# Measuring current in RLC circuit

1. Oct 28, 2012

### dannyR

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. Oct 28, 2012

### 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.

3. Oct 28, 2012

### dannyR

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