@Guest3123 I'm very happy to see that you have chosen to take this learning opportunity seriously.
But don't jump ahead too quickly without completing the current exercise. I have a lot more planned for you along this path.
Treat the learning experience as a journey and not a goal. Or as someone famously said "Life is a journey, not a destination".

Your two graphs are correct.

Now describe in words what you see as being different between the two graphs.

Trust me, I'm far from trying to take it too fast, too seriously.. I honestly would like to take it in baby steps if I had the choice, but I'm afraid nobobody wants to walk me threw it. Regardless of the equations everyone keeps spitting out, I honestly don't understand anything what anyone is saying. The only fun I've had so far was making that little graph in InkScape, but that's about it. I did try to spend a little money last night and watch another video, but ended up ending my subscription after what seemed to me as a very short video.

This was the video I ended up watching after the one I posted about in the above post..

Yeah, it seems to be about the letter e, and a bunch of other stuff, that I honestly don't have an inkling of a clue about. I understand what V, I, W, and Ω means.. I understand that tidle or T means time.. etc. I did look up e, and it's something to do with nature, and it's a very popular number, something like 2.14718 etc. I don't know.. but every time I try to look at posts or try to teach myself and try to pay attention to what is being talked about on a dedicated YouTube video, even when almost every video doesn't include an LED in the circuit, it's still too difficult. I honestly am starting to question if I'm really stupid or something else, I honestly don't know. I want to learn about all kinds of stuff, but it seems alot of the stuff I try learning about seems difficult, and foreign to me. I underderstand about the first thing.. Which was how to find a capacitor to discharge current threw the LED for a given time..

if I wanted to use 5vdc, 2 seconds delay, or decay..

0.02A x 2s = 0.04
Vf = LED's Forward Voltage. (7vdc)
Capacitance = (0.02A x 2s) = 0.04A / 7Vf = 0.005714 or.. 571 Hundred Thousandths of a Farad..?

etc.

It took me like a long time to learn about the n-channel mosfet, but I'm pretty familiar with it, and use an n-Channel mosfter every single day in the custom box mod I built. In fact, I've built a few box mods using IXYS power FET's.

I know how to wire up LED's now. That's cool. I wanted to know how to do that for quite some time.

I'm just sitting back now.. If you guys want to unfollow this post now, or just let it die, that's cool. I honestly don't "need" to learn about the discharging of a capacitor, it's something I simply just wanted to learn about for fun. It's not very fun, because alot of people it seems, have tried to help me, and I'm not ready yet. Maybe if I stayed in school or whatever and learned what I was supposed to learn in Math class, I probably wouldn't be in my situation.

Been working on the Chevy Silverado 1500 truck we bought in NJ, and that's basically what I've been up to all day since 8am this morning.

 

This stuff is usually taught alongside or following math prerequisites, and it becomes far easier if you've already learned calculus. Some things are tough to learn only at the practical level. It's easier to get the basic theory mastered (things like Ohm's law, RLC circuits, kirchoff's law) first and then add details as needed. Starting with the details obscures the forest for the trees.

 

Hello again,

We also can not ignore:
i=(e^(-t/RC)*(Vcc-vLED))/R

or in short:
i=(e^(-t/RC)*i0

where i0 is the initial current, and that is because we find that the current decreases to roughly 1/3 of it's initial value after only one time constant TC=R*C
and that means the brightness also follows that pattern. After that the current decreases roughly 1/2 of the previous current value for every additional time constant, and by the time 5 time constants have passed the current is down to less than 1 percent of the initial current.
This is a simpler way of looking at it too, but it also tells us that we need to select a final current value that we deem makes the LED too dim because many modern high brightness LED's stay visible way down in current, which may nor may not be enough brightness to fit the application at the end of some predefined time period. If we need sufficient light for say 10 seconds, then we have to make sure that the current level is still high enough to get that light at the end of the 10 second time period. This is very important. Without considering this final current value we may end up with only something like the first light brightness of the LED which probably wont be enough for most things.

