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Circuit question help?short

  1. Apr 26, 2013 #1
    1. The problem statement, all variables and given/known data
    I have to find i1 and i2 in the picture.


    2. Relevant equations

    The equations related to the division of the current.

    3. The attempt at a solution
    I found i1 by finding the equivalent resistance of the whole circuit which is 25 Ohm.So 25V/25 Ohm=1A.To find i2 I found the equivalent resistance of 3||6 which is 2 Ohm. So applying the law of division i2=6/(6+2)=3/4 ohm.I showed this to my teacher and he said that the current running through 6 Ohm isnt the same as the current running through 6 ohm and 3 ohm..how do I find i2? P19xTss.jpg
    EDIT : The equivalent resistance is 10 Ohm.so i =2.5 A.After this i find i1=[4/(4+6)]*i=(4*2.5)/(4+6)= 1A.I dont know how to find i2
     
    Last edited: Apr 26, 2013
  2. jcsd
  3. Apr 26, 2013 #2

    rl.bhat

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    Homework Helper

    Check the calculation of the equivalent resistance.
     
  4. Apr 26, 2013 #3
    I1 is right. I know for sure that i1=1 A,that is the answer in my book.I think my mistake is logical,rather than miscalc.
     
  5. Apr 26, 2013 #4

    gneill

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    Staff: Mentor

    Can you elaborate on the work you've done? Some of the numbers you're stating for things sound a bit odd to me. Could just be me misunderstanding your intent, so if you could show your steps that would help.
     
  6. Apr 26, 2013 #5
    Oh I miscalculated some things.I will edit it above.
     
  7. Apr 26, 2013 #6

    gneill

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    Staff: Mentor

    Knowing the total current and i1 you should be able to label the diagram with more currents and potentials. For example, what's the current through the second 4 Ω resistor? And thus the potential drop across it? Fill in as many currents and potentials as you can. Can you determine the current through the "upper" 6 Ω resistor in the branch of interest?
     
  8. Apr 26, 2013 #7
    That current is 3/4 Amper. because 3||6 Ohm has an equivalent resistance of 2 Ohm.Then So applying the law of division i 2 Ohm=6/(6+2)=3/4 ohm. Since 2 Ohm is in series with 6 Ohm,their current is 3/4 Amper
     
  9. Apr 26, 2013 #8

    gneill

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    Staff: Mentor

    I don't follow your logic here. Currents don't spring spontaneously from resistances, they control the flow of existing currents. Also, current is given in amps, resistance is in ohms; watch your units.

    If you're applying current division, then there must be an existing current to divide. How did you determine that current to begin with?
     
  10. Apr 26, 2013 #9
    Oh,that is i1. i1=1 A.
     
  11. Apr 26, 2013 #10

    gneill

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    Staff: Mentor

    :confused: But i1 is in a completely different branch.
     
  12. Apr 26, 2013 #11
    Oh Im completely lost.I dont even know what to apply here.Can you give me a hint? I feel very stupid and confused.
     
  13. Apr 26, 2013 #12

    gneill

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    Staff: Mentor

    Follow the advice I've given. Start with the total current from the source which you've already calculated, and start adding voltage drops and current values to your diagram. So, basic applications of Ohm's law and KCL.
     
  14. Apr 26, 2013 #13
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