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Need help finding current in a branch within a circuit

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data

    0KxaB.png

    2. Relevant equations

    KCL and KVL. V=IR

    3. The attempt at a solution

    I tried to combine all the resistors, and I ended up with an equivalent resistance of about .44444 mA. Is my value for equivalent resistance right? Also, does this mean that the resistance passing through the voltage source is .444 mA?
     
  2. jcsd
  3. Sep 25, 2011 #2
    Resistance is not measured in mA.

    If 0.444 mA was your answer for Is (or the voltage source current) try checking that.

    For example, find the voltage on every resistor and see if your solution satisfies KVL.

    Make sure all the currents going into the bottom node sum to Is.
     
  4. Sep 25, 2011 #3
    I'm sorry, but I meant that I found the total current going through the voltage source to be .444 mA.
     
  5. Sep 25, 2011 #4

    phinds

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    You need to show your work. You have expressed resistance in ma which makes no sense. Show WHERE you have what values.
     
  6. Sep 25, 2011 #5

    phinds

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    How could you possible have less current going through the voltage source than is going through the 10ohm resister?
     
  7. Sep 25, 2011 #6
    Some of my work. Where did I mess up?

    xrDQs.jpg
     
  8. Sep 25, 2011 #7
    The 6k Ω and 12 kΩ resistances are not in series.

    They are in the non-intersecting parts of two different meshes!

    I have noticed you seem to be having significant problems with circuit topology.
     
    Last edited: Sep 25, 2011
  9. Sep 25, 2011 #8

    gneill

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    Start with what you're given and see how far you can go finding voltages and currents without writing equations.

    You're given Io, so you should be able to use Ohm's Law to find the voltage across the 10K resistor. That voltage must also appear across the 15K resistor. Thus you should be able to find the current through that 15K resistor.

    That voltage also tells you how to find the voltage where Is meets the 8K resistor (what component is in between?). What more can you determine from that?
     
  10. Sep 25, 2011 #9

    phinds

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    Exactly.
     
  11. Sep 26, 2011 #10
    Ok, so doing the work...

    I got the voltage across the 10 kΩ resistor to be -40V and then the voltage across the 15 kΩ resistor to be 40V with a current of 8/3 mA.

    So that means, at the node right beneath Io, the current is 20/3 mA.

    Examining the outer loop and applying KVL, I get something like:

    -8V + 40V - [12kΩ * i] = 0

    Solving for i => i=(32V)/(12kΩ) => i=2.66666 mA;

    So if i=8/3 mA, and we know that the node beneath Io is 20/3 mA... we know that the current for Is is 12/3 or 4 mA.

    Is this logic correct? Did I mess up somewhere? Please help
     
  12. Sep 26, 2011 #11
    You've got to be disciplined!

    Label the 40 V on your diagram, with +/- symbols.

    Now draw an arrow going all the way around the loop.

    Pick a starting point and follow the loop around in the direction of an arrow.
    Each time it passes by an element's minus symbol first, add that voltage.
    If the arrow passes by the plus symbol first, subtract that voltage.
    When you've made a whole loop the sum should be zero.

    Also be disciplined when applying Ohm's law.
    If positive current flows into a resistor, that end of the resistor gets + voltage polarity.
     
  13. Sep 26, 2011 #12
    I'm so sorry. I have been working on this pset this whole weekend. It's not even in my major, I have to take it for a requirement :(

    I really can't get this question. I have already tried my best for it. Can someone please work this out for me? I mean, if the answer's not 4 mA, then I don't know what to do with this problem any more :(
     
  14. Sep 26, 2011 #13

    NascentOxygen

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    See that arrow for current, Io? The current is shown as directed "down" through that 10k resistor, so this means that the voltage at the "top" of that resistor will be positive with respect to the "bottom" of the resistor. You don't get a choice here, it's not your call. The moment the current in the resistor is marked with an arrow, that also defines the polarity of the voltage across that resistor. The only way you can get current to flow "down" through the 10k resistor is by having the "top" of that resistor more positive than the "bottom" of the resistor. For any individual element in a circuit, its current's direction and voltage's polarity go hand in glove.

    You have established that the voltage at the top of the 10k resistor is 40volts. But the 15k resistor is connected in parallel with the 10k, so the voltage across the 15k is also 40v and with the same polarity as the voltage across the 10k. So you can now calculate what current is going through the 15k resistor.

    Current does not materialize out of thin air, so the current that flows through that 10k, added together with that through the 15k, must be flowing through the 8v battery. Mark it in with a correctly-directed arrow.

    It's convenient to call the bottom node "ground" since it extends across the whole circuit.
     
  15. Sep 26, 2011 #14
    Working with your hints has gotten me this answer. Remotely close?

    wr5L1.jpg
     
  16. Sep 26, 2011 #15

    NascentOxygen

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    The figures look right.

    I suggest you always place the arrow for current right on the piece of wire. (Unless you are drawing it as a complete circle, as for mesh analysis.) Arrows for voltage are drawn to the side of the element that experiences that voltage, with the head of the arrow denoting the end of the element which has the greater voltage, i.e., more positive. (Note, -5 volts is "more positive" than -12 volts.)

    Neatness, and a large diagram that you can write on without clutter, are beneficial when analyzing schematics.
     
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