I also wonder if the end application was given yet, like what is this going to be used for. If it is just an indicator of some sort the current can be allowed to go lower for example, but if used for some medical purpose we probably want much more light at the end of that 10 second period.

 

Hi Guest,

Back on topic...


A linear approximation is probably the best place for you to start because it's the simplest view and we can only approximate anyway.

For a capacitor the definition is often given as:
dv/dt=i/C

So, ok bye
(Just kidding)

Let's look at this little equation a little bit...

dv/dt means the change that the voltage is making with time. We like to know this because this tells us how much it changes in a given time, like 1 second. Then we might estimate how long it will take to reach a certain voltage level where the LED does not light anymore.

What we really want to know though is how much time it takes, so we solve for the "dt" in that equation:
dv/dt=i/C
or:
dt/dv=C/i
or:
dt=C/(i*dv)

All we did was solve for "dt" algebraically.

Now that we have that, we need to know the starting current "i". We can estimate this by looking at the circuit for an LED in series with a resistor driven by a constant voltage source. The current is:
i=(Vcc-vLED)/R

The reason for this is because the LED drops a roughly constant voltage and Vcc is constant, and so when we subtract the two we get the voltage across the resistor R and then we use Ohm's Law:
I=V/R

and that gives us the current.

Now that we have the current, we can substitute that into the original equation:
dt=C/(i*dv)

and get:
dt=C/(dv*(Vcc-vLED)/R)

and multiply top and bottom by R and get:
dt=R*C/(dv*(Vcc-vLED))

and we are partly done. Next we need to determine what "dv" is. This is the decrease in cap voltage where the LED gets too dim to see much, but is kind of subjective to whatever we think is too dim. We'll set a limit on the current through the LED just for an example, let's say 0.2ma is low enough to make the LED dim by our standards of what dim is. That means when the cap discharges, the current will drop to:
i2=(Vcc-dv)/R

where i2 is the dim level set point current we choose for our liking. If we dont know this yet, we test for it with a small circuit.

Solving that for dv, we get:
R*i2=Vcc-dv
dv+R*i2=Vcc
dv=Vcc-R*i2

Next we substitute that into the previous equation:
dt=R*C/(dv*(Vcc-vLED))

and get:
dt=R*C/((Vcc-R*i2)*(Vcc-vLED))

and now we have an equation that gives us the time before the LED gets too dim, by our own standard of what dim is.
The Vcc here is the voltage that the cap is first charged up to. i2 is the LED current level we choose by previous experiment which is the current level where we think the LED is too dim. vLED is the characteristic voltage of the LED, often 1.8v to 3.5v depending on the particular LED part being used, and this can also be tested for via previous experiment. R is the series resistor, C is the capacitance.

If there is anything you dont understand here, just ask about that particular thing to stat with. We can take it one step at a time too if you like.

Once you see how this works if you would like to move on to the more exact solutions we can look at that next. That involves a little more math but not too much more.

What we really want to know though is how much time it takes, so we solve for the "dt" in that equation:
dv/dt=i/C
or:
dt/dv=C/i
or:
dt=C/(i*dv)

All we did was solve for "dt" algebraically.

Have you learned anything about the RC circuit yet?

There are two basic approaches but they do require a little algebra.

The DC approach and the AC approach.

The DC approach is when the RC circuit is driven with a step input that goes from 0 to some voltage in nearly zero time, then the cap starts to charge.

The AC approach treats the RC as a filter, with a constant AC output amplitude for constant input AC amplitude, and a constant phase shift for constant frequency input. This circuit is used as either a filter or a phase shift network. Because your question was originally related to a triac circuit, you will be eventually interested in the way the RC can shift the phase of the input AC wave.

We can look at these circuit in more detail, but i am not sure how much math you have had so far. It helps to know that so i know what kind of explanation would suite you best.


I'd like to start again, from the beginning. Working with DC Circuits first. I would like to request to start simple, and work my way into AC.
I understand that MrChips and a few other people, were also extremely helpful, but I've seemed to have lost my way on the whole Capacitors, resistors subject.

