when a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?

And whence the number 63%?

Alan Horowitz wrote:

when a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?

And whence the number 63%?

1-(1/e). Crack your textbook.

http://www.iop.org/EJ/S/UNREG/aS0aOQcupwxj4CNeiM5vaQ/article/-featured=jnl/0143-0807/23/1/304/ej2104.pdf
"Demonstration of the exponential decay law using beer froth"

Google
"exponential decay" 63 20,500 hits

Uncle Al gotta think of everything.

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf

On Mon, 11 Oct 2004 16:18:07 -0700, Uncle Al
wrote:

http://www.iop.org/EJ/S/UNREG/aS0aOQcupwxj4CNeiM5vaQ/article/-featured=jnl/0143-0807/23/1/304/ej2104.pdf
"Demonstration of the exponential decay law using beer froth"

Hi Al,

Good link. Here is one that is Tau intensive from my own work:
http://www.cybernalysis.com/tau/index.htm

73's
Richard Clark, KB7QHC

"Alan Horowitz" wrote in message
om...
when a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?

Definition of time-constant period.

John Lowry
Flight Physics


And whence the number 63%?

"Alan Horowitz" wrote in message
om...
when a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?

And whence the number 63%?

The voltage knows nothing about how it's "supposed" to behave. It just does
its thing without a care in the world.
The thing it does though, will always result in exactly the same voltage
shape, because with a fixed R and C and supply voltage it can do no other.
As the C voltage grows, the voltage across the R must drop. If the R voltage
drops then the charging current must drop. If the charging current drops,
then the C voltage must rise at a slower rate, ... and so on and so on ...
Everything slows down more and more as time goes on. A bit of thought and
you'll notice that the C can never actually charge exactly to the supply
voltage.
As this RL RC (dis)charging process must always result in this particular
shape or curve and this quite 'natural' curve turns up across all branches
of science, engineering and finance, it wasn't long before the
mathematicians found they could usefully model, or describe the curve
accurately, using an equation based on the 2.718 "e" value used for working
out 'natural' logarithms.
Hence the maths numbers and formulae that are taught are a good descriptive
model or analogue of what's happening in the circuit but have nothing to do
with the actual circuit workings.
Be wary when relying purely on maths models. They confer 'expertise' into
how something works, without offering 'understanding' of how something
works. The difference can be crucial.
regards
john

Alan Horowitz wrote:

when a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?

And whence the number 63%?

This all goes back to the solution of the differential equation for
the RC or RL system. e is a natural constant that has some very sweet
properties in many applications of mathematics, and simplifying
differential equations is one of them. Read through this tutorial and
see how the rate constant k in this tutorial is an example of a time
constant.
http://www.ugrad.math.ubc.ca/coursed...eqs/intro.html
--
John Popelish

On 11 Oct 2004 16:11:16 -0700, (Alan Horowitz) wrote:

when a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?

And whence the number 63%?

I haven't had a chance to read other responses, but here's mine:

Take the case of an RC:

,---,
| |
V| \
--- / R
- \
--- |
- +-----
| |
o --- C
/ ---
o |
'---+-----

Assume C is discharged and V has just been applied by closing the switch...

The current through R is based on V, less the voltage on C (which counters V),
so:

I(R) = ( V - V(C) ) / R

The above is a function of time, because V(C) is a function of time. So, what's
V(C)? Well, that needs to be arrived at more slowly.

First, we know that this is true:

Q = C*V

Well, actually, that's an average statement. More exactly, it's:

dQ = C * dV

In other words, the instantaneous change in Coulombs is equal to the capacitance
times the instantaneous change in voltage. Both sides can now be divided by an
instant of time to give:

dQ / dt = C * dV / dt

Since dQ/dt is just current (I), for the above capacitor this becomes:

I(C) = C * dV(C) / dt

So how does this help? Well, we know that the current from R must accumulate on
C. So, we know that:

I(C) = I(R) = ( V - V(C) ) / R

so, combining, we get:

C * dV(C) / dt = ( V - V(C) ) / R

Rearrangement of this gives:

dV(C) / dt + V(C) / (R*C) = V / ( R*C )

Which is the standard form for ordinary differential equations of this type.

The standard form with general terms looks like: dy/dx + P(x)*y = Q(x). In our
case, though, y = V(C), x = t, P(x) = 1 / (R*C), and Q(x) = V / (R*C).

