exponential equation with constant
- From: quiasmox@xxxxxxxxx
- Date: 2 Jun 2005 11:51:30 -0700
I had a problem in an electrical circuits class, which was to find the
point at which the voltage on two exponential curves is equal. These
were the decaying voltage on an inductor and the rising voltage on a
capacitor, the curves starting from a fixed voltage and from zero
(respectively) at the same instant. The equated formulas were
VL = 2* exp(-2000t) = 1 - exp(-1000t) = VC
By luck, I found I could solve it for t by turning it into a quadratic-
let y = exp(-1000t)
2y^2 + y - 1 = 0 = (2y-1)(y+1)
Then, using (2y-1),
2 exp(-1000t) = 1,
ln(2) = 1000t
t = 693.2 us.
This solution depends on the coincidence that one exponent is exactly
twice the other.
This brings me to the question -- is there a way, other than computer
simulation, to solve a*exp(-ct) = 1 - b*exp(-dt) for t?
--
john
.
- Follow-Ups:
- Re: exponential equation with constant
- From: Robert Israel
- Re: exponential equation with constant
- From: David W . Cantrell
- Re: exponential equation with constant
- From: achava
- Re: exponential equation with constant
- Prev by Date: Re: mathematician salaries
- Next by Date: Re: closed form of solution?
- Previous by thread: Splitting fields and Symmetric groups.
- Next by thread: Re: exponential equation with constant
- Index(es):
Relevant Pages
|