Can you please lead MrAl, let's start from the beginning. I'll be extremely patient, and even if it takes a year to learn these Capacitors and Resistors, at least I'll have a good understanding, and I'll understand how capacitors and resistors play a role in circuits.

I have a very solid understanding of Ohm's law. Resistors in series and parallel, and I also know that capacitors in parallel are added, and capacitors in series are more difficult. Kinda like how resistors are in parallel.

 

What's dt/dv?

d * time / d * voltage?

What's d ?

 

If you understand the basic physics, your hurdle is the math. I am assuming you have no comfort with differential equations or calculus? That's OK. Those tools make it easier but you can certainly get where you want without them.

The simplest possible approach is to assume the LED current is constant until the capacitor runs out of charge. Let's say the LED and resistor are such that your initial current is 20mA, or 0.02A. Each second of light requires 0.02 amp•seconds of charge.

The charge in your capacitor is measured in farads. A large 1F capacitor will develop a voltage of 1V for each coulomb it stores. A coulomb is a measure of quantity of charge (like gallons of water) and is equal to the charge that moves at a current of 1A for 1 second. So 1 coulomb = 1 amp•sec. That big 1F capacitor charged to 9V stores 9 amp•seconds.

Keeping with our simplifying assumption of constant voltage and current, we estimate the LED will light for 9 A•s/0.02A = 450 seconds = 7.5 minutes.

Unfortunately, we know the voltage on the capacitor declines as it discharges. That means the current through the LED will fall, slowing the discharge rate. Differential equations are how problems like this are solved, but there are other ways.

Before we move on, are you OK with this so far?

 

What's dt/dv?

d * time / d * voltage?

What's d ?

dV/dt is shorthand for ∆V/∆t at the limit where both approach zero. It's the instantaneous slope of V versus t. Assigning physical meaning and units to V and t requires knowing the context. In this context they are voltage in volts and time in seconds.

 

If you understand the basic physics, your hurdle is the math. I am assuming you have no comfort with differential equations or calculus? That's OK. Those tools make it easier but you can certainly get where you want without them.

I can learn the math, if someone is willing to teach me.
I enjoy math. Didn't get a proper education, which is kinda obvious, but now that I'm a little older, I regret that. In turn, I have found myself in the past couple of years learning different things, simple things about math, which I found to be very powerful.

The simplest possible approach is to assume the LED current is constant until the capacitor runs out of charge. Let's say the LED and resistor are such that your initial current is 20mA, or 0.02A. Each second of light requires 0.02 amp•seconds of charge.

Ok ! I'm listening. Each second of light requires 0.02 amps of current per second. Got it.

The charge in your capacitor is measured in farads. A large 1F capacitor will develop a voltage of 1V for each coulomb it stores.

Got it. So then 2V is 2 Coulombs, or.. 6,241,509,750,000,000,000 x 2 = 12,483,019,500,000,000,000 electrons. 12.483 Quintillion electrons.

A coulomb is a measure of quantity of charge (like gallons of water) and is equal to the charge that moves at a current of 1A for 1 second. So 1 coulomb = 1 amp•sec. That big 1F capacitor charged to 9V stores 9 amp•seconds.

So... hmm.. what's amp•seconds. that's.. 9 amps per second, per second? Or 9 amps, every second?

Keeping with our simplifying assumption of constant voltage and current, we estimate the LED will light for 9 A•s/0.02A = 450 seconds = 7.5 minutes.

So easier, it's 9,000mA per second / 0.02A = 450000 seconds.

Umm.. I think I got it wrong. 450,000 seconds is 7500 minutes.

So, that means that it's not 9A per second. 9 Amps per Millisecond?

7500 / 1000ms = 7.5 Minutes

I've don't think I understand 9 A•s.

That dot looks like a multiplication symbol. So, 9A x 1000ms ?

