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Measuring current in RLC circuit

  1. Oct 28, 2012 #1
    1. The problem statement, all variables and given/known data
    I've recently completed a experiment involving a Driven series RLC circuit. The potential difference across R " a variable resistor >> internal resistance of inductor" was measured.

    When calculating the current from the PD across R, should I include the internal resistance of L " the inductor"?

    2. Relevant equations

    V=IR


    3. The attempt at a solution

    If I include L's resistance:

    I think this would give a incorrect answers because the PD across R would be less.

    So the overall effect would be that the amplitude of current in the components will be reduced, and the measured effect of storing energy in C or L would come out as being very good quality components. Since numerically they would be storing a larger percentage of the supplied energy.

    If I don't include L's resistance:

    I'm not sure on the affect in the formula when calculating values from this current. Since L is assumed to be a perfect inductor, i.e. all the energy is stored within the magnetic field and not dissipated as heat.

    Thanks Danny
     
    Last edited: Oct 28, 2012
  2. jcsd
  3. Oct 28, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    The current I flowing through the resistor R produces the potential drop V across that resistance. Ohm's law applies individually to each resistance. While it is true that the same current flows through all the components in a series circuit, other resistances in the string will not affect the potential across that particular R due to that current.

    Suppose that you could somehow separate the internal resistance of the inductor from the inductor itself and make it a separate resistance r in series with an ideal inductance. Then, if you were so able to do so, you could measure the potential drop v across this r and deduce the same current I that you did for the external resistor R.

    The bottom line is, in order to find the current I that is flowing in the series circuit, it is enough to measure the potential drop across R and apply Ohm's law for that resistance alone: I = V/R.

    When deducing other characteristics of the circuit's behavior, however, you will have to take into account that the total resistance is R+r.
     
  4. Oct 28, 2012 #3
    Thank you very much, I think I need that push of clarity which you provided. :approve:
     
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