The solution to this includes multiplying by what is called "the integrating
factor", which is:

u(x) = e^(integral (P(x)*dx)) = e^(integral (dt/RC)) = e^(t/(R*C))

(This is a VERY POWERFUL method to learn, by the way, and it is probably covered
in the first few chapters of any ordinary differential equations book.) So,
going back to look at the general form and multiplying both sides:

u(x)*dy/dx + u(x)*P(x)*y = u(x)*Q(x)

But the left hand side is just d(u(x)*y)/dx, so:

d(u(x)*y)/dx = u(x)*Q(x)

or,

d(u(x)*y) = u(x)*Q(x) * dx

In our case, this means:

d( e^(t/(R*C)) * V(C) ) = V / ( R*C ) * e^(t/(R*C)) * dt

Taking the integral of both sides, we are left with:

e^(t/(R*C)) * V(C) = integral [ V / ( R*C ) * e^(t/(R*C)) * dt ]
= V / ( R*C ) * integral [ e^(t/(R*C)) * dt ]

setting z = t/(R*C), we have dz = dt/(R*C) or dt = R*C*dz, thus:

e^(t/(R*C)) * V(C) = V / ( R*C ) * R*C * [e^(t/(R*C)) + k1]
= V * [e^(t/(R*C)) + k1]
= V * e^(t/(R*C)) + V * k1
V(C) = V + V * k1 / e^(t/(R*C))
= V + V * k1 * e^(-t/(R*C))
= V * [ 1 + k1*e^(-t/(R*C)) ]

From initial conditions, where V(C) = 0V at t=0, we know that k1=-1, so:

V(C) = V * [ 1 - e^(-t/(R*C)) ]

Time constants are usually taken to be:

e^(-t/k)

with (k) being the constant. In our V(C) case, this means that k=R*C. So
that's the basic constant and it's in units of seconds.

So, what's the voltage after one such constant of time? Well:

V(C) = V * [ 1 - e^(R*C/(R*C)) ] = V * [ 1 - 1/e ] = V * .63212

Ah! There's that 63% figure. Actually, more like 63.212%.

Two time constants would be:

1 - 1/e^2 = .864665

and so on....

Oh... and there are other methods you can use to solve the simple RC formula,
but the method I chose is a very general and powerful one worth learning well.

Jon

**** Y I K E S !!!



--
Steve N, K,9;d, c. i My email has no u's.


"Jonathan Kirwan" wrote in message
...
On 11 Oct 2004 16:11:16 -0700, (Alan Horowitz) wrote:

when a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?

And whence the number 63%?

I haven't had a chance to read other responses, but here's mine:

Take the case of an RC:

,---,
| |
V| \
--- / R
- \
--- |
- +-----
| |
o --- C
/ ---
o |
'---+-----

Assume C is discharged and V has just been applied by closing the
switch...

The current through R is based on V, less the voltage on C (which counters
V),
so:

I(R) = ( V - V(C) ) / R

The above is a function of time, because V(C) is a function of time. So,
what's
V(C)? Well, that needs to be arrived at more slowly.

First, we know that this is true:

Q = C*V

Well, actually, that's an average statement. More exactly, it's:

dQ = C * dV

In other words, the instantaneous change in Coulombs is equal to the
capacitance
times the instantaneous change in voltage. Both sides can now be divided
by an
instant of time to give:

dQ / dt = C * dV / dt

Since dQ/dt is just current (I), for the above capacitor this becomes:

I(C) = C * dV(C) / dt

So how does this help? Well, we know that the current from R must
accumulate on
C. So, we know that:

I(C) = I(R) = ( V - V(C) ) / R

so, combining, we get:

C * dV(C) / dt = ( V - V(C) ) / R

Rearrangement of this gives:

dV(C) / dt + V(C) / (R*C) = V / ( R*C )

Which is the standard form for ordinary differential equations of this
type.

The standard form with general terms looks like: dy/dx + P(x)*y = Q(x).
In our
case, though, y = V(C), x = t, P(x) = 1 / (R*C), and Q(x) = V / (R*C).

