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Find Vo in the circuit / current direction / potential direction

  1. Jan 13, 2013 #1
    1. The problem statement, all variables and given/known data


    http://imageshack.us/a/img29/9056/homeworkprob10.jpg [Broken]


    Given: Va = 8V. Find Vo in the circuit.


    2. Relevant equations
    V = IR,

    KVL, KCL

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


    Current division (only for two resistors in parallel):
    (current through parallel resistor) = [ (OTHER resistance)/(sum of resistors) ](total incoming current)


    3. The attempt at a solution

    I think you can redraw it like this:

    http://img24.imageshack.us/img24/8245/homeworkprob10redraw.jpg [Broken]

    And since 8V = Va across the 4Ω resistor, V = IR means the current in the whole right node means that I = V/R, and then I = 8V / 4Ω

    I = 2A in the right branch right?


    Then I = 2A also in the 8Ω resistor in series with it,

    so then V = IR for the 8Ω resistor means, V = (2A)(8Ω), V = 16V over the 8Ω resistor

    Adding these, V = 24V for the entire right branch.


    Left branch resistors is simplified to 4Ω and then you can say V = 24V for it because of being in parallel, right?

    Then I = 6A for the left because I = V/R --> 24V/4Ω

    then that means 8A is supplied from the middle to both,


    but then that means V = (3Ω)(8A) means V = 24V through the middle 3Ω resistor

    It can't be more than 24V because it's in parallel right? There was no direction given for Va in the first place and that's confusing. The answer for Vo is given so the only thing that would make sense is if V = -24V for the middle 3Ω resistor and then Vo = 48V (which is also the answer, and just thought of this now).


    Is there a better way to tell if either the current or potential in the middle branch is + or - then given only Va first? Again, I just figured it out for the middle branch just now but it's still kind of confusing so, thanks anyway.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 13, 2013 #2

    gneill

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    Staff: Mentor

    Without being given the direction of the potential drop Va, you can't tell which direction the current is flowing. So there will be two possible answers, on for each choice of direction of Va.

    Remember that if the central resistor is dropping a certain amount of potential, the voltage source Vo will have to make up the difference so that their sum matches the required branch voltages.
     
  4. Jan 13, 2013 #3

    CWatters

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    Cyan..

    Your approach isn't correct. The 2A doesn't flow through the 8 Ohms. I redrew it like this...

    PS Vo is on the right.
     

    Attached Files:

  5. Jan 13, 2013 #4

    CWatters

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    PPS: Two of the resistors don't effect Vo.
     
  6. Jan 13, 2013 #5

    gneill

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    Staff: Mentor

    Hmm. The way I interpret the problem Va is not a voltage source, it's the potential drop across the 4Ω resistor due to the current flowing though it. The only independent voltage source in the circuit is Vo.
     
  7. Jan 13, 2013 #6

    gneill

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    Staff: Mentor

    If you're given a potential drop without its direction being specified, then all you can do is look at the circuit and make deductions/assumptions. Like, looking at the indicated polarity of Vo, if it is assumed that Vo is a positive value then the current can only flow one way through it since it's the only source in the circuit. That in turn tells you which way the current must flow in the branches, hence the polarity of Va.

    When these situations crop up on assignments or exams, be sure to explicitly state your assumptions along with your solution.
     
  8. Jan 13, 2013 #7
    Okay, thanks.


    And yes, Va was the potential drop across that resistor, and neither the drop direction or current direction were given. It's just given from being positive and then the direction of Vo.
     
  9. Jan 14, 2013 #8

    CWatters

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    Insert red face.

    That makes much more sense now I look at it again. I was fooled by the subscript on Vo.
     
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