RadioBanter - View Single Post - physical/intuitive understanding of RL/RC time constants?

October 13th 04, 10:30 AM

(Alan Horowitz) wrote in message . com...
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%?

First, I should make a few corrections to what you've said. The
voltage doesn't increase 63% during each time-constant period. If it
did, it would become 1.63 times its original value after the first
period, then 1.63*1.63 = 2.66 times, then 1.63^3 = 4.33 times, etc.
The voltage would increase forever, and the circuit would fry itself.

I'm assuming you're talking about a battery connected to a resistor
and capacitor (or inductor) in series, or similar circuits. In the
case of the capacitor, the voltage across the capacitor starts from
zero, and in the first time-constant period, the voltage across the
capacitor rises to 63% of the voltage across the battery. In each
time-constant period thereafter, it increases by 63% of the difference
between the battery voltage and its previous voltage.

Sorry if you knew this already, but it's hard to tell from just
reading a post.

It also isn't exactly 63%. The number predicted by theory is
63.21205588285576784044762298385 ... %. But that number will not be
exact either, due to the fact that the models which predict it assume
perfect wires, perfect resistors, etc., which is never the case. At
some point one has to round.

Let's examine such a circuit at at some time after the switch has been
flipped. We'll call the voltage of the capacitor at this point V and
the voltage of the battery E, the value of the resistor R and the
value of the capacitor C.

The voltage across the resistor is E-V, making the current through it
(E-V)/R. The charge on the capacitor is CV. If the capacitor were
fully charged, it would have a charge CE. That means the capacitor
needs CE-CV more charge to be fully charged. If the current through
the resistor continued to flow at its current rate, the time it would
take to supply this charge would be CR. This time is called the time
constant.

It's crucial to notice that the time constant we just computed does
not depend on how far the capacitor has been charged. At any time, it
is correct to say that if the current through the capacitor would only
stay the same, it would be fully charged after an amount of time later
given by the time constant.

Of course, the current through the resistor will not continue to flow
at its current rate. As capacitor is charged, the voltage across the
resistor decreases, and with it, the current.

Let's pretend the current really does stay constant for a small
fraction 1/n of the time constant, that is, for the short time CR/n.
In this time, the capacitor's voltage would increase by 1/n of the
difference between the battery's voltage E and its present voltage V.
The voltage E-V across the resistor would decrease by the same amount,
making it (1-1/n) of what it was before. After one time-constant, the
voltage across the resistor would have decreased by the same ratio n
times, making the voltage across the resistor (1-1/n)^n times its
previous value.

This all still an approximation; the current doesn't really stay
constant for any length of time. But if CR/n is very small, the drop
in current is insignificant. If we choose larger and larger values of
n, the answer we get becomes better and better. If we choose an
extremely large n, the error will go away with rounding, and we will
get (1-1/n)^n = 0.3678794... . And if the difference between the
battery voltage E and the capacitor voltage V is 37% of what it was
before, that means that the voltage of the capacitor increased by 63%
of that difference.

The reciprical of 0.3678794..., 1/0.3678794... = 2.7182818..., appears
in the answers to a lot of problems, and so it has been given a
special name: "e." It is near [n/(n-1)]^n, for large n. We can
simplify this formula even further by replacing n by m+1, making it
(1+1/m)^(m+1), and dividing by (1+1/m), which is very near 1, to get
(1+1/m)^m, for large m.

While in principle one could compute e by choosing a very large n, in
practice you need such a large n to get a correct answer that the
calculation that ensues becomes tedious. But, using the binomial
theorem, we can find:

(1+1/m)^m = 1 + m(1/m) + [m(m-1)/2!](1/m)^2
+ [m(m-1)(m-2)/3!](1/m)^3 + ...

(1+1/m)^m = 1 + 1 + (1-1/m)/2! + (1-1/m)(1-2/m)/3! + ...

(1+1/m)^m ~ 1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + ... = e

We can do a similar calculation to find a series for 1/e from "near
(1-1/n)^n for large n." It all works out the same, except that we're
dealing with powers of (-1/n) instead of (1/n). Terms that had odd
powers have their sign flipped:

1/e = 1 - 1 + 1/2! - 1/3! + 1/4! - 1/5! + 1/6! - 1/7! + ...

or

1/e = 3/3! - 1/3! + 5/5! - 1/5! + 7/7! - 1/7! + 9/9! - 1/9! + ...

1/e = 2/3! + 4/5! + 6/7! + 8/9! + 10/11! + 12/13! + ...

1/e = 1/1!/3 + 1/3!/5 + 1/5!/7 + 1/7!/9 + 1/9!/11 + 1/11!/13 + ...

With just the first two terms, which you can add in your head, you can
get 1/e accurate to two significant figures.

--
Jim Black

"Within the philosophy system, it is quite correct. Let's try this: if
it was in single quotes it would 'mean' "chaos". As it is not, there
'is some form'." -- Peter Kinane