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Diode two dc sources

  1. May 2, 2016 #1
    1. The problem statement, all variables and given/known data:Find the current that go across the diode ans=0.86mA
    zvZG2.png

    2. Relevant equations
    V=IR

    3. The attempt at a solution
    The Vt(v total) is 30v
    so,must i ignore the diode as off first?.In that case the current is 30/20=1.5 mA.
    i also know the that diode is foward bias because 20v>10V
    i know that the voltage drop for this diode is 0.7
    (20-0.7)/10=1.93,this step is correct.
    but i really need help for this qn
    mods pls dont delete.. :)
     
    Last edited by a moderator: May 2, 2016
  2. jcsd
  3. May 2, 2016 #2

    NascentOxygen

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    That is the current for the 20V source. Also determine the current for the 10V source.
     
  4. May 2, 2016 #3
    hey but how can i just split the circuit into two?? lol, isist that weird?? like i used to learn that i should find the current and so on...
     
  5. May 2, 2016 #4

    NascentOxygen

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    You have made the assumption that the diode is conducting....which means you are saying its cathode must be at a potential of -0.7V.

    So with this assumption, you can then determine the current in each resistor.
     
  6. May 2, 2016 #5
    http://imgur.com/l40uOZ2
    http://imgur.com/l40uOZ2
    the red arrow shows the current flow from the 10v source.No current from the 10V pass am I right??
     
  7. May 2, 2016 #6

    NascentOxygen

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    What does Ohm's Law tell you for the resistor on the left, given your assumption that the diode's cathode is being held at -0.7V?
     
  8. May 2, 2016 #7
    hey thats is the part i dont understand.. i really dont :(
     
  9. May 2, 2016 #8

    NascentOxygen

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    You have a resistor, you know how many ohms, and you also know the voltage at either end of it ....
     
  10. May 2, 2016 #9
    but shouldn't it be like (30-0.7 )/20????
     
  11. May 2, 2016 #10

    gneill

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    nathanxx, why don't you make things simpler for yourself? ... If you are studying diode circuits you very likely have already covered Thevenin equivalents. If you consider the diode as the load of the network, what would be the Thevenin equivalent of the circuit that is driving that load?

    upload_2016-5-2_15-58-30.png
     
  12. May 2, 2016 #11

    NascentOxygen

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    The current in each resistor will be what Ohm's Law says it is.

    ⏩ From their difference you can deduce how much current must be going through the diode.
     
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