maths question to do with electronics?

Question

i want to draw the graph of a capacitor charging and discharging the equation for working out the value to plot the graph for current is

Ic = Io e -t/cr

-t = minus time cr = time constant

Io = voltage divided by resistance

in my case the voltage is 8v and the resistance is 150,000ohms

the time constant is always 15 and the time is in intervals at
2, 10, 18, 25, 35, 45, 55, 65, 75, and 80

i would like to know the answers for each interval

Answer 1

Ic = (8/150000) e^(-t/15)

Get a calculator that does exponentiation.
Store the value (8/150000) in memory
START
Plug in the value for the time.
Divide it by -15
Hit the e^x button
Multiply by the recalled memory value of (8/150000)
Write it down so you can plot it later
Clear the calculator screen
Go back to START and do the next

Answer 2

You have all the info there, but can't plug and chug it?
Instantaneous current = max current (1 - e^(-t/cr))

Instantaneous current = V/R (1 - e^(-t/15))

Instantaneous current = 8/150K (1 - 2.718281828459045^(-t/15))

Now, in the above equation, replace the t with all of the interval numbers you have, one at a time. For each answer, plot the current vs. time. You should have a curve that starts at a relatively high current, shoots down to 3db above 0 current, then asymptotes towards zero current.

Answer 3

Your formula is missing the ^ .

You need a calculator with an e^x key.
Io = .053 mA e^ -t/rc for first interval is e^ -2/15 = Ic= .046mA.
2nd interval .053 (e^-10/15) = .027mA

The initial charge current is a surge, decreasing as the cap charges and becoming negligable after 5 time constants.
The same is true of discharge...surge becoming negligable after 5 rc

Answer 4

Ic =Io e^-t/15
Ic = 5.33*10^-5e^-t/15
Ic = 4.664*10^-5 for t =2
Ic = 2.734*10^-5 for t =10
Ic = 1.604*10^-5 for t = 18
Ic = 1.007*10^-5 for t = 25
Ic = 5.165*10^-6 for t =35
Ic = 2.654*10^-6 for t =45
Ic = 1.364*10^-6 for t =55
Ic = 6.982*10^-7for t =65
Ic = 3.592*10^-7 for t =75
Ic = 2.574*10^-7 for t =80

Answer 5

As I said..............