Unfortunately, we know the voltage on the capacitor declines as it discharges. That means the current through the LED will fall, slowing the discharge rate. Differential equations are how problems like this are solved, but there are other ways.

Before we move on, are you OK with this so far?

 

I've don't think I understand 9 A•s.

An amp is a flow rate, like gallons-per-minute. It's coulombs per second. A flow rate of 9A or 9 coulombs per second, collected and summed for one second, is a quantity of 9 amp•seconds = 9 coulombs.

 

What we really want to know though is how much time it takes, so we solve for the "dt" in that equation:
dv/dt=i/C
or:
dt/dv=C/i
or:
dt=C/(i*dv)

All we did was solve for "dt" algebraically.

Have you learned anything about the RC circuit yet?

There are two basic approaches but they do require a little algebra.

The DC approach and the AC approach.

The DC approach is when the RC circuit is driven with a step input that goes from 0 to some voltage in nearly zero time, then the cap starts to charge.

The AC approach treats the RC as a filter, with a constant AC output amplitude for constant input AC amplitude, and a constant phase shift for constant frequency input. This circuit is used as either a filter or a phase shift network. Because your question was originally related to a triac circuit, you will be eventually interested in the way the RC can shift the phase of the input AC wave.

We can look at these circuit in more detail, but i am not sure how much math you have had so far. It helps to know that so i know what kind of explanation would suite you best.


I'd like to start again, from the beginning. Working with DC Circuits first. I would like to request to start simple, and work my way into AC.
I understand that MrChips and a few other people, were also extremely helpful, but I've seemed to have lost my way on the whole Capacitors, resistors subject.

Can you please lead MrAl, let's start from the beginning. I'll be extremely patient, and even if it takes a year to learn these Capacitors and Resistors, at least I'll have a good understanding, and I'll understand how capacitors and resistors play a role in circuits.

I have a very solid understanding of Ohm's law. Resistors in series and parallel, and I also know that capacitors in parallel are added, and capacitors in series are more difficult. Kinda like how resistors are in parallel.

Hi,

Ok.

First, the equation dv/dt=i/C is a different view of the capacitor that by itself does not include a resistor. It is an expression with a constant current source 'i' that is charging the capacitor. In a circuit with a resistor, the current 'i' is changing al the time so we cant use that equation directly because once dv/dt changes so does 'i' and that means we have to set up a differential equation, which i dont think you want to get into just yet.

The equation for a capacitor charging through a resistor and driven with a voltage source is actually kind of simple:
Vc=Vs*(1-e^(-t/RC))

where
Vc is the cap voltage after time t,
t is time,
Vs is source voltage (like a battery),
RC is R*C is the resistance times the capacitance,
e is the number 2.718281828.... the base of the natural log system.

By pluggin in R*C and the value for t and the value for the voltage source, you can calculate the capacitor voltage that will exist after that time t.

So for example, With R=2 and C=1 and a source voltage of 10 volts, we have:
Vc=10*(1-e^(-t/2))

and after 2 seconds (t=2) we have:
Vc=10*(1-e^(-2/2))=10*(1-e^(-1))=10*(1-0.367879)=10*0.632121=6.32 volts approximately.

So after 2 seconds we see that the cap voltage went from 0 to 6.32 volts.

Does that make sense now?

e^(-x) is also sometimes written as exp(-x)

 

Keeping with our simplifying assumption of constant voltage and current, we estimate the LED will light for 9 A•s/0.02A = 450 seconds = 7.5 minutes.

Before we move on, are you OK with this so far?

I took another look at it, yes, it really was that simple. I honestly don't know how I misinterpreted that..

Is it also like 0.02A per second, if the capacitor has a charge of 9 Coulombs (9A•s)?
The Capacity or coulombs doesn't use up the current in the LED, the LED uses up the Coulombs in the capacitor..
So shouldn't it be 0.02A / 9A•s = 0.0022

Seems backwards, but anyways, I understand 9A•s ÷ 0.02A = 450 seconds.
You're right. I'm ready to continue.

