Measuring current in RLC circuit

AI Thread Summary
In a driven series RLC circuit, when calculating current from the potential difference across a variable resistor R, the internal resistance of the inductor L should not be included in the immediate calculation. Including L's resistance would reduce the measured potential difference across R, leading to an inaccurate representation of the current and energy storage efficiency. Ohm's law applies to each resistance individually, meaning the current can be determined solely from the voltage across R. However, for a complete analysis of the circuit's behavior, the total resistance, including both R and the internal resistance of L, must be considered. Accurate measurements and calculations are crucial for understanding the circuit's performance.
dannyR
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Homework Statement


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"?

Homework Equations



V=IR

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
 
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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.
 
Thank you very much, I think I need that push of clarity which you provided. :approve:
 
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