Wow. That was a really helpful response. Fantastic, really. Thank

you for taking the time to write that.

No problem.

After reading your post, I took my multimeter and measured the voltage

across the capacitor when the circuit was open, and sure enough, there

was a transient voltage of ~1.65V left in the capacitor from previous

charges.

Yep, thats what a capacitor(and inductors too) does... it stores charge. It

charges and discharges in a special way(not really but it looks "special").

Capacitors can be used as batteries in some sense too(but batteries tend to

have constant voltage for the majority of there life and last much longer).

One thing that struck me about what you were saying is that it seems

impossible for the capacitor to reach 9V of potential, because that

would mean the LED's branch would also have a 9V drop, leaving nothing

for the 10K resistor to drop. After thinking about that for a moment,

I measured the voltage across the capacitor when the switch was closed,

and I noticed that the voltage would rise quickly to about 2.53V, and

stay there. I left the circuit closed for two hours just to see if it

would rise any higher, and it stayed at 2.53V. I started wondering if

it would be possible to figure out the maximum voltage the capacitor

could reach in this setup.

Sure, its similar to the analysis I did for the simple circuits but involes

a little more algebra(the calculus part is pretty much identical but just

looks more complicated).

Anyway, this is what I was thinking: (this is long and could be totally

incorrect)

If you replaced the capacitor and LED branches in the original circuit

with a short, you would get a circuit that looked like this:

+-------------------+

| |

| |

| |

---- 9V Battery | Removed capacitor and LED branches

-- |

| |

| |

| |

+-------VVVVV-------+

10K

In this case, the current through the modified circuit would be 9 V/10

000 ohms = 0.0009A = 0.9 mA.

Now, at their lowest, the resistance (impedance?) of the capacitor and

LED branches in the original circuit would be 0 Ohms (a short). So the

0.9 mA could essentially be considered an upper bound on the current

flowing through the original circuit.

Impedence is the total restriction of flow. Resistance is basicaly what a

resistor does and reactive is like a resistance but it is not due to the

same mechanisms that resistors use and it usually depends on frequency(maybe

always). Impedence means exactly what it means... it impedes the flow of

current. Capacitors impead the flow too but in different way than

resistors... there "resistance" is actually changing over time(even the

ideal capacitor has this) and is also dependent on frequency(unlike an ideal

resistor). Basicaly impedence covers everything while reactance is used for

something that is "reactive"(capacitors and inductors) that has no

resistance... resistance is things like resistors. Its kinda circular the

way I gave it but the point I'm trying to make is impedence = reactance +

resistance. Ideal capacitors and inductors have reactance only and ideal

resistors have resistance only... you combine a circuit with them and you

get an impedence.... although you can refer to resistance and reactance as

impedence if you wish since impedence implies each of them too.

Yes, in the circuit above it is the maximum current that will ever flow when

you add more stuff(unless you add a power source)... adding passive elements

can only reduce the total current and stuff.

your 0.9mA is at t = 0.. at t = 0.0001, say, it might be 0.88998mA or

something like that.

(put capacitor back into circuit for what follows but leave out the other

branch with the diode)

the ideal is given then t = 0 here

Vc(t) = exp(-t/(RC))*(Vc0 - Vb) + Vb

Ic(t) = (Vb - Vc(t))/R

or with Vc0 = 0, Vb = 9v, R = 10k, and C = 1uF

hence R*C = 1/100 and

Vc(t) = 9v*(1 - exp(-100*t))

Ic(t) = 9v*exp(-100*t)/10k = 0.9mA*exp(-100*t)

these are the equations that tell you exactly(in the ideal case) what the

voltage and current are through the cap at any time t.

-------------------------

if you are not familiar with the exponential function it is not a difficult

concept. You just have to know a few properties of it.

exp(anything) >= 0

exp(0) = 1

exp(some really large positive number) ~= some much larger positive number

exp(some really large negative number) ~= 0

so

exp(0/192883 - 0*9843) = exp(0) = 1

exp(39849389289) = something really big and much larger than 39849389289

exp(-39849389289) ~= 0

exp(-t) starts at t = 0 with exp(0) = 1 and decays to exp(-10) in a special

way but basicaly you just need to know that as time increases the thing gets

smaller but does it at a slower and slower rate.

another thing to note is that exp(anything) ~= 2.7^(x)

check out

http://en.wikipedia.org/wiki/Exponential_function
or

http://ourworld.compuserve.com/homepages/g_knott/elect38.htm
for how it looks and a better explination.