 

Is it also like 0.02A per second, if the capacitor has a charge of 9 Coulombs (9A•s)?
The Capacity or coulombs doesn't use up the current in the LED, the LED uses up the Coulombs in the capacitor..
So shouldn't it be 0.02A / 9A•s = 0.0022

Seems backwards, but anyways, I understand 9A•s ÷ 0.02A = 450 seconds.

Keep an eye on units. There is no such thing as "0.02A per second", because amps are already a rate, expressed in units of charge (coulombs) per unit of time. Saying amps per second is like saying gpm per second. It would only make sense if you're talking about acceleration of the rate. And so in your last calculation, the units work out to inverse seconds. Since you want seconds, take the inverse.

I'm off to bed, so perhaps someone else will move forward with you in the meanwhile.

 

Before I turn in, here's where I would go next. The simple case of constant current gave us an estimate of 450 seconds. Now, instead of holding current constant over all that time, break the decay into intervals and see what happens. Start with the first 100 seconds and assume the current is constant at 0.02A over that period.

You should be able now to calculate the drop in charge on the capacitor during that time, and therefore the drop in voltage on the capacitor. With the new lower voltage, you can recalculate the current at the start of the next 100-second interval. It will be something less than 20mA. The ∆V across the LED is roughly constant, so you just need Ohm's law to calculate the new current.

Keep stepping forward in this manner. You can guess at the length of last interval, to get the charge on the capacitor as close to zero as you like.

When you've worked through that, you might try using 50-second intervals to see how the total time works out.

Calculus is the mathematical tool you use to step forward in an infinite number of infinitely small steps, i.e. continuously instead of stepwise. But even a handful of steps will get you close to an accurate answer. At 10 steps, you're probably getting more precise than the tolerances of the various components.

 

The equation for a capacitor charging through a resistor and driven with a voltage source is actually kind of simple:
Vc=Vs*(1-e^(-t/RC))

Looks complicated. But I'll give it a shot, thanks for participating in helping me.

Where
Vc is the cap voltage after time t,
t is time,
Vs is source voltage (like a battery),
RC is R*C is the resistance times the capacitance,
e is the number 2.718281828.... the base of the natural log system.

Vc is the capacitor voltage after time t, where t is in seconds.
t is time, in seconds.
Vs is source voltage (like a battery).
RC is Resistance in Ohm's & C is capacitance in Farads. So, Ω * F, resistance times capacitance. Got it.
e is Euler's number, 2.7182818284590452353602874713527 etc. , which is the base of the natural log system.

By plugging in R*C and the value for t and the value for the voltage source, you can calculate the capacitor voltage that will exist after that time t.

So for example, With R=2 and C=1 and a source voltage of 10 volts, we have: Vc=10*(1-e^(-t/2))

Vc = 10Ω * (1-e^(t/2))
Alright, so I'm lost. when it got to 1-e^(t/2)) I got messed up.

I don't know what 1-e^(t/2)) is..

It's 1-2.71828182^(t/2))..
Which is 1-2.71828182 to the power of (t/2), which is seconds divided by 2?

and after 2 seconds (t=2) we have:
Vc=10*(1-e^(-2/2))=10*(1-e^(-1))=10*(1-0.367879)=10*0.632121=6.32 volts approximately.
So after 2 seconds we see that the cap voltage went from 0 to 6.32 volts.

Does that make sense now?

No it doesn't.

e^(-x) is also sometimes written as exp(-x)

It's mostly foreign to me. The last Quote, e^(-x) is also complicated. It would be so much easier, if it were written out in simpler form..? If that's possible?

 

Here's where I would go next. The simple case of constant current gave us an estimate of 450 seconds. Now, instead of holding current constant over all that time, break the decay into intervals and see what happens. Start with the first 100 seconds and assume the current is constant at 0.02A over that period.

I don't understand what you want me to do, I'll wait until the next post to work on it. I did try to do charts, because I've done charts before on here, and it seemed to be fun, so I did it anyways, trying to write and draw the charts out like what I thought you were asking, but I think it's wrong. Simply because I don't understand what you were asking. But here they are anyways.

