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Finding Resistance in a Parallel Circuit

  1. Apr 2, 2013 #1
    So there is a parallel circuit with two resistors, R1 and R2, and an ammeter that are connected to a constant 30-volt source. The equivalent resistance of the circuit is 6.0-ohms. The resistance of R1 is 10.-ohms. The resistance of R2 is equal to
    a) 6.0 Ω
    b) 2.0 Ω
    c) 15 Ω
    d) 4.0 Ω



    1/Req = 1/R1 + 1/R2



    1/6Ω = 1/10Ω + 1/R2, solve for R2
    I plugged in the values and I got 15 Ω as my answer, but my teacher marked it as wrong in my homework.
     
    Last edited by a moderator: Apr 2, 2013
  2. jcsd
  3. Apr 2, 2013 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF.

    I get 15 [STRIKE](well 14.99999...)[/STRIKE] too. Is there more to the question? Does the ammeter have some shunt resistance that you need to take into account?
     
    Last edited: Apr 2, 2013
  4. Apr 2, 2013 #3

    PeterO

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    Homework Helper

    When resistors are connected in parallel, the effective resistance is always smaller than the smallest of the individual resistors. That means both of these two resistors has to have a resistance in excess of 6Ω.

    The ONLY option to meet those conditions is 15Ω.

    Your accurate calculation shows that to be the correct answer.

    It appears your teacher has not recognised your correct answer.

    EDIT: In case your teacher is being very technical ... Did you say R2 was equal to 15 Ohm, or did you say (c)
     
  5. Apr 2, 2013 #4
    Are you sure there is nothing special about the circuit? Why are voltage source and the ammeter mentioned?

    If the circuit is just two parallel resistors, your result is correct because 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6.
     
  6. Apr 2, 2013 #5

    PeterO

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    Homework Helper

    Indeed: where is the ammeter connected in this circuit?
     
  7. Apr 2, 2013 #6
    The other values given in the question that aren't useful are meant for other questions. I just really wanted to know if 15-ohms was the answer. My physics teacher makes a lot of mistakes. Thank you guys!
     
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