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### Author Topic: The bifilar pancake coil at its resonant frequency  (Read 372263 times)

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: The bifilar pancake coil at its resonant frequency
« Reply #420 on: April 07, 2017, 02:06:28 AM »
The maximum current into the capacitor will occur when it is at zero charge. The maximum current out of your capacitor will occur when it is at full charge. This is what the graph is showing you.

Say you are using 10 volts from a voltage-regulated supply. You first short the capacitor to assure that it is fully discharged, then you apply your ten volts. The current will be at its maximum value as soon as you apply the voltage and will taper off according to the curve until the voltage on the capacitor is ten volts. Actually it will asymptote to ten volts. Discharge is the reverse: Apply a resistor across the charged capacitor and look at the current through the resistor. It will be at its maximum value as soon as you make contact and will taper off from there as the capacitor "drains" down to zero voltage.

#### Free Energy | searching for free energy and discussing free energy

##### Re: The bifilar pancake coil at its resonant frequency
« Reply #420 on: April 07, 2017, 02:06:28 AM »

#### synchro1

• Hero Member
• Posts: 4491
##### Re: The bifilar pancake coil at its resonant frequency
« Reply #421 on: April 07, 2017, 09:13:04 AM »
The maximum current into the capacitor will occur when it is at zero charge. The maximum current out of your capacitor will occur when it is at full charge. This is what the graph is showing you.

Say you are using 10 volts from a voltage-regulated supply. You first short the capacitor to assure that it is fully discharged, then you apply your ten volts. The current will be at its maximum value as soon as you apply the voltage and will taper off according to the curve until the voltage on the capacitor is ten volts. Actually it will asymptote to ten volts. Discharge is the reverse: Apply a resistor across the charged capacitor and look at the current through the resistor. It will be at its maximum value as soon as you make contact and will taper off from there as the capacitor "drains" down to zero voltage.

You're right, unless the capacitor is charging and discharging through a resistor. I used those figures because the chart I posted includes them. I used .67 and .33 wrongly instead of .63 and .37. My comparison to the jug includes the neck as a resistor, while an open bucket would be better for yours. Thanks for helping clear that up.

"As with the previous RC charging circuit, in a RC Discharging Circuit, the time constant ( τ ) is still equal to the value of 63%. Then for a RC discharging circuit that is initially fully charged, the voltage across the capacitor after one time constant, 1T, has dropped to 63% of its initial value which is 1 – 0.63 = 0.37 or 37% of its final value.

So now this is given as the time taken for the capacitor to discharge down to within 37% of its fully discharged value which will be zero volts (fully discharged), and in our curve this is given as 0.37Vc.

As the capacitor discharges, it loses its charge at a declining rate. At the start of discharge the initial conditions of the circuit, are t = 0, i = 0 and q = Q. The voltage across the capacitors plates is equal to the supply voltage and Vc = Vs. As the voltage across the plates is at its highest value maximum discharge current flows around the circuit".

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: The bifilar pancake coil at its resonant frequency
« Reply #422 on: April 07, 2017, 10:48:04 AM »
No, I am right even if the charge-discharge is through a resistor. Read your quoted passage again, and look at your graph again. In the charging case, the slope of the graph is steepest right where it takes off from zero and becomes progressively flatter until the charging voltage is asymptotically reached. In the discharging case the slope of the graph is steepest right when discharging commences and becomes progressively flatter until zero voltage is again asymptotically reached. The slopes of the lines indicates the time rate of charge/discharge, and the sign of the slopes indicates charging (positive slope, going up from left to right) or discharging (negative slope, going down from left to right). Zero slope (asymptotically horizontal line) means that "steady state" has been reached, either zero or fully charged to the charging voltage. The lines represent charge vs. time and the slopes of the lines represent d(charge)/dt. That is, the change in charge over the change in time. Where the slope of the line is steepest, you have the greatest change in charge during the smallest time, that is, the fastest charging rate. Adding more resistance stretches the graph horizontally (takes more time for a given amount of charge or discharge) but does not change the qualitative slopes; the fastest rates (corresponding to the greatest currents) will still be at the beginnings of the charge or discharge. This is elementary differential calculus.