 

Looks complicated. But I'll give it a shot, thanks for participating in helping me.




Vc is the capacitor voltage after time t, where t is in seconds.
t is time, in seconds.
Vs is source voltage (like a battery).
RC is Resistance in Ohm's & C is capacitance in Farads. So, Ω * F, resistance times capacitance. Got it.
e is Euler's number, 2.7182818284590452353602874713527 etc. , which is the base of the natural log system.
View attachment 112412





Vc = 10Ω * (1-e^(t/2))
Alright, so I'm lost. when it got to 1-e^(t/2)) I got messed up.

I don't know what 1-e^(t/2)) is..

It's 1-2.71828182^(t/2))..
Which is 1-2.71828182 to the power of (t/2), which is seconds divided by 2?




No it doesn't.




It's mostly foreign to me. The last Quote, e^(-x) is also complicated. It would be so much easier, if it were written out in simpler form..? If that's possible?

Hi,

No problem, take your time.

The equation was:
Vc=Vs*(1-e^(-t/RC))

-t/RC is the exponent of e. We are taking 'e' up to the power of -t/RC, and note that the 't' is negative.
So for t=1 second the exponent is -1/RC, and for t=2 seconds the exponent is -2/RC and for t=3 seconds the exponent is -3/RC. If RC is 2 in all three of these, then the exponent is -1/2, -2/2, and -3/2 respectively.

When we raise the number e up to a power (the exponent) we usually use a calculator.
You can approximate to start with if you like to make it a little simpler too. For example:

Vc=Vs*(1-2.7^(-t/RC))

and since 2.7^(-t/RC) is equal to 1/(2.7^(t/RC)) you can do it that way if you like.

So looking at the 2 second solution again, we have:
Vc=Vs*(1-2.7^(-t/RC))
Vc=Vs*(1-2.7^(-2/RC)), stick the 2 for time t in there,
and with RC=2 and Vs=10 we have:
Vc=10*(1-2.7^(-2/2)), stick the RC and the Vs in there,
Vc=10*(1-2.7^(-1)), reduce the -2/2 part,
Vc=10*(1-0.37), reduce the 2.7^(-1) part,
Vc=10*(0.63), reduce the 1-0.37 part,
Vc=6.3 volts., multiply 0.63 times 10.

See if you can follow that and see if you can get the same result.
Remember that when we raise a number to a power like 2.7^(-1) we use a calculator, punching in the 2.7 first, then the "y^x" button, then the -1, then 'equals'. "y^x" is taking the number y up to the power of x, so 2.7 is y and -1 is x.
Try that and see how far you get.

I can assure you that once you do a few of these examples you'll get the hang of it really quick. It just takes a little repetitive practice, which is true for many things that are being learned for the first time. You seem to have plenty of patience and that is an advantage when learning new things.

If you have trouble doing the powers, try these first:
3^2
3^3
3^-1
3^-2

using a calculator. Many calculators have the "y^x" button but some may be different so you have to find the way to do it on the calculator you want to use.
The "y^x" button often looks like a lower case 'y' with a lower case 'x' up above and to the rigth of the lower case 'y'.

 

Vc=Vs*(1-e^(-t/RC))

Vc=Vs*(1-e^(-t/RC))

Vc = Capacitor voltage.
Vs = 10Vdc
R = 10Ω
C = 2 Farads ?
t = = -1 second, or.. (one negative second)

10v * (1-e^(-1/10Ω * 2F)) = 1.81269246922 volts..?

Because -1 is minus 1 second right?

I did it on Google.


Going to try it with two seconds now..

Vc=Vs*(1-e^(-t/RC))

Vc = Capacitor voltage.
Vs = 10Vdc
R = 10Ω
C = 2 Farads ?
t = = -2 seconds, or.. (negative two seconds)

10v * (1-e^(-2/10Ω * 2F)) = 3.29679953964 volts..?

