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Electric circuits problems, finding current.

  1. May 19, 2013 #1
    1. A series circuit initially consists of a cell of unknown voltage, a resistor R2 of 4.0 Ω, and another resistor R2 of 6.0 Ω. A student measures the current through R2 to be 3.0 A. When a wrench that has a small resistance is dropped on the circuit, the current through R2 is reduced to 1.0 A. What is the current flowing through the wrench? (Assume the supply voltage remains constant.)

    a) 1.0 A
    b) 2.0 A
    c) 5.0 A
    d) 7.5 A


    To summarize:

    R1=4.0 Ω
    R2= 6.0Ω
    I2 initially= 3.0 A
    I2 with wrench= 1.0 A
    Find I (current) going through the wrench.

    2. Relevant equations

    V=IR

    3. The attempt at a solution

    My answer is a) 1.0 A, but it is wrong. The correct answer is c) 5.0 A but I don't understand why.

    I was told that in a series circuit, the current should be the same throughout. So in the circuit (with wrench), if the current is measured after a resistor to have a current of 1.0 A, shouldn't the current through the wrench be the same?
     
  2. jcsd
  3. May 19, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    Dropping the wrench onto the circuit is unlikely to insert the wrench into the series scenario. Instead, it will likely lie across some component, providing a parallel path...
     
  4. May 19, 2013 #3
    Initially:

    R1 = 4.0 Ω
    I1 = 3.0 A
    V1 = 12.0 V

    R2 = 6.0 Ω
    I2 = 3.0 A
    V2 = 18.0 V

    So VT = 30.0 V

    With wrench:

    VT = 30.0 V

    R1 = 4.0 Ω
    I1 = 1.0 A
    V1 = 4.0 V

    R2 = 6.0 Ω
    I2 = 1.0 A
    V2 = 6.0 V

    Vwrench = 20 V

    But I still don't know R of the wrench so I can't figure out I.
     
  5. May 19, 2013 #4
    Gneill correctly states, Dropping the wrench onto the circuit is unlikely to insert the wrench into the series scenario. Instead, it will likely lie across some component, providing a parallel path...

    Draw a diagram of the initial circuit. Then figure out where you could drop a wrench that would reduce the current through R2. Then work the circuit.
     
  6. May 19, 2013 #5
    Okay, I got it! (plus the three other questions I posted as well).

    Thanks!
     
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