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Current Sources in Parallel when Simplifying Circuits

  1. Jan 17, 2012 #1
    When simplifying a circuit, can a parallel branch with a current source and a resistor in series be considered as just the current source?


    Also, if two parallel current sources 'point upwards' i.e. toward the positive rail, can I add them together as one source even if there are other branches in between them?


    Thanks everyone!!
     
  2. jcsd
  3. Jan 17, 2012 #2
    If you can ignore the resistance - yes.
    Yes add them together even if other branches are present.
     
  4. Jan 17, 2012 #3

    sophiecentaur

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    Yes, of course. They will both end up with the same voltage across them, each one supplying the current it's been set to deliver. (There are practical limits to that statement, of course - but only along the same lines as connecting Voltage sources in series).
     
  5. Jan 17, 2012 #4
    Thanks!

    How do I know if I can ignore the resistance?

    The question I am trying to do asks for the current in a particular wire of the circuit. Looking at my answer, It seems that i cannot combine things in parallel if they are on opposite sides of this wire. is that correct?
     
  6. Jan 17, 2012 #5
    If resistance is very small (close to zero ohm) ignore it like a short circuit.

    You should be specific about what kind of branch you want to combine.
     
  7. Jan 17, 2012 #6

    sophiecentaur

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    An ideal current source doesn't care about series resistance but in a 'real' circuit, you would need to know the details of actual resistances involved.
    Post a diagram to give us a better idea.
     
  8. Jan 17, 2012 #7
    Hi, Ive attached the circuit picture. To find current in xx I convert the 2 voltages to norton equivalents. Can I then combine the 2A with the left hand current source, ignoring its 5 ohm series resistance??

    I then think that I cannot combine the left hand and right hand norton equivalents into one, as I 'lose' the link xx??

    Thanks.
     

    Attached Files:

  9. Jan 17, 2012 #8
    Convert the voltage sources to Norton. Then combine everything except the current source branch to one current source and one parallel resistance. Then convert it back to Thevenin. Then the analysis is easy. Dont ignore the resistance.
     
  10. Jan 17, 2012 #9

    sophiecentaur

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    That 2A stays as 2A - no worries there. Just do Kirchoff on the circuit and you should get the right answer. You have currents at x adding to give zero (K1) and you have three loops that you can do K2 on. I think that's enough for a solution with four equations and four unknowns.
    @KS that's another approach. I don't see why mine shouldn't work too- not as clever as yours though.
     
  11. Jan 17, 2012 #10
    Thanks Guys,

    @SC, that approach worked for me when I tried, but the answers gave this as an equivalent circuit, and I can't work out how they got there...!
     
  12. Jan 17, 2012 #11
    sorry file didn't attach!
     

    Attached Files:

  13. Jan 19, 2012 #12
    It seems to me that the solution should look like this
     

    Attached Files:

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