Measuring current in RLC circuit

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SUMMARY

The discussion centers on measuring current in a driven series RLC circuit, specifically addressing whether to include the internal resistance of the inductor (L) when calculating current through a variable resistor (R). It is established that the potential difference (PD) across R can be used to calculate current using Ohm's Law (I = V/R) without factoring in L's internal resistance for that specific measurement. However, for a complete analysis of the circuit's behavior, the total resistance, including both R and L's internal resistance, must be considered.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Basic knowledge of RLC circuit components
  • Familiarity with series circuit behavior
  • Concept of internal resistance in inductors
NEXT STEPS
  • Study the effects of internal resistance in RLC circuits
  • Learn about energy storage in inductors and capacitors
  • Explore advanced circuit analysis techniques, including Thevenin's theorem
  • Investigate the impact of component quality on circuit performance
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing or experimenting with RLC circuits will benefit from this discussion.

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
 
Last edited:
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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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