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Homework Help: Current in a Complicated Circuit

  1. Aug 29, 2008 #1
    1. The problem statement, all variables and given/known data
    In the figure, the current of the circuit in the 8 ohm resistor is 0.5 A. What is the current in the 2 ohm resistor?

    Figure http://img181.imageshack.us/img181/5576/helpfk7.jpg [Broken]

    I'm trying to solve for the current in a complicated circuit. I'm given R1 = 16ohms, R2 = 8ohms, R3 = 20ohms, R4 = 6ohms, and R5 = 2ohms. Also, the question for the problem states that the current in R2 = 0.5 A

    2. Relevant equations
    Ohms Law: Potential Difference = Current(Resistance)
    Formula for Resistors in Series: Rt = R1 + R2 + R3...
    Formula for Resistors in Paralell: 1/Rt = 1/R1 + 1/R2 + 1/R3

    3. The attempt at a solution

    I'm not sure how to go about finding the current through R5 without knowing the voltage of the battery. But I think I may be able to figure this out by trying to compare R2's current with the entire circuit.

    Idea 1:
    R1 is twice R2, and being in parallel, R1 receives half R2's current.
    So R1 has a current of .25 A.
    Then from here somehow figure out how to find the current in R5.

    Idea 2:
    Using Ohm's law I can find the voltage in R2 V=0.5(8), V=4v.
    Then applying it to R5, 4v=I(2), I=8 A.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Aug 29, 2008 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    You know the current through R2, it's given. You know the current through R1, as in your Idea 1. You thus know the total current through R3 (it's the sum of the currents through R1 and R2).

    You can use Ohm's law to find the voltage across R2, and the voltage across R3. The sum of those two voltage drops is the voltage from the battery. From there, it's pretty straightforward.

    Your idea 2 is incorrect, because the same voltage is not applied to R2 as to R5 -- these resistors have different nets at their right hand terminals! Instead, the voltage applied to the entire "subcircuit" of R1-3 is the same as that applied to the entire "subcircuit" of R4-5, because each of these "subcircuits" is wired in parallel to the battery.

    - Warren
  4. Aug 29, 2008 #3
    Alright so my reworked solution is as follows:

    Current of R1 = 1/2 Current of R2 so, Current of R1 = .25 A
    Current Through R3 = Current through R1 + R2 so, Current of R3 = .75 A
    Voltage through R2 = .5(8) = 4v
    Voltage through R3 = .75(20) = 15v
    Battery Voltage = 4v+15v = 19v

    V = IA
    19v = I(2)
    9.5 A = I
  5. Aug 29, 2008 #4


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    Homework Helper

    That looks right. Good job.
  6. Aug 29, 2008 #5
    Thank you both for the help. :approve:
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