What Is the Maximum Current Through a 1.33 µF Capacitor?

[nooobie]

Homework Statement

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.

Homework Equations

i = C(dV/dt)

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 can't 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 don't 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 don't 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

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.