 

Simply because I don't understand what you were asking.

Very simple.

• In the simple case above where we assumed the current was constant at 0.02A, choose a resistor and starting voltage that give 20mA in the LED.
• Calculate the charge on the capacitor after 100 seconds of 0.02A discharge.
• From that remaining charge, calculate the new voltage on the capacitor.
• Adjust the current according to Ohm's law. It will be less than 0.02A.

Repeat the steps above, moving forward in increments of 100 seconds.
Guestimate the number of seconds in the final interval to take the capacitor near to zero charge.

Add up all the intervals and see how it compares to the first estimate.

If you're doing this in a spreadsheet, it'll be a lot easier!

 

Vc=Vs*(1-e^(-t/RC))

Vc = Capacitor voltage.
Vs = 10Vdc
R = 10Ω
C = 2 Farads ?
t = = -1 second, or.. (one negative second)

10v * (1-e^(-1/10Ω * 2F)) = 1.81269246922 volts..?

Because -1 is minus 1 second right?

I did it on Google.
View attachment 112421

Going to try it with two seconds now..

Vc=Vs*(1-e^(-t/RC))

Vc = Capacitor voltage.
Vs = 10Vdc
R = 10Ω
C = 2 Farads ?
t = = -2 seconds, or.. (negative two seconds)

10v * (1-e^(-2/10Ω * 2F)) = 3.29679953964 volts..?
View attachment 112423

Hello again,

You are definitely on the right track now. Just one small detail we have to clear up.

In the line:
"10v * (1-e^(-1/10Ω * 2F)) = 1.81269246922 volts..?"

it is clear that when you or your calculator calculated the part "-1/10Ω * 2F" you or it multiplied -1/10 by 2, which gave you or it -2/10, which is correct for that writing, but is not correct for the formula that has -t/RC in it because the implication is that R*C is to be performed first. That is partly my fault for writing it that way as the better way is:
-t/(R*C)

where it is more clear that we have to multiply R*C first.

So looking at the entire formula again:
Vc=Vs*(1-e^(-t/(R*C)))

and since R*C has to be done first we can do that right away, and with R=10 and C=2 this turns into:
Vc=Vs*(1-e^(-t/20))

You could try that again and see what you get, as well as the second example you gave.

I can assure you that this will become much much easier once you do a few more examples, and i see that you have this formula almost mastered already after only a few replies which is a good sign

Just a couple more small notes...
1. We are calculating the charge time of a capacitor. We can do the discharge time next if you wish.
2. In the formula with -t, it's not really negative time, it's just time entered into a formula that happens to have a minus sign in it which makes the whole exponent negative. So it's not really -t alone, it's -(t/(R*C)). This is just a small point though. The main thing is to be able to get the calculation right every time.

 

Very simple.
• In the simple case above where we assumed the current was constant at 0.02A, choose a resistor and starting voltage that give 20mA in the LED.

Resistor : 600Ω
Voltage : 12Vdc
Current : 0.02A
Wattage : 0.24W

Understood.

• Calculate the charge on the capacitor after 100 seconds of 0.02A discharge.

It draws 0.02A per second multiplied by 100 seconds = 2A are drawn, or 2Ah of capacity are taken after 100 seconds.

• From that remaining charge, calculate the new voltage on the capacitor.

The remaining charge after 2Ah are taken, the new voltage on the capacitor is...? I'm drawing a blank. It's 9Ah total right..? On the capacitor?

• Adjust the current according to Ohm's law. It will be less than 0.02A.

I'm drawing a blank. I don't understand.

Repeat the steps above, moving forward in increments of 100 seconds.
Guestimate the number of seconds in the final interval to take the capacitor near to zero charge.

Add up all the intervals and see how it compares to the first estimate.

If you're doing this in a spreadsheet, it'll be a lot easier!

I'll do it on a spreadsheet, once I understand.

I got the first part when you asked me to make up a voltage, and resistance for the LED that draws 0.02A, that was very easy.