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Finding the current in resistors

  1. Oct 2, 2009 #1
    1. The problem statement, all variables and given/known data
    In the figure, R1 = 99 . R2 = R3 = 49.5 . R4 = 75.0 , and the ideal battery has emf script e = 6.00 V.

    Re-GLORY.gif

    What is i in R1?

    2. Relevant equations
    V=IR
    Junction Rule
    Loop Rule?
    i = E / (R+r)


    3. The attempt at a solution
    The Loop Rule states that for any loop in a circuit, the sum of the voltages across the things in it will be zero. So by combining R2 R3 R4 into Req of resistance 18.609 ohms, I have a loop, so I add the voltages:

    E - iR1 - iReq = 0
    E = 6 as given in the problem. (Right? Maybe I'm wrong here.)
    By the Junction Rule, the current i at R1 and Req would be the same.
    So,
    6 - i(R1 + Req) = 0
    But this just gives me the equivilant resistance for all the resistors, and let's call that plain R.
    So I have i = E/R, which is an equation I already had, so I felt good about coming across it via application of the two rules, but when I use this to find the current I don't get the right answer. My numerical answer to the question is 0.51016494 A.

    I don't exactly know where I'm going wrong here...

    Thank you for looking at this.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 2, 2009 #2

    Delphi51

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    Homework Helper

    I get the 18.6 ohms, too. Adding R1 in series, we have 117.6 ohms.
    I = V/R = 6/117.6 = .051 A. Looks like you misplaced the decimal point. Quite forgivable at this time of night.
     
  4. Oct 2, 2009 #3

    hotvette

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    Homework Helper

    What is the right answer?
     
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