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Homework Help: Finding current in a curcuit

  1. Mar 12, 2008 #1
    1. The problem statement, all variables and given/known data

    I just need help with part B, i'm pretty sure part A is useless but I found Req to be 5 ohm

    http://img146.imageshack.us/img146/4499/physicsrh2.jpg [Broken]

    2. Relevant equations


    3. The attempt at a solution

    Well basically I found the voltage change going through the 5ohm to be 10v but then I don't understand how to find the current through anywhere else.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Mar 12, 2008 #2
    if the current through the [itex]5 \Omega[/itex] resistor is 2A, then the current through the assembly of the 3 resistors i.e. the assembly of the [itex]6 \Omega[/itex], [itex]3 \Omega[/itex] and [itex]2 \Omega[/itex] is also 2A as both these assemblies are in series.

    So, for the assembly of three resistors, you have the incoming current as 2A. When the current splits in more than two arms, the current in each arm is inversely proportional to current. This follows from the formula V = IR, as all three arms are in parallel, they are across the same potential difference and I becomes inversely proportional to R.

    So, if the arms have resistances in ratio 1:2:3, the current in each arm will split in the ratio 3:2:1 i.e. the arm with the lowest resistance will have the highest current through it.. however the ratio still holds.

    In your example, the resistances are in the ratio 6:3:2. Hence the current will be in the ratio 2:3:6. How does this divide into 2A of current? Construct a linear equation and solve for it. You shall have your answer...
  4. Mar 12, 2008 #3

    Thanks for the help, I don't really understand to show the science but basically the 6ohm resistor will get .33A, 3ohm will get .66A and the 2ohm will get 1A? How do I show that?
  5. Mar 12, 2008 #4
    well.. yes.. you have got it right. What exactly do you need help with showing? The principle behind this method or the mathematical solution of the answer?
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