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Simplyfing Parallel/Series Resistors in a Circuit

  1. Feb 4, 2007 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    V = IR
    Rseries = R1 + R2 + ... + Rn
    Rparallel = (1/R1 + 1/R2 + ... + 1/Rn)^-1

    3. The attempt at a solution

    I simplified the circuit, to find I = 3mA. What I did was this (in order starting from far right):

    6||3 = 2kohm, which is in series with 16kOhm, so 18kOhm.

    The 18kOhm is in || with 9kOhm, so 6kOhm.

    6kOhm is in || with 6kOhm, so total of 12kOhm.

    The 12kOhm is in || with 4kOhm, as well as the 6kOhm. So I got 2kOhm for that part.

    Then, then 1kOhm and the 2kOhm are in series, so that equals 3kOhm total equivalent resistence.

    Then, I found that I = 3mA.

    Now, it says to find I_0. I have no clue as to how to do this. I think my simplifying is correct.
  2. jcsd
  3. Feb 4, 2007 #2


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    Homework Helper

    Use Ohms law.

    You know the current through the 1K resistor is 3 mA, so you can find the voltage across that resistor.

    So, you can find the voltage across the 6K resistor. Then you can find the current through it.

    Another way would be to reduce the circuit to the 1K resistor, the 6K resistor, and another resistor is parallel. You have done most of the work already, to find what the other resistor value is!
  4. Feb 4, 2007 #3
    V across the 1k resistor = IR = 3mA * 1kOhm = 3V

    See, this is where I'm stumped. I can just do this?:

    V = IR
    (9 - Vacross 1k resistor) 6V = I_0 * 6kOhm
    I_0 = 1mA?

    I was thinking of doing it that way, keeping the 6kOhm and another resistor parallel to it and use the current division method. However, what should I do about the 1kOhm resistor in the way?
    Last edited: Feb 4, 2007
  5. Feb 4, 2007 #4


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    Imagine you are measuring the voltages at different points in the circuit from the "-" end of the battery. The + end of the battery measures obviously 9V. You know the voltage across the 1K resistor is 3V, and one end of it is at 9V. So what is the voltage at the other end?

    You know the total current flowing through the parallel resistors is 3 mA. So you can forget about the battery and the 1K resistor from now on - they are just providing that 3ma current. Work out how the 3mA is split between the two resistors in parallel and you are done.
  6. Feb 4, 2007 #5
    When I do these current divider type deals, I find it helpful to label the equivalent resistances as I'm reducing them, and draw an arrow to the circuit showing where it applies. This way when I need to do a current divider, or maybe multiple current dividers, then I already have the "2nd resistor" to plug into the equation.
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