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Find This Output Voltage

  1. Jun 17, 2012 #1
    Vo=? In terms of the given. This is the attached figure
     

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  2. jcsd
  3. Jun 17, 2012 #2

    psparky

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    I X R?

    Tough one.
     
  4. Jun 17, 2012 #3

    vk6kro

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    If the head of the Vo arrow is supposed to be the positive end, then Vo = minus I times little r = -(I * r)

    This is because the current is flowing upwards in the right hand side of the loop, so the bottom of the resistor r is positive.
     
  5. Jun 17, 2012 #4
    -I*r = Vo.
     
  6. Jun 17, 2012 #5

    psparky

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    I see what you guys are saying.....but if you remove little r.....the voltage is still the same!

    Little r has no bearing on v out in my opinion.
     
  7. Jun 17, 2012 #6

    jim hardy

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    looks to me like if you open little r you have opened the load on a current source, so the voltage will become the maximum the source can produce.
    Infinite, for "ideal" source .

    So it's little r not big R determines Vo in that circuit.
     
  8. Jun 17, 2012 #7

    vk6kro

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    It helps to put some real sizes on the components:

    Let R = 20 ohms, r = 30 ohms and the current source = 1 amp.

    Like this:
    http://dl.dropbox.com/u/4222062/current%20gen.PNG [Broken]

    So, the voltage across the resistor "r" = Vo = (1 amp * 30 ohms) = 30 volts (Negative 30 volts relative to the arrow)
    So, the voltage across the resistor "R" = (1 amp * 20 ohms) = 20 volts

    With this load, the voltage across the current source is 50 volts (1 amp * (30 ohms + 20 ohms)) and you can get the same result if you regard the resistors as part of a voltage divider.
     
    Last edited by a moderator: May 6, 2017
  9. Jun 18, 2012 #8

    psparky

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    Oh ya.....I see what you are saying once again!

    The circuit appears to be in parallel at first glance.....but apparentlly is not!
    The voltage across is clearly not the same. (edit....yes, voltage across is the same)

    Still though, If I slide a volt meter across one branch to the other branch I get two different voltages?

    How can this be?

    Perhaps I can answer this.....I can see how the current device has 50 volts across it.
    I guess it drops 20 volts accross R.....leaving 30 volts across each branch.

    Must be since the voltmeter is not going to change branch to branch.
     
    Last edited: Jun 18, 2012
  10. Jun 18, 2012 #9

    vk6kro

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    Perhaps I can answer this.....I can see how the current device has 50 volts across it.
    I guess it drops 20 volts across R.....leaving 30 volts across each branch.


    Yes, that's right, the 50 volts across the current source is partially cancelled by the 20 volts across the 20 ohms resistor (because they have opposite polarities in the left hand leg of the loop), so this leaves 30 volts between the top and bottom of the diagram.
     
  11. Jun 19, 2012 #10
    vk6kro thank you!!
     
  12. Jun 19, 2012 #11

    psparky

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    This is a great problem. I have stumped 5 out of 6 elec engineers with this problem at work....including myself.

    It is much easier to see when when you re-arrange the circuit in a basic series arrangement. Add the two resistors in series and you get 50 ohms....or 50 volts. Obviously there is 50 volts across the 50 ohm resistor.....or current source.

    Lesson learned.....always re-arrange circuits in a more familiar form....regardless if you are in college....taking state exams or just fooling around on this forum or whatever.

    Psparky out....thank you for schooling me as well.

    Excellent question M. next....keep em coming.
     
  13. Jun 20, 2012 #12
    psparky,

    Well, I just don't get it. It looks to me like a problem whose solution can be discerned almost as fast as it can be read. I mean, a current source in series with two resistors? What's the big deal?

    Ratch
     
  14. Jun 22, 2012 #13
    Are those resistance really in series ?
     
  15. Jun 22, 2012 #14

    psparky

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    Yes! Redraw the circuit leaving out the vout tails.

    You have a current source in series with two resistors.....simple as that.
     
  16. Jun 22, 2012 #15
    i think answer should be IR !
     
  17. Jun 22, 2012 #16
    suppose there are terminals in at the ends of r resistance , that would'nt be in series then and it did not asked for the voltage drop at r it asked for V(0) which would be equal to current in small r when small r and capital R are parallel!
     
  18. Jun 22, 2012 #17

    NascentOxygen

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    I'm not able to picture what you mean. Can you draw it?
     
  19. Jun 22, 2012 #18
    here it is
     

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  20. Jun 22, 2012 #19

    vk6kro

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    Here is the same circuit as above but redrawn in a more familiar way:

    http://dl.dropbox.com/u/4222062/series%202.PNG [Broken]

    The question asked for the voltage ACROSS the 30 ohm resistor, "r". That is, what voltage would you measure with a voltmeter?
     
    Last edited by a moderator: May 6, 2017
  21. Jun 22, 2012 #20
    ya i got it ! i was confused there ! thanks !
     
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