current through capacitor

nooobie

1. The problem statement, all variables and given/known data
find the greatest current through the capacitor with a capacitance of C=1.33uF, where the applied voltage is given by V = 250t^2 - 200t^3 volts.


2. Relevant equations
i = C(dV/dt)


3. The attempt at a solution
so i take the derivative of V:V'= 500t-600t^2...
and find values for time and i get

0= 100t(5 - 6t)
then t = 0, and 5/6.
testing these values in the second derivative V" = 500-1200t...I find 5/6 to be maximum...

plugging into the equation for i = (1.33*10^(-6))(500(5/6) - 600(5/6)^2)
which equates to zero...bah, this can be right...what am i doing wrong...am i taking the wrong derivative or something

pooface

you got the max point correct.

but why are you plugging the max x coordinate into V prime?

nooobie

the equation for I is C*V' , and i have no value for t except 5/6, do i have to plug into the orig?..but that would give me a constant, and i cant get the derivative of the constant, it would give me zero again..

Astronuc

One wants di/dt = 0 => d²V/dt² = 0 to find time at maximum current.

Using dV/dt = 0, one will find time of max or min voltage.


V" = 500-1200t = 0, then find t, which is the time of max current.

pooface

the equation means that Voltage is varying by time. you found the time when the voltage is at it's peak(max value). you need to find the voltage at that point in time to get the max value for I.

nooobie

thanks Astronuc
you were very helpful

pooface

I dont think astronucs method will work properly.

V'' is a straight line not a parabola.

My method suggested that since in i=C*V, C is constant and both C,V are directly proportional, the greatest value for V, which is the maximum point, would yield the greatest current.

EDIT: moreover, getting 500-1200t=0 and solving for t will give the x coordinate for the inflection point of V=250t^2-200t^3

learningphysics

i isn't C*V. i is Cdv/dt... ie: i = C*V'.... so to get the max of i you need to set i' = 0 ie: C*V'' = 0 => V'' = 0.

max voltage does not necessarily imply max current.