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Homework Help: Finding an Unknown Resistance of circuit

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data

    In the circuit in the figure below, the reading of the ammeter is the same when both switches are open and when both switches are closed. What is the unknown resistance R? (Let R1 = 103 Ω, R2 = 279 Ω, and R3 = 46.0 Ω.)


    2. Relevant equations

    Adding Resistance in Series: R1 + R2 = Req
    Adding Resistance in Parallel: 1/R1 + 1/R2 = 1/Req
    V = IR ... I = V/R

    3. The attempt at a solution

    I believe you have to find the resistance in both cases, switch open and closed, and then set them equal because I = V/R is equal in both cases, so R must be equal, since V doesn't change (always 1.5).

    For switch OPEN, I got: Req,open = R2 + R1 + R3
    For switch CLOSED, I got: Req, closed = R2 + (1/R1 + 1/Runknown)^-1

    I set these equal and solved for Ruknown and got 316.8, but that isn't right. Any tips? Thank you!!!
  2. jcsd
  3. Feb 2, 2012 #2


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    Staff: Mentor

    What remains the same in both cases is the current through R1 (as measured by the ammeter). So I'd suggest first determining what that current is when the switches are both open. Also determine the voltage that will appear across R1.

    Next draw a sketch of the circuit that results when the switches are closed (are any components bypassed and hence removed from the circuit?). Add the voltage and current you determined above to the sketch -- the current through the ammeter and hence the voltage across R1 should remain the same as before. Remember that an ideal ammeter has no resistance, hence no voltage drop will appear across it. Can you find the remaining currents?
  4. Feb 2, 2012 #3


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    Science Advisor
    Homework Helper

    Hi SMA777! :smile:
    No, the Req you found (for switches closed) gives you I = V/Req for the current next to the battery …

    that current will split before going through the ammeter, won't it? :wink:
  5. Feb 2, 2012 #4
    Oh, I see what you mean about how it splits up! Ok, got it. My question is, how do I know... how it splits up? IE how much goes to Runknown and how much to R1? Is the sum of their currents = total?
  6. Feb 3, 2012 #5


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    (just got up :zzz: …)
    Yes, from KCL.

    Then use KVL for the loop containing just R1 and R. :smile:
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