# Inductance coil with self-inductance and resistance

If a coil has $L$ henry self-inductance, resistance of $R$ ohms and potential deferens of $V$ volts what will be the current through it?
In my opinion this case should be considered as resistance and inductance in series and to derive the current from the following equation: $$V=Ri+L\frac{di}{dt}$$ What bothers me in this formula is that it considers the resistance in the coil as subjected to potential difference ($Ri$), different than that in the inductance ($L\frac{di}{dt}$), but both the resistance and the inductance span the ends of the coil and they should experience the same potential difference.
If I consider the case of connecting them parallel the same problem will occur but for the current rather than voltage.

• I think it should be the case of series connection because of energy considerations, i.e. both the resistance and the inductor will dissipate the energy of the charge so the total energy dissipated will be the sum of them. – Pekov Feb 12 '16 at 7:06

If you have a stationary circuit then Pekov's comment is the correct approach. The energy stored in an inductance in the quasistatic limit is approximately $\frac{1}{2}LI^2$ and the power dissipated through a stationary resistor is $RI^2$ so we get the total rate of energy loss to be $$RI^2+LI\frac{dI}{dt}.$$

And expressing that in a per current fashion we get

$$\frac{P}{I}=RI+L\frac{dI}{dt}.$$

And we know the powers add because the energy loss of the resistor is thermal so has no phase coupling to the inductive fields.

Consider the inductance as a conductor + inductance effect. So you have a superposition effect. Physically it is the same object, and the voltage across the inductance will be V=Ri+Ldi/dt .

I thought originally that a justification for splitting an non-ideal inductor into an ideal resistor in series with an ideal inductor was that the E-field in an ideal resistor is due to electric charges whereas the E-field in an ideal inductor is due to a changing magnetic flux. However that separation of E-fields cannot be done for a non ideal capacitor which cam also be modelled as an ideal resistor and ideal capacitor.

So perhaps the justification for modelling a non-ideal inductor as a ideal resistor in series with an ideal inductor is that in practice predictions using this model are found to be correct?

• actually even in an ideal inductor the non-conservative electric field produces a conservative one, thus we can define voltage across an inductor. Check this video youtu.be/EYYNRubHIno?t=39m – Tonylb1 Feb 12 '16 at 11:37
• @Tonylb1 Many thanks for the link to the video. I liked the lecturer's style and the use of a mutual inductor to explain his point which was really to transfer the $-L\frac{di}{dt}$ term from the right hand side to the left hand side as $+L\frac{di}{dt}$ and so enable the use Kirchhoff's voltage law which sums the voltages in a loop to give zero. – Farcher Feb 12 '16 at 12:13