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Homework Help: Dependent current source, find Vo

  1. Feb 4, 2013 #1
    1. The problem statement, all variables and given/known data

    http://imageshack.us/a/img26/4592/homeworkprintprob.jpg [Broken]

    Find the voltage V0 in the network.

    2. Relevant equations

    V = IR

    Voltage Division:
    (Voltage across series resistor) = [(resistance) / total series resistance)](total input V)

    Current Division (for 2 parallel resistors):
    (current across parallel resistor) = [(other resistor) / (sum of parallel resistors)](total incoming current)]

    3. The attempt at a solution

    I can simplify the 1k and 3k resistors to 4k no matter what.

    I'm not sure if I can simplify the 6k and 12k on the bottom as parallel though, but they will be 4k.

    4k + 4k + 2k in series becomes 10k, assuming you can do that with the current source.

    Then 10k + 6k makes

    R total = 16k ohm, and then I tot = 24 mA

    V = IR means Vtot = 384V.

    Then voltage division would make

    V0 = [ (6k)/(6k+10k) ](384V) = 144V

    would this be correct?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 4, 2013 #2


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    Staff: Mentor

    Yes, okay so far.
    Nope. Can't do that. There are separate paths with separate currents; there's a loop to the left of the current source and a loop to the right of the current source. The current will be different in the two loops. You might think about using the current division rule to see how much of the 24mA from the source flows in each loop...
  4. Feb 4, 2013 #3


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    Staff: Mentor

    Why don't you draw a simpler schematic?

    You said the 12 and 6 make 4. The 3 and 1 make 4. The 2 and 6 make 8. So draw it with those 3 resistances and see what you can make of it.

    BTW, it's not a dependent current source, it's a constant current source.
  5. Feb 4, 2013 #4


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    Science Advisor
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    Gold Member

    What others have said. OK so far.

    If you redraw the circuit at this point you should see that a further simplification can be made.

    Should be possible to treat it as a current divider at that point.
  6. Feb 4, 2013 #5
    I got confused and thought you could simplify the resistors in series like that since it was a constant current source, heh. Never mind then. 384V seemed way too high too.

    So then: :

    http://img835.imageshack.us/img835/7570/homeworkprintprobedit.jpg [Broken]

    and then doing current division:

    Current through right branch = (4k/16k)(24 mA) = 6 mA

    and then V knot is just V = IR, since current is the same through parallel elements

    V knot = (6 mA)(6 kΩ) = 36V

    Last edited by a moderator: May 6, 2017
  7. Feb 4, 2013 #6


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    That's what I make it.
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