The bifilar pancake coil at its resonant frequency

[TinselKoala]

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.

[synchro1]

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]

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".


 

[synchro1]

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".

*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.

[TinselKoala]

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.