-------------------------

Back to the circuit and the equations we can see that

Ic(t) = 0.9mA*exp(-t/100)

then Ic(0) = 0.9mA*exp(-0100) = 0.9mA*exp(0) = 0.9mA

then at t = 100s

Ic(100) = 0.9mA*exp(-100/100) = 0.9mA*exp(-1) = 0.9mA/e ~= 0.9mA/2.71 ~=

0.3mA

so after 100 seconds with a 1uF capacitor and 100k resistor and the 9V

battery there will be a 0.3mA current going through the capacitor

you can also see that if t = n*RC then

Ic(n*RC) = 0.9mA*exp(-n*RC/RC) = 0.9mA*exp(-n) = 0.9mA/exp(n)

so after one time constant(the time constant though depends on the circuit

and isn't always R*C but could be much more complicated) one has

Ic(RC) = 0.9mA*exp(-1) = 0.9mA/e ~= .331

so we can find out the percentage it dropped in 1 time constant by taking

the ratio of Ic before and after

Ic(RC)/Ic(0) = (0.9mA/e)/(0.9mA) = 1/e ~= 36% hence it dropped by 1 - 36% =

64%.

there for the current is 64% smaller after 1 time constant... you can do the

math to get what it is at after n time constants and its just 1/e^n. If you

want it at 50%, say, then you have to do a little more math though and you

would be dealing with fractions of time constants and it wouldn't be so

nice. The reason time constants are important is that they are independent

of the actual circuit... if you know the time constant of the circuit then

you know after one time constant whatever you are measuring will be 64%

smaller.

just so you know,

1/e ~= 36%

1/e^2 ~= 14%

1/e^3 ~= 5%

1/e^4 ~= 2%

1/e^5 ~= 0.7%

so you can see how the rate gets slower and slower. between the t = 0 and t

= RC it drops 64% and between the next time constant t = RC and t = 2RC it

drops only about 22% more. This is why, I suppose, they have the 5 time

constant rule thing... you would actually, theoretically, have to wait

forever to get a drop of 100%.

Back in the original circuit, by Kirchoff's law, if (I <= 0.9mA) then

(I_c + I_led <= 0.9mA). So at most, the current through the LED's

branch can be 0.9mA. Here I'm assuming I_c isn't negative... Maybe

that's not a safe assumption.

right, because the cap acts like short at t = 0 we have the simple circuit

with just Vb and R in series and nothing else... so it will be supplying

0.9mA's at that moment you flip the switch. I_c will never be negative cause

it will never be able to out do that battery(else it would charge it and

hence it would be supplying current... that is though only if the cap

doesn't have a larger voltage across it from being previously charged... but

its still ok long as the voltage across it isn't larger than the

battery(else it would be stronger and be able to force current through it).

So you wouldn't want to charge the cap up with a 100V battery say then stick

it in the circuit with the 9v battery cause it might screw up that battery).

Normally one can assume current is going in the wrong direction without

causing to many problems as long as one is consistant. If you do this, say,

mathematically and assume the wrong direction you always will get the a

negative in front of your current when you solve for it... i.e., if you

assume I goes one way you will eventually get I = -|I| which means you

guessed wrong. Even if you measure the current by putting the leads in the

wrong way you will get just a - of what you thought. While its important to

get the right sign in a qualitative analysis so it makes sense it always

happens to come out right in the math unless you make a mistake somewhere.

In this case though you just have to think how the 'electrons' are flowing

to see that the cap will not be able to change the direction of current when

the battery is hooked up. Remember, just think about the two extreme cases

of a capacitor when it is completely discharged(a short) to when it is

completely charged(open)... all the stuff inbetween is some a smooth

transition from one to the other.

i.e., say I have some complicated circuit with tons of resistors and

capacitors. If the circuit is in equalibrium(for DC) then it means that all

the capacitors are either acting as a short or an open circuit. Its just a

simple matter of figuring out which and then simplifying the circuit to get

the new circuit. Note that not all circuits will be in equalibrium because

say there will be a switch like you have that will keep on switching on and

off causing the capacitor to charge and discharge... or who knows what else

could be happening.

So if the maximum 0.9mA was running through the LED's branch (the

branch includes both the LED and the 1 Kohm resistor), it that would

mean that, at most, 0.9mA is running through the 1 Kohm resistor.

Using this you could calculate the maximum voltage drop across the

resistor:

V_s : resistor voltage drop

R_s : resistor resistance

I_s : resistor current

I_s <= 0.9mA

V_s/R_s <= 0.9mA

V_s <= 0.9mA * R_s

V_s <= 0.9mA * (1000ohm)

V_s <= 0.9V

Now if you add the voltage drop of the LED, or about 2.0V, you get 2.0V

+ 0.9V = 2.9V dropped on the LED's branch. This could be considered

the maximum possible voltage drop, because the LED's voltage drop

wouldn't change much for this current range.

