Current Division or Voltage Division

Discussion in 'Homework Help' started by M_bee, Dec 7, 2014.

  1. M_bee

    Thread Starter New Member

    Dec 7, 2014
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    I am having trouble deciding when to use current division or voltage division. I thought you could use either when in a case where there are resistors in both parallel and series.

    For this problem, I was to find the power dissipated in the 5 ohm resistor, given the source current is 15 A.

    For my first approach, I tried to use current division to find the current in the 6 ohm resistor using the Resistance equivalent of the entire system. Then I found the current in the 5 ohm resistor and solved for power. This resulted in an incorrect result.

    My second attempt used voltage division, and I found the correct answer, but I don't understand why the first method didn't work. IMG_1220.JPG
     
  2. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
    2,502
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    hi,
    Look at this clip from your image.
    E
     
  3. M_bee

    Thread Starter New Member

    Dec 7, 2014
    2
    0
    Thank you! Spotting that helpful similarity makes sense when the resistance in each branch is equal and they are in parallel, then their currents should also be equal (and half of the source current). I still am confused as to where my error would be in calculating the current in R6 using R-equivalen. I'm curious to find this so I am prepared for cases when the resistances in the parallel paths are not equal. Would you be able to help me with that?
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
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    Well simply, first use a current divider rule to find current in the 6 ohm resistor.
    http://en.wikipedia.org/wiki/Current_divider

    I6 = 15A* 10Ω/(10Ω + Rx )
    Where
    Rx = ( 6Ω + (5Ω||(8Ω+12Ω))) = 6Ω + 5Ω||20Ω = 6Ω + 4Ω = 10Ω

    So we have

    I6 = 15A * 10Ω/20Ω = 15A*0.5 = 7.5A

    And now we can use current divider rule to find current in the 5 ohm resistor.

    I5 = I6 * (8Ω + 12Ω)/(5Ω + (8Ω + 12Ω) ) = 7.5A * 20Ω/25Ω = 6A

    And P = 6A^2 * 5Ω = 36*5 = 180W
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    You correctly found the equivalent circuit as consisting of two 10Ω resistors in parallel. But then when you went to find the current in the 6Ω resistor you performed the calculation for a as though the resistance of the parallel branch you are interested ONLY consists of a 6Ω resistance. Remember that in order for two resistances to be in parallel, they must have the exact same voltage across them. The 6Ω resistor does NOT have the same voltage across it as the 10Ω resistor. The entire equivalent resistance of the right hand part of the circuit, which has an equivalent resistance of 10Ω, as the same voltage across it as the left-hand 10Ω resistance.
     
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