
  21. Mar 14, 2014 #20

    gneill

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    Staff: Mentor

    Yes. So what strategy does that suggest to you?
     
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