Since the capacitor's branch and the LED's branch must drop the same

voltage, the capacitor has the same maximum voltage drop as the LED

branch, or 2.9V.

Anyway, 2.9V seems to be pretty close to the actual value of 2.53V.

Is anything I wrote plausible? At the moment, most other kinds of

analysis are way over my head for RC circuits.

yeah, it makes sense. You are looking at the maximum(final) case which only

occurs after a long time but it is useful to know because you need to make

sure you don't burn anything up.

that is, initially the current all goes through the "shorted' capacitor but

it will slowly become an open circuit causing more and more current to go

through the LED.. after an infinite amount of time all the current will go

through the LED and the capacitor will be an open circuit... to calculate

the current we need just to analyze the circuit without the capacitor(i.e.,

it opened),

I-> R1

+-------(switch)-------+----------VVVV--------+

| |

| |

---- V Battery |

-- (LED) VL

| |

| |

| |

+----------VVVV--------+----------------------+

R2

so V - I*R1 - I*R2 - VL = 0

or

I = (V - VL)/(R1 + R2)

R1 = 1k

R2 = 10k

VL = 2

V = 9

so

I = (9 - 2)/(11k) = 7/11k ~= .63mA

the voltage drop then across the R1 is

..636mA*1kOhm = .636V

the voltage drop across R2 is

..636mA*11kOhm = 6.36V

the voltage drop across the LED is 2V

hence the voltage drop across the cap is

2V + 0.63V = 2.63V

(the reason you got 2.9 is you took into account the wrong current through.

You took the initial current instead of the "final")

i.e., we have three circuits we are dealing with

for very large time t(say 5*RC)

+----------------------+----------VVVV--------+

| |

| |

---- V Battery |

-- (LED) VL

| |

| |

| |

+----------VVVV--------+----------------------+

R2

and for very short time t(t = 0)

+----------------------+

| |

| |

---- 9V Battery --- 1 mF Cap

-- ---

| |

| |

| |

+----------VVVV--------+

R2

(we can forget the switch as it is used in the discharge part which gives

you the circuit

R1

+----------VVVV--------+

| |

| |

--- 1 mF Cap |

--- (LED)

| |

| |

| |

+----------------------+

but this is an easy circuit to analyze to some degree and only depends on

the initial charge of the capacitor)

The "mistake" you made was to to take the current in the time t = 0 circuit

and use it in the time t = infinity one... the problem is that its not the

same current... its not close but. Its true that the current will never get

above 0.9mA in the LED but it doesn't even get close to 0.7mA(well, 0.63

might be close to 0.7). So your upper bound is a bit exaggerated but might

be fine if you just need an approximation.

i.e., as I mentioned before, basicaly what you do in situations like this is

that you short the cap and reduce the circuit and figure out the values of

the quantities you want to know then you open the cap and do the same and

get the new values. What you then know, atleast in a simple circuit like

this, that there is a "smooth" decay or growth from one value to the other.

If V(0) = 10 and V(100) ~= 0

then I know that V(t) smoothly(exponentially) decays from 10 to 0 over that

time between 0 and 100 seconds.. we could linearly say something like this

then as a first approximation

V(t) = 10 - 10*t/100 for t between 0 and 100 else v(t) = 0

ofcourse we already know it decays exponentially but this is just a quick

way to get some idea. in actuality it would be something like

V(t) = 10*exp(-t/TC) for TC = the time constant which we could find a upper

bound for(if its to large, say TC = 100 then we know it will not be 0 but

only 3.67 after t = 100 but its suppose to be 0, hence the TC for this

hypothetical circuit is probably at most about TC = 10 since exp(-100/10) ~=

0.0004 which is close enough to zero for me)

Thank you in advance for wading through that.

I hope that makes some sense and is helpful. It seems like you got the gist

of how it works which is probably good enough for most basic things. Just

remember that many circuits depend on the the way a capacitor works to do

some "cool" stuff. They usually use the way a capacitor charges and

discharges to accomplish this. Another aspect of this charging/discharging

ability is that it can act as a filter for high and low

frequencies(depending on how its used)... among other things.

Hence, if you wanted to know the exact time the LED would be "on" you would

need to know how the capacitor is charging and discharging. This can be done

relatively easy by finding the equations and and figuring out when the

voltage drop across it when the switch was opened and also when there was

enough voltage across it so that there would be enough current going through

the LED(here this is an simply relationship... the higher the voltage across

the capacitor means the higher the current going through the LED)... then we

would have to know when the switch was opened which means the capacitor will

discharge through the LED... it will take a certain length of time for hte

current to drop below the amount needed to power the LED. (the nice thing

about the mathematical way is that you would know it very precisely for any

R's and C's instead of having to guess how it depends on them).

Anyways,

AD