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Find the voltage V in the circuit

  1. Sep 13, 2011 #1
    1. The problem statement, all variables and given/known data

    we need to find V in this circuit. -see attachment

    2. Relevant equations

    how do I begin to simplify the circuit?

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Sep 13, 2011 #2
    ok your given a 24 volt source in the middle.

    by definition, any circuit component parallel to each other share the same voltage. this will mean that 24 volts will appear across the 12 ohm resistors in the middle.

    from there you can try and determine the current and use Kirchoff current law to determine the loop currents. Hopefully you can then get the voltage you need

    hope this helps
     
  4. Sep 13, 2011 #3
    funny u say that, my professor doesnt want us to use the kcl and kvl laws, basically no nodal or mesh analysis, which is why im stuck, otherwise i can use those laws and figure it out in minutes.
     
  5. Sep 14, 2011 #4

    Zryn

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    Gold Member

    The diagram has been drawn so as to appear tricky, so redraw it with the source and each resistor laying vertical only for starters. You should only need the voltage divider equation (Ohms Law) to solve this.
     
  6. Sep 14, 2011 #5

    rude man

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    Homework Helper
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    1. Simplify the circuit: four of the resistors can immediately be reduced to two.
    2. Compute the equivalent resistance of the two R's in series & the one in parallel with those two.
    3. You now have 1 voltage source in series with two resistors.
     
  7. Sep 14, 2011 #6
    The 6 ohm on extreme left and 12 ohm resistor on extreme right are parallel to each other.

    Finding the combined parallel resistance would be a good start
     
  8. Sep 15, 2011 #7

    NascentOxygen

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    Staff: Mentor

    You haven't noticed that 4 ohm resistor on the source's -ve terminal?
     
  9. Sep 15, 2011 #8
    yes i have but the 6 ohm and 12 ohm on the extreme left and right of the circuit still share two common nodes on the circuit. By definition this makes them parallel
     
  10. Sep 15, 2011 #9

    Zryn

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    Gold Member

    NascentOxygen was commenting in regards to the bolded bit, which is incorrect, as the source is in series with the 4R resistor and this entire branch is then in parallel with the 12R resistors on either side of it, meaning that 24 volts will NOT appear across the 12R resistors as stated.
     
  11. Sep 15, 2011 #10

    gneill

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    Staff: Mentor

    Sometimes it can be helpful to simply redraw the circuit, moving things about a bit (but not connections!) in order to bring out more familiar patterns. No analysis is required for this, just a bit of spacial imagination and manipulation.

    For example, here's a version of the same circuit with things pushed about. Note that all the same component connections and nodes exist in this version; it is electrically identical to the original.

    attachment.php?attachmentid=38876&stc=1&d=1316094169.gif

    Parallel resistors are now obvious, so the obvious simplifications can be made there.

    Those of you who know about Thevenin equivalents will recognize the opportunity to replace the part of the circuit left of red X's with its single source and resistor equivalent. Then you'll end up with a voltage source and three resistances in series. Easy-peasy.

    Without Thevenin, KCL, or KVL, I suppose you can slog through the current calculations. Start by the net resistance seen by the voltage source and thus finding the total current that the source must drive through its 4 Ohm resistor, hence the voltage drop across it. This will tell you what the voltage is across the other two (parallel) branches. More current and voltage calculations... slog, slog, crank,... result.
     

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