"As the capacitor discharges, it loses its charge at a declining rate. At the start of discharge the initial conditions of the circuit, are t = 0, i = 0 and q = Q. The voltage across the capacitors plates is equal to the supply voltage and Vc = Vs. As the voltage across the plates is at its highest value maximum discharge current flows around the circuit".

#### Free Energy | searching for free energy and discussing free energy

##### Re: The bifilar pancake coil at its resonant frequency
« Reply #422 on: April 07, 2017, 10:48:04 AM »

#### synchro1

• Hero Member
• Posts: 4491
##### Re: The bifilar pancake coil at its resonant frequency
« Reply #423 on: April 07, 2017, 02:55:18 PM »
No, I am right even if the charge-discharge is through a resistor. Read your quoted passage again, and look at your graph again. In the charging case, the slope of the graph is steepest right where it takes off from zero and becomes progressively flatter until the charging voltage is asymptotically reached. In the discharging case the slope of the graph is steepest right when discharging commences and becomes progressively flatter until zero voltage is again asymptotically reached. The slopes of the lines indicates the time rate of charge/discharge, and the sign of the slopes indicates charging (positive slope, going up from left to right) or discharging (negative slope, going down from left to right). Zero slope (asymptotically horizontal line) means that "steady state" has been reached, either zero or fully charged to the charging voltage. The lines represent charge vs. time and the slopes of the lines represent d(charge)/dt. That is, the change in charge over the change in time. Where the slope of the line is steepest, you have the greatest change in charge during the smallest time, that is, the fastest charging rate. Adding more resistance stretches the graph horizontally (takes more time for a given amount of charge or discharge) but does not change the qualitative slopes; the fastest rates (corresponding to the greatest currents) will still be at the beginnings of the charge or discharge. This is elementary differential calculus.

"As the capacitor discharges, it loses its charge at a declining rate. At the start of discharge the initial conditions of the circuit, are t = 0, i = 0 and q = Q. The voltage across the capacitors plates is equal to the supply voltage and Vc = Vs. As the voltage across the plates is at its highest value maximum discharge current flows around the circuit".

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".

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.

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: The bifilar pancake coil at its resonant frequency
« Reply #424 on: April 07, 2017, 04:18:29 PM »
No, THIS is what you said:

@evostars,

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 you presented a graph that clearly shows 70 percent of charge leaving the capacitor in the FIRST RC time constant, a little under 20 percent of the charge leaving in the SECOND time constant, a fraction of 10 percent leaving in the THIRD time constant and the rest ( practically ) leaving in the FOURTH time constant and the tiny portion of the remainder leaving in the FIFTH time constant. After five time constants it is considered to have attained the Steady State (but in fact asymptotes to it.) The maximum DISCHARGE RATE occurs immediately after making the discharge connection, as your quoted passage says and as your graph shows, and the maximum CHARGE RATE occurs immediately after connecting the power source to the capacitor. The graph is right and continues to be right. Your verbal description was wrong and continues to be wrong.

THE SLOPE OF THE LINE INDICATES THE TIME RATE OF CHANGE OF THE LINE. THE RATE OF CHANGE (rate of charge or discharge) IS MAXIMUM WHERE THE SLOPE IS STEEPEST. IF THE SLOPE IS POSITIVE, going up from left to right, THE CAPACITOR IS CHARGING, AND IF THE SLOPE IS NEGATIVE, going down from left to right, THE CAPACITOR IS DISCHARGING. When the slope is zero (horizontal line) the "steady state" is reached.

And I can see that you either never took Differential Calculus in school, or you flunked miserably. Give it up AB, you are simply wrong. YET AGAIN.

#### Free Energy | searching for free energy and discussing free energy

##### Re: The bifilar pancake coil at its resonant frequency
« Reply #424 on: April 07, 2017, 04:18:29 PM »

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: The bifilar pancake coil at its resonant frequency
« Reply #425 on: April 07, 2017, 04:21:06 PM »
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".

