Quote from: poynt99 on April 13, 2011, 06:53:44 PM
No problem Laurent.

Sometimes it is better to think about inductors not in terms of AC or DC, but in terms of what the current is doing.

Ask yourself this question:

"Is the inductor current changing?"

If the answer is "YES", then you treat the inductor as a resistance that changes depending on how fast the current is changing through it. The faster the current changes, the higher the resistance (impedance).

So, in the case of the capacitor energy transfer experiment, the current is changing, and therefore the inductor will offer some impedance (similar to resistance, but frequency dependent and with a phase shift) to the circuit. If the inductance is large enough in comparison to the total wire resistance, most of the energy will be absorbed by the inductance rather than being burned in the total wire resistance. Therefore, more of the energy will be transferred to the second capacitor rather than being lost in the resistive wiring.

Make sense?

.99

Another way to say it is, as long as the currents are changing, it is considered A/C.   So in our circuit, from start to finish, we can say we are dealing with AC, as varying DC can be considered AC, as in an AC signal with dc bias ;] 

As for the energy is expended and stored  into the inductor, I fail to see how it is that the resistances in the circuit are ignored by the current in any situation. And virtually no heat is produced and losses by way of producing heat are seemingly nil? Current did flow through the resistances, didnt it? More will be in my reply. ;]


Mags

Mags,

Yes, current still flows through all the resistive wiring, but it will be a much lower current compared to that if the circuit consisted of the wiring and caps alone.

P = I²R correct?

So, since the R of the wiring has not changed, but the current has been reduced by a lot (when we utilize an inductor), there will be less power dissipated in the R of the wiring.

.99

Quote from: poynt99 on April 13, 2011, 06:53:44 PM
The faster the current changes, the higher the resistance (impedance).

.99

Is this what you really meant to say?

If it is, I can not even come close to grasping that concept and I need references to re-educate myself that a higher impedance speeds current flow.

mikemogo 

Quote from: mikemongo on April 13, 2011, 07:34:49 PM
Is this what you really meant to say?

If it is, I can not even come close to grasping that concept and I need references to re-educate myself that a higher impedance speeds current flow.

mikemogo

Yes, that is what I meant to say.

However, you've interpreted it backwards; A higher rate of change of current through the inductor produces a higher impedance across it's terminals.

Inductive Reactance: X_L = 2Ï€FL

.99

Quote from: Magluvin on April 13, 2011, 07:16:15 PM
Another way to say it is, as long as the currents are changing, it is considered A/C.   So in our circuit, from start to finish, we can say we are dealing with AC, as varying DC can be considered AC, as in an AC signal with dc bias ;] 

Mags

If you want to be technically correct, then you must not think of pulsed DC as AC. "Alternating Current" literally means that the current alternates direction i.e. a voltage going both above and below ground or the reference level, or a current flowing away from, then to the source.

Pulsed DC is a varying current that becomes lower or higher in magnitude over time, but the direction or polarity of the current is always either positive or negative.

.99