Re-arranging an Equation

[Denton91]

I have been given the following equation:

v=240(1-e^-Rt/L)

Where R=1ohm, L=200mH. Calculate the time(t) when the voltage(v) is expected to be 140V

I have already worked out the equation, which I will quickly explain below. I was hoping someone would be able to point me in the right direction of another method i can use to solve this equation.

The steps of the method I have used(missed some out on purpose):

(v/240)+e^-(Rt/L) =1

e^-(Rt/L) =1-(v/240)

Then using Napiers Law (lnA^B=AlnB)

lne^-(Rt/L)=ln(1-(v/240))

-(Rt/L)=ln(1-(v/240))

t=(-L/R)ln(1-(v/240))

After converting 200mH to 0.2H I was left with

t=(0.2/1)ln(1-(140/240))

which left me with an answer of 0.175 seconds.

Like I said, it isn't the answer I am after, I have been assured that is right by another tutor, but I am just looking for and hoping to work out another method for solving this equation.

 

[HallsofIvy]

I have no idea why that "isn't the answer I am after", but it is the correct answer! It is the only value of x (to three decimal places) that satisfies that equation.

 

[Denton91]

I have no idea why that "isn't the answer I am after", but it is the correct answer! It is the only value of x (to three decimal places) that satisfies that equation.

Sorry, my mis-wording I think, I meant it to say that I am not looking to be given the answer, just a push in the right direction of another method of solving this, I cannot see anything anywhere. But thanks for confirming it is right :-D

And that is if there is another method of doing it, as I have seriously racked my brin and looked at it non-stop for a hour, and cannot think of anything!!

 

[HallsofIvy]

Denton, are you simply looking for a less "wieldy way"?
If so, here's my "lazy" way:

GIVENS: R = 1, L = 200, v = 140

v=240(1-e^-Rt/L) : solve for t

Let f = R/L

v = 240(1 - e^(-tf))
v = 240 - 240/e^(tf)

e^(tf) = 240 / (240 - v) : let that = k

e^(tf) = LOG(k)

You mean e^(tf)= k so that tf= LOF(k)

tf = LOG(k) / LOG(e) = LOG(k) : since LOG(e) = 1

t = LOG(k) / f

t = 175.093747.....

 

[Ishuda]

Sorry, my mis-wording I think, I meant it to say that I am not looking to be given the answer, just a push in the right direction of another method of solving this, I cannot see anything anywhere. But thanks for confirming it is right :-D

And that is if there is another method of doing it, as I have seriously racked my brin and looked at it non-stop for a hour, and cannot think of anything!!

Just for grins and giggles, my looooooooooooooong way. Decide on the accuracy wanted, i.e.
|correct t - estimate of t| ~ 10^-m
let N = 200 (10 x)^m and compute
E = \(\displaystyle \Sigma_{j=1}^{j=N}\frac{x^j}{j}\)
where x = v/240. Then
t ~ 200 * E within (approximately) the wanted error.

EDIT: Added value of x and changed estimate of N. Note this is good for only a limited set of v's.