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Finding Current

  1. Sep 17, 2009 #1
    http://img33.imageshack.us/img33/7354/29346556.png [Broken]

    1/R = 1/R1 + 1/R2

    1/R = 1/38 + 1/37 = .053 But this is not right and I know I'm missing something just dont know what to do after this
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 17, 2009 #2
    What are you trying to do with the equations you provided? It looks like you're adding resistors in parallel... this is not what you want to do (what is your thought process? - maybe we can help put you back on track)... what other equations might be useful here?

  4. Sep 17, 2009 #3
    It seems that you have some problems in distinguishing the parallel and series configuration.

    Two resistors are parallel if their ends are connected together or sometimes it is said that if they have the same voltage. Look at R1 and R2; you assumed that they are paralle. What do you think; are they parallel? Now look at R2 and R3...are they parallel?
  5. Sep 17, 2009 #4
    Well what I dont understand is what happens to the current once it reaches this point

    http://img19.imageshack.us/img19/3268/98989815.png [Broken]

    Yes, isnt r2 r3 parallel since they are not the same current
    Last edited by a moderator: May 4, 2017
  6. Sep 17, 2009 #5
    At the point you pointed to it, the total current (red) is splited to two components. Ix and the other component (green).

    http://img246.imageshack.us/img246/7354/29346556.png [Broken]

    I don't understand what do you mean, but the diferrent current not always means that we have parallel configuration.

    Look at R2 and R3. Their ends are connected together as well as there are equal voltage across them.

    Try to obtain the equivalent (total) resistance of the circuit. Then obtain the total current of the circuit. Afterward use current divider to obtain Ix.
    Last edited by a moderator: May 4, 2017
  7. Sep 17, 2009 #6
    So the current flowing through r1 is 6/37 so wouldn't that mean that the current flowing across r2 and to r3 is is 12/37?
  8. Sep 17, 2009 #7
    Don't bother yourself. I understand. But don't be dissappointed. As soon as you practice these concepts will become very simple for you and in the future courses you will use these concepts as a tool to solve other problems. I'm sure.

    No. Let's advance step by step through the solution.

    1) What is the given information? We have the values of all components.
    V= 6 V; R1 = 37 Ohm; R2 = 38 Ohm; R3 = 34 Ohm.

    2) What is the purpose? We want to obtain the value of Ix.

    Begin the solution:

    3) Assume the currents as I pointed on my post. Let call them:
    Ir (red), Ig (green), Ix ( what to be find)

    4) Obtain the total resistance of the circuit.
    Assume that you are sitting in place of the voltage source. Look to the circuit. What do you see? OK, you see R1, R2, and R3.
    BEGIN from the END. (WHAT??!!). I mean look at the end of the circiut; remember that you are sitting on the place of the voltage source. What do you see??
    Let see. I see R2 and R3 at the end of the circiut. Do you see them? And I also see that they are in parallel ( their end connected togethrer). So, combine them and call the result Rp.

    Rp = R2 // R3
    1/Rp = 1/R2 + 1/R3
    Rearrange this:
    Rp = (R2 * R3)/(R2 + R3) = 19.95 Ohm

    Ok. Replace the two resistances R2 and R3 with Rp ( a single resistance)
    Again sit in place of voltage source and see. Now what do you see?
    I see R1 in series with Rp. Do you see it? Great.
    Calculate the total resistance of the circiut (call it Rt).

    Rt = R1 + Rp = 54.95 Ohm

    5) Calculate the total current using Ohm's law (call it It). (Ir and It are the same).

    It = V/Rt = 0.11 Amp [ ow, I forgot, you can now stand up. Give back the voltage source its place :) ]

    6) All things are going well. The last thing is to obtain Ix. Back to the first arrangement of the circiu ( With R1, R2, and R3; no Rp).

    7) Use the current divider to obtain Ix:

    Ix = It * R2/(R2 + R3) = 0.057 Amp

    That's all.
    Please do the steps yourself so that you never forget them. Check my calculations to be sure that there are no mistakes ( I think there won't)
    Last edited: Sep 17, 2009
  9. Sep 17, 2009 #8
    I got it! Thank you so much!
    Last edited: Sep 17, 2009
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