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Finding current using current dividers

  1. Feb 1, 2013 #1
    1. The problem statement, all variables and given/known data
    find i3?


    2. Relevant equations
    i1=i(R3/(R3+R6))
    i2=i(R6/(R3+R6))
    R(parallel)=R6(R3)/(R6+R3)
    v=iR


    3. The attempt at a solution
    i=V/R4=(12sin(t))/(4 ohm)= 3sin(t) A then using R parallel I get 2 ohms as the resistor then again using v=iR. I get 1.333sin(t) A. Did I do this wrong or would this be acceptable
     

    Attached Files:

  2. jcsd
  3. Feb 1, 2013 #2
    I tried this way using v=iR4 finding i=3sin(t) A. Then using the equation to find i1 I get i1=sin(t) A and using the equation to find i2=2sin(t). i3=i1+i2 so I get i3= 3sin(t) A which is incorrect. By the way i1 is the current on the leg with the 6 ohm resistor and i2 is the current on the leg with the 3 ohm resistor.
     
  4. Feb 1, 2013 #3
    Ugg no answers did I make it confusing again I always do that bad grammar and what not anyways please help!
     
  5. Feb 3, 2013 #4

    CWatters

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    Science Advisor
    Homework Helper

    Perhaps it's me but I don't follow that at all.

    I did it this way...

    Let the source current be Is and the source voltage Vs =12 Sin(t)

    Then

    Is = Vs/(4 + 6//3)
    Is = Vs/6................................(1)

    By inspection 2/3rds of Is goes through the 3 Ohm and 1/3rd through the 6 Ohm resistor (they must sum to 3/3rds Is due to KCL)

    so

    I3 = 2/3 * Is

    Using eqn 1...

    I3 = 2/3 * Vs/6
    I3 = Vs/9

    Then substitute Vs with 12Sin(t)

    I3 = 12/9 Sin(t)
    I3 = 1.333 Sin(t) Amps
     
  6. Feb 3, 2013 #5
    Oh yah well would this work as well what I did is combine the 6 ohm and 3 ohm resistor into a parallel and got 2 ohms. I then put it in series with the 4 ohm resisor and get 6 ohms. then using v=ir i get i=2sin(t) A then using my formula for i2= 2sin(t)(6/9)=(4/3)sin(t) A which is correct. My question is what I did correct as well?
     
  7. Feb 4, 2013 #6

    NascentOxygen

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    Staff: Mentor

    You have it right now. But your first post wasn't correct.
    There isn't a current of 3 anywhere.
     
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