The solution to this includes multiplying by what is called "the
integrating
factor", which is:

u(x) = e^(integral (P(x)*dx)) = e^(integral (dt/RC)) = e^(t/(R*C))

(This is a VERY POWERFUL method to learn, by the way, and it is probably
covered
in the first few chapters of any ordinary differential equations book.)
So,
going back to look at the general form and multiplying both sides:

u(x)*dy/dx + u(x)*P(x)*y = u(x)*Q(x)

But the left hand side is just d(u(x)*y)/dx, so:

d(u(x)*y)/dx = u(x)*Q(x)

or,

d(u(x)*y) = u(x)*Q(x) * dx

In our case, this means:

d( e^(t/(R*C)) * V(C) ) = V / ( R*C ) * e^(t/(R*C)) * dt

Taking the integral of both sides, we are left with:

e^(t/(R*C)) * V(C) = integral [ V / ( R*C ) * e^(t/(R*C)) * dt ]
= V / ( R*C ) * integral [ e^(t/(R*C)) * dt ]

setting z = t/(R*C), we have dz = dt/(R*C) or dt = R*C*dz, thus:

e^(t/(R*C)) * V(C) = V / ( R*C ) * R*C * [e^(t/(R*C)) + k1]
= V * [e^(t/(R*C)) + k1]
= V * e^(t/(R*C)) + V * k1
V(C) = V + V * k1 / e^(t/(R*C))
= V + V * k1 * e^(-t/(R*C))
= V * [ 1 + k1*e^(-t/(R*C)) ]

From initial conditions, where V(C) = 0V at t=0, we know that k1=-1, so:

V(C) = V * [ 1 - e^(-t/(R*C)) ]

Time constants are usually taken to be:

e^(-t/k)

with (k) being the constant. In our V(C) case, this means that k=R*C. So
that's the basic constant and it's in units of seconds.

So, what's the voltage after one such constant of time? Well:

V(C) = V * [ 1 - e^(R*C/(R*C)) ] = V * [ 1 - 1/e ] = V * .63212

Ah! There's that 63% figure. Actually, more like 63.212%.

Two time constants would be:

1 - 1/e^2 = .864665

and so on....

Oh... and there are other methods you can use to solve the simple RC
formula,
but the method I chose is a very general and powerful one worth learning
well.

Jon

On Tue, 12 Oct 2004 16:20:32 -0500, "Steve Nosko"
wrote:

**** Y I K E S !!!



Jon

Alan Horowitz wrote:
when a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?

And whence the number 63%?

Suppose you are trying to fill up a box with balls. However, for some
strange reason, you've decided that each time you throw in balls, you'll
throw in 1/2 of the balls that will fit in the remaining space.

At the first second, you have 1/2 the balls. Next second, you'll have
that plus 1/2 of the remaining space, which is 1/2 + 1/4 = 3/4. The
third second, you'll have that plus 1/2 the remaining space, ie, 1/2 +
1/4 + 1/8 = 7/8...

So, the number of balls at any time t will be:

B(t) = 1 - (1/2)^t

Thus, after 3 seconds, there will be B(3) = 1 - (1/2)^3 = 1 - 1/8 = 7/8,
just like above.

Now, apply that same reasoning, only instead of using the ratio 1/2, use
the ratio 1/e (since we are applying arbitrary rules)

Then

B(t) = 1 - (1/e)^t

After the first second, you'll have

B(1) = 1 - (1/e)^1 = 1 - 1/e = 0.632 (that is, 63%)

Strange coincidence, isn't it? It happens because when you are charging
a capacitor through a resistor, you are throwing balls, in the form of
charges, into a box (the capacitor), and the number of charges you throw
at any given time (the current) depends on how many charges are already
on the capacitor (the voltage).

Each step of the formula above is one time constant, RC. By dividing out
the RC, you can get the answer given seconds, ie

B(t) = 1 - (1/e)^(t/RC) = 1 - e^(-t/RC)

Where B is the percentage 'filled' the capacitor is (ie, what percentage
it is of the input voltage).

Why is 1/e used instead of 1/2? That has to do with the fact that we
must have a continuous solution, not a solution based on ratios of
existing values; the rate of change of the current (ie, how many balls
we throw in per unit time) is proportional to the voltage remaining,
which is continuously changing. Using 1/e instead of 1/2 allows us to
generalize to this, in the same way as the compound interest formula
allows us to compute 'continuously compounding' interest.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.