# Reducing non-linear laws to linear fractions

• Nov 17th 2011, 07:16 AM
batman121
Reducing non-linear laws to linear fractions
current in mA flowing through a capacitor is discharging over time ms, as shown
i 203 61.14 22.49 6.13 2.49 0.615
t 100 160 210 275 320 390

the graph follows the law of, i = Ie^-t/T (e refers to log, and its to the power of -t/T)

find values for I and T

my solution

log i = log I e ^ -t/T

log i = log I + log e ^ -t/T

log i = -t / T log I
Y = M X + C

i then plot the graph of log of Y against x (shown in attachment)

so - t should = change in log Y / change in x

-t = 3.2 / 392 = 8.16*10^-3

log I = point of intersect 3.2

I = e3.2

I = 24.53
t = 8.16*10^-3
i do this and then put the values into the forumla

24.53e^((8.16*10^-3)/t) = i

but am not getting the correct answers if t is 100 it should be 203 but i get 24.53

any help on where iv gone wrong would be helpful
• Nov 17th 2011, 02:16 PM
skeeter
Re: Reducing non-linear laws to linear fractions
I graphed $i$ vs. $\log(t)$ in my TI-83

linear regression yielded the equation $\log(i) = -0.200 t + 7.313$

$I = i(0) = e^{7.313} \approx 1500$

$i(t) = I \cdot e^{-.02t}$

$i(100) = 203$