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Finding current through resistor (Kirchhoff's Laws)

  1. Mar 21, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider the circuit shown in the diagram below, for R1 = 5 Ω, R2 = 8 Ω, R3 = 8 Ω, R4 = 8 Ω, and V0 = 8.0 V. Calculate the current through R4.

    lICBmYQ.gif

    2. Relevant equations

    Loop rule: The sum of all potential changes around a closed loop is zero

    Junction rule: The currents entering and leaving a junction are equal

    V = IR

    3. The attempt at a solution

    I guess I'm kind of lost with where to begin. The examples in class were a little more simplified but I'll say what my thought process is to solving it.

    First I found the equivalent resistance through the whole circuit

    1/R4 + 1/R3 = 1/Req34
    1/8 + 1/8 = 1/Req34

    Req34 = 4 Ω

    Then that resistance + R2 = the equivalent resistance on the far left loop

    4 + 8 = 12 Ω

    Once more, to find the equivalent resistance between that and R1, I went

    1/12 + 1/R1 = 1/Req

    Req for the circuit = 3.53 Ω

    Then to find the current through R4, I've been using V = IR where V = 8.0 V and Req = 3.53

    I = 2.27 A

    This is incorrect, I'm just not quite sure how else to approach this problem. Am I on the right track?

    Thanks!
     
  2. jcsd
  3. Mar 21, 2014 #2
    Yes, you are definitely on the right track. Right now, it looks like you have calculated the circuit current (current through the battery). If this was a series circuit, then everything would also have the same current, but because there are some parallel components, you must calculate how much current goes through each "fork" (use junction rule).

    Remember that the lower the resistance, the more current that's going to go through that wire when it splits, so you will have to calculate the Req of each wire coming from each fork. (You are doing very well, so I think this is enough guidance for now. If you are still stumped, let us know ^_^)
     
  4. Mar 21, 2014 #3
    Potential difference across R4 is not 8.0 V. Voltage of the battery divides itself between Req34 and R2. Think how will 8.0V divide itself.

    [Hint: Take the potential difference as V1 and V 2 for potential difference across Req34 and R2 respectively. After that find out the ratio [itex]\frac{V1 }{V2 }[/itex] , using the relation V=IR]
     
  5. Mar 21, 2014 #4
    Okay, so I have the total circuit current and all of the individual resistances and the battery's voltage. I'm a little confused on where to go from here to be honest!
     
  6. Mar 21, 2014 #5
    I'd say that since R4 is pretty much the last resistor you get to, you might as well take it leg by leg. For example, have you noticed that R1 branch is parallel to the whole equivalent R234 branch? What do you know about the voltage across parallel branches?

    After that, it's just Junction Rule to find out how much current remains after the R1 branch, and you're on your way.
     
  7. Mar 22, 2014 #6
    It looks like you have two branches of wire that share the same two nodes. This means they must be equipotential, right?

    On the second branch, notice again that your third and fourth resistor share nodes, so these are also equipotential.

    Perhaps if you could find the voltage drop over R2, and subtract that from the total voltage drop over that length of wire, you could find the drop over your two parallel resistors, R3 and R4
     
  8. Mar 22, 2014 #7

    NascentOxygen

    User Avatar

    Staff: Mentor

    Using a coloured pen, trace the complete circuit taken by the battery current passing through R4.
     
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