# Inductor Rise time in resonance

Discussion in 'General Electronics Chat' started by wes, Jun 19, 2012.

1. ### wes Thread Starter Active Member

Aug 24, 2007
242
2
I asked a question sort of like this one a while ago but I still don't think I really understand it. So here it is, pretty straight forward.

If a inductor is connected to a circuit where there is an appropriate valued capacitor in series with the inductor so they are in resonance, they cancel each others reactance. Does this mean that the current in the inductor will rise faster because the reactance is no longer there or least not as great as what the reactance would normally be for that inductors inductance.

To make it simpler, when a inductor is in resonance, does that mean the time constant for the current decreases because the reactance is less which means a higher current in the same amount of time for the same voltage and inductance?

2. ### #12 Expert

Nov 30, 2010
16,705
7,358
No. The vanishing act is only apparent to the outside world. The inductor and the capacitor each act normally within the confines of their own little piece of the world.

In other words, you can't repeal what's-his-names laws by putting something in series with an inductor.

Edit: Actually, that's a fun part of learning to do the math. You examine the different points of reference and understand what's really happening in each part.

3. ### wes Thread Starter Active Member

Aug 24, 2007
242
2
alright, I just wasn't sure, I was thinking of it in a different way and it got me a little confused, but I see now just as i did the last time I talked about this, that it does not change it.