QUESTION about current divider

Discussion in 'Homework Help' started by SUN SHINE, Jan 21, 2017.

  1. SUN SHINE

    Thread Starter New Member

    Jan 6, 2017
    13
    0
    Hi

    i want to ask a question about the attached photo
    finding the current in the 3-ohm resistor by this way isn't it wrong?? because i think that the 3-ohm resistor is connected in paralel with the the 2 and 4-ohm resistors
    so i think we should first sum 4+2=6 and the equivalent parallel resistor between 6-ohm and 3 -ohm
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    4,868
    1,364
    Yes , you can use the current divider rule but the current I that is flowing through 3kΩ resistor is equal to :
    I = 6mA * (4kΩ+2kΩ)/(3kΩ +(4kΩ+2kΩ)) = 4mA.

    Also notice that if I = 2mA then V = 2mA *3kΩ = 6V and (4kΩ+2KΩ) current is 6V/6kΩ = 1mA ----->2mA + 1mA is not equal to 6mA.
    And this is why it mus be wrong.
     
  3. RBR1317

    Senior Member

    Nov 13, 2010
    463
    87
    You are correct in stating that the 2KΩ & 4KΩ resistors in series make an equivalent 6KΩ resistor which is in parallel with the 3KΩ resistor.

    In the general form, the current division rule is stated in terms of conductance:

    In=It∙(Gn/(G1+G2+...+Gk)) which in the particular case of two parallel resistors reduces to I1=It∙R2/(R1+R2)

    So your solution of i=2mA is correct. To verify the solution calculate the voltage across the 3KΩ and 6KΩ resistors which must be equal.
     
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