The bifilar pancake coil at its resonant frequency

[TinselKoala]

The discharge curve shows 6 units of charge dissipating over 1 unit of time down to .37. Then it shows 4 units of charge dissapating over 3 units of time.

Here's what I said:

"Discharging a capacitor is like decanting water from a five gallon jug; Slow starting, followed by a strong gush, (then*) at .67, tapering off to a slow flow: Charging exactly the reverse, max charge rate at .33 capacity". That places the maximum discharge at 4 units of dissipation over 1/2 unit of time. Take another look at the graph "Mister Differential Calculus".

*Addition to original.

What I'm implying is that the first 1/3 of the capacitor charge spills most forcefully from the jug; This is with a resistor. Shorting the capacitor results in an explosion.

Naturally, the charge curve is the inverse; The first 2/3 of the charge the most rapid then the remaining 1/3 equally time consuming as the last 1/3 of the discharge lag.

Now you are trying to "spin" what you said and claimed at first. You have been confusing RATE with QUANTITY, apparently.

"What you are implying" you are now stating nearly correctly. The first 1/3 of the capacitor charge or discharge is happening at a FASTER RATE than the remaining charge or discharge. And the FASTEST rate, corresponding to the highest current, occurs during the first time constant immediately after making the connection.

Again, this is what you said at first:

Discharging a capacitor is like decanting water from a five gallon jug; Slow starting, followed by a strong gush  at .67, tapering off to a slow flow: Charging exactly the reverse, max charge rate at .33 capacity:

And the parts I have emphasized in your original statement are wrong.

[nelsonrochaa]

Yes Evostars, the bogeyman is going to speak to you again.  I am going to make good sense, so the question for you is are you going to have enough character to respond to me?  Am I allowed to eat at the same lunch counter as you?

The power going into the capacitor is the product of the voltage times the current.  Both are functions of time.

p(t) = i(t) * v(t).

And of course the energy is the summation of the power over the charging cycle.  We can also say that the energy is the integral of the current times the voltage with respect to time from the start to the end of the cycle.

Let's go back to p(t) = i(t) * v(t).

We are charging a capacitor, and we know that the voltage on a capacitor v = q/C.

And we also know that q(t) is the integral of i(t) with respect to time.  So we can do some substitutions:

p(t) = i(t) * [integral i(t)dt/C]

And there you go.  If you have a DSO you can record the current waveform and the capacitor voltage waveform and have the DSO do the multiplication for you.

Alternatively you can record the current waveform only and then put the file on a USB drive.   Then you can load the current waveform into a spreadsheet and do a multiply-accumulate function and get your answer like that.  This is the more "profound" way to do this because as long as you know the current waveform and the capacitance you have enough information to get your answer.  It's like Spice running through your veins.

MileHigh

Thanks by explanation .

cheers

[synchro1]

Now you are trying to "spin" what you said and claimed at first. You have been confusing RATE with QUANTITY, apparently.

"What you are implying" you are now stating nearly correctly. The first 1/3 of the capacitor charge or discharge is happening at a FASTER RATE than the remaining charge or discharge. And the FASTEST rate, corresponding to the highest current, occurs during the first time constant immediately after making the connection.

Again, this is what you said at first:
And the parts I have emphasized in your original statement are wrong.

Here's what you had to say:

"No, I am right even if the charge-discharge is through a resistor"!

You're wrong!

This is right:

"The bigger the value of RC the slower the rate at which the capacitor discharges".

[TinselKoala]

Try and get a grip, AB.

You said,

You're right, unless the capacitor is charging and discharging through a resistor.

--referring back to my statement that the maximum charge rate happens when the cap is first connected to the power source, and the maximum discharge rate is when the cap is first connected to the discharge circuit. Whether it be a resistor or the negligible resistor formed by a straight bit of wire.

And I said, I'm right even when the capacitor is charging and discharging through a resistor. Yes, charging and discharging through a resistor will be slower overall-- that is what is meant by "stretching the graph horizontally". But the qualitative shape of the charge-discharge curve remains the same: It will still have the greatest slope, positive or negative, at the very beginning of the charging or discharging, meaning that the fastest RATE of charge/discharge is at those points, meaning that the greatest CURRENT occurs at those points. Anyone who can interpret a graph will tell you the same thing.

Do you actually understand what the word RATE means? It means quantity per time. Miles per hour, charge per RC period, coulombs per second, etc. These are all RATES, or to use the calculus term, time rate of change. The most quantity of charge per the least amount of time happens at the very beginning of the charge or discharge curve. The greatest currents occur at these points, NOT one third or two thirds down the graph. As your quoted passage and your graph CLEARLY STATE AND SHOW.

What exactly is your problem here? You made a statement that is clearly wrong, you posted information from other sources that clearly demonstrate that you are wrong, I have tried over and over to explain it to you but you persist in your error!  No wonder nobody bothers to respond to you at EF any more.

This is right:

"The bigger the value of RC the slower the rate at which the capacitor discharges".

Yes, that is right, but that is not what we have been talking about. Your claim was that the MAXIMUM RATE happens 1/3 or 2/3 down the graph, which is wrong, no matter what resistance is used.

Discharging a capacitor is like decanting water from a five gallon jug; Slow starting, followed by a strong gush  at .67, tapering off to a slow flow: Charging exactly the reverse, max charge rate at .33 capacity:

Those are your words, clearly stating your claim, and that claim is wrong.


 

[synchro1]

Tinselkola,

There you go again with your "Freddy the Freeloader" cigarbut stuck on a toothpick routine.

My point in your over quoted statement by me was merely to describe how the charge and discharge curve are equal and opposite. Nothing more!

I choose to compare the capacitor discharge to a jug of wáter to phase into a civil discussion of the RC formula, which is a necessary corolary. The diameter of the bottle neck analogous to the level of resistance.

I caught you using ambiguous language too. Everyone knows zero resistance accross the capacitor electrodes would result in a "Pulse", not a slope. This would appear as a near vertical line on the graph.

I can start to mistreat you as though you were a moron, even though I know you know better. I can't argue with anything you've had to say up till now. All I can assure you of, is that I have nothing to learn from you about this subject.