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Finding current with parallel resistors

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data



    R1 = 2 Ω R2 = 5 Ω R3 = 11 Ω R4 = 10 Ω V = 7 V

    What is the current through R2?

    2. Relevant equations

    Req for parallel resistors: 1/Req = 1/R1 + 1/R2...
    Req for series resitors: Req = R1 + R2...
    Electric potential V = current I x resistance R
    I have found that Req is 1.755 and the current supplied by the battery is 3.988.


    3. The attempt at a solution

    I tried taking the 3.988 x 3.33 (the Req for R2 and R4) which is 13.29V and since the voltage drop is the same for parallel resistors I then divided by the 5 ohms to get 2.658 amps but that's not right.
     

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    Last edited: Feb 15, 2009
  2. jcsd
  3. Feb 15, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi cm846! Welcome to PF! :smile:

    How did you get the 3.988? :confused:
     
  4. Feb 15, 2009 #3
    I just took the 7V divided by the Req of 1.755 to get 3.988
     
  5. Feb 15, 2009 #4

    tiny-tim

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    But that would require the battery and the R2-R4 combination to be the only things in the loop. :wink:
     
  6. Feb 15, 2009 #5
    3.99 is the answer to the question "What is the current supplied by the battery?" so I guess I'm not sure how I'd find it any other way than the way I showed?
     
  7. Feb 15, 2009 #6

    tiny-tim

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    Sorry, I mistook the 1.755 it for the R2-R4 Req.

    I should have asked, where did the 1.755 come from?
     
  8. Feb 15, 2009 #7
    Thats ok. I got the 1.755 for the Req for the entire circuit: (R2) 1/5 + (R4) 1/10 = 1/Req which is 3.33 then I added R3. 11 + 3.33 = 14.33 then 1/14.33 + (R1) 1/2 = 1/Req which is 1.755.
     
  9. Feb 15, 2009 #8

    tiny-tim

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    ah, got it …

    that's the current through the battery itself …

    but you need the current through the loop containing the battery and R2 R3 and R4. :wink:
     
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