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Voltage/Current Division with central Voltage Source

  1. Jan 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Find V0 using voltage/current division

    2. Relevant equations
    V=(VsR)/Req
    I=(IsIeq)/I

    3. The attempt at a solution
    I used general circuit rules (parallel and series) to get the total resistance as 14.775 ohms{?}. And I'm stuck with what to do next . . .
     

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  3. Jan 20, 2015 #2

    gneill

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    Show your work. How did you calculate "total resistance"?
     
  4. Jan 20, 2015 #3
    Ah yes. I was going to do that but was in a hurry.

    [(10+3)//(15+12)] + 4 + 2 = 14.775

    That's part one that's tripping me up. Part two is using the those division equations to find the source voltage . . .
     
  5. Jan 20, 2015 #4

    gneill

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    So, by "total resistance" you mean the Thevenin resistance as seen at the open terminals. Okay. Be sure to include units on all results. All the resistors in your circuit have units of kΩ, so what should be the units of your result by the method you've used?

    As for the current division, what are the resistances of the two branches that the 18 mA sees?
     
  6. Jan 20, 2015 #5
    So yes, I'm in K-Ohms. The 18mA sees a 13 K-Ohm and 27 K-Ohm branch, correct? We haven't quite got to Thevenin in this class, or just ignored it maybe, but sure. From the vague recollection I have of it from last semester, learning from these forums.
     
  7. Jan 20, 2015 #6

    gneill

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    Right.
    Nope. Trace the paths that the current can flow from the current source, around a loop, and back to the current source. Remember that the Vo terminals are open: no current can flow across an open path.
    Well, you don't really need Thevenin or Norton here (although they could be used). Which is why I was curious about your total resistance calculation; I didn't see how it was going to be useful if you're just looking to find Vo. Current division and voltage division is the way to go here.
     
  8. Jan 20, 2015 #7
    So it would be going through +15 and +10 K-Ohm resistors (they + is just saying it's going through the positive terminal to negative . . . )? So do that with all 3 loops (well, I guess meshes in this case)?

    And is it legal to substitute V0R in for "i" across the resistors? I'm just remembering my notes and we did something similar though the context has been lost. But anyways, do 3 loops?
     
  9. Jan 20, 2015 #8

    gneill

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    You don't need any loop equations if you're going to use current and voltage division. You just need to identify the two branches in parallel with the current source that the current will flow through. You've identified one so far (the 10 k +15 k branch).
    No, there's no current associated with Vo as it's an open circuit.

    Just identify the two branches that parallel the current source and use current division o determine the current in each of them. Then you can play with voltage division to find the potential on either side of the open terminals, hence Vo.
     
  10. Jan 20, 2015 #9
    You say 2 loops but I see 3? Why would you ignore the 2k and 4k Ohm resistors on the inside (I'm assuming the second loop doesn't include them)? Or does it not matter which loop I choose for number 2?
     
  11. Jan 20, 2015 #10

    gneill

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    Vo is an open circuit. There is no path for current there, hence no loop current, hence not a loop. You cannot make a loop that includes an open circuit.

    Here are the three possible loops for your circuit. Any two of them will produce linearly independent equations, the third will produce redundant information. Hint: You only every need enough loops so that each current-carrying component in the circuit is included in the path of at least one loop.

    Here are the three possible loops for your circuit:
    Fig1.gif
     
  12. Jan 20, 2015 #11
    So is the first loop equation essentially 18mA + V15/15K + V10/10K = 0?

    And then doing the entire loop gives 18mA + V12/15K + V3/10K = 0?
     
  13. Jan 20, 2015 #12

    gneill

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    You seem to be introducing a lot of unknowns and setting yourself up for a lot of extra work.

    Why not just use current division to find the branch currents, then voltage division to finds the potentials at the resistor junctions? Very straight forward and you have all the information you require.
     
  14. Jan 20, 2015 #13
    So 18mA + i15 + i10 = 0? And same with the entire loop. Then how do I find voltage with that? Don't you need a "V" in there somewhere?
     
  15. Jan 20, 2015 #14

    gneill

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    Okay, current division.

    Consider the following circuit:
    Fig2.gif

    The current from the source, ##I##, will be divided into the two parallel branches as ##i_1## and ##i_2##. The branches have total resistances R1 and R2 respectively. One can determine the values of those currents from the total current ##I## and the branch resistances. Namely:
    $$i_1 = I \frac{R2}{R1 + R2}$$
    $$i_2 = I \frac{R1}{R1 + R2}$$
    So you're in a position to determine the currents in the branches of your circuit. With those currents you can determine the potential drops across the individual resistors in those branches.....
     
  16. Jan 20, 2015 #15
    So i1 = 18mA(25KΩ/40KΩ) and i2 = 18mA(15KΩ/40KΩ)?
     
  17. Jan 20, 2015 #16

    gneill

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    No, you didn't pay attention to which resistance forms the numerator of the ratio. Revisit post #14 and look closely at the expressions.
     
  18. Jan 20, 2015 #17
    I ended up just getting a solutions manual since I realized it'll be a long semester . . . and unless I make it click my way, it never will.

    So it solved each branch first. I understand that, using the parallel knowledge (and that's circuit division, yes?). IE [(10+15)//(3+12)]/(10+15) * 18

    The left most was 6.75 mA, and the right most 11.25 mA. The numerators were Req, and the denominators were just the branch resistances in series.

    Then they calculated voltage across each resistor towards the top, which I also understand. IE V15 = -(6.75)(15) because of the direction of the current source.

    The final equation they got was V = -101.25- -135. I understand everything up until the end . . . Wouldn't that entire "lower wire" (is node appropriate?) be one voltage? Why the subtraction?
     
  19. Jan 20, 2015 #18

    gneill

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    No, the numerators were the resistances of the other branch, the denominators were the total resistance of both branches. The current divider is an important concept, so it'll crop up in exams. Be sure to remember the procedure: The current in one branch is equal to the total current multiplied by the ratio of the resistance of the other branch to the total resistance of both branches.
    The calculation is for the potential across the terminals where Vo is indicated. According to the figure, Vo is the potential at the left terminal minus the potential at the right terminal (the diagram puts the + of the voltage is on the left, the - on the right). So they calculated the potential at each terminal with respect to some reference node (the bottom node) and then calculated the difference in the potentials at the terminals.

    The potentials at the terminals are the same as the potentials at the junctions of the resistors of the branches (since no current can flow through the 2 K or 4 K resistors). Choosing a common reference node (the bottom rail), they found the potential drops across the 15k and 12k resistors using the branch currents, and thus determined the potentials at those junctions with respect to that reference node.
     
  20. Jan 20, 2015 #19
    But the numerators had the parallel notation?

    I understand the last part now. So using general knowledge of the equal voltages across parallel branches, where the "top" and "bottoms" are the same voltage. But now that first part is confusing me since apparently they did it wrong? How did they get parallel notation in the numerator???
     
  21. Jan 20, 2015 #20

    gneill

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    Sorry, I don't follow. Perhaps I haven't directly seen their work.

    There are two parallel branches supplied by a current source. Current division using the branch resistances give you the branch currents. Each branch is also a voltage divider, since there are two resistors in each branch. Using the branch current and the resistance values the potential difference at the junction of the resistors can be determined with respect to a reference node (choose the top or bottom rail as the reference and the resistor between the reference node and the junction as the device across which the potential with respect to the reference node is to be determined). Once the potentials at the junctions are established you can determine the potential difference between the junctions.
     
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