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:
Quote
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

• Hero Member
• Posts: 665
##### Re: The bifilar pancake coil at its resonant frequency
« Reply #426 on: April 07, 2017, 04:56:53 PM »
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

#### Free Energy | searching for free energy and discussing free energy

##### Re: The bifilar pancake coil at its resonant frequency
« Reply #426 on: April 07, 2017, 04:56:53 PM »

#### synchro1

• Hero Member
• Posts: 4491
##### Re: The bifilar pancake coil at its resonant frequency
« Reply #427 on: April 07, 2017, 05:11:29 PM »
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

• Hero Member
• Posts: 13968
##### Re: The bifilar pancake coil at its resonant frequency
« Reply #428 on: April 07, 2017, 06:43:33 PM »
Try and get a grip, AB.

You said,
Quote
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.

Quote
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.

Quote
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.

#### Free Energy | searching for free energy and discussing free energy

##### Re: The bifilar pancake coil at its resonant frequency
« Reply #428 on: April 07, 2017, 06:43:33 PM »

#### synchro1

• Hero Member
• Posts: 4491
##### Re: The bifilar pancake coil at its resonant frequency
« Reply #429 on: April 07, 2017, 06:51:39 PM »
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.
« Last Edit: April 07, 2017, 09:07:50 PM by synchro1 »

#### starcruiser

• Hero Member
• Posts: 693
##### Re: The bifilar pancake coil at its resonant frequency
« Reply #430 on: April 07, 2017, 08:02:41 PM »
Time for remedial Electronics guys?

try this for a refresher
http://www.electronics-tutorials.ws/rc/rc_1.html

#### Free Energy | searching for free energy and discussing free energy

##### Re: The bifilar pancake coil at its resonant frequency
« Reply #430 on: April 07, 2017, 08:02:41 PM »

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: The bifilar pancake coil at its resonant frequency
« Reply #431 on: April 07, 2017, 08:08:07 PM »

Quote
Everyone knows zero resistance accross the capacitor electrodes would result in a "Pulse", not a slope. This would appear as a vertical line on the graph.

Wrong again! All real wires have resistance and ALL REAL CAPACITORS do too. Look up "ESR" wrt capacitors. For this claim of yours to be true you would have to have a capacitor with zero resistance and a connection with zero resistance, both impossible in reality, and the current on charge and discharge would be _infinite_, taking no time, which is what a vertical line means. Infinite slope! A change in charge over +no change in time+ !!!! Ridiculous on the face of it, as everyone _actually_ knows.

Good luck with that.

Anyone with a scope and the wit to use it can prove you wrong in fifteen minutes.

You want to keep moving goalposts and constructing your strawman arguments instead of admitting that you were wrong, go ahead. You'll wind up here just like you wound up at EF: nobody pays attention to you since they all know that you are full of yourself, you misrepresent and misunderstand and prevaricate, and you will never admit that you are wrong.

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: The bifilar pancake coil at its resonant frequency
« Reply #432 on: April 07, 2017, 08:11:48 PM »
Time for remedial Electronics guys?

try this for a refresher
http://www.electronics-tutorials.ws/rc/rc_1.html

Perhaps you would like to try to explain to Synchro just where the maximum charge RATE and the maximum discharge RATE are shown on that graph.

Do you believe this very clear and unambiguous statement that he made is correct, or not?
Quote
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:

#### synchro1

• Hero Member
• Posts: 4491
##### Re: The bifilar pancake coil at its resonant frequency
« Reply #433 on: April 07, 2017, 08:26:29 PM »
Perhaps you would like to try to explain to Synchro just where the maximum charge RATE and the maximum discharge RATE are shown on that graph.

Do you believe this very clear and unambiguous statement that he made is correct, or not?

Why and the hell do you keep reposting that comment when I made it clear that it was an attempt by me to compare the charge and discharge curves as equal and opposite and nothing more?

You're a compulsive psychopath and need help.

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: The bifilar pancake coil at its resonant frequency
« Reply #434 on: April 07, 2017, 08:37:56 PM »
SO when you say, "the max charge rate is at .33 capacity" you didn't actually _mean_ that the maximum charge rate is at .33 capacity? Or when you say "slow starting followed by a gush at .67" you didn't actually _mean_ slow starting or that there is a gush at .67?

I see. Well, you will have to excuse me for misunderstanding then, since I can only go by what you SAY, not what you claim to have meant.

(insert facepalm character here)