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A multi loop curcuit- Resistors in parallel and series

  1. Feb 18, 2009 #1
    1. The problem statement, all variables and given/known data

    The battery in the figure below has negligible internal resistance. Find
    l_51ef2b6852a62056a601d62c5837c0c9.jpg

    (a) the current in each resistor:
    (b) the power delivered by the battery:


    2. Relevant equations

    V = IR

    Resistor Connected in series

    Req = R1 + R2 + R3

    Resistor Connected in Parallel

    1/Req = 1/R1 + 1/R1 + 1/R3

    3. The attempt at a solution

    I think that the resistors 3omega and vertical-2omega are connected in series, and the other 2 are connected in parallel.

    So the ones in Parallel..

    1/Req1 = 1/2 + 1/4
    1/Req1 = 3/4
    Req1 = 1.333

    The ones in Series

    Req2 = 3 + 2 = 5

    so...

    Req = Req1 +Req2 =1.333 + 5 = 6.333


    But to find the Current, I you just use Ohm's law

    I = V/r

    so the 3.0-omega resistor would be

    I = 6/3 = 2

    ...but thats wrong and Im at a loss to know why?? thanks for the help!
     
  2. jcsd
  3. Feb 18, 2009 #2

    LowlyPion

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    Homework Helper

    Unfortunately the 2Ω, the 2Ω and the 4Ω are all in parallel. They share a common node on each side. That's what || means.

    Find the equivalent of those 3 together and then you can add in series.
     
  4. Feb 18, 2009 #3
    The problem that you are having comes from the diagram you are using. When you see something like this you should redraw it in an easier way to see it. Start with the battery, and connect it to the first resister. Then notice that the connection splits in three, so this means you can redraw those three resisters in parallel. Finally they reconnect at the end, and the single wire returns to the battery.

    When you draw it this way, you will see that all the 2 ohm resistors and the 4 ohm are in parallel, and the 3 ohm resistor is in series with them.
     
  5. Feb 18, 2009 #4
    So would the current for the first resistor be I = V/R = 6 V/ 2 = 3 ?
     
  6. Feb 18, 2009 #5

    LowlyPion

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    You have to determine the Req for the network, in order to determine I.
     
  7. Feb 18, 2009 #6
    I got Req= 3.8

    parallel = 1/Req = 1/2 + 1/2 + 1/4 = 5/4
    Req-p = 0.8

    so Req = 0.8 + 3 = 3.8 A

    Then do I use this in V = IR?
     
  8. Feb 18, 2009 #7

    LowlyPion

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    That would be the way to do it.

    Except it's 3.8Ω
     
  9. Feb 18, 2009 #8
    But be careful, since the current isn't nesseceraly the same in every resistor.
     
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