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Voltage and current division

  1. Sep 11, 2012 #1
    1. The problem statement, all variables and given/known data
    http://i.imgur.com/XbMaK.png

    Use voltage and current division to determine [itex]V_{0}[/itex] in the circuit given that [itex]V_{out}[/itex] = 0.2V.

    2. Relevant equations

    [itex]v_{i}[/itex] = [itex]\frac{R_{i}}{R_{eq}}[/itex] * [itex]v_{s}[/itex]
    [itex]i_{n}[/itex] = [itex]\frac{R_{eq}}{R_{n}}[/itex] * [itex]i_{s}[/itex]

    3. The attempt at a solution

    I got [itex]\frac{236}{23}[/itex] for the [itex]R_{eq}[/itex]

    0.2 = [itex]\frac{1}{\frac{236}{23}}[/itex] * [itex]v_{s}[/itex]

    2.05 = [itex]v_{s}[/itex]
     
  2. jcsd
  3. Sep 11, 2012 #2

    gneill

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    The result doesn't look right. What are those formulas under Relevant Equations? What do Ri and Req represent? How did you calculate the Req that you used in your attempt?
     
  4. Sep 11, 2012 #3

    phinds

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    I have no idea how you are going about this but it seems to be an extremely straightforward problem and your answer is WAY off.

    Just think logically for a minute. What you have is a stream of voltage dividers. How could you possible have a source voltage that is comparable to the output voltage the way you do? Just at a glance you should see that it has to be at least an order magnitude higher
     
  5. Sep 11, 2012 #4
    My book lists those formulas under voltage and current division.
    Any individual resistor is Ri,Rn
    Req is the equivalent resistance in the circuit

    I started out by combining the 1 and 2 ohm resistors to make a 2/3 resistor. Then I combined that with the 2 ohm resistor in parallel to make a 1/2 ohm resistor. I combined that with the 4 ohm resistor in series to make a 4.5 ohm resistor and then combined that result with the 4 ohm resistor in parallel to make a 8.5 ohm resistor. Finally I combined the 8 ohm resistor in series to get 16.5 ohms.


    I guess I added wrong in my previous answer.


    16.5 = [itex]R_{eq}[/itex]

    0.2 = [itex]\frac{1}{16.5}[/itex] * [itex]v_{s}[/itex]

    3.3 = [itex]v_{s}[/itex]
     
    Last edited: Sep 11, 2012
  6. Sep 11, 2012 #5
    I just did what I read in my book. I found the equivalent resistance so I could find the proportionate fraction of the voltage for the 1 ohm resistor. So since 0.2v is given I solved for Vs
     
  7. Sep 11, 2012 #6

    gneill

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    Ah. Unfortunately that technique will not work here. You might determine the net resistance in order to calculate the load on Vo (the diagram designates the source voltages as Vo), perhaps to determine the current it supplies as I = Vo/Req, but it won't allow you to find Vo given Vout.

    I can see several possible approaches, two of which are as follows.

    1. Reduce the network behind the 1 Ohm resistor to a single resistor and voltage supply. That is, find the Thevenin equivalent of the network up to but not including the 1 Ohm resistor, leaving the Vo as an unknown. Then apply the voltage divider formula to the resulting simplified circuit to extract a value for Vo. Reduction of the circuit can be accomplished stepwise by working from left to right, forming successive Thevenin models as you go.

    2. Starting at the output, note that with Vout specified you can find the current in the 1 Ohm resistor. Where does that current come from? It must come from the 2 Ohm resistor. So what's the voltage drop across that resistor? What then is the potential at the next junction (where the 4 Ohm resistor meets the two 2 Ohm resistors)? Continue to work back in a similar manner until you reach Vo.
     
  8. Sep 11, 2012 #7

    phinds

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    Yeah, that's what I did. It's VERY simple and as quick as any method you're likely to find.
     
  9. Sep 11, 2012 #8
    The 2ohm resistor has a current of 0.2A so it's Voltage is 0.4V. At the node where the 4 ohm and two 2 ohm resistors meet is what confuses me.

    If I use KCL, then I1+I2 = 0.2A. So how can I solve for the two unknown currents?
     
  10. Sep 11, 2012 #9

    gneill

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    What's the node voltage?
     
  11. Sep 11, 2012 #10
    I don't think I can use the node voltage method. My professor says that we can't use information we haven't learned yet to solve our current homework problems.
     
  12. Sep 11, 2012 #11

    gneill

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    You can't add together two potential changes? :surprised:
    If so, I don't see how you can do any problems at all! :smile:
     
  13. Sep 11, 2012 #12
    Well, there is a section in the next chapter called the Node-Voltage method so I thought you were referring to that.

    Is the voltage 0.4V?
     
  14. Sep 11, 2012 #13

    gneill

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    No, just the voltage (or potential) at the node.
    That's just the potential increase as you cross over the 2Ω resistor.

    You want the potential with respect to the common rail that runs along the bottom (it's a convenient reference point for potentials for this circuit).
     
  15. Sep 11, 2012 #14
    I'm still confused, am I supposed to add the voltages of the 8 ohm resistor 4 ohm resistor to find the voltage at that node?
     
  16. Sep 11, 2012 #15

    gneill

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    No, you do a KVL "walk" from the bottom of the 1Ω resistor, through the 2Ω resistor to the node. You're building on the values that you're calculating along the way...
     
  17. Sep 11, 2012 #16
    is this correct?

    -0.2V + 0.4V +2I = 0

    i = -0.2
     
  18. Sep 11, 2012 #17

    gneill

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    You just want to "walk" from the reference node to the node in question, adding up the potential changes along the way:

    attachment.php?attachmentid=50705&stc=1&d=1347423983.gif
     

    Attached Files:

  19. Sep 11, 2012 #18
    Does this mean the potential across the 2 ohm resistor is 0.6 volts?
     
  20. Sep 11, 2012 #19

    gneill

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    It does indeed! So now you can work out what current it must be flowing through it. And you already know what current flows out through the other 2Ω resistor... a little KCL at the node should then tell you the current through the 4Ω resistor... carry on.
     
  21. Sep 12, 2012 #20
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