Current Divider Law

Discussion in 'Homework Help' started by atomtm, Nov 1, 2012.

  1. atomtm

    Thread Starter New Member

    Aug 13, 2012
    24
    0
  2. bertus

    Administrator

    Apr 5, 2008
    15,634
    2,342
    Hello,

    How did you calculate the current?

    Bertus
     
  3. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,787
    We are NOT mind readers! How can you POSSIBLY expect us to know how you got a particular wrong answer unless you show us HOW you went about coming up with that answer?
     
  4. panic mode

    Senior Member

    Oct 10, 2011
    1,318
    304
    hello atomtm,

    welcome to the forum. as you have noticed by now, there are some expectations when it comes to the questions - you need to give us enough of information.

    the result of 5mA is incorrect for current through R1 when powered from 6V. this can be verified directly using Ohms law. I=V/R=6V/1k=6mA.
    but when using current divider, we normally do not know the voltage, we know current.
    in this example, total current is I=I1+I2+I3=6+2+3=11mA
    if we didn't know voltage but did know that total current is 11mA and resistor values are 1k 3k and 2k, we could still deduce individual currents through each resistor.

    to use current division rule we need to simplify circuit. if R1 is the branch of interest, we need to find equivalent resistance of rest of the circuit (all other branches). in this case that would be R2 and R3 which are in parallel so

    Rt= R2 || R3

    this can be calculated by solving
    1/Rt = 1/R2+ 1/R3

    or by remembering simplification:
    Rt=R2*R3/(R2+R3)
    Rt=3*2/(3+2)=6/5 kOhm

    I=11mA, R1=1k, Rt=1.2 k;

    I1=I*Rt/(R1+Rt)
    I1=11mA * 1.2k /(1k +1.2k)= 11mA * 1.2/2.2 = 6mA

    it does not matter which method you use, if you don't make any mistake result will always be the same.

    if you wanted to apply this to find current through R2, then rest of the circuit is parallel connection of R1 and R3

    Rt=R1*R3/(R1+R3)=1*2/(1+2)=2/3 kOhm

    I=11mA, R2=3k, Rt= (2/3) k;

    I2= I*Rt/(R2+Rt)=11mA * (2/3)/(3 + 2/3)=11mA * (2/3)/(9/3 + 2/3)= 11mA * (2/3) / (11/3)= 11mA * 2/11 = 2mA

    since we do know that V=6V we can do direct check,

    I2= 6V/R2= 6V/3k = 2mA
     
    Last edited: Nov 1, 2012
  5. atomtm

    Thread Starter New Member

    Aug 13, 2012
    24
    0
    Well thank you all for your answers and I 'm really sorry for the dumb question i finally got the right answer!!

    Thank you again
     
Loading...