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Adding a wrench to a series circuit ?

  1. Feb 27, 2012 #1
    If someone could also clarify how to go about this one.

    Does the wrench added make it a parallel circuit?
    I found the voltage of the initial circuit by simply doing V = IR = (3)(10) = 30 V.

    The current flowing through the 4 ohm resistor must be the same as before... so the current for that is still 3.0 A?

    Help please :s
     

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  3. Feb 27, 2012 #2

    SammyS

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    attachment.php?attachmentid=44481&d=1330405295.png

    Well, in a sense the wrench makes it a parallel circuit, with the wrench being in parallel with R2. But the resistance of the wrench is small. (I assume that it's much smaller than the resistance of either resistor.) The effective resistance of two resistors in parallel is less than the resistance of either one.
     
  4. Feb 27, 2012 #3

    ehild

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    the wrench is parallel with R2.
    Correct, for the emf of the battery.

    NO! It is the emf that is the same as before, 30 V. The currents will change. What is the voltage across R2 if 1 A current flows through it? What is then the voltage across R1? And what current flows through it? And then apply Kirchhoff's current Law.
     
  5. Feb 27, 2012 #4
    I don't know how to go about finding the current through the wrench though... especially if it has negligible resistance.
     
  6. Feb 27, 2012 #5
    oooh okay let me do that then...
     
  7. Feb 27, 2012 #6
    Voltage across R2
    V = IR = (1)(6) = 6 V

    Is the voltage across R1 the EMF minus the voltage parallel?
    So voltage across R1 = 30 - 6 = 24 V?
    Current through R1 would be I=V/R = 24/4 = 6 A.

    Oh! And then if there's only 1 A through R2 then there must be 5 A through the wrench due to Kirchoff's current law...
     
  8. Feb 28, 2012 #7

    ehild

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    Excellent!:smile:

    ehild
     
  9. Feb 28, 2012 #8
    Thanks! Finally get it, awesome.
     
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