Vi = E.e ^-t/cr
if having been fully charged to voltage e=10v c=250uf it is allowed to discharge through a 3.3k resistor determine the values of capacitor voltage after time; t=0, t=1, up to t=0
b) plot the voltage values against time values
c) draw a tangenrt to the urve at t=2s. calculate from the slope the value of rate of change of voltage in V/s
d) use differential Calculus to find the gradient of the curve at t=2s
e) compare your values from c) and d) on the accuracy.
Vi = E.e-t/cr (I believe you left out the negative in the exponent !)
You are given E = 10 v c = 250 x 10-6 FARADS r = 3.3 ohms (? dimensions left off in your Q)
Plugging into the given equation
V0 = 10 e-0/(250 x10-6 * 3.3) this is the value of V at t= 0
V1 = 10 e-1 /(250 x 10-6 * 3.3) this is the value of V at t= 1
etc etc for other 't' values
Start with the graph and show us what you find.....we'l go from there !
Calculate the v at points t = 0 1 2 3 4 ...... to whatever edpoint is given to you (your question did not have an end point)
t will be on the x axis the calculated v will be the y axis
at t= 0 v = 10 first point (0,10) at t = 2 v = ? second point would be (2, ?) etc etc
Here is my example......it may or may not be correct...I used 3.3 kOhm instead of 3.3 ohm ...I just graphed the equation....your question wants you to plot the points and draw a curve......