Does current flow through an inductor when charging and how is it calculated?

[Lowrie]

My engineering course has told me inductors are the dual of capacitors. One of the key things it has left out so far is whether, while charging, any current passes through the inductor, and if so how to calculate the voltage or current on the other side. It would make sense to me if no current passed through the inductor to the other side until 5 time constants have completed, however I also realize inductors and capacitors charge exponentially. Meaning that it will never be fully charged.

So my main questions are:

• Does current pass through an inductor when it is charging?
• If so how do you calculate the voltage/current on the other side of the inductor?

Any help is greatly apprechiated.
I hope i conveyed my questions appropriately.

 

[nivaOne]

Hi Lowrie,

Inductors do not like change. Which means they will try to block current upon applying power to the circuit. ( DC power that is ). A capacitor does the opposite, it acts as a short circuit in the beginning and ends as an extreme high value resistor. An inductor will act as extreme high value resistor in the beginning and end as - an almost - 0 Ohm resistor. ( neglectable in a lot of cases I mean )

Current does flow however in every state. Check the site below to understand it better:

http://webphysics.davidson.edu/physlet_resources/bu_semester2/c19_RL.html

The formulas are also included.

I hope that helps. Do not worry about the fact that theoretically an inductor is never fully charged. the 5 times constant may be considered as 'fully'

remember in an LR circuit (DC powered ), the inductor initially is equivalent to an extreme high value resistor and in the end, after 5 \Gamma it represents an almost 0 Ohm resistor. Initially a micro small current will start to flow and in the end the current will be determined by the value of R.

( I hope that I have clicked the correct greek sign, I'm not so familiar with those ;)

 

[sophiecentaur]

It is the fact that current is flowing and its value is increasing that causes a changing magnetic field -and hence a 'back emf' is induced which acts against the applied PD. The limit to the increase in value of current is set by the DC resistance of the rest of the circuit. Hence, the (ideal) inductor will start with almost the full PD across it and end up with zero voltage drop. Theoretically, it will take for ever but, as nivaOne says, after five or so "time constants' (=RL or 1/C) things have near enough got as far as they're going to.
The value of this current flow follows the same exponential curve as the voltage across a charging capacitor.
When used in an AC circuit, there will be a difference in phase between V across and I through both components.